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Original questions in the style of each exam boardFull worked mark schemes with method and accuracy marksCreated by Evan Kotronis with AI assistanceUpdated 16 October 2026 · new questions added continuouslyFree access until 31 December 2026How every question is checked →
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1IBMI-033-00-01Sampling and descriptive statisticsPaper 1Easy5 marks
A fitness centre has 840 members. 336 members are aged under 30, 364 are aged 30 to 50 and 140 are aged over 50. The manager wants a sample of 60 members to answer a questionnaire about opening hours.
(a)
The manager uses a stratified sample by age group. Find the number of members aged over 50 in the sample.
(2)
(b)
The manager instead lists all 840 members alphabetically, chooses a random starting number from 1 to 14 and then selects every 14th member. Name this sampling technique.
(1)
(c)
The questionnaire asks each member for the number of visits they made last month. State, with a reason, whether this data is discrete or continuous.
(2)
Show mark scheme
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Mark
Notes
Part (a)
(140 ÷ 840) × 60 M1
✓ 1
10 members A1
✓ 1
Part (b)
Systematic sampling A1
✓ 1
Part (c)
Discrete A1
✓ 1
Number of visits is a count, so it can only take whole-number values R1
✓ 1
Exam tip: In a stratified sample each stratum's share of the sample equals its share of the population: (stratum size ÷ population size) × sample size.
2IBMI-033-00-02Sampling and descriptive statisticsPaper 1Easy6 marks
The times, t minutes, that 80 employees of a company take to travel to work are recorded in the table.
0 < t ≤ 10 : 6 10 < t ≤ 20 : 18 20 < t ≤ 30 : 27 30 < t ≤ 40 : 19 40 < t ≤ 60 : 10
(a)
Write down the modal class.
(1)
(b)
Find an estimate of the mean travel time.
(3)
(c)
Find an estimate of the standard deviation of the travel times.
(2)
Show mark scheme
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Mark
Notes
Part (a)
20 < t ≤ 30 A1
✓ 1
Part (b)
Uses midpoints 5, 15, 25, 35, 50 M1
✓ 1
Σfx ÷ 80 = 2140 ÷ 80 M1
✓ 1
26.8 minutes (26.75) A1
✓ 1
Part (c)
Midpoints and frequencies entered into GDC, or √(69350 ÷ 80 − 26.752) M1
✓ 1
12.3 minutes A1
✓ 1
Common error: The last class is 20 minutes wide, so its midpoint is 50, not 45. Always check that classes have equal widths before writing midpoints.
3IBMI-033-00-04Sampling and descriptive statisticsPaper 1Easy4 marks
A weather station records the maximum daily temperature, in °C, for every day in June. The mean is 18.4 °C and the standard deviation is 3.5 °C. The temperatures are converted to degrees Fahrenheit using F = 1.8C + 32.
(a)
Find the mean maximum daily temperature in °F.
(2)
(b)
Find the standard deviation of the maximum daily temperatures in °F.
(2)
Show mark scheme
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Mark
Notes
Part (a)
1.8 × 18.4 + 32 M1
✓ 1
65.1 °F (65.12) A1
✓ 1
Part (b)
Multiplies by 1.8 only (adding 32 does not change spread) M1
✓ 1
6.30 °F A1
✓ 1
Common error: Adding a constant shifts every value equally, so it changes the mean but not the standard deviation. Do not add 32 to the standard deviation.
4IBMI-033-00-08Sampling and descriptive statisticsPaper 1Easy5 marks
A town council wants to know adult residents' opinions on a proposed cycle lane. A council worker questions the first 50 adults who arrive at a town-centre car park at 9 a.m. on a Monday.
(a)
Name the sampling technique used by the council worker.
(1)
(b)
Explain why this sample is likely to be biased.
(1)
(c)
The town has 18 000 adults: 7200 in North district, 6300 in Central district and 4500 in South district. A stratified sample of 400 adults is taken by district. Find the number of adults sampled from each district.
(2)
(d)
Describe how the 160 adults from North district could be chosen so that the sample within the district is random.
People arriving at a car park are drivers, so cyclists and non-drivers are under-represented. Also accept: only people in the town centre at 9 a.m. on a Monday (e.g. commuters) can be chosen, so others such as those working elsewhere or at home are excluded. R1
✓ 1
Part (c)
Use (district size ÷ 18 000) × 400 M1
✓ 1
North 160, Central 140, South 100 A1
✓ 1
Part (d)
Number every adult in North district from 1 to 7200 and use a random number generator to select 160 different numbers. Accept any equivalent simple random sampling method, e.g. drawing names from a hat. R1
✓ 1
Exam tip: When explaining bias, name the specific group that is over- or under-represented in this context rather than just saying 'not random'.
5IBMI-033-07-r2-01Probability and the binomial distributionPaper 1Easy5 marks
A garden centre sells packets of tomato seeds. The probability that any one seed germinates is 0.85, independently of all other seeds. A gardener plants 20 seeds from one packet. Let X be the number of seeds that germinate.
(a)
Find the expected number of seeds that germinate.
(1)
(b)
Find the probability that exactly 18 seeds germinate.
(2)
(c)
Find the probability that at least 17 seeds germinate.
(2)
Show mark scheme
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Mark
Notes
Part (a)
E(X) = 20 × 0.85 = 17 A1
✓ 1
Part (b)
Recognises X ~ B(20, 0.85) and uses P(X = 18) M1
✓ 1
P(X = 18) = 0.229 A1
✓ 1
Part (c)
Uses P(X ≥ 17) = 1 − P(X ≤ 16) M1
✓ 1
= 0.648 A1
✓ 1
Exam tip: "At least 17" includes 17. On a GDC, the cumulative binomial function gives P(X ≤ k), so use 1 − P(X ≤ 16), not 1 − P(X ≤ 17).
6IBMI-033-07-r2-02Probability and the binomial distributionPaper 1Easy6 marks
A city council surveyed 240 commuters about how they travel to work. Of these, 132 use the bus, 90 use the train and 38 use both the bus and the train. Let B be the event that a randomly chosen commuter from the survey uses the bus and T the event that they use the train.
(a)
Find the number of commuters who use neither the bus nor the train.
(2)
(b)
Determine, with a reason, whether the events B and T are independent.
(2)
(c)
A commuter from the survey is chosen at random and is known to use the bus. Find the probability that this commuter also uses the train.
Since these are not equal, B and T are not independent A1
✓ 1
Part (c)
P(T | B) = 38/132 M1
✓ 1
= 0.288 A1
✓ 1
Common error: For P(T | B) the denominator is the number who use the bus (132), not the whole survey (240). Given information reduces the sample space.
7IBMI-033-08-r2-01Normal distribution and chi-squared testsPaper 1Easy6 marks
The battery life, X hours, of a particular model of smartphone on a full charge can be modelled by a normal distribution with mean 18.4 hours and standard deviation 1.6 hours.
(a)
Find the probability that a randomly chosen phone of this model has a battery life of less than 16 hours.
(2)
(b)
Only 10% of phones of this model have a battery life longer than k hours. Find the value of k.
(2)
(c)
A retailer sells 250 phones of this model. Find the expected number of these phones that have a battery life of less than 16 hours.
(2)
Show mark scheme
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Mark
Notes
Part (a)
Recognises P(X < 16) with X ~ N(18.4, 1.62), e.g. normalcdf with correct bounds
✓ 1
0.0668
✓ 1
Part (b)
Uses P(X > k) = 0.1, i.e. inverse normal with area 0.9 to the left (or 0.1 to the right)
✓ 1
k = 20.5 hours (20.450…)
✓ 1
Part (c)
250 × 0.0668…
✓ 1
16.7 phones
✓ 1
Exam tip: An expected number does not have to be a whole number. Leave it as 16.7 rather than rounding to 17, because rounding loses the accuracy mark.
8IBMI-033-08-r2-02Normal distribution and chi-squared testsPaper 1Easy7 marks
A school surveys 200 students about how they usually travel to school. The results, classified by year group, are shown below.
Year 11: Bus 38, Car 22, Walk/cycle 30 (total 90) Year 12: Bus 34, Car 46, Walk/cycle 30 (total 110)
A χ2 test for independence is carried out at the 5% significance level.
(a)
State the null hypothesis for this test.
(1)
(b)
Show that the expected frequency for Year 12 students who travel by car is 37.4.
(2)
(c)
Use your graphic display calculator to find the p-value for this test.
(2)
(d)
State the conclusion of the test, giving a reason for your answer.
(2)
Show mark scheme
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Notes
Part (a)
H0: method of travel to school is independent of year group
✓ 1
Part (b)
Car total = 22 + 46 = 68 and uses (row total × column total) ÷ grand total
✓ 1
\(\frac{110 \times 68}{200}\) = 37.4
✓ 1
Part (c)
χ2calc = 6.76
✓ 1
p-value = 0.0340
✓ 1
Part (d)
0.0340 < 0.05, so reject H0
✓ 1
There is evidence that method of travel and year group are not independent
✓ 1
Common error: Writing 'travel method depends on year group' as the null hypothesis is wrong: H₀ always states independence. The conclusion must also refer back to travel method and year group to earn the context mark.
9IBMI-033-05-r1-02Sampling and descriptive statisticsPaper 1Easy5 marks
A council surveys 50 commuters about their daily journey time, t minutes. The results are grouped as follows: 0 < t ≤ 10: 6 commuters 10 < t ≤ 20: 14 commuters 20 < t ≤ 30: 18 commuters 30 < t ≤ 40: 9 commuters 40 < t ≤ 50: 3 commuters
(a)
Write down the modal class.
(1)
(b)
Use mid-interval values to find an estimate of the mean journey time.
(2)
(c)
Find an estimate of the standard deviation of the journey times.
(2)
Show mark scheme
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Mark
Notes
Part (a)
A1: 20 < t ≤ 30
✓ 1
Part (b)
M1: Use of midpoints 5, 15, 25, 35, 45 with frequencies, e.g. \(\frac{1140}{50}\)
✓ 1
A1: 22.8 minutes
✓ 1
Part (c)
M1: Use of GDC with midpoints and frequencies, or \(\sqrt{\frac{31650}{50}-22.8^2}\)
✓ 1
A1: 10.6 minutes
✓ 1
Common error: Entering the class upper boundaries (10, 20, 30, …) instead of the midpoints into the calculator shifts the mean up by 5 minutes.
10IBMI-033-03-r2-01Normal distribution and chi-squared testsPaper 1Easy7 marks
An orchard grades its apples by mass. The mass, X grams, of an apple is modelled by a normal distribution with mean 152 g and standard deviation 11 g.
(a)
Find the probability that a randomly chosen apple has a mass less than 140 g.
(2)
(b)
Find the probability that a randomly chosen apple has a mass between 145 g and 165 g.
(2)
(c)
Apples are packed in boxes of 40. Find the expected number of apples in a box with mass between 145 g and 165 g.
(1)
(d)
Apples heavier than m grams are sold as 'premium'. Exactly 5% of apples are premium. Find m.
(2)
Show mark scheme
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Notes
Part (a)
Correct use of normal cdf with lower bound −∞ (or a very large negative number), upper 140, μ = 152, σ = 11 (M1)
✓ 1
P(X < 140) = 0.138 (A1)
✓ 1
Part (b)
Correct use of normal cdf with bounds 145 and 165 (M1)
✓ 1
P(145 < X < 165) = 0.619 (A1)
✓ 1
Part (c)
40 × 0.61909… = 24.8 apples (A1)
✓ 1
Part (d)
Use of inverse normal with area 0.95 (or upper-tail area 0.05), μ = 152, σ = 11 (M1)
✓ 1
m = 170 g (170.09…) (A1)
✓ 1
Exam tip: An expected number is a mean, so it does not have to be a whole number; give 24.8 rather than rounding to 25.
11IBMI-033-02-r2-01Probability and the binomial distributionPaper 1Easy5 marks
A garden centre sells packets of tomato seeds. Each seed in a packet germinates with probability 0.82, independently of the other seeds. A packet contains 15 seeds. Let \(X\) be the number of seeds in a packet that germinate.
(a)
Find the probability that exactly 12 seeds in a packet germinate.
(2)
(b)
Find the probability that at least 13 seeds in a packet germinate.
(2)
(c)
Find the expected number of seeds that germinate in a packet.
(1)
Show mark scheme
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Notes
Part (a)
M1: recognises \(X \sim B(15, 0.82)\) and attempts \(P(X=12)\)
Exam tip: For 'at least 13', use the complement \(1 - P(X \le 12)\) on your GDC. Using \(P(X \le 13)\) instead is a frequent slip.
12IBMI-033-02-r2-02Probability and the binomial distributionPaper 1Easy6 marks
In a survey of 200 commuters in a city, 120 said they regularly use the train, 75 said they regularly use the bus and 40 said they regularly use both. One commuter is selected at random from the survey.
(a)
Find the number of commuters who regularly use neither the train nor the bus.
(2)
(b)
Given that the commuter regularly uses the train, find the probability that they also regularly use the bus.
(2)
(c)
Determine whether the events 'uses the train' and 'uses the bus' are independent. Justify your answer.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1: \(120 + 75 - 40 = 155\) use at least one
✓ 1
A1: \(200 - 155 = 45\)
✓ 1
Part (b)
M1: \(\frac{40}{120}\)
✓ 1
A1: \(\frac{1}{3}\) or \(0.333\)
✓ 1
Part (c)
M1: compares \(P(T)\times P(B) = 0.6 \times 0.375 = 0.225\) with \(P(T \cap B) = 0.2\) (or compares \(P(B|T)=0.333\) with \(P(B)=0.375\))
✓ 1
R1: the values are not equal, so the events are not independent
✓ 1
Common error: Adding 120 and 75 without subtracting the 40 who use both counts those commuters twice and gives a negative 'neither' region.
13IBMI-033-02-r1-01Probability and the binomial distributionPaper 1Easy6 marks
A garden centre sells packets of tomato seeds. Each seed in a packet germinates with probability 0.85, independently of the other seeds. A packet contains 20 seeds. Let \(X\) be the number of seeds in a packet that germinate.
(a)
Find the expected number of seeds in a packet that germinate.
(2)
(b)
Find the probability that exactly 17 seeds in a packet germinate.
(2)
(c)
The garden centre advertises that a packet produces "at least 18 plants". Find the probability that a randomly chosen packet meets this claim.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1: uses \(E(X)=np=20\times0.85\)
✓ 1
A1: 17 seeds
✓ 1
Part (b)
M1: recognises \(X\sim B(20,\,0.85)\) and attempts \(P(X=17)\), e.g. \(\binom{20}{17}(0.85)^{17}(0.15)^3\)
✓ 1
A1: 0.243
✓ 1
Part (c)
M1: attempts \(P(X\ge18)=1-P(X\le17)\) or \(P(18)+P(19)+P(20)\)
✓ 1
A1: 0.405
✓ 1
Exam tip: For 'at least' probabilities, write the inequality first, e.g. \(P(X\ge18)=1-P(X\le17)\), before using the calculator's cumulative binomial function. This earns the method mark even if the final value is mistyped.
14IBMI-033-02-r1-02Probability and the binomial distributionPaper 1Easy5 marks
At a railway station café, 60% of customers order a hot drink. Of the customers who order a hot drink, 35% also buy a pastry. Of the customers who do not order a hot drink, 20% buy a pastry.
(a)
Find the probability that a randomly chosen customer buys a pastry.
(2)
(b)
A customer is seen buying a pastry. Find the probability that this customer also ordered a hot drink.
(3)
Show mark scheme
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Mark
Notes
Part (a)
M1: adds the two branch products, \(0.6\times0.35+0.4\times0.2\)
✓ 1
A1: 0.29
✓ 1
Part (b)
M1: uses conditional probability \(\frac{P(H\cap P)}{P(P)}\)
✓ 1
A1: correct numerator \(0.6\times0.35=0.21\)
✓ 1
A1: \(\frac{0.21}{0.29}=0.724\)
✓ 1
Common error: In part (b), writing 0.35 confuses \(P(\text{pastry}\mid\text{hot drink})\) with \(P(\text{hot drink}\mid\text{pastry})\). The condition is what you divide by, and here that is buying a pastry.
15IBMI-033-07-r1-01Probability and the binomial distributionPaper 1Easy5 marks
A factory produces smartphone screens. Each screen independently has a probability of 0.04 of being defective. A quality inspector selects a random sample of 25 screens from the production line. Let X be the number of defective screens in the sample.
(a)
Find the expected number of defective screens in the sample.
(1)
(b)
Find the probability that exactly 2 screens in the sample are defective.
(2)
(c)
The production line is stopped for checking if 2 or more defective screens are found in the sample. Find the probability that the line is stopped.
(2)
Show mark scheme
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Mark
Notes
Part (a)
E(X) = 25 × 0.04 = 1 A1
✓ 1
Part (b)
Recognises X ~ B(25, 0.04), e.g. uses binomial pdf with n = 25, p = 0.04 M1
✓ 1
P(X = 2) = 0.188 A1
✓ 1
Part (c)
P(X ≥ 2) = 1 − P(X ≤ 1) = 1 − 0.7358… M1
✓ 1
= 0.264 A1
✓ 1
Common error: For P(X ≥ 2), using 1 − P(X ≤ 2) instead of 1 − P(X ≤ 1) loses the value at X = 2 itself. Write the complement out in words before reaching for the calculator.
16IBMI-033-07-r1-02Probability and the binomial distributionPaper 1Easy6 marks
Priya travels to work each day either by bus or by bicycle. She takes the bus on 70% of days and cycles on the remaining days. When she takes the bus, the probability that she is late is 0.15. When she cycles, the probability that she is late is 0.05.
(a)
Find the probability that Priya is late on a randomly chosen day.
(2)
(b)
Given that Priya is late on a particular day, find the probability that she took the bus.
(2)
(c)
Assume that whether Priya is late is independent from day to day. Find the probability that, in a 5-day working week, she is late on exactly one day.
Exam tip: In a conditional probability, the denominator is the probability of the event you are told has happened (being late), not the probability of taking the bus.
17IBMI-033-01-r1-01Correlation and regressionPaper 1Easy5 marks
A water company records the maximum daily temperature, T °C, and the volume of water supplied, W megalitres, in a town on 40 summer days. On these days the temperatures ranged from 14 °C to 31 °C. The regression line of W on T is \(W = 1.84T + 23.6\), and Pearson's product-moment correlation coefficient is r = 0.873.
(a)
Describe the correlation between T and W.
(1)
(b)
Interpret, in context, the value 1.84 in the equation of the regression line.
(1)
(c)
A manager wants to use the line to estimate water supplied on a day when the maximum temperature is 38 °C. Comment on the reliability of this estimate.
(1)
(d)
Estimate the volume of water supplied on a day when the maximum temperature is 25 °C.
(2)
Show mark scheme
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Mark
Notes
Part (a)
Strong, positive (linear) correlation, since r = 0.873 is close to 1
✓ 1
Part (b)
For each 1 °C increase in maximum daily temperature, the water supplied increases by about 1.84 megalitres
✓ 1
Part (c)
Unreliable, because 38 °C lies outside the range of temperatures in the data (14–31 °C), so this would be extrapolation
✓ 1
Part (d)
M1: substitutes T = 25 into the line, W = 1.84 × 25 + 23.6
✓ 1
A1: W = 69.6 megalitres
✓ 1
Exam tip: When interpreting a gradient, name both variables and their units, and say 'for each 1 unit increase in x'. A bare number with no context does not earn the mark.
18IBMI-033-01-r1-02Correlation and regressionPaper 1Easy5 marks
A gardener investigates how the mass of fertiliser applied, x grams per plant, affects the yield of tomatoes, y kg per row. Five rows were tested.
The regression line of y on x for these data is \(y = 0.6x + 4.44\).
(a)
Find \(\bar{x}\), the mean mass of fertiliser applied.
(1)
(b)
Using the fact that the regression line passes through the point \((\bar{x}, \bar{y})\), find \(\bar{y}\).
(2)
(c)
Hence find the value of k.
(2)
Show mark scheme
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Mark
Notes
Part (a)
A1: \(\bar{x}\) = 90 ÷ 5 = 18 g
✓ 1
Part (b)
M1: substitutes x = 18 into the line, 0.6 × 18 + 4.44
✓ 1
A1: \(\bar{y}\) = 15.24 kg
✓ 1
Part (c)
M1: total of y-values = 5 × 15.24 = 76.2, so 10.2 + 13.1 + k + 17.5 + 20.0 = 76.2
✓ 1
A1: k = 15.4
✓ 1
Exam tip: The least-squares regression line of y on x always passes through the mean point. This is a quick route to a missing mean or a missing data value.
19IBMI-033-06-r1-01Correlation and regressionPaper 1Easy5 marks
A teacher records the number of hours, x, that each of 15 students spent revising for a chemistry test and their test score, y, out of 100. The revision times range from 1 hour to 9 hours. The Pearson's product-moment correlation coefficient is r = 0.87. The equation of the regression line of y on x is y = 4.2x + 31.5.
(a)
Describe the correlation between revision time and test score.
(1)
(b)
Interpret the value 4.2 in the context of the question.
(1)
(c)
Explain why it would not be appropriate to use the regression line to estimate the score of a student who revised for 20 hours.
(1)
(d)
Use the regression line to estimate the score of a student who revised for 6.5 hours.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Strong positive (linear) correlation
✓ 1
Part (b)
Each additional hour of revision is associated with an increase of about 4.2 marks in the test score
✓ 1
Part (c)
20 hours is outside the range of the data (1–9 hours), so this would be extrapolation and the estimate is unreliable
✓ 1
Part (d)
Substitutes: y = 4.2 × 6.5 + 31.5
✓ 1
y = 58.8
✓ 1
Exam tip: When interpreting a gradient, always give both the size of the change and the units of both variables, e.g. 'marks per hour of revision'.
20IBMI-033-06-r1-02Correlation and regressionPaper 1Easy6 marks
A café owner records the maximum daily temperature, T °C, and the number of iced drinks sold, N, on six days.
Find the value of Pearson's product-moment correlation coefficient, r, for these data.
(2)
(b)
Write down the equation of the regression line of N on T, in the form N = aT + b.
(2)
(c)
Use your equation to estimate the number of iced drinks sold on a day when the maximum temperature is 27 °C.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Correct use of GDC with the data entered as two lists
✓ 1
r = 0.998
✓ 1
Part (b)
a = 2.95
✓ 1
b = −1.36, so N = 2.95T − 1.36
✓ 1
Part (c)
Substitutes T = 27: N = 2.9545… × 27 − 1.3636…
✓ 1
N = 78.4, so about 78 drinks
✓ 1
Common error: Using the rounded coefficients 2.95 and −1.36 gives 78.29, not 78.4. Store the full GDC values and use those when making a prediction.
21IBMI-033-01-r2-02Correlation and regressionPaper 1Easy7 marks
A food blogger records the price of a cappuccino, P (in dollars), and the average customer rating, R (out of 10), at six coffee shops.
Shop: A, B, C, D, E, F P: 3.20, 4.10, 3.80, 2.90, 4.50, 3.50 R: 7.8, 8.6, 8.1, 6.9, 8.4, 7.2
(a)
Calculate Spearman's rank correlation coefficient, rs, for these data.
(3)
(b)
Interpret your answer to part (a) in context.
(1)
(c)
The owner of shop E says that the result proves that raising prices will improve customer ratings. Comment on this claim.
(1)
(d)
A seventh shop, G, is added. It charges $5.00 and has an average rating of 8.9. Find the new value of rs.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Correct ranks, e.g. P: 2, 5, 4, 1, 6, 3 and R: 3, 6, 4, 1, 5, 2 (A1)
✓ 1
Σd² = 4, or Pearson's r applied to the ranks (M1)
✓ 1
rs = 0.886 (A1)
✓ 1
Part (b)
Strong positive association: shops with higher cappuccino prices tend to have higher customer ratings (R1)
✓ 1
Part (c)
Correlation does not imply causation; another factor such as coffee quality or location could affect both price and rating (R1)
✓ 1
Part (d)
G ranks 7th for both variables, so d = 0 and Σd² stays 4 with n = 7, or uses the GDC on the new ranks (M1)
✓ 1
rs = 0.929 (A1)
✓ 1
Common error: Rank both variables in the same direction, either both smallest-first or both largest-first. Mixing directions reverses the sign of rs.
22IBMI-033-06-r2-01Correlation and regressionPaper 1Easy6 marks
A kiosk owner records the maximum daily temperature, x °C, and the number of iced drinks sold, y, on 20 summer days. The recorded temperatures range from 14 °C to 31 °C. The regression line of y on x is \(y = 12.4x - 38.6\) and Pearson's product-moment correlation coefficient is r = 0.912.
(a)
Describe the correlation between temperature and iced drink sales.
(1)
(b)
Interpret the value 12.4 in the context of the question.
(1)
(c)
Use the regression equation to estimate the number of iced drinks sold on a day when the maximum temperature is 25 °C.
(2)
(d)
The owner wants to use the equation to estimate sales on a day when the maximum temperature is 40 °C. State, with a reason, whether this estimate would be reliable.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Strong positive correlation
✓ 1
Part (b)
For each 1 °C rise in maximum temperature, about 12.4 more iced drinks are sold on average
✓ 1
Part (c)
M1: substitutes 25 into the equation: 12.4(25) − 38.6
✓ 1
A1: 271.4, so about 271 drinks
✓ 1
Part (d)
Not reliable
✓ 1
R1: 40 °C lies outside the data range of 14 °C to 31 °C, so this is extrapolation and the linear trend may not continue
✓ 1
Exam tip: When you interpret a gradient, give the rate of change and the units of both variables, for example 'about 12.4 more drinks per 1 °C'.
23IBMI-033-06-r2-02Correlation and regressionPaper 1Easy7 marks
A teacher records the number of hours, x, that six students spent revising for a test and their test scores, y.
x: 2, 4, 5, 7, 8, 10 y: 41, 50, 58, 63, 70, 79
(a)
Find Pearson's product-moment correlation coefficient, r.
(2)
(b)
Interpret your value of r in context.
(1)
(c)
Find the equation of the regression line of y on x.
(2)
(d)
Use your regression equation to estimate the test score of a student who revised for 6.5 hours.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: correct data entered into the GDC, or a valid formula attempt
✓ 1
A1: r = 0.994
✓ 1
Part (b)
Very strong positive linear correlation: students who revised for longer tended to score higher
✓ 1
Part (c)
A1: gradient 4.69
✓ 1
A1: y = 4.69x + 32.0
✓ 1
Part (d)
M1: substitutes 6.5 into their equation
✓ 1
A1: 62.5
✓ 1
Exam tip: Substitute the unrounded gradient and intercept stored in your GDC. If you use rounded values, your final answer may differ in the third significant figure.
24IBMI-033-03-r1-01Normal distribution and chi-squared testsPaper 1Easy6 marks
A roastery fills bags of coffee. The mass, M grams, of coffee in a bag is normally distributed with mean 502 g and standard deviation 4.5 g.
(a)
Find the probability that a randomly chosen bag contains less than 495 g of coffee.
(2)
(b)
The roastery fills 600 bags in one shift. Find the expected number of these bags that contain less than 495 g.
(2)
(c)
90% of bags contain more than k grams. Find the value of k.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: uses M ~ N(502, 4.52) to find P(M < 495), e.g. with lower bound −∞ and upper bound 495 on the GDC
✓ 1
A1: 0.0599
✓ 1
Part (b)
M1: 600 × 0.0599…
✓ 1
A1: 35.9 bags (accept 36)
✓ 1
Part (c)
M1: uses inverse normal with lower-tail area 0.1, i.e. P(M < k) = 0.1
✓ 1
A1: k = 496 g (496.23…)
✓ 1
Common error: In part (c), '90% exceed k' means a lower-tail area of 0.1. Entering 0.9 into inverse normal gives 508 g, which is on the wrong side of the mean.
25IBMI-033-03-r1-02Normal distribution and chi-squared testsPaper 1Easy7 marks
A gym surveys 200 members about their preferred class. The results are shown below.
A χ2 test for independence is carried out at the 5% significance level.
(a)
State the null and alternative hypotheses.
(1)
(b)
Show that the expected frequency of members under 30 who prefer Yoga is 30.
(1)
(c)
Write down the number of degrees of freedom for this test.
(1)
(d)
Find the χ2 statistic and the p-value for this test.
(2)
(e)
State the conclusion of the test, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1: H0: preferred class is independent of age group; H1: preferred class is not independent of age group
✓ 1
Part (b)
A1: \(\frac{100 \times 60}{200} = 30\)
✓ 1
Part (c)
A1: (2 − 1)(3 − 1) = 2
✓ 1
Part (d)
A1: χ2 = 6.21 (6.2095…)
✓ 1
A1: p = 0.0448
✓ 1
Part (e)
R1: 0.0448 < 0.05
✓ 1
A1: reject H0; there is sufficient evidence that preferred class depends on age group
✓ 1
Exam tip: Give the reason (comparing the p-value with the significance level) and then state the conclusion in the context of the question. The R mark depends on the comparison being shown.
26IBMI-033-05-r2-01Sampling and descriptive statisticsPaper 1Easy4 marks
A secondary school has 1200 students: 480 in Year 1, 420 in Year 2 and 300 in Year 3. The head teacher wants to find out how long students take to travel to school. She plans to collect data from a sample of 60 students.
(a)
State whether the time taken to travel to school is discrete or continuous data.
(1)
(b)
A teacher suggests surveying the first 60 students to arrive at school on Monday morning. Give one reason why a stratified sample by year group, with students chosen at random within each year group, would be better.
(1)
(c)
A stratified sample of 60 students is taken, stratified by year group. Find the number of Year 3 students in the sample.
(2)
Show mark scheme
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Mark
Notes
Part (a)
Continuous A1
✓ 1
Part (b)
Any one valid reason that says why the first-60 method is unrepresentative and how the stratified sample improves on it. Accept either of these:
—
• The first 60 arrivals might come mostly from one or two year groups, whereas a stratified sample includes each year group in proportion to its size.
—
• The first arrivals are a convenience sample and are likely to have unusual journeys (e.g. living close by or being driven), so they are biased; random selection within each year group avoids this bias.
—
R1
✓ 1
Part (c)
\(\frac{300}{1200}\times 60\) M1
✓ 1
15 students A1
✓ 1
Exam tip: In a stratified sample, each stratum's share of the sample equals its share of the population. Check that your numbers for all strata add up to the total sample size: here 24 + 21 + 15 = 60.
27IBMI-033-01-r2-01Correlation and regressionPaper 1Easy7 marks
A website lists used cars of one model. For a sample of 20 cars aged between 1 and 9 years, the age, x years, and the advertised price, y US dollars, were recorded. The regression line of y on x is \(y = -1850x + 21400\) and the Pearson product-moment correlation coefficient is \(r = -0.912\).
(a)
Interpret, in context, the value −1850 in the regression equation.
(1)
(b)
Interpret the value 21400 in context, and state why this interpretation may not be reliable.
(2)
(c)
Use the regression line to estimate the price of a 6-year-old car of this model.
(2)
(d)
State, with two reasons, whether your estimate from the previous part is likely to be reliable.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
For each additional year of age, the advertised price is predicted to fall by about 1850 US dollars
✓ 1
Part (b)
It is the predicted advertised price (21400 US dollars) of a car aged 0 years, i.e. a new car
✓ 1
Age 0 lies outside the data range of 1 to 9 years, so this is extrapolation
✓ 1
Part (c)
Substitutes x = 6: y = −1850(6) + 21400
✓ 1
y = 10300, so the estimated price is 10 300 US dollars
✓ 1
Part (d)
Accept either reason, in any order; the estimate is likely to be reliable because:
—
r = −0.912 shows a strong (negative) linear correlation
✓ 1
x = 6 lies within the data range of 1 to 9 years, so this is interpolation
✓ 1
Exam tip: To judge whether a regression estimate is reliable, check two things: how strong the correlation is, and whether the x-value lies inside the range of the data.
28IBMI-033-05-r1-01Sampling and descriptive statisticsPaper 1Easy5 marks
A secondary school has 1200 students: 300 in Year 10, 420 in Year 11 and 480 in Year 12. The head of student welfare wants to survey students about the length of the lunch break.
(a)
One teacher suggests taking an alphabetical list of all 1200 students, choosing a random starting point among the first 15 names, and then selecting every 15th name. State the name of this sampling method.
(1)
(b)
Instead, a stratified sample of 80 students is chosen, proportional to the size of each year group. Calculate the number of Year 11 students in the sample.
(2)
(c)
A third teacher proposes asking the first 80 students who walk into the canteen at lunchtime. State the name of this sampling method and give one reason why it may give biased results in this context.
R1: Valid reason in context, e.g. students who skip the canteen (who bring packed lunches or leave early) are excluded, so their views on the lunch break are not represented / one year group may arrive first and be over-represented
✓ 1
Exam tip: For proportional stratified sampling, check that your stratum sizes add up to the total sample: here 20 + 28 + 32 = 80.
29IBMI-033-04-r1-01T-tests, confidence intervals and Markov chainsPaper 1Easy5 marks
A sports scientist wants to know whether regular video-game players have a different mean reaction time from non-players. She records the reaction times, in milliseconds, of a random sample of 12 regular players and an independent random sample of 10 non-players.
Sample mean for players: 248 ms Sample mean for non-players: 271 ms
She carries out a two-tailed two-sample t-test at the 5% significance level. Her GDC gives a p-value of 0.0232.
(a)
State the null and alternative hypotheses for this test.
(2)
(b)
State, with a reason, the conclusion of the test in context.
(2)
(c)
State one assumption that is needed for this test to be valid.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1: H0: the population mean reaction time of regular players equals the population mean reaction time of non-players, \(\mu_P = \mu_N\)
✓ 1
A1: H1: the population mean reaction times are different, \(\mu_P \neq \mu_N\)
✓ 1
Part (b)
R1: p-value 0.0232 < 0.05
✓ 1
A1: reject H0; there is sufficient evidence at the 5% level that the mean reaction times of regular players and non-players are different
✓ 1
Part (c)
A1: any one of: the reaction times in both populations are normally distributed; the two populations have equal variances (for a pooled test); the two samples are independent
✓ 1
Exam tip: Write hypotheses in terms of population means and say in words what the parameters represent. A conclusion of just 'reject H₀' does not earn the context mark; say what it means for reaction times.
30IBMI-033-04-r2-01T-tests, confidence intervals and Markov chainsPaper 1Easy7 marks
A market gardener grows tomato plants in two greenhouses. Fertiliser A is used in the first greenhouse and fertiliser B in the second. She records the yield, in kg, of six randomly chosen plants from each greenhouse.
She wants to test, at the 5% significance level, whether plants given fertiliser B have a higher mean yield than plants given fertiliser A.
(a)
State suitable null and alternative hypotheses.
(2)
(b)
Use a pooled two-sample t-test to find the p-value for this test.
(2)
(c)
State the conclusion of the test, in context, giving a reason.
(2)
(d)
State one assumption needed for this test to be valid.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
H0: \(\mu_A=\mu_B\), where \(\mu\) is the population mean yield per plant
✓ 1
H1: \(\mu_B>\mu_A\)
✓ 1
Part (b)
Attempt at a one-tailed pooled two-sample t-test using the data (for reference, t = 3.28 with 10 degrees of freedom)
✓ 1
p = 0.00412 (accept 0.0041)
✓ 1
Part (c)
0.00412 < 0.05
✓ 1
Reject H0. There is significant evidence that the mean yield of plants given fertiliser B is higher than that of plants given fertiliser A
✓ 1
Part (d)
Any one of the following earns the mark: the yields for each fertiliser are normally distributed; OR the two populations have equal variances; OR the two samples are random and independent of each other
✓ 1
Exam tip: The direction of H₁ comes from the question wording ('higher'), not from the data. A one-tailed test needs the one-tailed p-value from the GDC.
31IBMI-033-08-r1-02Normal distribution and chi-squared testsPaper 1Easy7 marks
A café owner records the drink ordered by 180 customers, together with their age group. The results are shown below.
Under 30: coffee 38, tea 12, juice 30 (total 80) 30 and over: coffee 52, tea 28, juice 20 (total 100)
A \(\chi^2\) test for independence is carried out at the 5% significance level.
(a)
State the null hypothesis for this test.
(1)
(b)
Show that the expected frequency for customers under 30 ordering tea is 17.8, correct to 3 significant figures.
(1)
(c)
Write down the number of degrees of freedom.
(1)
(d)
Find the \(\chi^2\) statistic and the p-value.
(2)
(e)
State the conclusion of the test, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
H0: the drink ordered is independent of the customer's age group
✓ 1
Part (b)
Tea total = 12 + 28 = 40, so expected = \(\frac{80 \times 40}{180} = 17.78 \approx 17.8\)
Conclusion: there is sufficient evidence that the drink ordered depends on age group
✓ 1
The conclusion mark requires a correct reason.
—
Exam tip: Always write the conclusion in context: say which variables appear to be related, not just 'reject H₀'. Quote the p-value and the significance level in the reason.
32IBMI-033-05-r2-02Sampling and descriptive statisticsPaper 1Easy6 marks
A takeaway restaurant records the delivery time, t minutes, for 40 orders.
10 ≤ t < 20: 8 orders 20 ≤ t < 30: 15 orders 30 ≤ t < 40: 12 orders 40 ≤ t < 50: 5 orders
(a)
Write down the modal class.
(1)
(b)
Find an estimate of the mean delivery time.
(2)
(c)
Find an estimate of the standard deviation of the delivery times.
(2)
(d)
Explain why your values for the mean and the standard deviation are only estimates.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
20 ≤ t < 30 A1
✓ 1
Part (b)
Use of midpoints 15, 25, 35, 45, e.g. \(\frac{8(15)+15(25)+12(35)+5(45)}{40}\) M1
✓ 1
28.5 minutes A1
✓ 1
Part (c)
Midpoints with frequencies entered into GDC, or \(\sqrt{\frac{36000}{40}-28.5^2}\) M1
✓ 1
9.37 minutes A1
✓ 1
Part (d)
The actual delivery times are not known (data are grouped); each time is assumed to be at the midpoint of its class R1
✓ 1
Common error: Do not use the class widths or upper boundaries as the x-values. Grouped data calculations always use the class midpoints.
33IBMI-033-08-r1-01Normal distribution and chi-squared testsPaper 1Easy6 marks
The masses of apples from an orchard are modelled by a normal distribution with mean 152 g and standard deviation 9 g.
(a)
Find the probability that a randomly chosen apple has a mass greater than 165 g.
(2)
(b)
A crate contains 200 of these apples. Find the expected number of apples in the crate with a mass less than 140 g.
(2)
(c)
Find the mass that is exceeded by 10% of the apples.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Correct set-up, e.g. P(X > 165) with X ~ N(152, 92), or z = 13/9 = 1.444
✓ 1
0.0743 (accept 0.07429)
✓ 1
Part (b)
P(X < 140) = 0.0912
✓ 1
200 × 0.0912 = 18.2 apples (accept 18.24)
✓ 1
Part (c)
Use of inverse normal with area 0.9 to the left (or 0.1 to the right), e.g. z = 1.2816
✓ 1
m = 152 + 1.2816 × 9 = 164 g (accept 163.5 or 163.53 g)
✓ 1
Common error: In part (c), 10% of the apples exceed the mass, so the area to the left is 0.9. Entering 0.1 into the inverse normal function with a left tail gives about 140 g. That mass is below the mean, which should look wrong.
34IBMI3S-4-07-01Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 1Easy5 marks
A hospital employs 480 nurses, 300 doctors and 220 technicians. A manager wants to survey a stratified sample of 50 staff about shift patterns.
(a)
Show that the sample should contain 24 nurses.
(2)
(b)
Find the number of doctors and the number of technicians in the sample.
(2)
(c)
Give one reason why stratified sampling is more suitable than simple random sampling here.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to find the proportion of nurses, \(\frac{480}{1000}\times 50\)
✓ 1
A1 24 (shown, with the 1000 total seen)
✓ 1
Part (b)
A1 doctors: 15
✓ 1
A1 technicians: 11
✓ 1
Part (c)
R1 it guarantees every staff group is represented in proportion to its size (a simple random sample might miss or under-represent a group)
✓ 1
Common error: Round only at the end; if the proportions do not give whole numbers, the sample sizes must still add up to the required total.
35IBMI3S-4-07-02Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 1Easy6 marks
A courier company records the delivery time, in minutes, for twelve parcels: 18, 22, 25, 27, 29, 31, 33, 34, 38, 41, 44, 67
(a)
Find the median and the interquartile range of the delivery times.
(3)
(b)
A delivery time is an outlier if it is more than 1.5 × IQR above the upper quartile. Determine whether 67 minutes is an outlier.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 median = 32
✓ 1
A1 lower quartile 26 and upper quartile 39.5
✓ 1
A1 IQR = 13.5
✓ 1
Part (b)
M1 attempt to find \(Q_3+1.5\times\text{IQR}\)
✓ 1
A1 \(39.5+1.5\times13.5=59.75\)
✓ 1
R1 67 > 59.75, so 67 minutes is an outlier
✓ 1
Exam tip: When a question tells you the outlier rule, show the boundary value you calculate and compare it explicitly with the data value.
36IBMI3S-4-07-03Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 1Easy7 marks
A wind-farm engineer records the average wind speed, v m s−1, and the power output, P kW, of a turbine on six days.
Find the Pearson's product-moment correlation coefficient, r.
(1)
(b)
Find the equation of the regression line of P on v.
(2)
(c)
Interpret the gradient of the line in this context.
(1)
(d)
Estimate the power output when the wind speed is 9 m s−1.
(2)
(e)
Explain why the equation should not be used to estimate the power output at 30 m s−1.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 r = 0.996
✓ 1
Part (b)
A1 gradient 9.80
✓ 1
A1 \(P=9.8v-18.5\)
✓ 1
Part (c)
A1 for each extra 1 m s−1 of wind speed the power output increases by about 9.8 kW
✓ 1
Part (d)
M1 substitute v = 9 into the regression equation
✓ 1
A1 69.7 kW
✓ 1
Part (e)
R1 30 m s−1 is far outside the data range (extrapolation), so the linear model may not hold
✓ 1
Common error: Commenting that a strong correlation proves a model works everywhere is not enough; always refer to whether the value lies inside the range of the data.
37IBMI3S-4-07-04Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 1Easy5 marks
A company packs phone chargers. Each charger is independently faulty with probability 0.04. A box contains 25 chargers.
(a)
Write down the expected number of faulty chargers in a box.
(1)
(b)
Find the probability that a box contains no faulty chargers.
(2)
(c)
Find the probability that a box contains at least two faulty chargers.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 1 (or 1.00)
✓ 1
Part (b)
M1 \(X\sim B(25,0.04)\) and attempt to find \(P(X=0)\), e.g. \(0.96^{25}\)
✓ 1
A1 0.360
✓ 1
Part (c)
M1 \(1-P(X\le1)\)
✓ 1
A1 0.264
✓ 1
Exam tip: 'At least two' means \(X\ge2\), so use the complement \(1-P(X\le1)\) rather than adding many terms.
38IBMI3S-4-07b-01Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 1Easy5 marks
A survey telescope has stored 1200 images, classified as 540 galaxies, 420 nebulae and 240 star clusters. An astronomer wants a stratified sample of 60 images, to be checked by hand for processing errors.
(a)
Calculate how many images of each type should be in the sample.
(3)
(b)
State one advantage of stratified sampling over simple random sampling for this task.
(1)
(c)
Explain why the astronomer cannot simply check the first 60 images stored.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use proportion (stratum size/1200) × 60
✓ 1
A1 27 galaxies and 21 nebulae
✓ 1
A1 12 star clusters
✓ 1
Part (b)
R1 every type of image is guaranteed to be represented in proportion to its frequency
✓ 1
Part (c)
R1 the first images may come from one night or one type, so the sample would not be representative
✓ 1
Exam tip: In stratified sampling each sample size is (stratum size ÷ population) × total sample, and the sizes must add up to the sample total.
39IBMI3S-4-07b-02Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 1Easy7 marks
A sports scientist records the time, in seconds, taken by ten athletes to run a 30 m flying sprint: 3.42, 3.51, 3.38, 3.47, 3.55, 3.44, 3.49, 3.40, 3.46, 3.94
(a)
Find the mean and the sample standard deviation of the times.
(2)
(b)
Find the median and the interquartile range.
(2)
(c)
Use the 1.5 × IQR rule to determine whether the slowest time is an outlier.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 mean = 3.51 s
✓ 1
A1 standard deviation = 0.161 s
✓ 1
Part (b)
A1 median = 3.465 s
✓ 1
A1 Q1 = 3.42, Q3 = 3.51, so IQR = 0.09 s
✓ 1
Part (c)
M1 upper fence = Q3 + 1.5 × IQR
✓ 1
A1 3.51 + 1.5(0.09) = 3.645
✓ 1
R1 3.94 is greater than 3.645, so it is an outlier
✓ 1
Common error: The upper fence uses Q₃ plus 1.5 × IQR, not the mean plus 1.5 × IQR.
40IBMI3S-4-07b-03Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 1Easy6 marks
A study of club cyclists compares the weekly training time, x hours, with the maximal oxygen uptake, y mL kg−1 min−1. For the 25 cyclists studied, the product-moment correlation coefficient was r = 0.91 and the regression line of y on x was y = 1.8x + 38.2
(a)
Describe the correlation between training time and oxygen uptake.
(1)
(b)
Interpret the value 1.8 in this context.
(2)
(c)
Estimate the oxygen uptake of a cyclist who trains for 12 hours each week.
(2)
(d)
A coach uses the line to predict the oxygen uptake of a person who trains 40 hours each week. Comment on the reliability of this prediction.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 strong positive (linear) correlation
✓ 1
Part (b)
A1 each extra hour of weekly training is associated with
✓ 1
R1 an average increase of 1.8 mL kg−1 min−1 in oxygen uptake
✓ 1
Part (c)
M1 substitution x = 12 into the equation
✓ 1
A1 59.8 mL kg−1 min−1
✓ 1
Part (d)
R1 40 hours is probably outside the data range, so this is extrapolation and unreliable
✓ 1
Common error: Describing a correlation as causation: say 'is associated with', not 'causes'.
41IBMI3S-4-07b-04Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 1Easy7 marks
A tram on a city route arrives on time with probability 0.85 on any journey, independently of other journeys. Mira makes 12 journeys on this route in a month. Let X be the number of journeys on which the tram is on time.
(a)
State the distribution of X and find the expected number of journeys on which the tram is late.
(2)
(b)
Find P(X = 10).
(2)
(c)
Find the probability that the tram is on time for at least 11 of the 12 journeys.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 X ~ B(12, 0.85)
✓ 1
A1 12 × 0.15 = 1.8 journeys late
✓ 1
Part (b)
M1 attempt to use the binomial probability formula or GDC
✓ 1
A1 0.292
✓ 1
Part (c)
M1 P(X ≥ 11) = P(11) + P(12)
✓ 1
A1 0.301 + 0.142
✓ 1
A1 0.443
✓ 1
Exam tip: 'At least 11' means 11 or 12, so add two terms, or use 1 − P(X ≤ 10).
42IBMI3S-4-01r7-01Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy5 marks
The mass of sulfur dioxide released by a volcano each day is modelled by a normal distribution with mean 420 tonnes and standard deviation 65 tonnes.
(a)
Find the probability that more than 500 tonnes are released on a given day.
(2)
(b)
On 10% of days, the mass released exceeds k tonnes. Find k, correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to find P(X > 500) using N(420, 652)
✓ 1
A1 0.109
✓ 1
Part (b)
M1 use of inverse normal with upper tail 0.1, giving z = 1.2816
✓ 1
M1 k = 420 + 1.2816 × 65
✓ 1
A1 k = 503 tonnes
✓ 1
Exam tip: Read whether the question asks for an upper or lower tail before using the inverse normal function.
43IBMI3S-4-01r7-02Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy4 marks
A volcano observatory has 240 seismic sensors in four regions: A (96 sensors), B (72), C (48) and D (24). The observatory selects a stratified sample of 60 sensors for a maintenance check, with strata based on region.
(a)
Calculate the number of sensors chosen from region B.
(2)
(b)
A technician suggests a simple random sample of 60 sensors instead. Explain one reason why this could give a less representative sample than the stratified sample.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 72⁄240 × 60
✓ 1
A1 18
✓ 1
Part (b)
M1 simple random sampling does not guarantee sensors from every region
✓ 1
R1 region D, being small, could have no sensors chosen (or regions could be over- or under-represented)
✓ 1
Common error: In stratified sampling, the sample from each stratum is proportional to the size of that stratum.
44IBMI3S-4-01r7-03Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy5 marks
A theatre sells 20 tickets for a studio performance. Each ticket holder independently fails to attend with probability 0.12. Let X be the number of ticket holders who fail to attend.
(a)
State the distribution of X, and find P(X = 3), correct to 3 significant figures.
(3)
(b)
Find the probability that at least two ticket holders fail to attend, correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 X ~ B(20, 0.12)
✓ 1
M1 P(X = 3) = C(20,3) × 0.123 × 0.8817
✓ 1
A1 0.224
✓ 1
Part (b)
M1 P(X ≥ 2) = 1 − P(X ≤ 1)
✓ 1
A1 0.711
✓ 1
Exam tip: “At least two” is easiest as 1 minus P(X ≤ 1) rather than adding many terms.
45IBMI3S-4-01r7-04Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy6 marks
The times, in hours, taken to unload 12 cargo ships at a port are listed in order. 6.2, 6.8, 7.1, 7.4, 7.9, 8.3, 8.8, 9.1, 9.6, 10.0, 10.9, 14.5 Take the lower quartile as the median of the lowest six values and the upper quartile as the median of the highest six values.
(a)
Find the median, the lower quartile and the upper quartile.
(3)
(b)
A time is an outlier if it is more than 1.5 × IQR above the upper quartile. Show whether 14.5 hours is an outlier.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 median = 8.55
✓ 1
A1 lower quartile = 7.25
✓ 1
A1 upper quartile = 9.8
✓ 1
Part (b)
A1 IQR = 2.55
✓ 1
M1 upper fence = 9.8 + 1.5 × 2.55
✓ 1
A1 fence = 13.625 and 14.5 > 13.625, so 14.5 is an outlier
✓ 1
Exam tip: Write out the outlier boundary as a number before comparing it with the largest value.
46IBMI3S-4-01r14-01Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 1Easy4 marks
A health clinic has 2400 registered patients: 480 are aged under 18, 1320 are aged 18 to 64 and 600 are aged 65 or over. The clinic plans a survey about winter vaccination and takes a stratified sample of 120 patients, with each age group represented in proportion to its size.
(a)
Give one reason why a stratified sample is more suitable here than a simple random sample.
(1)
(b)
Calculate the number of patients aged 18 to 64 in the sample.
(2)
(c)
Find the number of patients aged 65 or over in the sample.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 e.g. it guarantees every age group is represented / the sample reflects the population structure / vaccination uptake is likely to differ between age groups
✓ 1
Part (b)
M1 \(\frac{1320}{2400}\times 120\) or sampling fraction \(\frac{120}{2400}=\frac{1}{20}\) seen
✓ 1
A1 66
✓ 1
Part (c)
A1 \(\frac{600}{2400}\times120=30\)
✓ 1
Common error: Do not divide the sample size by the number of groups: proportional stratification uses each group's share of the population.
47IBMI3S-4-01r14-02Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 1Easy6 marks
The energy output of a wind turbine (in MWh) was recorded on each of 12 days: 17.0, 25.6, 13.8, 15.9, 12.4, 18.3, 14.5, 8.1, 16.4, 13.0, 17.8, 15.1
(a)
Find the median daily output.
(2)
(b)
Find the lower quartile, the upper quartile and the interquartile range.
(2)
(c)
A value is an outlier if it is more than 1.5 × IQR above the upper quartile or below the lower quartile. Determine whether 25.6 and 8.1 are outliers.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 ordered data / average of 6th and 7th values: \(\frac{15.1+15.9}{2}\)
✓ 1
A1 15.5 MWh
✓ 1
Part (b)
A1 \(Q_1=13.4\) and \(Q_3=17.4\)
✓ 1
A1 IQR = 4.0
✓ 1
Part (c)
M1 attempt at fences: \(Q_3+1.5\times IQR=23.4\) and \(Q_1-1.5\times IQR=7.4\)
✓ 1
A1 25.6 > 23.4 so 25.6 is an outlier; 8.1 > 7.4 so 8.1 is not an outlier
✓ 1
Exam tip: Write the fences as numbers before comparing; a clear comparison for each value is what earns the final mark.
48IBMI3S-4-01r14-03Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 1Easy6 marks
At a supermarket, the waiting time T minutes of a customer at the checkout is modelled by a normal distribution with mean 4.2 minutes and standard deviation 1.1 minutes.
(a)
Find the probability that a customer waits more than 6 minutes.
(2)
(b)
Find the probability that a customer waits between 3 and 5 minutes.
(2)
(c)
On a busy day, 300 customers use the checkout. Using the unrounded probability from part (a), find the expected number of customers who wait more than 6 minutes, to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 normal cdf with correct parameters / \(z=\frac{6-4.2}{1.1}\) used
✓ 1
A1 0.0509 (accept 0.0509)
✓ 1
Part (b)
M1 \(P(3
✓ 1
A1 0.629
✓ 1
Part (c)
M1 \(300\times P(T>6)\)
✓ 1
A1 15.3
✓ 1
Common error: Check that the probability asked for is the upper tail: P(T > 6) is not the value the calculator gives for P(T < 6).
49IBMI3S-4-01r14-06Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 1Easy5 marks
An online shop finds that each order is returned with probability 0.12. In one hour the shop receives 25 orders. Let X be the number of these orders that are returned; assume X ~ B(25, 0.12).
(a)
State one assumption needed for this binomial model.
(1)
(b)
Find the probability that at least 5 orders are returned.
(2)
(c)
Find the probability that exactly 3 orders are returned.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 orders are returned independently of each other / the probability of return is constant for every order
✓ 1
Part (b)
M1 \(1-P(X\le4)\) or \(P(X\ge5)\) attempted
✓ 1
A1 0.173
✓ 1
Part (c)
M1 \(\binom{25}{3}(0.12)^3(0.88)^{22}\) or binomial pdf with correct parameters
✓ 1
A1 0.239
✓ 1
Exam tip: "At least 5" means 5 or more, so use 1 − P(X ≤ 4), not 1 − P(X ≤ 5).
50IBMI3S-4-01r15-01Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy5 marks
A fish farm keeps its stock in four ponds, A, B, C and D, which hold 240, 360, 180 and 120 trout respectively. The farm manager wants a sample of 75 trout to weigh. She decides to take a sample that has the same proportion of trout from each pond.
(a)
State the name of this type of sampling.
(1)
(b)
Calculate the number of trout that should be taken from each pond.
(3)
(c)
Give one reason why this method is likely to give a more representative sample than a simple random sample taken from all 900 trout.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 stratified sampling
✓ 1
Part (b)
M1 uses the sampling fraction 75/900 (or 1/12)
✓ 1
A1 A: 20 and B: 30
✓ 1
A1 C: 15 and D: 10
✓ 1
Part (c)
R1 every pond is guaranteed to be represented in proportion to its size, so the sample cannot by chance over-represent one pond
✓ 1
Common error: Sample sizes from the strata must be whole numbers that add up to the total, so check that your values sum to the overall sample size.
51IBMI3S-4-01r15-02Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy7 marks
A robotics club times how long its robot arm takes to sort a tray of components. The times, in seconds, for 12 trials are 14.2, 15.1, 15.6, 15.9, 16.3, 16.4, 16.8, 17.0, 17.5, 17.9, 18.4, 24.6. In this question, take the lower quartile and upper quartile to be the medians of the lower six and upper six values respectively.
(a)
Use your calculator to find the mean and the standard deviation of these times, treating the 12 trials as the whole population.
(2)
(b)
Find the median and the interquartile range.
(2)
(c)
An observation is classed as an outlier if it is more than 1.5 × IQR above the upper quartile. Show that the time 24.6 s is an outlier.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 mean = 17.1 s
✓ 1
A1 standard deviation = 2.52 s
✓ 1
Part (b)
A1 median = 16.6 s
✓ 1
A1 Q1 = 15.75, Q3 = 17.7, so IQR = 1.95 s
✓ 1
Part (c)
M1 attempts upper boundary = Q3 + 1.5 × IQR
✓ 1
A1 upper boundary = 20.625 s
✓ 1
R1 24.6 > 20.625, so 24.6 is an outlier
✓ 1
Exam tip: State which quartile rule you are using when the rule is not given; here the lower and upper halves each contain exactly six values, so there is no ambiguity.
52IBMI3S-4-01r15-08Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy6 marks
A research vessel lands 120 fish in one day. The table shows how each fish was caught and its species.
Cod
Haddock
Other
Trawl net
36
28
16
Line
14
12
14
One fish is selected at random from the 120.
(a)
Find the probability that the fish was caught by trawl net.
(1)
(b)
Find the probability that the fish is a cod that was caught on a line.
(1)
(c)
Given that the fish is a cod, find the probability that it was caught on a line.
(2)
(d)
Determine whether the events 'the fish was caught on a line' and 'the fish is a cod' are independent. Justify your answer.
R1 the probabilities are different, so the events are not independent
✓ 1
Exam tip: To test independence, compare P(A | B) with P(A), or P(A and B) with P(A) × P(B); one equality is enough to check.
53IBMI3S-4-01r15-13Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy4 marks
Two judges score eight school robots in a competition out of 10. The scores are
Robot
A
B
C
D
E
F
G
H
Judge 1
7.8
6.1
8.4
5.2
9.0
4.7
6.9
7.3
Judge 2
6.5
5.9
8.8
5.0
7.7
5.6
6.2
7.9
(a)
Write down the rank of each robot in the scores of Judge 1, giving rank 1 to the highest score.
(1)
(b)
Use your calculator to find Spearman's rank correlation coefficient, rs, between the two sets of scores.
(2)
(c)
Interpret your value of rs in context.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 ranks A to H: 3, 6, 2, 7, 1, 8, 5, 4
✓ 1
Part (b)
M1 ranks both sets (Judge 2: 4, 6, 1, 8, 3, 7, 5, 2) and finds the correlation of the ranks
✓ 1
A1 rs = 0.857
✓ 1
Part (c)
R1 strong positive association: the judges tend to rank the robots in a similar order
✓ 1
Exam tip: Spearman's coefficient is the Pearson coefficient of the ranks, so it measures how consistently the order agrees, not whether the relationship is a straight line.
54IBMI3S-4-01r16-01Sampling and descriptive statistics; T-tests, confidence intervals and Markov chains; Probability and the binomial distribution; Normal distribution and chi-squared testsPaper 1Easy5 marks
A coastal fishery holds 1680 fish in a holding tank: 840 trout, 560 perch and 280 pike. A biologist wants a stratified sample of 60 fish, with the number from each species proportional to the number of that species in the tank.
(a)
Calculate the number of fish of each species that should be in the sample.
(3)
(b)
Explain why stratified sampling is more suitable here than simple random sampling.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use (species number ÷ 1680) × 60
✓ 1
A1 trout 30 and perch 20
✓ 1
A1 pike 10
✓ 1
Part (b)
R1 guarantees every species is represented in the sample, including the smallest group (pike)
✓ 1
R1 a simple random sample could by chance contain very few or no pike, so would be less representative of the tank
✓ 1
Exam tip: In a proportional stratified sample, each stratum's sample size is (stratum size ÷ population size) × sample size. Check that your stratum samples add up to the total.
55IBMI3S-4-01r16-05Sampling and descriptive statistics; T-tests, confidence intervals and Markov chains; Probability and the binomial distribution; Normal distribution and chi-squared testsPaper 1Easy5 marks
A warehouse operator wants to know whether robot Model X has a longer mean battery life than robot Model Y. Battery lives are assumed to be normally distributed with equal variances. A one-tailed two-sample t-test is carried out and gives a p-value of 0.0312.
(a)
State the null and alternative hypotheses.
(2)
(b)
State the conclusion of the test at the 5% significance level. Give a reason for your answer.
(2)
(c)
State the conclusion if the test is carried out at the 1% significance level.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: the mean battery lives of Models X and Y are equal (μX = μY)
✓ 1
A1 H1: the mean battery life of Model X is greater than that of Model Y (μX > μY)
✓ 1
Part (b)
M1 compare 0.0312 with 0.05
✓ 1
A1 0.0312 < 0.05, so reject H0: there is evidence that Model X has a longer mean battery life
✓ 1
Part (c)
A1 0.0312 > 0.01, so do not reject H0: insufficient evidence that Model X has a longer mean battery life
✓ 1
Common error: Do not write that H₀ has been "proved" when it is not rejected. Say there is insufficient evidence to reject it, and state the conclusion in the context of the question.
56IBMI3S-4-01r16-08Sampling and descriptive statistics; T-tests, confidence intervals and Markov chains; Probability and the binomial distribution; Normal distribution and chi-squared testsPaper 1Easy6 marks
At a clothing shop, 45% of customers buy shoes (event S), 30% buy a bag (event B), and 60% buy at least one of these two items.
(a)
Find the probability that a customer buys both shoes and a bag.
(2)
(b)
Find the probability that a customer buys shoes, given that they buy a bag.
(2)
(c)
Determine whether S and B are independent.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(S ∪ B) = P(S) + P(B) − P(S ∩ B) with values substituted
✓ 1
A1 P(S ∩ B) = 0.15
✓ 1
Part (b)
M1 P(S | B) = 0.15 ÷ 0.30
✓ 1
A1 0.5 (ECF from (a))
✓ 1
Part (c)
M1 P(S) × P(B) = 0.45 × 0.30 = 0.135
✓ 1
R1 0.135 ≠ 0.15, so the events are not independent
✓ 1
Exam tip: To test independence, compare P(A ∩ B) with P(A) × P(B), or P(A | B) with P(A). Show the numbers you compare.
57IBMI3S-4-01r16-12Sampling and descriptive statistics; T-tests, confidence intervals and Markov chains; Probability and the binomial distribution; Normal distribution and chi-squared testsPaper 1Easy6 marks
The length, L metres, of a roll of fabric from a mill is normally distributed with mean 25.0 m and standard deviation 0.4 m. A roll shorter than 24.5 m is rejected.
(a)
Find the probability that a roll is rejected.
(2)
(b)
The mill guarantees that 10% of rolls are longer than a length d. Find d to 3 significant figures.
(2)
(c)
In a delivery of 500 rolls, find the expected number that are rejected, using the unrounded probability from (a). Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(L < 24.5) with correct normal distribution, or z = −1.25
✓ 1
A1 0.106
✓ 1
Part (b)
M1 inverse normal with area 0.9 (or z = 1.2816) and 25.0 + 1.2816 × 0.4
✓ 1
A1 d = 25.5 m
✓ 1
Part (c)
M1 500 × P(L < 24.5) = 500 × 0.10565
✓ 1
A1 52.8
✓ 1
Exam tip: Sketch the normal curve and shade the required area before using the calculator. For "exceeded by 10%" the area to the left is 0.9.
58IBMI3S-4-02r16-01Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regressionPaper 1Easy6 marks
The time, T hours, taken by a climber to reach a high camp on a mountain is normally distributed with mean 5.2 hours and standard deviation 0.6 hours.
(a)
Find the probability that a climber reaches the camp in less than 4.5 hours.
(2)
(b)
Find the probability that a climber takes between 4.5 and 6.0 hours.
(2)
(c)
On one day 150 climbers set out. Find the expected number who take more than 6.0 hours, using the unrounded probability. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(T < 4.5) with correct normal distribution (or z = −1.167)
✓ 1
A1 0.122
✓ 1
Part (b)
M1 P(4.5 < T < 6.0) attempted
✓ 1
A1 0.787
✓ 1
Part (c)
M1 150 × P(T > 6.0) = 150 × 0.09121
✓ 1
A1 13.7
✓ 1
Exam tip: Write down the distribution, for example T ~ N(5.2, 0.6²), before using your calculator. Note that the second parameter is the variance, so enter the standard deviation 0.6, not 0.36.
59IBMI3S-4-02r16-05Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regressionPaper 1Easy6 marks
A mountaineering club has 200 members, numbered 1 to 200 on its membership list. The secretary wants a sample of 25 members to complete a survey.
(a)
Calculate the sampling interval needed for a systematic sample of 25 members.
(2)
(b)
Describe how the secretary could select the systematic sample.
(2)
(c)
Instead, the secretary asks the first 25 members who arrive at the clubhouse. Name this sampling method and give one reason why the sample may be biased.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 200 ÷ 25
✓ 1
A1 8
✓ 1
Part (b)
A1 choose a starting member at random from members 1 to 8
✓ 1
A1 then select every 8th member on the list after that
✓ 1
Part (c)
A1 convenience sampling
✓ 1
R1 e.g. early arrivals may be more active or keen than other members, so they may not represent the whole club
✓ 1
Exam tip: For systematic sampling, the interval is the population size divided by the sample size. A random start within the first interval keeps the sample unbiased.
60IBMI3S-4-02r16-12Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regressionPaper 1Easy5 marks
A telecoms company compares the mean download speeds of two network types, using a two-sample t-test with pooled variance. The hypotheses are H0: μ1 = μ2 and H1: μ1 ≠ μ2. The test statistic is t = 2.31 with 18 degrees of freedom.
(a)
Explain why this is a two-tailed test.
(1)
(b)
Find the p-value for this test.
(2)
(c)
State the conclusion at the 5% significance level, in context.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 H1 states only that the means differ, not which is larger, so both tails are used
✓ 1
Part (b)
M1 2 × P(T > 2.31) with 18 degrees of freedom
✓ 1
A1 p = 0.0330
✓ 1
Part (c)
M1 compare 0.0330 with 0.05
✓ 1
A1 reject H0: evidence that the mean download speeds of the two network types differ
✓ 1
Exam tip: For a two-tailed test, the p-value is twice the area in one tail of the t-distribution beyond |t|. Compare it with the full significance level.
61IBMI3S-4-02r16-14Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regressionPaper 1Easy5 marks
A telecoms company records the number of subscribers, s thousand, to a new network at the end of each month, m, for the first 12 months (m = 1, 2, …, 12). The regression line of s on m is s = 12.4m + 85.2, and the correlation coefficient is r = 0.973.
(a)
Interpret the gradient 12.4 in this context.
(1)
(b)
Interpret the value of r.
(1)
(c)
Estimate the number of subscribers at the end of month 9, to 3 significant figures.
(2)
(d)
The company uses the line to predict the number of subscribers at the end of month 60. Comment on the reliability of this prediction.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 the number of subscribers increases by about 12.4 thousand per month
✓ 1
Part (b)
R1 strong positive linear correlation between the month and the number of subscribers
✓ 1
Part (c)
M1 12.4 × 9 + 85.2
✓ 1
A1 197 thousand
✓ 1
Part (d)
R1 month 60 is far outside the range of the data (extrapolation), so the prediction is unreliable
✓ 1
Common error: A high value of r shows a strong linear relationship within the data, but it does not justify predicting far outside the range of the data.
62IBMI3S-4-02r18-01Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared testsPaper 1Easy5 marks
A fisheries officer wants to survey 40 of the 400 registered fishing boats in a region. The boats are based at three harbours: harbour A has 120 boats, harbour B has 200 boats and harbour C has 80 boats.
(a)
The officer uses stratified sampling, with the harbours as the strata and the sample proportional to the number of boats at each harbour. Calculate the number of boats to be sampled from each harbour.
(3)
(b)
State one advantage of stratified sampling over taking a simple random sample of 40 boats from all 400.
(1)
(c)
The officer considers surveying only the first 40 boats to arrive at the nearest harbour. Give one reason why this would be a poor sampling method.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sample size = (boats in harbour ÷ 400) × 40
✓ 1
A1 harbour A: 12 and harbour B: 20
✓ 1
A1 harbour C: 8
✓ 1
Part (b)
R1 Each harbour is guaranteed to be represented in proportion to its size (so the sample is more representative of the whole population)
✓ 1
Part (c)
R1 It is convenience sampling: boats at one harbour (or those arriving early) may differ from the rest, so the sample is likely to be biased
✓ 1
Common error: In stratified sampling, each stratum's sample size is its share of the population multiplied by the total sample size, not an equal share for every stratum.
63IBMI3S-4-02r18-02Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared testsPaper 1Easy7 marks
A robotics team tests 12 rechargeable batteries and records the running time, in hours, of each battery before it goes flat: 4.8, 5.1, 5.3, 5.4, 5.6, 5.8, 5.9, 6.0, 6.2, 6.3, 6.5, 8.4
(a)
Find the median, the lower quartile and the upper quartile of the running times.
(3)
(b)
Show that the running time of 8.4 hours is an outlier, using the rule that an outlier is more than 1.5 × IQR above the upper quartile.
(2)
(c)
Calculate the mean running time and the population standard deviation, each to 3 significant figures.
Exam tip: State the numerical value of the fence (here 7.6) before making the comparison, so the reasoning mark is clear.
64IBMI3S-4-02r18-04Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared testsPaper 1Easy6 marks
On any given day, let A be the event that a fishing cooperative lands more than 500 kg of fish, and let B be the event that the sea is rough. It is known that P(A) = 0.45, P(B) = 0.30 and P(A ∩ B) = 0.12.
(a)
Find the probability that the cooperative lands more than 500 kg of fish or the sea is rough (or both).
(2)
(b)
Determine whether events A and B are independent. Justify your answer.
(2)
(c)
Find the probability that the cooperative lands more than 500 kg of fish given that the sea is rough.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(A ∪ B) = 0.45 + 0.30 − 0.12
✓ 1
A1 0.63
✓ 1
Part (b)
M1 P(A) × P(B) = 0.45 × 0.30 = 0.135
✓ 1
R1 0.135 ≠ 0.12 = P(A ∩ B), so the events are not independent
✓ 1
Part (c)
M1 P(A | B) = P(A ∩ B) ÷ P(B) = 0.12 ÷ 0.30
✓ 1
A1 0.4
✓ 1
Common error: To test independence, compare P(A)P(B) with P(A ∩ B); showing that P(A ∪ B) is less than 1 says nothing about independence.
65IBMI3S-4-02r18-07Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared testsPaper 1Easy6 marks
The mass of fish landed at a small harbour is normally distributed with mean 1.80 kg and standard deviation 0.25 kg.
(a)
Find the probability that a randomly chosen fish has a mass greater than 2.20 kg. Give your answer to 3 significant figures.
(2)
(b)
Find the probability that a randomly chosen fish has a mass between 1.50 kg and 2.00 kg. Give your answer to 3 significant figures.
(2)
(c)
The heaviest 10 % of fish are sold as premium grade. Find the minimum mass of a premium-grade fish. Give your answer to 3 significant figures.
M1 inverse normal with P(X < m) = 0.90 (z = 1.2816)
✓ 1
A1 2.12 kg
✓ 1
Exam tip: For "the heaviest 10 %", the area below the required value is 0.90, not 0.10.
66IBMI3S-4-02r20-01Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy6 marks
The daily energy output, X MWh, of a wind farm is modelled by a normal distribution with mean 48 MWh and standard deviation 6 MWh.
(a)
Find the probability that the output on a randomly chosen day is less than 40 MWh.
(2)
(b)
Find the probability that the output on a randomly chosen day is greater than 55 MWh.
(2)
(c)
On 10% of days the output is greater than k MWh. Find the value of k.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 X ~ N(48, 62), P(X < 40) by GDC or standardising
✓ 1
A1 0.0912
✓ 1
Part (b)
M1 P(X > 55)
✓ 1
A1 0.122
✓ 1
Part (c)
M1 P(X < k) = 0.9 and use of inverse normal
✓ 1
A1 k = 55.7
✓ 1
Exam tip: Sketch the normal curve and shade the required area first; this shows whether you need a lower tail, an upper tail or an inverse normal.
67IBMI3S-4-02r20-02Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy6 marks
A supplier knows that 4% of the solar panels it manufactures are defective, independently of each other. A pallet contains 25 panels chosen at random from the supplier's output. Let X be the number of defective panels on the pallet.
(a)
State the distribution of X, including its parameters.
(1)
(b)
Find the probability that exactly 2 panels on the pallet are defective.
(2)
(c)
Find the probability that at least one panel on the pallet is defective.
(2)
(d)
Find the expected number of defective panels on a pallet.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 X ~ B(25, 0.04)
✓ 1
Part (b)
M1 P(X = 2) using the binomial distribution
✓ 1
A1 0.188
✓ 1
Part (c)
M1 1 − P(X = 0)
✓ 1
A1 0.640
✓ 1
Part (d)
A1 25 × 0.04 = 1 (1.00)
✓ 1
Exam tip: For "at least one" use the complement: 1 − P(X = 0) is much quicker than adding up many probabilities.
68IBMI3S-4-02r20-03Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy7 marks
The number of units of a product sold by a small shop on each of 12 days is: 142, 155, 138, 160, 171, 149, 166, 152, 145, 158, 163, 210.
(a)
Find the mean and the median number of units sold per day.
(2)
(b)
Find the sample standard deviation of the number of units sold.
(1)
(c)
An outlier is defined as a value that is more than 1.5 × IQR above the upper quartile or below the lower quartile. Show that 210 is an outlier.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 mean = 159 (159.08...)
✓ 1
A1 median = 156.5
✓ 1
Part (b)
A1 18.8 (18.84...)
✓ 1
Part (c)
A1 Q1 = 147 and Q3 = 164.5, so IQR = 17.5
✓ 1
M1 upper boundary = 164.5 + 1.5 × 17.5
✓ 1
A1 upper boundary = 190.75
✓ 1
R1 210 > 190.75, so 210 is an outlier
✓ 1
Exam tip: The mean is pulled towards an outlier but the median is not, which is why the median is 3 units below the mean here.
69IBMI3S-4-02r20-04Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy7 marks
A kiosk records the maximum temperature, x °C, and the number of cold drinks sold, y, on eight days.
x
12
15
18
21
24
27
30
33
y
85
102
118
140
151
178
190
214
(a)
Find the Pearson correlation coefficient, r, and describe the correlation.
(2)
(b)
Find the equation of the regression line of y on x.
(2)
(c)
Interpret the gradient of the regression line in context.
(1)
(d)
Use the regression line to estimate the number of drinks sold on a day with maximum temperature 20 °C. Give your answer to the nearest whole number.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 r = 0.998
✓ 1
A1 very strong positive correlation
✓ 1
Part (b)
A1 gradient 6.09 (6.087...)
✓ 1
A1 y = 6.09x + 10.3
✓ 1
Part (c)
A1 for each 1 °C increase in maximum temperature, about 6 more drinks are sold
✓ 1
Part (d)
M1 substitution of x = 20
✓ 1
A1 132 drinks
✓ 1
Exam tip: Quote the regression line to 3 significant figures only after calculating with the full GDC values.
70IBMI3S-4-01r22-01Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regressionPaper 1Easy6 marks
A volcano observatory operates 12 seismometers on the flanks of a volcano. During a tremor, each seismometer independently detects the event with probability 0.85. Let X be the number of seismometers that detect a particular tremor.
(a)
Find P(X = 10).
(2)
(b)
Find P(X ≥ 11).
(2)
(c)
The observatory issues a warning only when at least 11 seismometers detect a tremor. Over 40 independent tremors, find the expected number of tremors that lead to a warning, giving your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of X ~ B(12, 0.85), e.g. C(12,10)(0.85)10(0.15)2
✓ 1
A1 0.292
✓ 1
Part (b)
M1 P(X = 11) + P(X = 12), or 1 − P(X ≤ 10)
✓ 1
A1 0.443
✓ 1
Part (c)
M1 40 × P(X ≥ 11), using the unrounded probability from (b) or 0.443
✓ 1
A1 17.7 (40 × 0.443 = 17.72 also gives 17.7)
✓ 1
Common error: For 'at least 11' out of 12, the complement is P(X ≤ 10), not P(X ≤ 11) - check which end of the distribution you need.
71IBMI3S-4-01r22-04Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regressionPaper 1Easy6 marks
The running time, T minutes, of a touring theatre production is modelled by a normal distribution with mean 142 and standard deviation 6.
(a)
Find the probability that a performance lasts longer than 150 minutes.
(2)
(b)
Find the probability that a performance lasts between 135 and 148 minutes.
(2)
(c)
The longest 10% of performances last more than t minutes. Find t, to the nearest 0.1 minute.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of the normal distribution with mean 142 and standard deviation 6 (e.g. z = 1.33)
✓ 1
A1 0.0912
✓ 1
Part (b)
M1 P(T < 148) − P(T < 135), or equivalent
✓ 1
A1 0.720
✓ 1
Part (c)
M1 inverse normal with probability 0.9 (z = 1.2816)
✓ 1
A1 t = 149.7
✓ 1
Exam tip: For 'the longest 10%' use the 0.9 cumulative probability, not 0.1, when using an inverse normal function.
72IBMI3S-4-01r22-06Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regressionPaper 1Easy7 marks
A port handles 1200 standard containers, 450 refrigerated containers and 150 tank containers. An inspector selects a stratified sample of 60 containers.
(a)
Find the number of containers of each type in the sample.
(2)
(b)
The masses, in tonnes, of 11 of the standard containers are: 18.2, 21.5, 19.8, 22.1, 20.4, 18.9, 21.0, 35.6, 19.5, 20.9, 22.4 Find the median, the lower quartile and the upper quartile. (Take each quartile as the median of the lower or upper five values.)
(3)
(c)
A mass is an outlier if it is more than 1.5 × IQR above the upper quartile. Show whether 35.6 tonnes is an outlier.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sample size proportional to type, e.g. (1200 / 1800) × 60
Exam tip: Sort the data first, then find the quartiles; the outlier test needs the IQR (Q3 − Q1), not the range.
73IBMI3S-4-01r22-14Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regressionPaper 1Easy7 marks
A port authority believes that ships are equally likely to arrive on each weekday from Monday to Friday. A random sample of 150 ship arrivals gives these frequencies.
Day
Mon
Tue
Wed
Thu
Fri
Arrivals
39
22
28
35
26
A χ2 goodness-of-fit test is carried out at the 5% significance level.
(a)
State the null and alternative hypotheses.
(2)
(b)
Write down the number of degrees of freedom.
(1)
(c)
Calculate the χ2 statistic, to 3 significant figures.
(2)
(d)
The critical value at the 5% level for this number of degrees of freedom is 9.49. State the conclusion, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: arrivals are equally likely on each weekday (uniform distribution)
✓ 1
A1 H1: arrivals are not equally likely on each weekday
✓ 1
Part (b)
A1 5 − 1 = 4
✓ 1
Part (c)
M1 expected frequency 30 for each day and Σ(O − E)2/E
✓ 1
A1 6.33
✓ 1
Part (d)
R1 6.33 < 9.49
✓ 1
A1 do not reject H0: no significant evidence that arrivals differ across weekdays
✓ 1
Common error: In a goodness-of-fit test, degrees of freedom are (number of categories - 1), not the number of observations.
74IBMI3S-4-02r25-01Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 1Easy6 marks
The turnaround time, T minutes, of a short-haul aircraft between landing and the next departure is normally distributed with mean 42 minutes and standard deviation 6 minutes.
(a)
Find the probability that a turnaround takes longer than 50 minutes.
(2)
(b)
In a week, 200 turnarounds are recorded. Find the expected number that take between 38 and 46 minutes.
(2)
(c)
The airline sets a target time, k minutes, that is met by 90% of turnarounds. Find k, correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 T ~ N(42, 62), P(T > 50) attempted
✓ 1
A1 0.0912 (0.0912)
✓ 1
Part (b)
M1 P(38 < T < 46) = 0.4950 and multiplied by 200
✓ 1
A1 99 turnarounds (99.0)
✓ 1
Part (c)
M1 P(T < k) = 0.9 (inverse normal, or z = 1.2816)
✓ 1
A1 k = 49.7
✓ 1
Exam tip: State the distribution, for example T ~ N(42, 6²), before using your calculator; it shows the method even if a key is mis-pressed.
75IBMI3S-4-02r25-06Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 1Easy5 marks
A farm has 480 experimental plots: 240 on clay soil, 160 on sandy soil and 80 on loam. A researcher wants a stratified sample of 60 plots to test a new seed variety.
(a)
Find the number of plots that should be sampled from each soil type.
(3)
(b)
Give one reason why a stratified sample is more suitable here than a simple random sample.
(1)
(c)
Describe how the 30 clay plots could be chosen.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sample fraction 60 ÷ 480 = 1/8 applied to each stratum
✓ 1
A1 clay 30 and sandy 20
✓ 1
A1 loam 10
✓ 1
Part (b)
R1 it guarantees every soil type is represented in proportion to its size, whereas a simple random sample could by chance contain few or no loam plots
✓ 1
Part (c)
R1 number the 240 clay plots 1 to 240 and select 30 distinct numbers using a random number generator
✓ 1
Common error: Round only at the end: sample size in a stratum = (stratum size ÷ population size) × sample size, and the stratum sizes must add up to the total sample.
76IBMI3S-4-02r25-13Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 1Easy6 marks
A music teacher records the weekly practice time, x hours, and the exam score, y (out of 100), of her students. The regression line of y on x is y = 3.8x + 41.6, and the correlation coefficient is r = 0.89.
(a)
Describe what the value of r tells you about the relationship between practice time and exam score.
(1)
(b)
Interpret the gradient 3.8 in context.
(1)
(c)
Use the regression line to predict the score of a student who practises for 9 hours a week.
(2)
(d)
A student who practises for 9 hours a week scores 82. Find the residual and state what it shows.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 strong positive linear correlation: students who practise more tend to score higher
✓ 1
Part (b)
A1 each additional hour of weekly practice is associated with an increase of 3.8 marks in the exam score
✓ 1
Part (c)
M1 3.8 × 9 + 41.6
✓ 1
A1 75.8
✓ 1
Part (d)
M1 82 − 75.8
✓ 1
A1 residual = 6.2: the student scored 6.2 marks higher than the model predicts
✓ 1
Exam tip: A residual is actual minus predicted; a positive residual means the point lies above the regression line.
77IBMI3S-4-02r31-01Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy4 marks
Calculator
The lengths, L cm, of adult mackerel caught by a marine research vessel are normally distributed with mean 31.0 cm and standard deviation 2.8 cm.
(a)
Find the probability that a randomly chosen mackerel has a length between 30 cm and 35 cm.
(2)
(b)
The longest 10% of mackerel are classed as large. Find the minimum length of a mackerel that is classed as large. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to find P(30 < L < 35) using the normal distribution
✓ 1
A1 0.563
✓ 1
Part (b)
M1 inverse normal with P(L < k) = 0.9 (or the standardised value 1.2816)
✓ 1
A1 k = 34.6 cm
✓ 1
Exam tip: Sketch the normal curve and shade the required region before using your calculator; for the longest 10%, the area to the left of the cut-off is 0.9, not 0.1.
78IBMI3S-4-02r31-02Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy5 marks
Calculator
A logistics company records the delivery times, in minutes, for 11 parcels: 22, 25, 27, 28, 30, 31, 33, 34, 36, 41, 58 Quartiles are to be found by taking the median of the lower half and the median of the upper half, excluding the overall median from both halves.
(a)
Find the median and the interquartile range of the delivery times.
(3)
(b)
A delivery time is an outlier if it is more than 1.5 × IQR above the upper quartile. Show whether 58 minutes is an outlier.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 median = 31 minutes
✓ 1
A1 Q1 = 27 and Q3 = 36
✓ 1
A1 IQR = 36 − 27 = 9 minutes
✓ 1
Part (b)
M1 upper boundary = 36 + 1.5 × 9 (ECF from (a))
✓ 1
A1 49.5 minutes, and 58 > 49.5 so 58 is an outlier
✓ 1
Common error: The 1.5 × IQR rule is added to the upper quartile (and subtracted from the lower quartile); it is not added to the median.
79IBMI3S-4-02r31-09Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy6 marks
A survey of 120 building projects finds that 45 include solar panels (S), 38 include a green roof (G) and 15 include both.
(a)
Find the probability that a randomly chosen project includes solar panels or a green roof or both.
(2)
(b)
Find the probability that a project includes solar panels, given that it has a green roof.
R1 this does not equal P(S ∩ G) = 15/120 = 0.125, so the events are not independent
✓ 1
Common error: To test independence compare P(S) × P(G) with P(S ∩ G); having both events occur is not the same as them being independent.
80IBMI3S-4-04r33-02Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy4 marks
A fashion company has 750 employees: 120 in design, 450 in production and 180 in sales. The company surveys a sample of 50 employees and chooses the sample so that each department is represented in proportion to its size.
(a)
State the name of this sampling method.
(1)
(b)
Calculate the number of employees that should be sampled from production.
(2)
(c)
Give one reason why this method is preferred to sampling the first 50 employees to arrive at work.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 stratified sampling
✓ 1
Part (b)
M1 use of \(\frac{450}{750}\times50\)
✓ 1
A1 30 employees
✓ 1
Part (c)
R1 it gives a representative sample of every department (accept: removes bias from arrival time / every department represented)
✓ 1
Exam tip: In stratified sampling the sample size for each stratum is (stratum size ÷ population size) × total sample size.
81IBMI3S-4-04r33-03Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy5 marks
A robotics company tests its warehouse robots. Each robot independently passes the final inspection with probability 0.92. A batch of 15 robots is inspected. Let X be the number of robots in the batch that pass.
(a)
Find the probability that exactly 14 robots pass.
(2)
(b)
Find the probability that at least 14 robots pass.
(2)
(c)
Find the expected number of robots that pass in a batch.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 recognise \(X\sim B(15,0.92)\) and use \(P(X=14)\)
✓ 1
A1 0.373
✓ 1
Part (b)
M1 use of \(P(X=14)+P(X=15)\) (or \(1-P(X\le13)\))
✓ 1
A1 0.660
✓ 1
Part (c)
A1 \(15\times0.92=13.8\)
✓ 1
Exam tip: 'At least' means the stated value and everything above it, so \(P(X\ge14)=1-P(X\le13)\).
82IBMI3S-4-04r33-04Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy4 marks
The lengths of haddock caught in a fishery are normally distributed with mean 34 cm and standard deviation 4.5 cm.
(a)
Find the probability that a randomly chosen haddock is longer than 40 cm.
(2)
(b)
The longest 10% of haddock are classed as 'large'. Find the minimum length of a large haddock. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(L\sim N(34,4.5^2)\) and \(P(L>40)\)
✓ 1
A1 0.091
✓ 1
Part (b)
M1 use of inverse normal with \(P(L
✓ 1
A1 39.8 cm
✓ 1
Common error: For 'the top 10%' the inverse normal needs an area of 0.9 to the left, not 0.1.
83IBMI3S-4-02r34-01Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Probability and the binomial distributionPaper 1Easy4 marks
A cheese producer ages 900 wheels of cheese in three caves: 450 wheels in cave A, 300 in cave B and 150 in cave C. A quality team takes a stratified sample of 60 wheels, with the strata being the caves.
(a)
Find the number of wheels that should be sampled from each cave so that the sample is proportional.
(3)
(b)
Give one reason why stratified sampling is more suitable than simple random sampling here.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sample fraction \(\dfrac{60}{900}\) applied to each cave size, e.g. \(\dfrac{150}{900}\times60\)
✓ 1
A1 two correct sample sizes
✓ 1
A1 cave A: 30, cave B: 20, cave C: 10
✓ 1
Part (b)
A1 stratified sampling guarantees every cave is represented in proportion to its size (a simple random sample could, by chance, contain few or no wheels from cave C)
✓ 1
Common error: Rounding sample sizes independently can make the total different from the sample size. Check that the sizes add to the required total.
84IBMI3S-4-02r34-02Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Probability and the binomial distributionPaper 1Easy6 marks
The times, in hours, taken by 12 climbers to reach a summit are: 5.3, 5.6, 5.9, 6.1, 6.4, 6.6, 6.8, 7.1, 7.3, 7.7, 8.0, 10.4 The quartiles are found as the medians of the lower six and upper six values.
(a)
Calculate the mean time, to 3 significant figures.
(2)
(b)
Find the interquartile range.
(2)
(c)
A time is an outlier if it is more than 1.5 × IQR above the upper quartile. Show that 10.4 hours is an outlier.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sum of the times divided by 12 (\(83.2\div12\))
✓ 1
A1 6.93 hours (6.9333…)
✓ 1
Part (b)
M1 \(Q_1=\dfrac{5.9+6.1}{2}=6.0\) and \(Q_3=\dfrac{7.3+7.7}{2}=7.5\)
✓ 1
A1 IQR \(=1.5\) hours
✓ 1
Part (c)
M1 \(7.5+1.5\times1.5=9.75\)
✓ 1
A1 \(10.4>9.75\), so 10.4 is an outlier
✓ 1
Exam tip: For the outlier test, the fences are calculated from the quartiles, not from the mean. State the fence value before comparing.
85IBMI3S-4-02r34-06Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Probability and the binomial distributionPaper 1Easy5 marks
A mountain guide knows that each climber in her groups reaches the summit with probability 0.72, independently of the other climbers. A group has 10 climbers. Let X be the number of climbers who reach the summit.
(a)
Find \(P(X=8)\), to 3 significant figures.
(2)
(b)
Find the probability that at least 8 climbers reach the summit, to 3 significant figures.
(2)
(c)
Find the expected number of climbers in a group who reach the summit.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(X\sim B(10,0.72)\) and \(P(X=8)=\binom{10}{8}(0.72)^8(0.28)^2\)
✓ 1
A1 0.255 (0.2548…)
✓ 1
Part (b)
M1 \(1-P(X\le7)\) or \(P(X=8)+P(X=9)+P(X=10)\)
✓ 1
A1 0.438 (0.4378…)
✓ 1
Part (c)
A1 \(10\times0.72=7.2\)
✓ 1
Common error: 'At least 8' means 8 or more, so the complement is \(P(X\le7)\), not \(P(X\le8)\).
86IBMI3S-4-02r34-13Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Probability and the binomial distributionPaper 1Easy5 marks
A telecoms provider has 2400 customers, numbered 1 to 2400 on a list. It selects a sample of 80 customers using systematic sampling. The first customer selected is number 17.
(a)
Find the sampling interval.
(1)
(b)
Write down the number of the fifth customer selected.
(1)
(c)
Show that the last customer selected is number 2387.
(2)
(d)
The list is arranged in blocks of 30 customers, with the 17th customer in each block being a business account. State one problem with using this sample.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(2400\div80=30\)
✓ 1
Part (b)
A1 \(17+4\times30=137\)
✓ 1
Part (c)
M1 \(17+79\times30\)
✓ 1
A1 \(=2387\)
✓ 1
Part (d)
R1 every selected customer is a business account, so the sample is biased and not representative of all customers
✓ 1
Common error: Using 80 as the interval. The interval is the population size divided by the sample size.
87IBMI3S-4-02r36-01Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy7 marks
Each batch of roasted coffee produced by a small roastery is independently defective with probability 0.1. A buyer receives 15 batches. Let X be the number of defective batches.
(a)
State the distribution of X, including its parameters.
(1)
(b)
Find P(X = 2).
(2)
(c)
Find P(X ≥ 1).
(2)
(d)
Find the probability that more than 3 batches are defective.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 X ~ B(15, 0.1)
✓ 1
Part (b)
M1 use of the binomial distribution
✓ 1
A1 0.267
✓ 1
Part (c)
M1 1 − P(X = 0) = 1 − 0.915
✓ 1
A1 0.794
✓ 1
Part (d)
M1 1 − P(X ≤ 3)
✓ 1
A1 0.0556
✓ 1
Exam tip: 'At least one' is most efficiently found as 1 minus the probability of none.
88IBMI3S-4-02r36-06Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy6 marks
A strain sensor on a bridge records a reading every week for 120 weeks. The lower quartile of the readings is 21 and the upper quartile is 25 (in microstrain). A reading is an outlier if it is more than 1.5 × IQR above the upper quartile or below the lower quartile.
(a)
Find the interquartile range.
(1)
(b)
Determine whether a reading of 14 is an outlier.
(2)
(c)
Find the upper limit above which a reading is an outlier, and state whether a reading of 31 is an outlier.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 4
✓ 1
Part (b)
M1 lower fence = 21 − 1.5 × 4 = 15
✓ 1
R1 14 < 15, so it is an outlier
✓ 1
Part (c)
M1 25 + 1.5 × 4
✓ 1
A1 31
✓ 1
R1 31 is not above 31, so it is not an outlier
✓ 1
Exam tip: Show the fence calculation explicitly; a bare 'yes' or 'no' about outliers earns no reasoning mark.
89IBMI3S-4-02r36-13Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy6 marks
A bridge-maintenance company has 240 employees: 60 inspectors, 100 technicians, 40 office staff and 40 drivers. The company wants a sample of 48 employees for a survey.
(a)
Find the number of employees from each group in a stratified sample of size 48.
(3)
(b)
State one advantage of stratified sampling over simple random sampling.
(1)
(c)
Instead, a systematic sample of 48 employees is taken from an alphabetical list of all 240 employees. Find the interval between selected employees.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (group size ÷ 240) × 48
✓ 1
A1 inspectors 12 and technicians 20
✓ 1
A1 office staff 8 and drivers 8
✓ 1
Part (b)
R1 each group is represented in proportion to its size, so the sample is more representative
✓ 1
Part (c)
M1 240 ÷ 48
✓ 1
A1 every 5th employee
✓ 1
Common error: In stratified sampling the sample sizes must add up to the required total; check your numbers sum to 48.
90IBMI3S-4-02r36-14Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy6 marks
In a café, 45% of customers order an espresso (event E), 30% order a pastry (event P) and 12% order both.
(a)
Show that E and P are not independent.
(2)
(b)
Find P(E ∪ P).
(2)
(c)
Find the probability that a customer who orders a pastry also orders an espresso.
Exam tip: Independence is tested by checking whether P(A ∩ B) equals P(A) × P(B); mutually exclusive is a different idea.
91IBMI3S-4-04r37-02Sampling and descriptive statistics; T-tests, confidence intervals and Markov chains; Probability and the binomial distribution; Normal distribution and chi-squared tests; Correlation and regression; Normal distribution and chi-squared tests; Probability and the binomial distributionPaper 1Easy5 marks
A fashion company has 600 staff in four branches: Athens (90), Patras (150), Thessaloniki (210) and Heraklion (150). The company wants to survey a sample of 40 staff about their working conditions.
(a)
Use stratified sampling, with each branch proportionally represented, to find how many staff should be sampled from each branch.
(2)
(b)
A manager suggests surveying the first 40 staff who arrive at the head office on Monday. Name this type of sampling and give one reason why it may give a biased sample.
(2)
(c)
Give one advantage of stratified sampling over simple random sampling in this situation.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 branch size ÷ 600 × 40, for example \(\frac{210}{600}\times40\)
R1 staff arriving early may share a branch or job type, so not every staff member has a chance of being chosen and the sample may not represent all staff
✓ 1
Part (c)
R1 it guarantees that every branch is represented in proportion to its size, whereas a simple random sample could, by chance, miss a branch
✓ 1
Exam tip: Describe how a sampling method is chosen and how it could introduce bias; naming the method alone does not earn the reasoning mark.
92IBMI3S-4-04r37-05Sampling and descriptive statistics; T-tests, confidence intervals and Markov chains; Probability and the binomial distribution; Normal distribution and chi-squared tests; Correlation and regression; Normal distribution and chi-squared tests; Probability and the binomial distributionPaper 1Easy5 marks
An online fashion retailer sells through its website and through its shops. 60% of items are bought online and 40% are bought in shops. Of the items bought online, 25% are returned. Of the items bought in shops, 10% are returned.
(a)
Find the probability that a randomly chosen item is returned.
(2)
(b)
Given that an item is returned, find the probability that it was bought online. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(0.6\times0.25+0.4\times0.1\)
✓ 1
A1 0.19
✓ 1
Part (b)
M1 \(P(\text{online}\mid\text{returned})=\frac{P(\text{online and returned})}{P(\text{returned})}\)
✓ 1
A1 \(\frac{0.6\times0.25}{0.19}\)
✓ 1
A1 0.789
✓ 1
Exam tip: Draw a tree diagram with the first branches for online and shop. The conditional probability is the branch you want divided by the total probability of the condition.
93IBMI3S-4-04r37-09Sampling and descriptive statistics; T-tests, confidence intervals and Markov chains; Probability and the binomial distribution; Normal distribution and chi-squared tests; Correlation and regression; Normal distribution and chi-squared tests; Probability and the binomial distributionPaper 1Easy5 marks
The mass, M, of a mackerel landed at a port is normally distributed with mean 420 g and standard deviation 35 g.
(a)
Find the probability that a randomly chosen mackerel has a mass greater than 450 g. Give your answer to 3 significant figures.
(2)
(b)
The heaviest 10% of mackerel are sold as premium grade. Find the minimum mass of a premium grade mackerel. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(P(M>450)\) with \(\mu=420\), \(\sigma=35\) (or \(z=0.857\))
✓ 1
A1 0.196
✓ 1
Part (b)
M1 \(P(M>m)=0.1\), so \(P(M
✓ 1
A1 \(z=1.2816\)
✓ 1
A1 \(m=420+1.2816\times35=465\) g
✓ 1
Common error: For inverse normal questions, convert "the heaviest 10%" into the area to the left, 0.9, before using the inverse normal function.
94IBMI3S-4-02r38-01Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 1Easy7 marks
Calculator
Archaeologists excavate three layers of a settlement and recover 840 potsherds: 420 from layer A, 280 from layer B and 140 from layer C. They take a stratified sample of 60 sherds, in proportion to the number found in each layer. The masses, in grams, of ten of the sampled sherds are 42, 55, 38, 61, 47, 52, 44, 58, 50, 53.
(a)
Find the number of sherds sampled from each of the three layers.
(2)
(b)
Find the mean mass of the ten sherds.
(1)
(c)
Find the population standard deviation of the ten masses. Give your answer to 3 significant figures.
(2)
(d)
An eleventh sherd has mass 75 g. Determine whether it lies more than two standard deviations above the mean of the ten masses.
(2)
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Marking point
Mark
Notes
Part (a)
M1 proportion, e.g. 420 ÷ 840 × 60
✓ 1
A1 layer A 30, layer B 20, layer C 10
✓ 1
Part (b)
A1 50 g
✓ 1
Part (c)
M1 use of GDC or formula for σ with these ten values
✓ 1
A1 6.90 g
✓ 1
Part (d)
M1 mean + 2σ = 50 + 2 × 6.90 = 63.8 (accept 63.8)
✓ 1
A1 75 > 63.8, so it is more than two standard deviations above the mean
✓ 1
Common error: Using the sample standard deviation (sn−₁) when the question asks for the population standard deviation, which is the σn value on most calculators.
95IBMI3S-4-02r38-02Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 1Easy7 marks
Calculator
The times, in minutes, taken by a cyclist to complete the same training loop on twelve days are 34, 36, 37, 39, 41, 42, 44, 45, 47, 49, 52, 71. Quartiles are found as the medians of the lower six and upper six values.
(a)
Find the median, the lower quartile and the upper quartile.
(3)
(b)
Show that the time of 71 minutes is an outlier, using the rule that an outlier is more than 1.5 × IQR above the upper quartile.
(2)
(c)
The time of 71 minutes was caused by a puncture and is removed. Find the mean of the remaining eleven times, to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 median = 43
✓ 1
A1 lower quartile = 38
✓ 1
A1 upper quartile = 48
✓ 1
Part (b)
M1 IQR = 48 − 38 = 10 and 48 + 1.5 × 10
✓ 1
A1 63 < 71, so 71 is an outlier
✓ 1
Part (c)
M1 (sum of times − 71) ÷ 11 = 466 ÷ 11
✓ 1
A1 42.4 minutes
✓ 1
Exam tip: Write down the fences (Q1 − 1.5 × IQR and Q3 + 1.5 × IQR) explicitly, then compare each suspected outlier with the relevant fence.
96IBMI3S-4-02r38-08Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 1Easy6 marks
Calculator
In a data centre, the cooling system fails on a given day with probability 0.04. If the cooling system fails, a server shuts down with probability 0.7. If the cooling system does not fail, a server shuts down with probability 0.01.
(a)
Find the probability that the cooling system fails and the server shuts down.
(1)
(b)
Show that the probability that the server shuts down on a given day is 0.0376.
(2)
(c)
Given that the server shuts down, find the probability that the cooling system had failed. Give your answer to 3 significant figures.
(3)
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Marking point
Mark
Notes
Part (a)
A1 0.04 × 0.7 = 0.028
✓ 1
Part (b)
M1 0.04 × 0.7 + 0.96 × 0.01
✓ 1
A1 0.0376 (AG)
✓ 1
Part (c)
M1 conditional probability P(A ∩ B) ÷ P(B)
✓ 1
A1 0.028 ÷ 0.0376 (ECF from (a) and (b))
✓ 1
A1 0.745
✓ 1
Exam tip: For conditional probability, the denominator is the total probability of the 'given' event, found by adding all the branches that lead to it.
97IBMI3S-4-02r38-13Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 1Easy8 marks
Calculator
An excavation produces 80 finds. Of these, 33 are pottery (P), 24 are decorated (D), and 9 are both pottery and decorated.
(a)
Find the number of finds that are neither pottery nor decorated.
(2)
(b)
A find is chosen at random. Given that it is pottery, find the probability that it is decorated. Give your answer to 3 significant figures.
(2)
(c)
Determine whether the events P and D are independent. Justify your answer.
(2)
(d)
Two different finds are chosen at random, one after the other, without replacement. Find the probability that both are decorated. Give your answer to 3 significant figures.
(2)
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Marking point
Mark
Notes
Part (a)
M1 33 + 24 − 9 = 48 finds that are pottery or decorated (or Venn diagram regions 24, 9, 15)
R1 the values differ, so the events are not independent
✓ 1
Part (d)
M1 24/80 × 23/79
✓ 1
A1 0.0873
✓ 1
Common error: Writing P(D | P) as 9 ÷ 24; the conditional probability uses the 'given' event, pottery, as the denominator.
98IBMI3S-4-02r44-01Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy6 marks
A beekeeper records the mass of honey, in kg, harvested from each of 12 hives: 18, 22, 25, 27, 29, 30, 32, 33, 36, 38, 41, 58. In this question the lower quartile Q1 is the median of the lower six values and the upper quartile Q3 is the median of the upper six values.
(a)
Find the median mass.
(1)
(b)
Find Q1 and Q3.
(2)
(c)
A value is an outlier if it is more than \(1.5\times\text{IQR}\) above Q3. Show that 58 kg is an outlier.
(3)
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Marking point
Mark
Notes
Part (a)
A1 median \(=\frac{30+32}{2}=31\) kg
✓ 1
Part (b)
A1 \(Q_1=26\) kg
✓ 1
A1 \(Q_3=37\) kg
✓ 1
Part (c)
M1 IQR \(=37-26=11\) (ECF from (b)) and attempt to use \(Q_3+1.5\times\text{IQR}\)
✓ 1
A1 upper boundary \(=37+1.5(11)=53.5\)
✓ 1
R1 \(58>53.5\), so 58 kg is an outlier
✓ 1
Exam tip: Write the boundary value you calculate before comparing it with the data point; the comparison earns the reasoning mark.
99IBMI3S-4-02r44-02Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy6 marks
A hospital clinic books 12 patients for scans each morning. Each patient independently attends with probability 0.9. Let X be the number of booked patients who attend, so that \(X\sim B(12,\,0.9)\).
(a)
State one assumption that is needed for X to follow a binomial distribution.
(1)
(b)
Find the probability that all 12 patients attend.
(2)
(c)
Find the probability that at least 11 patients attend.
(2)
(d)
Find the expected number of patients who attend.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 each patient attends independently of the others (the probability 0.9 is the same for every patient)
✓ 1
Part (b)
M1 \(0.9^{12}\) (or use of binomial pdf with \(X=12\))
✓ 1
A1 0.282
✓ 1
Part (c)
M1 \(P(X=11)+P(X=12)\) or \(1-P(X\le 10)\)
✓ 1
A1 0.659
✓ 1
Part (d)
A1 \(12\times 0.9=10.8\)
✓ 1
Common error: 'At least 11' means 11 or 12; do not use \(P(X\le 11)\).
100IBMI3S-4-02r44-07Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy6 marks
A beekeeping cooperative has 120 hives at site A, 80 at site B and 200 at site C. A researcher wants a sample of 40 hives and uses stratified sampling, with the sites as strata.
(a)
Find the number of hives sampled from each site.
(3)
(b)
Explain why stratified sampling is more suitable here than choosing 40 hives from the whole list at random.
(1)
(c)
Instead, the researcher takes the 40 hives nearest the road. Name this type of sampling and give one reason it may be biased.
R1 it guarantees each site is represented in proportion to its size, whereas a simple random sample might miss or over-represent a site
✓ 1
Part (c)
A1 convenience sampling
✓ 1
R1 hives near the road may differ from others (for example in exposure or flowers nearby), so the sample may not represent all hives
✓ 1
Common error: In stratified sampling, the number from each stratum is proportional to the stratum's size, and must be a whole number.
101IBMI3S-4-02r44-14Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy6 marks
Of 80 patients on a hospital ward, 35 have hypertension (H), 28 have diabetes (D) and 10 have both conditions. A patient is chosen at random from the ward.
(a)
Find the number of patients who have neither condition.
(2)
(b)
Find the probability that the patient has hypertension, given that the patient has diabetes.
(2)
(c)
Determine whether H and D are independent events.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(35+28-10=53\) patients have at least one condition
R1 this is not equal to \(P(H\cap D)=\frac{10}{80}=0.125\), so H and D are not independent
✓ 1
Common error: To test independence, compare \(P(A)P(B)\) with \(P(A\cap B)\); events being non-overlapping is a different idea (mutually exclusive).
102IBMI3S-4-02r47-01Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy6 marks
The daily water use of a household in a town, X litres, is modelled by a normal distribution with mean 420 and standard deviation 35.
(a)
Find the probability that a household uses more than 480 litres in a day.
(2)
(b)
Find the probability that a household uses between 380 and 450 litres in a day.
(2)
(c)
The water company wants to supply enough water so that 90% of households use less than this amount on a given day. Find this amount, to the nearest litre.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use the normal distribution, for example P(X > 480) with μ = 420, σ = 35
✓ 1
A1 0.0432
✓ 1
Part (b)
M1 P(380 < X < 450) with correct parameters
✓ 1
A1 0.678
✓ 1
Part (c)
M1 use of inverse normal with probability 0.9 (z = 1.2816…)
✓ 1
A1 465 litres
✓ 1
Exam tip: Write the probability statement, for example P(X < k) = 0.9, before using the inverse normal function; it shows the method even if the calculator entry is wrong.
103IBMI3S-4-02r47-02Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy7 marks
Twelve players each recorded the time, in seconds, taken to complete a level of a video game. The times, in increasing order, are 34, 41, 45, 47, 52, 55, 58, 61, 64, 68, 72, 121.
(a)
Find the mean time. Give your answer to 3 significant figures.
(2)
(b)
Find the standard deviation of these twelve times, treating them as the whole population. Give your answer to 3 significant figures.
(1)
(c)
Find the lower quartile and the upper quartile, using the median of the lower half and of the upper half of the data.
(2)
(d)
A time is treated as an outlier if it is more than 1.5 × IQR above the upper quartile. Show that 121 seconds is an outlier, and state the upper boundary for outliers.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sum ÷ 12 (718 ÷ 12 = 59.83…)
✓ 1
A1 59.8 seconds
✓ 1
Part (b)
A1 21.3 seconds (from the calculator, population standard deviation σn)
✓ 1
Part (c)
A1 Q1 = 46
✓ 1
A1 Q3 = 66
✓ 1
Part (d)
M1 IQR = 20 and 1.5 × IQR = 30 (ECF)
✓ 1
A1 upper boundary 96, and 121 > 96 so it is an outlier
✓ 1
Common error: Quartiles from a calculator can differ slightly depending on the method; state the method you use. Use the form of standard deviation (population or sample) that the question names.
104IBMI3S-4-02r47-09Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy6 marks
On any given day the water level in a reservoir is high with probability 0.35. If the level is high, the probability of a supply interruption is 0.02. If the level is not high, the probability of a supply interruption is 0.11.
(a)
Find the probability that there is a supply interruption on a randomly chosen day.
(2)
(b)
Find the expected number of days with a supply interruption in a period of 100 days.
(1)
(c)
Given that there is an interruption, find the probability that the water level was high. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.35 × 0.02 + 0.65 × 0.11
✓ 1
A1 0.0785
✓ 1
Part (b)
A1 100 × 0.0785 = 7.85
✓ 1
Part (c)
M1 P(high | interruption) = P(high and interruption) ÷ P(interruption)
✓ 1
A1 0.007 ÷ 0.0785
✓ 1
A1 0.0892
✓ 1
Exam tip: For conditional probability, write the formula P(A | B) = P(A ∩ B) ÷ P(B) first; the denominator is the probability of the condition.
105IBMI3S-4-04r47-01Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy4 marks
A fisheries officer inspects the catch landed at a harbour in one morning. The catch contains 840 cod, 560 haddock and 280 plaice. The officer takes a stratified random sample of 60 fish, with the species as the strata.
(a)
Find the number of fish of each species in the sample.
(3)
(b)
Give one reason why stratified sampling is preferred to simple random sampling here.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sampling fraction 60/1680 (= 1/28) applied to each species
✓ 1
A1 30 cod
✓ 1
A1 20 haddock and 10 plaice
✓ 1
Part (b)
R1 it guarantees that each species is represented in proportion to its numbers in the catch (so the sample is more representative)
✓ 1
Exam tip: In stratified sampling, each stratum's sample size is (stratum size ÷ population size) × sample size.
106IBMI3S-4-04r47-07Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy6 marks
A fisheries survey of 120 boats records which species each boat caught on one day. 70 boats caught cod (C), 55 caught haddock (H) and 20 caught both. Each boat caught at least one of the two species or none at all.
(a)
Find the number of boats that caught neither cod nor haddock.
(2)
(b)
A boat is chosen at random. Given that it caught haddock, find the probability that it also caught cod.
(2)
(c)
Determine whether the events C and H are independent.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 70 + 55 − 20 = 105 boats caught at least one
✓ 1
A1 120 − 105 = 15 boats
✓ 1
Part (b)
M1 P(C | H) = 20/55
✓ 1
A1 0.364
✓ 1
Part (c)
M1 P(C) × P(H) = (70/120)(55/120) = 0.267
✓ 1
R1 P(C ∩ H) = 20/120 = 0.167 ≠ 0.267, so the events are not independent
✓ 1
Exam tip: To test independence compare P(A) × P(B) with P(A ∩ B); showing both numbers is needed for the reasoning mark.
107IBMI3S-4-04r47-14Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Easy5 marks
A sample of 100 cod caught in one net is grouped by mass, m grams.
Mass m (g)
100 ≤ m < 200
200 ≤ m < 300
300 ≤ m < 400
400 ≤ m < 500
500 ≤ m < 600
Frequency
8
22
35
25
10
(a)
Estimate the mean mass of the cod.
(2)
(b)
Estimate the population standard deviation of the masses, correct to 3 significant figures.
(1)
(c)
Write down the modal class.
(1)
(d)
Explain why the values in (a) and (b) are only estimates.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of mid-interval values 150, 250, 350, 450, 550
✓ 1
A1 357 g
✓ 1
Part (b)
A1 109 g
✓ 1
Part (c)
A1 300 ≤ m < 400
✓ 1
Part (d)
R1 the exact masses within each class are not known, so mid-interval values are used
✓ 1
Common error: Use the mid-interval value of each class, not the lower boundary, when estimating a mean from grouped data.
108IBMI3S-4-02r48-01Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 1Easy4 marks
Calculator
A pharmacist weighs a random sample of 10 tablets from a batch. The masses, in mg, are 498, 502, 505, 497, 501, 503, 499, 506, 500, 504.
(a)
Find the mean mass of the tablets.
(1)
(b)
Find the median mass.
(1)
(c)
Use your calculator to find the sample standard deviation \(s_{n-1}\) of the masses. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 501.5 mg
✓ 1
Part (b)
A1 501.5 mg
✓ 1
Part (c)
M1 use of the statistics function for the sample (unbiased) standard deviation, or \(\sqrt{\frac{\sum(x-\bar x)^2}{9}}\)
✓ 1
A1 3.03 mg
✓ 1
Common error: A calculator shows both the population (\(\sigma_x\)) and the sample (\(s_x\)) standard deviation. A sample taken from a larger batch needs \(s_{n-1}\).
109IBMI3S-4-02r48-02Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 1Easy5 marks
Calculator
The time taken by players to complete a level of a video game is normally distributed with mean 42 minutes and standard deviation 6.5 minutes.
(a)
Find the probability that a randomly chosen player completes the level in less than 35 minutes.
(2)
(b)
The slowest 15% of players receive a hint. Find the time above which a player receives a hint. Give your answer to the nearest 0.1 minute.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(T\sim N(42,6.5^2)\) and use of normal cdf for \(P(T<35)\)
✓ 1
A1 0.1408
✓ 1
Part (b)
M1 \(P(T>t)=0.15\), so \(P(T
✓ 1
M1 inverse normal: \(z=1.0364\) or \(t=42+1.0364\times6.5\)
✓ 1
A1 48.7 minutes
✓ 1
Exam tip: Sketch the normal curve and shade the region before using your calculator; for the slowest 15%, the area to the left of \(t\) is 0.85.
110IBMI3S-4-02r48-03Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 1Easy4 marks
Calculator
In a trial of a new drug, each patient independently has a probability of 0.35 of showing a response. A group of 12 patients is given the drug. Let \(X\) be the number of patients in the group who respond.
(a)
Find the probability that at most 2 patients respond.
(2)
(b)
Find the probability that exactly 5 patients respond.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(X\sim B(12,0.35)\) and use of binomial cdf for \(P(X\le2)\)
✓ 1
A1 0.1513
✓ 1
Part (b)
M1 \(\binom{12}{5}(0.35)^5(0.65)^7\) or binomial pdf
✓ 1
A1 0.2039
✓ 1
Common error: \(P(X\le2)\) includes \(X=0\), \(X=1\) and \(X=2\); use the cumulative function, not the single-value function.
111IBMI3S-4-02r48-04Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 1Easy6 marks
Calculator
A town's water supply uses two independent pumps, A and B. On any day pump A works with probability 0.92 and pump B works with probability 0.85.
(a)
Find the probability that both pumps fail on a given day.
(1)
(b)
Find the probability that at least one pump works on a given day.
(2)
(c)
Given that at least one pump works, find the probability that exactly one pump works. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(0.08\times0.15=0.012\)
✓ 1
Part (b)
M1 \(1-P(\text{both fail})\) or sum of the three working cases
Find the median, the lower quartile and the upper quartile of the battery lives.
(3)
(b)
An outlier is a value more than 1.5 × IQR above the upper quartile or below the lower quartile. Show that 21.7 is an outlier.
(3)
(c)
The consumer group discovers that the phone with 21.7 hours was faulty, so this value is removed. Find the mean battery life of the remaining phones.
(2)
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Notes
Part (a)
Median = 11.9 hours A1
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Lower quartile = 10.3 hours A1
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Upper quartile = 13.5 hours A1
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Part (b)
IQR = 3.2 and 13.5 + 1.5 × 3.2 M1
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Upper boundary = 18.3 A1
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21.7 > 18.3, so 21.7 is an outlier R1
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Part (c)
Sum of remaining 14 values = 162.6, divided by 14 M1
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11.6 hours A1
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Exam tip: For an outlier mark you need the comparison stated: give the boundary value and say clearly that the data value lies beyond it.
113IBMI-033-00-07Sampling and descriptive statisticsPaper 1Medium6 marks
A café records the number of cakes it sells each day for 8 days:
34, 41, 29, k, 38, 45, 36, 33
The mean number of cakes sold per day is 37.
(a)
Find the value of k.
(2)
(b)
Find the standard deviation of the number of cakes sold per day.
(2)
(c)
On day 9 the café sells exactly 37 cakes. State, with a reason, the effect this has on the mean and on the standard deviation.
(2)
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Notes
Part (a)
(256 + k) ÷ 8 = 37 M1
✓ 1
k = 40 A1
✓ 1
Part (b)
Uses all 8 values in GDC, or Σ(x − 37)2 = 180 M1
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4.74 cakes A1
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Part (c)
Mean remains 37 A1
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Standard deviation decreases, because the new value has zero deviation from the mean while the number of values increases R1
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Exam tip: Adding a value equal to the mean leaves the mean unchanged but always reduces the standard deviation, since the total squared deviation is shared among more values.
114IBMI-033-00-12Sampling and descriptive statisticsPaper 1Medium6 marks
A library surveys a random sample of students about the number of books, x, they read during the summer holiday.
x = 0 : 6 students x = 1 : 10 students x = 2 : p students x = 3 : 8 students x = 4 : 6 students
The mean number of books read is 1.95.
(a)
Find the value of p.
(3)
(b)
Write down the median number of books read.
(1)
(c)
Find the standard deviation of the number of books read.
(2)
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Notes
Part (a)
Σfx = 58 + 2p and Σf = 30 + p M1
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(58 + 2p) ÷ (30 + p) = 1.95 A1
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p = 10 A1
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Part (b)
2 books A1
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Part (c)
Uses p = 10 in GDC, or √(218 ÷ 40 − 1.952) M1
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1.28 books A1
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Exam tip: When a frequency is unknown, remember it appears in both the numerator (Σfx) and the denominator (Σf) of the mean.
115IBMI-033-07-r2-03Probability and the binomial distributionPaper 1Medium7 marks
A factory produces electronic components, 4% of which are defective. Every component passes through an automatic test. • If a component is defective, the test flags it with probability 0.95. • If a component is not defective, the test wrongly flags it with probability 0.03.
(a)
Find the probability that a randomly chosen component is flagged by the test.
(2)
(b)
A component is flagged. Find the probability that it is actually defective.
(2)
(c)
A batch of 50 components is tested. Outcomes for different components are independent. Find the probability that at most 2 components in the batch are flagged.
(3)
Show mark scheme
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Notes
Part (a)
0.04 × 0.95 + 0.96 × 0.03 M1
✓ 1
= 0.0668 A1
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Part (b)
(0.04 × 0.95) ÷ 0.0668 = 0.038 ÷ 0.0668 M1
✓ 1
= 0.569 A1
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Part (c)
Recognises binomial with n = 50 M1
✓ 1
and p = 0.0668, using P(Y ≤ 2) M1
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= 0.342 A1
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Exam tip: In part (c) the success probability is the overall flagging probability from part (a), not 0.95 or 0.04. Keep the unrounded value 0.0668 in your GDC.
116IBMI-033-07-r2-05Probability and the binomial distributionPaper 1Medium6 marks
A manufacturer of LED bulbs finds that 2% of its bulbs are faulty. Faults occur independently. Bulbs are packed into boxes.
(a)
Find the smallest number of bulbs a box must contain so that the probability of the box containing at least one faulty bulb is greater than 0.5.
(3)
(b)
A box contains 40 bulbs. Find the probability that it contains at least 2 faulty bulbs.
(3)
Show mark scheme
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Mark
Notes
Part (a)
1 − 0.98n > 0.5, i.e. 0.98n < 0.5 M1
✓ 1
n > 34.3 A1
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Smallest n = 35 A1
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Part (b)
X ~ B(40, 0.02) M1
✓ 1
P(X ≥ 2) = 1 − P(X ≤ 1) M1
✓ 1
= 0.190 A1
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Common error: When solving 0.98^n < 0.5, n = 34.3 must be rounded up to 35, since n must be an integer and 34 bulbs gives a probability below 0.5.
117IBMI-033-08-r2-04Normal distribution and chi-squared testsPaper 1Medium7 marks
The owner of a café claims that customers choose coffee, tea, hot chocolate and juice in the ratio 5 : 2 : 2 : 1. To test this claim, the drink choices of 160 randomly selected customers are recorded.
Coffee 72, Tea 38, Hot chocolate 34, Juice 16
A χ2 goodness of fit test is carried out at the 5% significance level.
(a)
Write down the null hypothesis for this test.
(1)
(b)
Find the expected frequency for each drink.
(2)
(c)
Find the value of the χ2 test statistic.
(2)
(d)
The critical value for this test is 7.815. State, with a reason, whether the data support the owner's claim.
(2)
Show mark scheme
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Mark
Notes
Part (a)
H0: customers choose coffee, tea, hot chocolate and juice in the ratio 5 : 2 : 2 : 1
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Part (b)
Uses proportions of 160, e.g. \(\frac{5}{10} \times 160\)
✓ 1
Coffee 80, tea 32, hot chocolate 32, juice 16
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Part (c)
Correct use of \(\sum \frac{(O-E)^2}{E}\) or GDC with correct observed and expected lists
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χ2calc = 2.05
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Part (d)
2.05 < 7.815 (or p-value 0.562 > 0.05), so do not reject H0
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There is insufficient evidence to doubt the owner's claim; the data are consistent with the 5 : 2 : 2 : 1 ratio
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Common error: Failing to reject H₀ does not prove the claim is true. Say the data are consistent with the claim, or that there is insufficient evidence against it.
118IBMI-033-05-r1-04Sampling and descriptive statisticsPaper 1Medium6 marks
Over 30 days in June, the daily maximum temperature in City A had a mean of 18.4 °C and a standard deviation of 3.5 °C. A travel website converts temperatures to degrees Fahrenheit using F = 1.8C + 32.
(a)
Find the mean daily maximum temperature for City A in °F.
(2)
(b)
Find the standard deviation of the daily maximum temperatures for City A in °F.
(2)
(c)
Over 20 days in the same June, City B had a mean daily maximum temperature of 70.2 °F. Find the mean daily maximum temperature, in °F, of all 50 recorded days from both cities combined.
(2)
Show mark scheme
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Notes
Part (a)
M1: 1.8 × 18.4 + 32
✓ 1
A1: 65.1 °F (65.12)
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Part (b)
M1: 1.8 × 3.5, with the +32 correctly ignored
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A1: 6.30 °F
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Part (c)
M1: \(\frac{30\times 65.12+20\times 70.2}{50}\)
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A1: 67.2 °F (67.152)
✓ 1
Common error: Adding a constant shifts the mean but not the spread, so the standard deviation is only multiplied by 1.8; adding 32 to it is wrong.
119IBMI-033-03-r2-04Normal distribution and chi-squared testsPaper 1Medium7 marks
The manager of a customer helpline records the number of complaints received on each weekday over one month. The totals are:
The manager tests, at the 1% significance level, whether complaints are spread evenly across the five weekdays.
(a)
State the null hypothesis for this test.
(1)
(b)
Write down the expected frequency for each day and the number of degrees of freedom.
(2)
(c)
Find the χ2 statistic and the p-value.
(2)
(d)
State the conclusion of the test in context, giving a reason.
(2)
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Notes
Part (a)
H0: complaints are uniformly distributed across the five weekdays (equally likely on each day) (A1)
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Part (b)
Expected frequency = 50 (A1)
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Degrees of freedom = 5 − 1 = 4 (A1)
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Part (c)
χ2 = 12.5 (12.52) (A1)
✓ 1
p = 0.0139 (A1)
✓ 1
Part (d)
0.0139 > 0.01 so do not reject H0 (R1)
✓ 1
There is insufficient evidence at the 1% level that complaints are unevenly spread across the weekdays (A1)
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Do not award A1 without R1.
—
Common error: A large-looking χ² value does not automatically mean rejection: compare the p-value with the stated significance level, here 1%, not 5%.
120IBMI-033-03-r2-05Normal distribution and chi-squared testsPaper 1Medium8 marks
Priya drives to work each day. Her journey time, T minutes, is modelled by a normal distribution with mean 38.5 minutes and standard deviation 6.2 minutes.
(a)
Find the probability that a journey takes longer than 45 minutes.
(2)
(b)
Priya wants to leave early enough to arrive on time on 90% of days. Find the journey time t such that P(T < t) = 0.9.
(2)
(c)
Priya records her journeys over 20 working days. State one assumption needed to model the number of journeys longer than 45 minutes by a binomial distribution.
(1)
(d)
Find the probability that more than 3 of the 20 journeys take longer than 45 minutes.
(3)
Show mark scheme
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Notes
Part (a)
Correct use of normal cdf with lower bound 45, μ = 38.5, σ = 6.2 (M1)
✓ 1
P(T > 45) = 0.147 (0.14723…) (A1)
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Part (b)
Use of inverse normal with area 0.9 (M1)
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t = 46.4 minutes (A1)
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Part (c)
Journey times on different days are independent (or the probability of a journey exceeding 45 minutes is the same each day) (R1)
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Part (d)
Recognising Y ~ B(20, 0.14723) (M1)
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P(Y > 3) = 1 − P(Y ≤ 3) (M1)
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= 0.339 (accept 0.338 from p = 0.147) (A1)
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Common error: 'More than 3' means 1 − P(Y ≤ 3), not 1 − P(Y ≤ 4) or 1 − P(Y ≤ 2).
121IBMI-033-02-r2-03Probability and the binomial distributionPaper 1Medium7 marks
A screening test is used for a condition that affects 3% of a population. If a person has the condition, the test gives a positive result with probability 0.95. If a person does not have the condition, the test gives a positive result with probability 0.08. A person is chosen at random and tested.
(a)
Find the probability that the test result is positive.
(3)
(b)
Given that the test result is positive, find the probability that the person has the condition.
(3)
(c)
Comment on the reliability of a positive result from this test.
(1)
Show mark scheme
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Mark
Notes
Part (a)
M1: multiplies along branches, e.g. \(0.03 \times 0.95\) or \(0.97 \times 0.08\)
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M1: adds the two positive branches \(0.0285 + 0.0776\)
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A1: \(0.106\)
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Part (b)
M1: numerator \(0.03 \times 0.95 = 0.0285\)
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M1: divides by their answer to (a), \(\frac{0.0285}{0.1061}\)
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A1: \(0.269\)
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Part (c)
R1: about 73% of positive results are false positives, so a positive result alone is unreliable and should be confirmed by a further test
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Exam tip: Draw a tree diagram first. Conditional probability questions become one division once both positive branches are labelled.
122IBMI-033-02-r1-04Probability and the binomial distributionPaper 1Medium8 marks
A basketball player scores each free throw with probability 0.72, independently of other throws. In a match she takes 15 free throws. Let \(X\) be the number she scores.
(a)
Find the mean and the standard deviation of \(X\).
(2)
(b)
In a separate training drill, the player takes \(n\) free throws. Find the least value of \(n\) for which the probability that she misses at least one throw is greater than 0.95.
(3)
(c)
Find the probability that, in the match, the number of free throws she scores is within one standard deviation of the mean.
(3)
Show mark scheme
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Notes
Part (a)
A1: mean \(=15\times0.72=10.8\)
✓ 1
A1: standard deviation \(=\sqrt{15\times0.72\times0.28}=\sqrt{3.024}=1.74\)
✓ 1
Part (b)
M1: forms the inequality \(1-0.72^n>0.95\), i.e. \(0.72^n<0.05\)
✓ 1
A1: \(n>9.12\) (by logarithms or GDC)
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A1: least \(n=10\)
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Part (c)
M1: interval \(10.8\pm1.74\), i.e. \(9.06
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A1: identifies \(X=10,\,11,\,12\), i.e. \(P(10\le X\le12)\)
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A1: 0.614
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Common error: In part (c), \(X\) only takes whole-number values. The interval 9.06 to 12.54 contains 10, 11 and 12 only. Including 9 or 13 is a frequent slip.
123IBMI-033-01-r1-03Correlation and regressionPaper 1Medium6 marks
A food blogger visits seven cafés and records the price of a standard coffee and gives each café an overall rating out of 10.
Café: A, B, C, D, E, F, G Price ($): 2.80, 3.10, 3.45, 2.60, 3.90, 3.25, 4.20 Rating: 6.8, 7.4, 7.1, 6.2, 8.5, 7.9, 8.1
(a)
Rank the cafés by price and by rating, giving rank 1 to the lowest value in each case.
(2)
(b)
Calculate Spearman's rank correlation coefficient, rs, for these data.
(2)
(c)
Interpret your value of rs in context.
(1)
(d)
The blogger concludes that raising the price of coffee will cause a café's rating to improve. Comment on this conclusion.
(1)
Show mark scheme
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Notes
Part (a)
Price ranks A–G: 2, 3, 5, 1, 6, 4, 7
✓ 1
Rating ranks A–G: 2, 4, 3, 1, 7, 5, 6
✓ 1
Part (b)
M1: correct use of ranks, e.g. GDC with ranked lists or differences 0, −1, 2, 0, −1, −1, 1 giving Σd² = 8
✓ 1
A1: rs = 0.857
✓ 1
Part (c)
Strong positive association: cafés with higher coffee prices tend to receive higher ratings
✓ 1
Part (d)
Not justified: correlation does not imply causation; a third factor such as quality of food or location may influence both
✓ 1
Common error: Ranking the data in opposite directions for the two variables reverses the sign of rs. Use the same direction (lowest = 1) for both variables.
124IBMI-033-05-r2-04Sampling and descriptive statisticsPaper 1Medium6 marks
The daily maximum temperatures in a city over 30 days in June had a mean of 18.4 °C and a standard deviation of 3.5 °C. A travel website converts the temperatures to degrees Fahrenheit using F = 1.8C + 32.
(a)
Find the mean of the temperatures in °F.
(2)
(b)
Find the standard deviation of the temperatures in °F.
(2)
(c)
On one day the maximum temperature was 24.0 °C. Find how many standard deviations above the mean this temperature is, using the °F values.
(2)
Show mark scheme
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Mark
Notes
Part (a)
1.8 × 18.4 + 32 M1
✓ 1
65.1 °F (65.12) A1
✓ 1
Part (b)
Recognising that adding 32 does not change the spread, so multiply by 1.8 only M1
✓ 1
6.3 °F A1
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Part (c)
24.0 °C = 75.2 °F and \(\frac{75.2-65.12}{6.3}\) M1
✓ 1
1.6 standard deviations (same as in °C) A1
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Common error: Adding a constant shifts the mean but leaves the standard deviation unchanged. Only the multiplier 1.8 affects the spread, so 1.8 × 3.5 + 32 is wrong.
125IBMI-033-06-r1-03Correlation and regressionPaper 1Medium7 marks
A consumer website compares seven smartphone models, A to G. The table gives the price, in dollars, and the mean customer satisfaction score out of 10 for each model.
Model: A, B, C, D, E, F, G Price ($): 180, 240, 310, 395, 450, 620, 890 Satisfaction score: 6.4, 6.1, 7.3, 7.0, 8.8, 7.9, 8.2
(a)
Find Spearman's rank correlation coefficient, rs, between price and satisfaction score.
(3)
(b)
Interpret your answer to part (a) in context.
(1)
(c)
Suggest why Spearman's rank correlation coefficient may be more appropriate than Pearson's product-moment correlation coefficient for these data.
(1)
(d)
The price of model G is increased to $1200. State, with a reason, whether the value of rs changes.
(2)
Show mark scheme
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Mark
Notes
Part (a)
Ranks both variables correctly, e.g. price ranks 1, 2, 3, 4, 5, 6, 7 and score ranks 2, 1, 4, 3, 7, 5, 6
✓ 1
Uses the ranks in the GDC or in \(1-\frac{6\sum d^2}{n(n^2-1)}\) with \(\sum d^2 = 10\)
✓ 1
rs = 0.821
✓ 1
Part (b)
There is a strong positive association: more expensive models tend to have higher satisfaction scores
✓ 1
Part (c)
The relationship need not be linear (e.g. scores level off at high prices) or the high price of G is an extreme value; Spearman's only requires a monotonic relationship
✓ 1
Part (d)
rs does not change
✓ 1
Reason: G is still the most expensive, so all the ranks are unchanged
✓ 1
Exam tip: Spearman's coefficient depends only on the ranks. Any change to a data value that keeps the order the same leaves rs unchanged.
126IBMI-033-06-r1-06Correlation and regressionPaper 1Medium6 marks
A researcher collects data from 10 towns on the number of public libraries, x, and the number of crimes reported in one year, y. The number of libraries ranges from 2 to 12. The Pearson's product-moment correlation coefficient is r = 0.91. The regression line of y on x is y = 38.6x + 12.4.
(a)
Explain why it is appropriate to use the regression line to estimate the number of crimes in a town with 7 libraries.
(2)
(b)
A town with 7 libraries reported 305 crimes. Find the residual for this town.
(2)
(c)
A councillor claims that building more libraries causes crime to increase. Comment on this claim.
(2)
Show mark scheme
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Notes
Part (a)
r = 0.91 shows strong linear correlation
✓ 1
x = 7 is within the range of the data (2 to 12), so this is interpolation
✓ 1
Part (b)
Predicted value 38.6 × 7 + 12.4 = 282.6
✓ 1
Residual = 305 − 282.6 = 22.4
✓ 1
Part (c)
Correlation does not imply causation, so the claim is not supported
✓ 1
A third variable such as town population is likely to increase both the number of libraries and the number of crimes
✓ 1
Common error: A residual is observed minus predicted. Working out predicted minus observed gives the wrong sign, so a point above the line would wrongly get a negative residual.
127IBMI-033-03-r1-04Normal distribution and chi-squared testsPaper 1Medium6 marks
A café owner records the number of loaves of sourdough sold on each weekday over a period of several weeks.
She tests at the 10% significance level whether sales are uniformly distributed across the five weekdays.
(a)
State the null hypothesis.
(1)
(b)
Write down the expected frequency for each day.
(1)
(c)
Find the χ2 statistic and the p-value.
(2)
(d)
State the conclusion of the test in context, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1: H0: sales of sourdough are uniformly distributed across the five weekdays
✓ 1
Part (b)
A1: 250 ÷ 5 = 50
✓ 1
Part (c)
A1: χ2 = 5.72
✓ 1
A1: p = 0.221 (4 degrees of freedom)
✓ 1
Part (d)
R1: 0.221 > 0.10
✓ 1
A1: do not reject H0; there is insufficient evidence that sourdough sales differ across weekdays
✓ 1
Common error: Failing to reject H₀ does not prove that sales are uniform. Say there is insufficient evidence of a difference.
128IBMI-033-02-r1-03Probability and the binomial distributionPaper 1Medium7 marks
A factory packs rechargeable batteries in boxes of 50. Each battery is defective with probability 0.04, independently of the others. A box is rejected if it contains more than 3 defective batteries.
(a)
Find the probability that a box contains no defective batteries.
(2)
(b)
Show that the probability that a box is rejected is 0.139, correct to 3 significant figures.
(2)
(c)
A shop receives a delivery of 12 boxes. Find the probability that at least 2 of these boxes would be rejected.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: \(X\sim B(50,\,0.04)\), \(P(X=0)=0.96^{50}\)
✓ 1
A1: 0.130
✓ 1
Part (b)
M1: \(P(X>3)=1-P(X\le3)=1-0.86087\ldots\)
✓ 1
A1: \(0.13913\ldots=0.139\)
✓ 1
Part (c)
M1: recognises \(Y\sim B(12,\,0.139)\) for the number of rejected boxes
✓ 1
M1: uses \(P(Y\ge2)=1-P(Y\le1)\)
✓ 1
A1: 0.513 using the unrounded \(p=0.13913\ldots\) (accept 0.512 from \(p=0.139\))
✓ 1
Exam tip: Carry the unrounded rejection probability (0.13913…) into part (c). Using the rounded 0.139 gives 0.512 rather than 0.513, and early rounding in a second binomial model can cost the final accuracy mark.
129IBMI-033-06-r2-03Correlation and regressionPaper 1Medium7 marks
An ecologist records the altitude, in metres, and the number of plant species found at eight mountain sites.
Site: A, B, C, D, E, F, G, H Altitude (m): 120, 340, 560, 410, 890, 1020, 730, 250 Number of species: 48, 45, 31, 39, 22, 12, 18, 41
(a)
Calculate Spearman's rank correlation coefficient, rs, for these data.
(3)
(b)
Interpret your answer to part (a) in context.
(1)
(c)
The ecologist notices that the number of species changes little between 120 m and 410 m but falls much more rapidly at higher altitudes. Explain why Spearman's coefficient is more appropriate than Pearson's coefficient for these data.
(1)
(d)
A ninth site, at an altitude of 1200 m, is found to have 9 plant species. Calculate the new value of rs for all nine sites.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: both variables ranked correctly, e.g. altitude ranks 1, 3, 5, 4, 7, 8, 6, 2 and species ranks 8, 7, 4, 5, 3, 1, 2, 6
✓ 1
M1: Pearson's r found for the ranks on a GDC, or Σd² = 164 substituted into \(1-\frac{6\sum d^2}{n(n^2-1)}\)
✓ 1
A1: rs = −0.952
✓ 1
Part (b)
Very strong negative association: higher sites tend to have fewer plant species
✓ 1
Part (c)
R1: the relationship appears to be monotonic (decreasing) but curved, not linear. Spearman's coefficient measures monotonic association, whereas Pearson's measures only linear correlation
✓ 1
Part (d)
M1: re-ranks all nine sites. The new site has altitude rank 9 and species rank 1, and Σd² = 236
✓ 1
A1: rs = −0.967
✓ 1
Common error: When a new data point is added, the existing ranks can change. Here every original species rank goes up by 1, so all the ranks must be redone, not just the new one.
130IBMI-033-07-r1-05Probability and the binomial distributionPaper 1Medium8 marks
In a sixth-form college, 120 students were asked which foreign languages they study. 65 study Spanish, 48 study French and 20 study both Spanish and French. A student is chosen at random. Let S be the event that the student studies Spanish and F the event that the student studies French.
(a)
Find the probability that the student studies neither Spanish nor French.
(2)
(b)
Given that the student does not study Spanish, find the probability that the student studies French.
(2)
(c)
Determine whether the events S and F are independent. Justify your answer.
(2)
(d)
Two different students are chosen at random. Find the probability that both of them study Spanish and French.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Number studying at least one = 65 + 48 − 20 = 93, so neither = 120 − 93 = 27 M1
✓ 1
P = \(\frac{27}{120}\) = 0.225 A1
✓ 1
Part (b)
French only = 48 − 20 = 28 and not Spanish = 120 − 65 = 55 M1
The values are not equal, so S and F are not independent R1
✓ 1
Part (d)
\(\frac{20}{120} \times \frac{19}{119}\) M1
✓ 1
= 0.0266 A1
✓ 1
Exam tip: When two different people are chosen, the second probability changes: use 19/119, not 20/120 again.
131IBMI-033-08-r1-04Normal distribution and chi-squared testsPaper 1Medium7 marks
A genetic model predicts that guinea pigs born to a certain pair of parent types will have black, brown and white coats in the ratio 2 : 1 : 1. A breeder records the coat colours of 200 such guinea pigs: 112 black, 41 brown and 47 white. A \(\chi^2\) goodness of fit test is carried out at the 10% significance level.
(a)
State the null and alternative hypotheses.
(2)
(b)
Write down the expected frequencies.
(1)
(c)
Find the \(\chi^2\) statistic and the p-value.
(2)
(d)
State, with a reason, whether the data support the genetic model.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
H0: coat colours occur in the ratio 2 : 1 : 1
✓ 1
H1: coat colours do not occur in the ratio 2 : 1 : 1
✓ 1
Part (b)
Black 100, brown 50, white 50
✓ 1
Part (c)
$\chi^2_{calc}$ = 3.24
✓ 1
p = 0.198 (3 s.f.)
✓ 1
Part (d)
0.198 > 0.10 (or 3.24 < critical value 4.605), so do not reject H0
✓ 1
There is insufficient evidence to reject the 2 : 1 : 1 model, so the data are consistent with it. This mark depends on a correct comparison.
✓ 1
Exam tip: 'Do not reject H₀' is not the same as proving the model correct. Phrase the conclusion as 'insufficient evidence to reject' or 'consistent with' the model.
132IBMI-033-01-r2-04Correlation and regressionPaper 1Medium7 marks
A biologist measures the wing length, x mm, and the mass, y g, of eight birds of the same species. The masses are 22.1, 25.4, 24.0, 27.8, 23.5, 26.9, 21.7, k. The wing lengths range from 40.5 mm to 49.5 mm, and the mean wing length is 45.0 mm. The regression line of y on x is \(y = 0.84x - 12.6\), and the Pearson correlation coefficient is r = 0.67.
(a)
Show that the mean mass of the eight birds is 25.2 g.
(2)
(b)
Find the value of k.
(2)
(c)
A student rearranges the regression equation to estimate the wing length of a bird with mass 28 g. Explain why this method may not be appropriate.
(1)
(d)
Use the regression line to estimate the mass of a bird with wing length 48 mm, and comment on the reliability of your estimate.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The regression line passes through the mean point \((\bar{x}, \bar{y})\) (R1)
Sum of masses = 8 × 25.2 = 201.6, and the seven known masses sum to 171.4 (M1)
✓ 1
k = 201.6 − 171.4 = 30.2 (A1)
✓ 1
Part (c)
The line of y on x is designed to predict y from x; estimating x from y requires the regression line of x on y (R1)
✓ 1
Part (d)
y = 0.84(48) − 12.6 = 27.7 g (A1)
✓ 1
Reasonably reliable, because 48 mm lies within the data range (interpolation) and r = 0.67 shows a moderate positive correlation (R1)
✓ 1
Exam tip: Any least-squares regression line passes through \((\bar{x}, \bar{y})\). This is often the fastest route to a missing data value.
133IBMI-033-07-r1-03Probability and the binomial distributionPaper 1Medium7 marks
A garden centre sells packets of 40 tomato seeds. Each seed independently has a probability of 0.85 of germinating. Let X be the number of seeds in a packet that germinate.
(a)
Write down the mean of X.
(1)
(b)
Find the standard deviation of X.
(2)
(c)
The garden centre advertises that “at least 35 seeds in every packet will germinate”. Find the probability that a randomly chosen packet meets this claim.
(2)
(d)
A customer buys 3 packets. Find the probability that exactly 2 of the packets meet the claim.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
E(X) = 40 × 0.85 = 34 A1
✓ 1
Part (b)
Var(X) = 40 × 0.85 × 0.15 = 5.1 M1
✓ 1
σ = √5.1 = 2.26 A1
✓ 1
Part (c)
P(X ≥ 35) = 1 − P(X ≤ 34), or uses the binomial cumulative distribution correctly M1
✓ 1
= 0.433 (0.43250…) A1
✓ 1
Part (d)
Uses B(3, 0.43250…), e.g. 3 × 0.43252 × 0.5675 M1
✓ 1
= 0.318 (accept 0.319 from using 0.433) A1
✓ 1
Exam tip: Carry the full calculator value of P(X ≥ 35) into the final part. Rounding it to 0.433 first can change the third significant figure of the answer.
134IBMI3S-4-07-05Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 1Medium7 marks
The length, L cm, of steel rods cut by a machine is normally distributed with mean μ and standard deviation σ. It is known that 10% of rods are shorter than 48.4 cm and 5% of rods are longer than 51.5 cm.
(a)
Find the standardised value (z-value) corresponding to 48.4 cm and the z-value corresponding to 51.5 cm.
(2)
(b)
Hence find the value of σ.
(3)
(c)
Find the value of μ.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(z=-1.28\) for 48.4 cm
✓ 1
A1 \(z=1.64\) for 51.5 cm
✓ 1
Part (b)
M1 forms \(\frac{48.4-\mu}{\sigma}=-1.2816\) and \(\frac{51.5-\mu}{\sigma}=1.6449\)
✓ 1
M1 subtracts (or eliminates μ) to get \(\sigma=\frac{3.1}{2.9265}\)
✓ 1
A1 σ = 1.06
✓ 1
Part (c)
M1 substitute the unrounded σ into one equation, e.g. \(\mu=48.4+1.2816\sigma\)
✓ 1
A1 μ = 49.8 cm
✓ 1
Common error: The lower z-value is negative: a probability below 0.5 to the left of the mean gives a negative z.
135IBMI3S-4-07-06Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 1Medium7 marks
A shop manager believes that customer numbers are equally likely on each of the five weekdays. Over one week, 240 customers were recorded:
Mon 52, Tue 38, Wed 45, Thu 41, Fri 64.
A χ2 goodness-of-fit test at the 5% significance level is used to test the manager's belief.
(a)
State the null and alternative hypotheses.
(1)
(b)
Find the expected frequency for each day and calculate the χ2 statistic.
(2)
(c)
Write down the number of degrees of freedom and the p-value.
(2)
(d)
State the conclusion of the test, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: customers are uniformly distributed over the five days; H1: they are not
✓ 1
Part (b)
M1 expected frequency \(240\div5=48\)
✓ 1
A1 χ2 = 8.96
✓ 1
Part (c)
A1 degrees of freedom = 4
✓ 1
A1 p-value = 0.0621
✓ 1
Part (d)
R1 0.0621 > 0.05
✓ 1
A1 do not reject H0: insufficient evidence that customer numbers differ between the weekdays
✓ 1
Exam tip: Compare the p-value with the significance level in a sentence, then state the conclusion in the context of the question.
136IBMI3S-4-07-09Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 1Medium6 marks
Eight tablet computers are tested in a laboratory. Each receives a durability score, d (out of 100), and an average customer rating, c (out of 5).
Give one reason why Spearman's coefficient is more suitable than Pearson's coefficient for customer ratings.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 durability ranks: 4, 3, 7, 1, 8, 2, 5, 6
✓ 1
A1 rating ranks: 4, 2, 8, 1, 7, 3, 5, 6
✓ 1
Part (b)
M1 use of ranks, e.g. \(\sum d^2=4\) or a correlation calculation on the ranks
✓ 1
A1 \(r_s=0.952\)
✓ 1
Part (c)
A1 strong positive association: tablets that score higher for durability tend to receive higher customer ratings
✓ 1
Part (d)
R1 ratings are subjective/ordinal, so only their order is reliable, and the relationship need not be linear
✓ 1
Exam tip: Spearman's coefficient is calculated from ranks; assign rank 1 consistently to the smallest (or largest) value in both lists.
137IBMI3S-4-07-10Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 1Medium6 marks
A delivery drone has a battery failure on any one flight with probability 0.03, independently of other flights.
(a)
Find the probability that exactly one failure occurs in 20 flights.
(2)
(b)
Find the smallest number of flights, n, for which the probability of at least one failure exceeds 0.9.
(3)
(c)
Find the expected number of failures in n flights, using your value of n.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(X\sim B(20,0.03)\), \(P(X=1)\)
✓ 1
A1 0.336
✓ 1
Part (b)
M1 \(1-0.97^n>0.9\)
✓ 1
M1 solve \(0.97^n<0.1\), e.g. by logarithms or a table
✓ 1
A1 n = 76
✓ 1
Part (c)
A1 \(76\times0.03=2.28\) (ECF from (b))
✓ 1
Exam tip: For 'smallest n' questions check both the answer and the value just below it (here 75 flights gives 0.898, which is not enough).
138IBMI3S-4-07-11Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 1Medium7 marks
The time, T hours, taken by amateur runners to complete a marathon is normally distributed with mean 4.2 and standard deviation 0.6.
(a)
Find the probability that a randomly chosen runner finishes in under 3.5 hours.
(2)
(b)
A club has 12 runners who enter the race. Find the probability that at least three of them finish in under 3.5 hours.
(3)
(c)
The fastest 10% of runners receive a medal. Find the slowest time that earns a medal.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(T\sim N(4.2,0.6^2)\) with \(P(T<3.5)\)
✓ 1
A1 0.122
✓ 1
Part (b)
M1 \(X\sim B(12,0.1217)\)
✓ 1
M1 \(1-P(X\le2)\)
✓ 1
A1 0.172
✓ 1
Part (c)
M1 \(P(T<t)=0.1\), \(z=-1.2816\)
✓ 1
A1 3.43 hours
✓ 1
Common error: Use the unrounded probability from part (a) in the binomial model; a value rounded to 0.12 gives a different answer.
139IBMI3S-4-07b-05Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 1Medium7 marks
The finishing times, T minutes, of runners in a city marathon are modelled by a normal distribution with mean 232 and standard deviation 24. There are 800 runners.
(a)
Find the probability that a randomly chosen runner finishes in under 210 minutes.
(2)
(b)
The fastest 15% of runners receive a medal. Find the slowest time that earns a medal.
(3)
(c)
Estimate how many of the 800 runners finish in under 210 minutes.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 normal cdf with mean 232, SD 24
✓ 1
A1 P(T < 210) = 0.18
✓ 1
Part (b)
M1 inverse normal with area 0.15
✓ 1
A1 z = −1.036 or equivalent GDC use
✓ 1
A1 207 minutes
✓ 1
Part (c)
M1 800 × 0.1797
✓ 1
A1 144 runners
✓ 1
Common error: Give the count of people as a whole number, and keep the unrounded probability in your calculator for the multiplication.
140IBMI3S-4-07b-08Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 1Medium8 marks
On a commuter rail line, the probability of rain at the start of a morning is 0.35. If it rains, the probability that the train is late is 0.4. If it does not rain, the probability that the train is late is 0.1.
(a)
Find the probability that the train is late.
(3)
(b)
Given that the train is late, find the probability that it rained.
(3)
(c)
Determine whether 'rain' and 'train late' are independent events. Justify your answer.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of the law of total probability or tree diagram
✓ 1
A1 0.35 × 0.4 + 0.65 × 0.1
✓ 1
A1 0.205
✓ 1
Part (b)
M1 conditional probability P(R ∩ L) / P(L)
✓ 1
A1 0.14 / 0.205
✓ 1
A1 0.683
✓ 1
Part (c)
M1 compares P(L | R) = 0.4 with P(L) = 0.205
✓ 1
R1 they are not equal, so the events are not independent
✓ 1
Common error: P(rain | late) is not P(late | rain): the order of the conditioning matters, so divide by P(late).
141IBMI3S-4-07b-09Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 1Medium7 marks
A footballer scores from a penalty with probability 0.78. In a training session she takes 15 penalties. Assume each penalty is independent. Let X be the number of penalties she scores.
(a)
State two assumptions needed for X to be modelled by a binomial distribution.
(2)
(b)
Find P(X = 12).
(2)
(c)
Find P(X ≤ 10).
(1)
(d)
Find the probability that she misses at least once.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 the probability of scoring is constant for every penalty
✓ 1
R1 penalties are independent of one another
✓ 1
Part (b)
M1 binomial model B(15, 0.78)
✓ 1
A1 0.246
✓ 1
Part (c)
A1 0.219
✓ 1
Part (d)
M1 1 − P(scores all 15) = 1 − 0.7815
✓ 1
A1 0.976
✓ 1
Exam tip: 'At least one' problems are usually quickest by the complement: 1 − P(none).
142IBMI3S-4-07b-14Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 1Medium7 marks
The distances to a sample of nearby stars have a mean of 26.0 light-years and a standard deviation of 8.5 light-years. One light-year is 0.3066 parsecs.
(a)
Find the mean and standard deviation of the distances in parsecs.
(2)
(b)
A calibration error is found. Every distance in parsecs must be increased by 0.4. State the new mean and the new standard deviation.
(2)
(c)
A star is measured at 52 light-years. Find how many standard deviations this is above the original mean, and state whether it is more than 2 standard deviations away.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 mean = 26.0 × 0.3066 = 7.97 parsecs
✓ 1
A1 standard deviation = 8.5 × 0.3066 = 2.61 parsecs
✓ 1
Part (b)
A1 new mean = 7.9716 + 0.4 = 8.37 parsecs
✓ 1
A1 standard deviation unchanged at 2.61 parsecs
✓ 1
Part (c)
M1 (52 − 26.0) / 8.5
✓ 1
A1 3.06
✓ 1
R1 3.06 > 2, so it is more than 2 standard deviations above the mean
✓ 1
Common error: Adding a constant changes the mean but not the standard deviation; multiplying by a constant scales both.
143IBMI3S-4-01r7-05Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium7 marks
A port authority records the day of the week on which each of 150 container ships arrives (Monday to Friday). It claims that arrivals are equally likely on each of the five days. The observed numbers are: Monday 38, Tuesday 24, Wednesday 31, Thursday 29, Friday 28. A χ2 goodness-of-fit test is carried out at the 5% significance level.
(a)
State the null and alternative hypotheses.
(2)
(b)
Write down the expected frequency for each day and calculate the χ2 statistic, correct to 3 significant figures.
(3)
(c)
The degrees of freedom are 4 and the p-value is 0.473. State, with a reason, whether the port authority’s claim is rejected.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: arrivals are equally likely on each of the five days (the uniform model fits)
✓ 1
A1 H1: arrivals are not equally likely on each day
✓ 1
Part (b)
A1 expected frequency 30
✓ 1
M1 Σ(O − E)2/E = 82+62+12+12+22⁄30
✓ 1
A1 χ2 = 3.53
✓ 1
Part (c)
A1 0.473 > 0.05
✓ 1
R1 do not reject H0; insufficient evidence that arrivals are not equally likely
✓ 1
Exam tip: The degrees of freedom for a goodness-of-fit test are the number of categories minus 1 when no parameters are estimated.
144IBMI3S-4-01r7-06Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium7 marks
A theatre records the type of show chosen by 250 customers and whether each customer is under 30 or aged 30 and over.
Drama
Musical
Comedy
Under 30
28
52
40
30 and over
62
48
20
A χ2 test for independence is carried out at the 5% significance level.
(a)
State the null and alternative hypotheses.
(2)
(b)
Show that the expected frequency for under-30 customers choosing comedy is 28.8.
(2)
(c)
Use your calculator to find the χ2 statistic and the p-value, each correct to 3 significant figures.
(2)
(d)
State the conclusion of the test, with a reason.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: show type and age group are independent
✓ 1
A1 H1: show type and age group are not independent (associated)
✓ 1
Part (b)
M1 120 × 60⁄250
✓ 1
A1 28.8 (shown)
✓ 1
Part (c)
A1 χ2 = 19.3
✓ 1
A1 p = 6.44 × 10−5
✓ 1
Part (d)
R1 p < 0.05, so reject H0; there is evidence of an association between age group and show type
✓ 1
Common error: Quote the comparison (p-value against 0.05) in words before giving the conclusion in context.
145IBMI3S-4-01r7-07Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium6 marks
The masses of shipping containers loaded at a port are normally distributed with mean μ tonnes and standard deviation σ tonnes. 10% of containers have a mass less than 12 tonnes and 20% of containers have a mass greater than 20 tonnes.
(a)
Write down two equations connecting μ and σ. Use z-values of −1.2816 and 0.8416, correct to 4 decimal places.
(2)
(b)
Find σ, correct to 3 significant figures.
(2)
(c)
Hence find μ, correct to 3 significant figures. Use the unrounded value of σ.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 (12 − μ)/σ = −1.2816
✓ 1
A1 (20 − μ)/σ = 0.8416
✓ 1
Part (b)
M1 subtract the equations: 8 = (1.2816 + 0.8416)σ
✓ 1
A1 σ = 3.77
✓ 1
Part (c)
M1 μ = 12 + 1.2816 σ
✓ 1
A1 μ = 16.8
✓ 1
Exam tip: Sketching the normal curve with the two tail areas shows which z-value is negative.
146IBMI3S-4-01r7-08Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium6 marks
Each of the 15 members of a drama group independently attends a rehearsal with probability p. The number attending, X, has mean 10.5.
(a)
Find the value of p.
(2)
(b)
Find P(X = 12), correct to 3 significant figures.
(2)
(c)
Find the probability that fewer than 10 members attend, correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 15p = 10.5
✓ 1
A1 p = 0.7
✓ 1
Part (b)
M1 C(15,12) × 0.712 × 0.33
✓ 1
A1 0.17
✓ 1
Part (c)
M1 P(X ≤ 9) using B(15, 0.7)
✓ 1
A1 0.278
✓ 1
Common error: “Fewer than 10” means X ≤ 9, not X ≤ 10.
147IBMI3S-4-01r7-09Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium6 marks
A cargo ship leaves a port during a storm with probability 0.35. If it leaves during a storm, the probability that it is delayed is 0.6. If it leaves when there is no storm, the probability that it is delayed is 0.15.
(a)
Find the probability that the ship is delayed.
(3)
(b)
Given that the ship is delayed, find the probability that it left during a storm. Give your answer correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.35 × 0.6 + 0.65 × 0.15 (sum of the two delayed branches)
✓ 1
A1 0.2100 + 0.0975
✓ 1
A1 0.3075
✓ 1
Part (b)
M1 P(storm | delayed) = P(storm and delayed)/P(delayed)
✓ 1
A1 0.21⁄0.3075
✓ 1
A1 0.683
✓ 1
Exam tip: For a conditional probability, the denominator is the probability of the condition, not of the whole sample space.
148IBMI3S-4-01r14-04Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 1Medium7 marks
A rooftop solar array records the number of hours of sunshine, x, and the energy produced, y kWh, on eight days. x | 3.2 | 4.5 | 5.1 | 6.0 | 7.2 | 8.4 | 9.1 | 10.3 y | 9.8 | 13.1 | 14.9 | 17.9 | 19.6 | 24.2 | 25.0 | 29.4
(a)
Write down Pearson's product-moment correlation coefficient, r, and describe the correlation.
(2)
(b)
Find the equation of the regression line of y on x, with coefficients to 3 significant figures.
(2)
(c)
Use your equation from part (b) with the coefficients as rounded in part (b) to estimate the energy produced on a day with 6.5 hours of sunshine.
(2)
(d)
Explain why the equation should not be used to estimate the output on a day with 14 hours of sunshine.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 r = 0.997
✓ 1
A1 strong positive correlation
✓ 1
Part (b)
A1 gradient 2.70
✓ 1
A1 y = 2.70x + 1.07
✓ 1
Part (c)
M1 substitution of \(x=6.5\)
✓ 1
A1 18.62 kWh
✓ 1
Part (d)
R1 14 is outside the range of the data (extrapolation) so the linear relationship may not hold
✓ 1
Exam tip: Interpolation (inside the data range) is reliable; extrapolation is not. Name which one is happening.
149IBMI3S-4-01r14-05Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 1Medium6 marks
In a screening programme, 2% of the population have a particular infection. A test gives a positive result for 95% of people who have the infection and for 4% of people who do not have it. A person is selected at random from the population and tested.
(a)
Find the probability that the test result is positive.
(2)
(b)
Given that the test result is positive, find the probability that the person has the infection.
(2)
(c)
In a screening round, 7500 people are tested. Find the expected number of people who do not have the infection but test positive.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(0.02\times0.95+0.98\times0.04\) (tree diagram or equivalent)
Common error: P(infection | positive) is not the same as P(positive | infection). The denominator must be the total probability of a positive result.
150IBMI3S-4-01r14-07Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 1Medium6 marks
A medical practice has 240 walk-in visits in a week, as follows. Day | Mon | Tue | Wed | Thu | Fri Visits | 61 | 40 | 38 | 47 | 54 The practice manager wants to test, at the 5% significance level, whether walk-in visits are equally likely on each weekday, using a χ2 goodness-of-fit test.
(a)
State the null and alternative hypotheses.
(2)
(b)
Show that each expected frequency is 48, and calculate the χ2 statistic.
(2)
(c)
The degrees of freedom are 4 and the p-value is 0.103. State the conclusion, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: walk-in visits are equally distributed across the five weekdays (uniform)
✓ 1
A1 H1: walk-in visits are not equally distributed across the weekdays
A1 do not reject H0: insufficient evidence that visits are unequally spread across weekdays
✓ 1
Common error: The degrees of freedom for a goodness-of-fit test are the number of categories minus 1 (when no parameters are estimated), not the number of categories.
151IBMI3S-4-01r15-03Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium8 marks
A fashion retailer tests eight different prices, p pounds (£), for the same style of scarf. For each price it records the number of scarves sold in a week, n.
p (£)
20
25
30
35
40
45
50
55
n
121
106
101
88
83
66
64
49
(a)
Find the Pearson's product-moment correlation coefficient, r, and describe the correlation.
(2)
(b)
Write down the equation of the regression line of n on p, with each coefficient given to 3 significant figures.
(2)
(c)
Interpret the gradient of the regression line in context.
(1)
(d)
Use the equation from part (b), with the coefficients to 3 significant figures, to estimate the number of scarves sold in a week when the price is £42. Give your answer to 3 significant figures.
(2)
(e)
The retailer proposes to use the equation to predict sales at a price of £90. Comment on the reliability of this prediction.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 r = -0.992
✓ 1
A1 strong negative linear correlation
✓ 1
Part (b)
A1 gradient = -1.96
✓ 1
A1 intercept = 158, so n = -1.96p + 158
✓ 1
Part (c)
R1 for every £1 increase in price, the number of scarves sold per week falls by about 1.96
✓ 1
Part (d)
M1 substitutes p = 42 into n = -1.96p + 158
✓ 1
A1 n = 75.7 (accept any value from 75.7 to 75.9)
✓ 1
Part (e)
R1 £90 lies far outside the data range (20 to 55), so this is extrapolation and unreliable; the model would even give a negative number of sales
✓ 1
Common error: Regression of n on p is used to predict n from p; do not swap the variables to predict p from n.
152IBMI3S-4-01r15-04Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium8 marks
On a fishing boat, 15% of the fish hauled up in a net are under the legal size and must be thrown back, independently of each other. A crew member takes a random group of 12 fish from the net. Let X be the number of these fish that are under the legal size.
(a)
Write down the distribution of X, including its parameters.
(1)
(b)
Find P(X = 2).
(2)
(c)
Find P(X ≥ 4).
(2)
(d)
Find the expected number of fish in the group that are under the legal size.
(1)
(e)
Find the probability that at least one fish in the group is under the legal size.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 X ~ B(12, 0.15)
✓ 1
Part (b)
M1 uses the binomial probability with n = 12, p = 0.15
✓ 1
A1 P(X = 2) = 0.292
✓ 1
Part (c)
M1 uses P(X ≥ 4) = 1 − P(X ≤ 3)
✓ 1
A1 P(X ≥ 4) = 0.0922
✓ 1
Part (d)
A1 E(X) = 12 × 0.15 = 1.8
✓ 1
Part (e)
M1 uses 1 − P(X = 0) = 1 − 0.8512
✓ 1
A1 probability = 0.858
✓ 1
Exam tip: For 'at least one' questions use the complement, 1 − P(X = 0), rather than adding many terms.
153IBMI3S-4-01r15-05Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium7 marks
The lengths of mackerel landed at a harbour are normally distributed with mean 36.0 cm and standard deviation 4.5 cm.
(a)
Find the probability that a randomly chosen mackerel is longer than 40 cm.
(2)
(b)
The longest 10% of the mackerel are sold as premium fish. Find the minimum length of a premium mackerel.
(2)
(c)
Mackerel shorter than 30 cm may not be sold. In a catch of 600 mackerel, estimate how many are shorter than 30 cm. Use P(L < 30) to 3 significant figures, and give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempts the normal probability P(L > 40)
✓ 1
A1 probability = 0.187
✓ 1
Part (b)
M1 uses inverse normal with cumulative probability 0.9
✓ 1
A1 length = 41.8 cm
✓ 1
Part (c)
A1 P(L < 30) = 0.0912
✓ 1
M1 expected number = 600 × 0.0912
✓ 1
A1 about 54.7 mackerel
✓ 1
Common error: For the top 10%, the cumulative area to the left of the boundary is 0.9, not 0.1.
154IBMI3S-4-01r15-09Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium7 marks
A company makes warehouse robots on two assembly lines. Line A produces 60% of the robots and line B produces the rest. A robot from line A has a sensor fault with probability 0.03, and a robot from line B has a sensor fault with probability 0.07. A robot is selected at random from the company's production.
(a)
Find the probability that the robot comes from line A and has a sensor fault.
(2)
(b)
Show that the probability that the robot has a sensor fault is 0.046.
(2)
(c)
Given that the robot has a sensor fault, find the probability that it came from line B. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.6 × 0.03
✓ 1
A1 0.018
✓ 1
Part (b)
M1 adds the two branches: 0.6 × 0.03 + 0.4 × 0.07
✓ 1
A1 0.018 + 0.028 = 0.046 (AG)
✓ 1
Part (c)
M1 conditional probability P(B | fault) = P(B and fault) ÷ P(fault)
✓ 1
M1 0.028 ÷ 0.046
✓ 1
A1 0.609
✓ 1
Common error: P(B | fault) is not 0.07: that is P(fault | B). Divide the joint probability by the probability of the condition.
155IBMI3S-4-01r15-11Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium6 marks
The masses, m grams, of 100 farmed salmon are summarised in the table.
Mass (g)
Frequency
200 ≤ m < 300
8
300 ≤ m < 400
21
400 ≤ m < 500
33
500 ≤ m < 600
24
600 ≤ m < 800
14
(a)
Write down the modal class.
(1)
(b)
Estimate the mean mass of the salmon.
(2)
(c)
Write down the class that contains the median mass.
(1)
(d)
Assuming that masses are spread evenly within each class, estimate the number of salmon with a mass of at least 550 g.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 400 ≤ m < 500
✓ 1
Part (b)
M1 uses midpoints 250, 350, 450, 550, 700
✓ 1
A1 mean = 472 g
✓ 1
Part (c)
A1 400 ≤ m < 500 (cumulative frequencies 8, 29, 62; the 50th and 51st values are in this class)
✓ 1
Part (d)
M1 half of the 500 ≤ m < 600 class: 24 × (50/100) = 12
✓ 1
A1 12 + 14 = 26 salmon
✓ 1
Common error: Use the midpoint of each class to estimate the mean; for the last class (600 to 800) the midpoint is 700, not 600 or 800.
156IBMI3S-4-01r15-14Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium6 marks
A fashion designer runs a stall at a weekend market. Her profit, X pounds (£), on a weekend has the probability distribution shown, where k is a constant.
x
−200
50
200
450
P(X = x)
0.20
3k
2k
0.10
(a)
Find the value of k.
(2)
(b)
Calculate the expected profit E(X) for a weekend.
(2)
(c)
The designer will keep the stall only if the expected profit is more than £75 per weekend. State, with a reason, whether she keeps it.
(1)
(d)
Estimate her total profit over 40 weekends of trading.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses the sum of the probabilities equal to 1: 0.30 + 5k = 1
✓ 1
A1 k = 0.14
✓ 1
Part (b)
M1 uses E(X) = Σ x P(X = x) with P(50) = 0.42 and P(200) = 0.28
✓ 1
A1 E(X) = −40 + 21 + 56 + 45 = £82
✓ 1
Part (c)
R1 £82 > £75, so she keeps the stall
✓ 1
Part (d)
A1 40 × 82 = £3280
✓ 1
Common error: Do not forget to use the value of k found in part (a) when writing P(X = 50) and P(X = 200) in the expected-value sum.
157IBMI3S-4-01r16-02Sampling and descriptive statistics; T-tests, confidence intervals and Markov chains; Probability and the binomial distribution; Normal distribution and chi-squared testsPaper 1Medium7 marks
A robotic arm on an assembly line takes a different time to complete each pick-and-place cycle. The times, in seconds, for 12 consecutive cycles are 4.8, 5.1, 5.3, 5.0, 4.9, 5.2, 5.4, 5.1, 5.0, 5.3, 4.9, 6.4
(a)
Use your calculator to find the mean cycle time and the population standard deviation, giving the standard deviation to 3 significant figures.
(2)
(b)
Find the median and the interquartile range of the cycle times.
(3)
(c)
A cycle time is classed as an outlier if it is more than 1.5 × IQR above Q3. Determine whether the longest cycle time is an outlier.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 mean = 5.20 s (5.2)
✓ 1
A1 standard deviation = 0.402 s
✓ 1
Part (b)
A1 median = 5.1 s
✓ 1
A1 Q1 = 4.95 and Q3 = 5.3
✓ 1
A1 IQR = 0.35 s
✓ 1
Part (c)
M1 upper boundary = 5.3 + 1.5 × 0.35 = 5.825
✓ 1
R1 6.4 > 5.825, so 6.4 s is an outlier (ECF from (b))
✓ 1
Common error: The mean and standard deviation are strongly affected by an outlier, but the median and IQR are not. Always compare the data value with the calculated boundary before calling it an outlier.
158IBMI3S-4-01r16-03Sampling and descriptive statistics; T-tests, confidence intervals and Markov chains; Probability and the binomial distribution; Normal distribution and chi-squared testsPaper 1Medium6 marks
A hatchery records the lengths, l cm, of 60 young salmon. Length (cm) | 20 ≤ l < 25 | 25 ≤ l < 30 | 30 ≤ l < 35 | 35 ≤ l < 40 | 40 ≤ l < 45 Frequency | 6 | 14 | 21 | 12 | 7
(a)
Write down the modal class.
(1)
(b)
Estimate the mean length of the salmon.
(2)
(c)
Use your calculator to estimate the standard deviation of the lengths, using the mid-interval values and treating the data as a population.
(2)
(d)
Explain why your answers to (b) and (c) are estimates.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 30 ≤ l < 35
✓ 1
Part (b)
M1 use of mid-interval values 22.5, 27.5, 32.5, 37.5, 42.5
✓ 1
A1 mean = 32.5 cm
✓ 1
Part (c)
M1 use of mid-interval values with frequencies in the calculator
✓ 1
A1 standard deviation = 5.70 cm
✓ 1
Part (d)
R1 the actual lengths within each class are not known, so mid-interval values are used
✓ 1
Exam tip: For grouped data, always use the mid-interval value of each class. State that the results are estimates when the exact data are not available.
159IBMI3S-4-01r16-06Sampling and descriptive statistics; T-tests, confidence intervals and Markov chains; Probability and the binomial distribution; Normal distribution and chi-squared testsPaper 1Medium7 marks
A fisheries scientist compares the masses, in kg, of adult sea trout caught in two lakes. Assume the masses in each lake are normally distributed with the same variance.
Lake A (8 fish): 1.42, 1.55, 1.38, 1.61, 1.47, 1.52, 1.44, 1.58 Lake B (10 fish): 1.35, 1.47, 1.30, 1.51, 1.37, 1.42, 1.36, 1.46, 1.39, 1.32
She tests, at the 5% significance level, whether the mean mass differs between the two lakes.
(a)
State the null and alternative hypotheses.
(2)
(b)
Carry out a two-sample t-test using a pooled variance and find the p-value.
(2)
(c)
State the conclusion of the test at the 5% significance level.
(2)
(d)
Would the conclusion change at the 1% significance level? Give a reason.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: μA = μB
✓ 1
A1 H1: μA ≠ μB
✓ 1
Part (b)
M1 use of two-sample pooled t-test on the calculator
✓ 1
A1 p-value = 0.0112 (t = 2.86, df = 16)
✓ 1
Part (c)
M1 compare 0.0112 with 0.05 (two-tailed, no halving needed)
✓ 1
A1 0.0112 < 0.05, so reject H0: evidence that the mean masses differ
✓ 1
Part (d)
R1 yes, 0.0112 > 0.01, so H0 would not be rejected
✓ 1
Exam tip: The calculator's two-tailed p-value already accounts for both tails, so compare it directly with the significance level. Mention both assumptions (normality and equal variance) if asked to justify a pooled test.
160IBMI3S-4-01r16-07Sampling and descriptive statistics; T-tests, confidence intervals and Markov chains; Probability and the binomial distribution; Normal distribution and chi-squared testsPaper 1Medium7 marks
A fisheries biologist claims that the catch from a large trawl contains cod, haddock, whiting and pollack in the ratio 4 : 3 : 2 : 1. A random sample of 200 fish from the catch contains 70 cod, 70 haddock, 38 whiting and 22 pollack. A χ2 goodness-of-fit test is used at the 5% significance level.
(a)
State the null hypothesis.
(1)
(b)
Find the expected frequency for each species.
(1)
(c)
Calculate the value of the χ2 statistic.
(2)
(d)
Write down the number of degrees of freedom.
(1)
(e)
The critical value of χ2 for 3 degrees of freedom at the 5% level is 7.815. State the conclusion of the test.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: the species in the catch are in the ratio 4 : 3 : 2 : 1
✓ 1
Part (b)
A1 cod 80, haddock 60, whiting 40, pollack 20
✓ 1
Part (c)
M1 Σ (O − E)2 ÷ E = 100/80 + 100/60 + 4/40 + 4/20
✓ 1
A1 χ2 = 3.22
✓ 1
Part (d)
A1 degrees of freedom = 3
✓ 1
Part (e)
M1 3.22 < 7.815
✓ 1
A1 do not reject H0: no evidence that the catch differs from the claimed ratio
✓ 1
Exam tip: For a goodness-of-fit test the degrees of freedom are the number of categories minus 1 (when no parameters are estimated). Compute expected frequencies from the claimed proportions, not from the observed data.
161IBMI3S-4-01r16-09Sampling and descriptive statistics; T-tests, confidence intervals and Markov chains; Probability and the binomial distribution; Normal distribution and chi-squared testsPaper 1Medium7 marks
Each fish caught by a trawler is independently undersized with probability 0.18. A sample of 12 fish is inspected.
(a)
Find the probability that exactly 3 of the 12 fish are undersized.
(2)
(b)
Another boat inspects n fish. Find the smallest value of n for which the probability that at least one fish is undersized exceeds 0.95.
(3)
(c)
Using your value of n from (b), find the expected number of undersized fish in a sample of n fish.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 X ~ B(12, 0.18) and P(X = 3) attempted
✓ 1
A1 0.215
✓ 1
Part (b)
M1 1 − 0.82n > 0.95
✓ 1
M1 solve, e.g. n > ln 0.05 ÷ ln 0.82 = 15.1 (or table of values: n = 15 gives 0.949, n = 16 gives 0.958)
✓ 1
A1 n = 16
✓ 1
Part (c)
M1 n × 0.18
✓ 1
A1 2.88 (ECF from (b))
✓ 1
Common error: "At least one" is best found as 1 − P(none). Remember that n must be a whole number, so round up, not to the nearest integer.
162IBMI3S-4-01r16-10Sampling and descriptive statistics; T-tests, confidence intervals and Markov chains; Probability and the binomial distribution; Normal distribution and chi-squared testsPaper 1Medium7 marks
A box holds 12 servo motors for a robot, of which 4 are faulty. Two motors are selected at random from the box without replacement.
(a)
Find the probability that both motors are faulty.
(2)
(b)
Find the probability that exactly one motor is faulty.
(2)
(c)
Given that at least one of the two motors is faulty, find the probability that both are faulty. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 4/12 × 3/11
✓ 1
A1 1/11 (0.0909)
✓ 1
Part (b)
M1 4/12 × 8/11 + 8/12 × 4/11
✓ 1
A1 16/33 (0.485)
✓ 1
Part (c)
M1 P(at least one faulty) = 1 − 8/12 × 7/11 = 19/33
✓ 1
M1 conditional probability = (1/11) ÷ (19/33)
✓ 1
A1 3/19 = 0.158
✓ 1
Common error: When items are selected without replacement, the second probability changes: both the numerator and the denominator reduce after the first selection.
163IBMI3S-4-01r16-13Sampling and descriptive statistics; T-tests, confidence intervals and Markov chains; Probability and the binomial distribution; Normal distribution and chi-squared testsPaper 1Medium7 marks
The lengths of sea bass raised on a fish farm are normally distributed with mean μ cm and standard deviation σ cm. It is found that 12% of the fish are shorter than 28 cm and 20% of the fish are longer than 41 cm.
(a)
Find the value of μ and the value of σ, each to 3 significant figures.
(5)
(b)
A fish is classed as large if its length is more than 1.5 standard deviations above the mean. Using the unrounded values of μ and σ, find the smallest length of a fish classed as large, to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 z for 28 cm is −1.175 (P(Z < z) = 0.12)
✓ 1
A1 z for 41 cm is 0.842 (P(Z < z) = 0.80)
✓ 1
M1 two equations: 28 = μ − 1.175σ and 41 = μ + 0.8416σ
✓ 1
A1 σ = 6.45
✓ 1
A1 μ = 35.6
✓ 1
Part (b)
M1 μ + 1.5σ = 35.57… + 1.5 × 6.446…
✓ 1
A1 45.2 cm
✓ 1
Common error: The 12% area gives a negative z-value because 28 cm is below the mean. Do not use the rounded μ and σ in later calculations if the question asks for unrounded values.
164IBMI3S-4-02r16-02Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regressionPaper 1Medium7 marks
A food factory fills jars of jam. The mass, M grams, of jam in a jar is normally distributed with standard deviation 6 g. The label on each jar states 450 g.
(a)
The mean mass μ is set so that 2% of jars contain less than 450 g. Find μ, to 1 decimal place.
(3)
(b)
Using the unrounded value of μ, find the probability that a jar contains more than 470 g of jam.
(2)
(c)
A new machine has mean 458 g and a different standard deviation σ. For this machine, 2% of jars contain less than 450 g. Find σ, to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 standardise: (450 − μ) ÷ 6 = z, with z = −2.054 (inverse normal for 0.02)
✓ 1
M1 μ = 450 + 2.054 × 6
✓ 1
A1 μ = 462.3 g
✓ 1
Part (b)
M1 P(M > 470) with M ~ N(462.32…, 62)
✓ 1
A1 0.100
✓ 1
Part (c)
M1 (450 − 458) ÷ σ = −2.0537
✓ 1
A1 σ = 3.90 g
✓ 1
Common error: A probability of 0.02 in the lower tail corresponds to a negative z-value. Keep the sign when forming (x − μ) ÷ σ = z, then rearrange carefully.
165IBMI3S-4-02r16-04Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regressionPaper 1Medium7 marks
A mountain guide records whether climbers who attempted three routes reached the summit. Route | Reached summit | Turned back North route | 14 | 16 South route | 22 | 18 West route | 11 | 29 A χ2 test for independence is carried out at the 5% significance level.
(a)
State the null hypothesis.
(1)
(b)
Write down the number of degrees of freedom.
(1)
(c)
Find the χ2 statistic and the p-value.
(3)
(d)
State the conclusion of the test, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: the route chosen and whether the summit is reached are independent
✓ 1
Part (b)
A1 (3 − 1)(2 − 1) = 2
✓ 1
Part (c)
M1 use of the χ2 test for independence on the calculator
✓ 1
A1 χ2 = 6.44
✓ 1
A1 p-value = 0.0399
✓ 1
Part (d)
M1 compare 0.0399 with 0.05 (or 6.44 with the critical value 5.991)
✓ 1
A1 reject H0: there is evidence that the success rate depends on the route
✓ 1
Common error: Do not give the number of degrees of freedom as the number of cells minus 1. For an r × c table it is (r − 1)(c − 1). Also write the conclusion in context, not just "reject H₀".
166IBMI3S-4-02r16-06Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regressionPaper 1Medium7 marks
A food scientist measures the sugar content, in grams per 100 g, of 16 breakfast cereals: 3.2, 4.1, 4.4, 4.8, 5.0, 5.2, 5.5, 5.9, 6.1, 6.4, 6.6, 7.0, 7.3, 7.8, 8.2, 14.5
(a)
Find the minimum, lower quartile, median, upper quartile and maximum of the data.
(3)
(b)
A value is an outlier if it is more than 1.5 × IQR above Q3 or below Q1. Determine whether any of the values is an outlier.
(3)
(c)
State, with a reason, whether the mean or the median is the better measure of the centre of these data.
R1 14.5 > 10.525 so 14.5 is an outlier; no other value lies outside the boundaries (ECF from (a))
✓ 1
Part (c)
R1 the median, because it is not affected by the outlier 14.5 (the mean is pulled upwards)
✓ 1
Common error: Calculate the outlier boundaries from Q₁ and Q₃, not from the median. Check both ends of the data, even when only one end looks unusual.
167IBMI3S-4-02r16-07Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regressionPaper 1Medium6 marks
A telecoms company records the monthly mobile data used, x GB, by a sample of customers. Data used (GB) | 0 ≤ x < 2 | 2 ≤ x < 4 | 4 ≤ x < 6 | 6 ≤ x < 8 | 8 ≤ x < 10 Frequency | 8 | 14 | k | 9 | 4 The estimated mean data used is 4.5 GB.
(a)
Find the value of k.
(3)
(b)
State the modal class.
(1)
(c)
Using the mid-interval values, estimate the standard deviation of the data used, treating the data as a population. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sum of (mid-interval value × frequency) = 149 + 5k
✓ 1
M1 (149 + 5k) ÷ (35 + k) = 4.5
✓ 1
A1 k = 17
✓ 1
Part (b)
A1 4 ≤ x < 6 (ECF from (a))
✓ 1
Part (c)
M1 use of mid-interval values 1, 3, 5, 7, 9 with frequencies 8, 14, 17, 9, 4 (ECF from (a))
✓ 1
A1 standard deviation = 2.28 GB
✓ 1
Exam tip: For an estimated mean from a grouped table with an unknown frequency, form an equation using the sum of (mid-value × frequency) over the total frequency, both in terms of the unknown.
168IBMI3S-4-02r16-09Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regressionPaper 1Medium7 marks
On each day at a high camp, the weather is suitable for a summit attempt with probability 0.35, independently of other days. A team waits at camp for 7 days.
(a)
Find the probability that exactly 2 of the 7 days are suitable.
(2)
(b)
Find the probability that at least one of the 7 days is suitable.
(2)
(c)
On a different mountain, a team waits 5 days and the probability that none of the 5 days is suitable is 0.0778. Find the probability p that a day is suitable on that mountain, to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 X ~ B(7, 0.35) and P(X = 2) attempted
✓ 1
A1 0.298
✓ 1
Part (b)
M1 1 − P(X = 0) = 1 − 0.657
✓ 1
A1 0.951
✓ 1
Part (c)
M1 (1 − p)5 = 0.0778
✓ 1
M1 1 − p = 0.07781/5 = 0.6000…
✓ 1
A1 p = 0.400
✓ 1
Common error: "At least one" is not the same as "exactly one". Use the complement: 1 − P(none). For an unknown probability, take the nth root of both sides of the equation.
169IBMI3S-4-02r16-10Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regressionPaper 1Medium7 marks
A food laboratory tests 120 snack products for allergens. Of these, 55 contain nuts, 48 contain gluten, and 30 contain neither.
(a)
Find the number of products that contain both nuts and gluten.
(2)
(b)
A product is chosen at random from those containing gluten. Find the probability that it also contains nuts.
(2)
(c)
Two different products are chosen at random from the 120 without replacement. Find the probability that exactly one of them contains nuts. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 products containing at least one allergen = 120 − 30 = 90, and 55 + 48 − 90
M1 correct substitution of the values, 2 × 55 × 65 ÷ (120 × 119)
✓ 1
A1 0.501
✓ 1
Exam tip: A Venn diagram helps: start with the overlap, then find the "only" regions by subtraction, and check that the regions add up to the total. Here nuts only = 42, gluten only = 35, both = 13, neither = 30.
170IBMI3S-4-02r16-11Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regressionPaper 1Medium7 marks
A food scientist tests whether a new packaging type increases the shelf life of a soup. Shelf lives, in days, are recorded for samples in each packaging type. Assume the shelf lives are normally distributed with equal variances.
State the null and alternative hypotheses for a one-tailed test.
(2)
(b)
Carry out a one-tailed two-sample t-test using a pooled variance and find the p-value.
(2)
(c)
State the conclusion at the 5% significance level, in context.
(2)
(d)
State why a pooled test is appropriate here.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: μnew = μold
✓ 1
A1 H1: μnew > μold
✓ 1
Part (b)
M1 use of the pooled two-sample t-test on the calculator
✓ 1
A1 p = 0.0196 (t = 2.29, df = 13)
✓ 1
Part (c)
M1 compare 0.0196 with 0.05
✓ 1
A1 reject H0: there is evidence that the new packaging gives a longer mean shelf life
✓ 1
Part (d)
R1 the test assumes the two populations have equal variances (and that both are normally distributed)
✓ 1
Exam tip: If the calculator gives a two-tailed p-value but the test is one-tailed, halve it, provided the sample means differ in the direction of H₁. Many calculators let you choose the tail directly.
171IBMI3S-4-02r18-05Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared testsPaper 1Medium7 marks
A factory assembles robot arms. Each arm independently passes the final quality test with probability 0.92. A batch of 15 arms is tested. Let X be the number of arms in the batch that pass.
(a)
State the distribution of X and write down the expected number of arms that pass.
(2)
(b)
Find the probability that exactly 14 arms pass. Give your answer to 3 significant figures.
(2)
(c)
Find the probability that at least 13 arms pass. Give your answer to 3 significant figures.
Exam tip: "At least 13" means 13 or more; with a calculator, use the cumulative function carefully (P(X ≥ 13) = 1 − P(X ≤ 12)).
172IBMI3S-4-02r18-06Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared testsPaper 1Medium8 marks
A fashion label finds that 6 % of its garments have a stitching fault, independently of each other. Garments are packed in boxes at random.
(a)
A box contains 20 garments. Find the probability that none of them has a stitching fault. Give your answer to 3 significant figures.
(2)
(b)
A buyer wants the probability that a box contains at least one faulty garment to be greater than 0.95. Find the smallest number of garments in a box for which this is true.
(3)
(c)
For a box of 20 garments, find the probability that at least 2 garments have a stitching fault. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.9420
✓ 1
A1 0.290
✓ 1
Part (b)
M1 1 − 0.94n > 0.95, so 0.94n < 0.05
✓ 1
A1 n > 48.4 (accept a table or solver giving 48.4)
✓ 1
A1 n = 49
✓ 1
Part (c)
M1 1 − P(0 faults) − P(1 fault)
✓ 1
A1 P(1 fault) = 20 × 0.06 × 0.9419 = 0.3703
✓ 1
A1 0.340
✓ 1
Common error: P(at least 1) = 1 − P(none); writing P(at least 2) as 1 − P(none) forgets to remove the case of exactly one fault.
173IBMI3S-4-02r18-11Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared testsPaper 1Medium8 marks
A motor in a delivery robot is run continuously for different lengths of time, x hours, and its temperature, y °C, is then measured.
x
0.5
1
1.5
2
3
3.5
4
5
y
24
29
31
36
42
47
49
58
(a)
Find the Pearson's product-moment correlation coefficient, r, to 3 significant figures.
(2)
(b)
Describe the correlation between x and y.
(1)
(c)
Find the equation of the regression line of y on x. Write the gradient and intercept to 3 significant figures.
(2)
(d)
Interpret the value of the gradient in this context.
(1)
(e)
Use the unrounded regression line to estimate the temperature of the motor after 2.5 hours. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use the regression/statistics function with the data
✓ 1
A1 r = 0.998
✓ 1
Part (b)
A1 strong positive linear correlation
✓ 1
Part (c)
A1 gradient 7.33
✓ 1
A1 y = 7.33x + 20.7
✓ 1
Part (d)
R1 For each extra hour of continuous running, the temperature rises by about 7.3 °C
✓ 1
Part (e)
M1 substitute x = 2.5 into the regression line
✓ 1
A1 39.0 °C
✓ 1
Exam tip: Only predict within the range of the data (interpolation); the line may not hold for times far outside 0.5 to 5 hours.
174IBMI3S-4-02r18-12Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared testsPaper 1Medium7 marks
Two judges each give a score out of 10 to eight robots in a design competition.
Robot
A
B
C
D
E
F
G
H
Judge 1
8.2
6.5
9.1
7.4
5.8
7.9
6.9
8.6
Judge 2
7.8
6.9
8.7
6.5
6.2
8.4
7.1
8.5
(a)
Rank the robots for each judge, giving rank 1 to the highest score.
(2)
(b)
Find the sum of the squares of the differences between the ranks, Σd2.
(2)
(c)
Calculate Spearman's rank correlation coefficient, rs, to 3 significant figures, and comment on the agreement between the judges.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 Judge 1: 3, 7, 1, 5, 8, 4, 6, 2 (for A to H)
✓ 1
A1 Judge 2: 4, 6, 1, 7, 8, 3, 5, 2 (for A to H)
✓ 1
Part (b)
M1 differences between ranks for each robot, squared
✓ 1
A1 Σd2 = 8
✓ 1
Part (c)
M1 rs = 1 − 6Σd2 ÷ (8 × (82 − 1))
✓ 1
A1 rs = 0.905
✓ 1
R1 strong positive agreement between the judges' rankings
✓ 1
Exam tip: Spearman's coefficient uses ranks, so it measures how well the orderings agree, not the size of the scores.
175IBMI3S-4-02r20-05Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium7 marks
In a study of 500 adults, each person's vaccination status (unvaccinated, one dose, boosted) and whether they caught a particular infection during the winter were recorded.
Infected
Not infected
Unvaccinated
18
142
One dose
35
165
Boosted
27
113
A χ2 test for independence at the 5% significance level is carried out.
(a)
Write down the null and alternative hypotheses.
(1)
(b)
Show that the expected frequency for the boosted adults who were infected is 22.4, and write down the number of degrees of freedom.
(3)
(c)
Find the χ2 statistic and the p-value.
(2)
(d)
State the conclusion of the test, giving a reason.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: vaccination status and infection are independent; H1: they are not independent (associated)
✓ 1
Part (b)
M1 80 × 140/500
✓ 1
A1 22.4 (must be shown)
✓ 1
A1 degrees of freedom = (3 − 1)(2 − 1) = 2
✓ 1
Part (c)
A1 χ2 = 4.15 (4.145...)
✓ 1
A1 p-value = 0.126
✓ 1
Part (d)
R1 p = 0.126 > 0.05 so do not reject H0: there is insufficient evidence of an association between vaccination status and infection
✓ 1
Exam tip: "Do not reject H₀" is not the same as proving independence; word the conclusion as a lack of evidence.
176IBMI3S-4-02r20-06Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium7 marks
Records over many years show that the prevailing wind direction at a site is north in 40% of days, west in 30%, south in 20% and east in 10%. After a new building is erected nearby, the prevailing direction is recorded on 250 days.
Direction
North
West
South
East
Frequency
125
60
45
20
A χ2 goodness of fit test at the 5% significance level is used to decide whether the historic proportions still apply.
(a)
Write down the null hypothesis.
(1)
(b)
Find the expected frequencies.
(1)
(c)
Find the number of degrees of freedom and the χ2 statistic.
(3)
(d)
The critical value at the 5% level is 7.815. State the conclusion, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: the wind directions follow the historic proportions (40%, 30%, 20%, 10%)
✓ 1
Part (b)
A1 100, 75, 50, 25
✓ 1
Part (c)
A1 degrees of freedom = 3
✓ 1
M1 Σ(O − E)2/E = 6.25 + 3 + 0.5 + 1
✓ 1
A1 χ2 = 10.75
✓ 1
Part (d)
M1 comparison of 10.75 with 7.815 (or p = 0.0132 < 0.05)
✓ 1
R1 reject H0: there is evidence that the wind direction proportions have changed
✓ 1
Exam tip: Expected frequencies are found from the proportions in H₀ and the total number of observations, not from the observed counts.
177IBMI3S-4-02r20-07Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium7 marks
In a vaccine trial, each of 30 volunteers independently has probability 0.12 of developing mild side effects. Let X be the number of volunteers who develop mild side effects.
(a)
Find P(X ≤ 3).
(2)
(b)
Find P(2 ≤ X ≤ 5).
(2)
(c)
The clinic keeps k recovery chairs ready. Find the smallest value of k such that the probability that there are enough chairs for all volunteers with side effects is at least 0.95.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 X ~ B(30, 0.12), use of cumulative binomial probability
✓ 1
A1 0.507
✓ 1
Part (b)
M1 P(X ≤ 5) − P(X ≤ 1)
✓ 1
A1 0.747
✓ 1
Part (c)
M1 find k such that P(X ≤ k) ≥ 0.95, using a table of cumulative probabilities
✓ 1
A1 P(X ≤ 6) = 0.939 and P(X ≤ 7) = 0.978
✓ 1
A1 k = 7
✓ 1
Exam tip: For "at least" or "at most" probabilities with k unknown, tabulate cumulative probabilities and check the values either side of the target.
178IBMI3S-4-02r20-08Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium6 marks
A screening test for an infection is given to a large population. 4% of the population have the infection. If a person has the infection, the test is positive with probability 0.92. If a person does not have the infection, the test is positive with probability 0.04.
(a)
Find the probability that a randomly chosen person tests positive.
(3)
(b)
A randomly chosen person tests positive. Find the probability that they have the infection.
(2)
(c)
Explain why fewer than half of the people who test positive have the infection, even though the test is accurate.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.04 × 0.92 (infected and positive)
✓ 1
M1 0.96 × 0.04 (not infected and positive)
✓ 1
A1 0.0368 + 0.0384 = 0.0752
✓ 1
Part (b)
M1 P(infected | positive) = 0.0368/0.0752 (ECF from (a))
✓ 1
A1 0.489 (0.4893...)
✓ 1
Part (c)
R1 the infection is rare (4%), so the many uninfected people produce more false positives than the few infected people produce true positives
✓ 1
Exam tip: Conditional probability: the numerator is the probability of both events; the denominator is the probability of the given event.
179IBMI3S-4-02r20-09Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium7 marks
The mass, M kg, of rice in a bag filled by a machine is normally distributed. 10% of bags contain less than 4.90 kg and 20% of bags contain more than 5.30 kg.
(a)
Write down the z-values corresponding to these two probabilities.
(2)
(b)
Find the mean and the standard deviation of M.
(3)
(c)
The machine fills 500 bags. Using the unrounded values of μ and σ, find the expected number of bags containing more than 5.00 kg.
Exam tip: Standardise each condition separately, then solve the two linear equations in μ and σ together; keep full calculator values until the end.
180IBMI3S-4-02r20-10Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium7 marks
A hospital employs 120 doctors, 340 nurses, 90 administrative staff and 50 technicians. A stratified sample of 60 staff is to be selected.
(a)
Find the number of nurses in the sample.
(2)
(b)
The waiting times, t minutes, of 100 patients in the emergency department are summarised below.
Waiting time
0 ≤ t < 10
10 ≤ t < 20
20 ≤ t < 30
30 ≤ t < 40
40 ≤ t < 60
Frequency
8
22
33
27
10
Write down the class that contains the median waiting time.
(1)
(c)
Estimate the mean waiting time.
(2)
(d)
Estimate the population standard deviation of the waiting times.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (340/600) × 60
✓ 1
A1 34
✓ 1
Part (b)
R1 20 ≤ t < 30, since the cumulative frequencies are 8, 30 and 63, so the 50th value lies in this class
✓ 1
Part (c)
M1 use of mid-interval values 5, 15, 25, 35, 50
✓ 1
A1 26.4 minutes
✓ 1
Part (d)
M1 use of GDC with mid-interval values and frequencies
✓ 1
A1 11.9 minutes (11.897...)
✓ 1
Exam tip: Grouped data give only estimates because every value in a class is replaced by its mid-interval value.
181IBMI3S-4-02r20-11Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium7 marks
A wind farm operator records the average wind speed, w m/s, and the power output, P kW, of eight turbines on one afternoon.
w
4.2
5.8
7.1
6.3
8.9
9.6
5.1
11.4
P
12
35
40
47
130
98
21
140
(a)
Rank the values of w and P, from lowest (rank 1) to highest (rank 8).
Give a reason why Spearman's coefficient may be preferred to Pearson's coefficient for this data.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 ranks of w: 1, 3, 5, 4, 6, 7, 2, 8
✓ 1
A1 ranks of P: 1, 3, 4, 5, 7, 6, 2, 8
✓ 1
Part (b)
M1 differences in rank 0, 0, 1, −1, −1, 1, 0, 0 and Σd2 = 4 (or use of Pearson's formula on the ranks)
✓ 1
M1 rs = 1 − 6 × 4/(8 × 63)
✓ 1
A1 0.952
✓ 1
Part (c)
A1 strong positive association: turbines with higher wind speed tend to have higher output
✓ 1
Part (d)
R1 the output does not rise linearly with wind speed (it increases faster than linearly), but the association is still monotonic
✓ 1
Exam tip: Spearman's coefficient measures how well the ranks agree, so it only needs the relationship to be consistently increasing or decreasing, not linear.
182IBMI3S-4-01r22-02Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regressionPaper 1Medium6 marks
Volcanologists record the daily sulfur dioxide output, x (thousand tonnes per day), and the number of tremors, y, detected on the same day at a volcano. Data for seven days are shown.
x
2.1
3.4
4.0
5.2
6.5
7.1
8.3
y
16
19
27
30
44
40
55
(a)
Find the Pearson product-moment correlation coefficient, r, for these data.
(1)
(b)
Find the equation of the regression line of y on x, giving the coefficients to 3 significant figures.
(2)
(c)
Interpret the gradient of the regression line in this context.
(1)
(d)
Use the regression line to estimate the number of tremors on a day when the output is 6.0 thousand tonnes per day. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 r = 0.973
✓ 1
Part (b)
A1 gradient 6.20
✓ 1
A1 intercept 0.602, so y = 6.20x + 0.602
✓ 1
Part (c)
R1 each extra thousand tonnes of sulfur dioxide per day is associated with about 6.2 more tremors per day
✓ 1
Part (d)
M1 substitute x = 6.0 into the regression equation
✓ 1
A1 37.8 (37.78 unrounded; 37.80 with rounded coefficients)
✓ 1
Exam tip: Use the regression line of y on x only to predict y from x, and only for x values inside the range of the data.
183IBMI3S-4-01r22-07Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regressionPaper 1Medium6 marks
An automatic alarm monitors a volcano. In any given week, an eruption occurs with probability 0.04. When an eruption occurs, the alarm sounds with probability 0.92. When no eruption occurs, the alarm sounds (falsely) with probability 0.07.
(a)
Find the probability that the alarm sounds in a given week.
(3)
(b)
Given that the alarm sounds, find the probability that an eruption occurs. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 identifies both routes: eruption and alarm, no eruption and alarm
✓ 1
A1 0.04 × 0.92 + 0.96 × 0.07 = 0.0368 + 0.0672
✓ 1
A1 0.104
✓ 1
Part (b)
M1 P(eruption | alarm) = P(eruption and alarm) / P(alarm)
✓ 1
A1 0.0368 / 0.104
✓ 1
A1 0.354
✓ 1
Common error: P(eruption | alarm) is not the same as P(alarm | eruption); the first needs the whole alarm probability as the denominator.
184IBMI3S-4-01r22-08Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regressionPaper 1Medium7 marks
Two stage crews, A and B, take different times to assemble a set between scenes. The times, in hours, for a sample of builds are:
Assume that both sets of times come from normal populations with equal variances. A two-tailed two-sample t-test is carried out at the 5% significance level to decide whether the mean assembly times differ.
(a)
State the null and alternative hypotheses.
(2)
(b)
Find the value of the test statistic and the p-value.
(3)
(c)
State the conclusion in the context of the question, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: the population mean assembly times of the two crews are equal
✓ 1
A1 H1: the population mean assembly times are not equal
✓ 1
Part (b)
M1 use of pooled two-sample t-test with a two-tailed alternative (13 degrees of freedom)
✓ 1
A1 t = −2.42 (accept 2.42)
✓ 1
A1 p-value = 0.0309
✓ 1
Part (c)
R1 0.0309 < 0.05
✓ 1
A1 reject H0: there is evidence that the mean assembly times of the two crews differ
✓ 1
Common error: For a two-tailed test the p-value must cover both tails; do not halve or double it a second time if your calculator already gives the two-tailed value.
185IBMI3S-4-01r22-09Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regressionPaper 1Medium7 marks
Geologists measure the width, w mm, of 100 ash particles taken from a deposit.
Width w (mm)
0 ≤ w < 5
5 ≤ w < 10
10 ≤ w < 15
15 ≤ w < 20
20 ≤ w < 25
25 ≤ w < 30
Frequency
6
19
35
24
11
5
(a)
Estimate the mean width of the particles, giving your answer to 3 significant figures.
(2)
(b)
Estimate the standard deviation of the widths, treating the 100 particles as the whole population and giving your answer to 3 significant figures.
(1)
(c)
Estimate the median width, assuming that widths are evenly spread within each class. Give your answer to 3 significant figures.
(2)
(d)
All 100 particles were collected from the side of the deposit closest to the access road. Explain why this may give a biased sample and suggest an improvement.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of mid-interval values 2.5, 7.5, ..., 27.5 with the frequencies
✓ 1
A1 14.0
✓ 1
Part (b)
A1 6.06
✓ 1
Part (c)
M1 cumulative frequencies 6, 25, 60, ... so the median lies in the class 10 ≤ w < 15; interpolate: 10 + (50 − 25)/35 × 5
✓ 1
A1 13.6
✓ 1
Part (d)
R1 particles near the road may differ from the rest of the deposit (convenience sample, not representative of the whole deposit)
✓ 1
A1 select sampling locations at random across the whole deposit (accept stratified or systematic sampling over the deposit)
✓ 1
Exam tip: Mid-interval values estimate grouped data, so quote means and standard deviations as estimates; the median needs the cumulative frequency, not the modal class.
186IBMI3S-4-01r22-10Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regressionPaper 1Medium6 marks
A theatre magazine gives each of eight productions a critic's score and an audience score, both out of 10.
Production
1
2
3
4
5
6
7
8
Critic
7.5
6.2
8.4
5.1
6.8
9.0
4.3
7.1
Audience
6.9
7.4
8.8
4.0
5.5
8.1
5.2
7.7
(a)
Rank the productions on each score, giving rank 1 to the lowest score.
(2)
(b)
Calculate Spearman's rank correlation coefficient, rs, to 3 significant figures.
(2)
(c)
Interpret your value of rs in context.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 critic ranks 6, 3, 7, 2, 4, 8, 1, 5
✓ 1
A1 audience ranks 4, 5, 8, 1, 3, 7, 2, 6
✓ 1
Part (b)
M1 correlation of the ranks, or 1 − 6Σd2/(n(n2 − 1)) with Σd2 = 14 (ECF from (a))
✓ 1
A1 rs = 0.833
✓ 1
Part (c)
A1 strong positive association between the two sets of scores
✓ 1
R1 productions that critics score higher tend to be scored higher by audiences (association, not proof of cause)
✓ 1
Common error: Rank both variables in the same direction before calculating; ranking one from high to low turns a positive association into a negative value.
187IBMI3S-4-01r22-13Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regressionPaper 1Medium7 marks
A shipping company records the age, a years, of seven vessels and the annual maintenance cost, C (in thousand euros).
a
3
5
6
8
10
13
15
C
48
70
66
98
107
140
150
(a)
Find the value of the Pearson correlation coefficient r and the equation of the regression line of C on a, giving all values to 3 significant figures.
(3)
(b)
Interpret the gradient of the regression line in context.
(1)
(c)
The company will retire a vessel when its predicted annual maintenance cost reaches 250 thousand euros. Use the regression line to estimate the age at which this happens, to 3 significant figures, and comment on the reliability of the estimate.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 r = 0.991
✓ 1
A1 gradient 8.74
✓ 1
A1 intercept 22.1, so C = 8.74a + 22.1
✓ 1
Part (b)
R1 each additional year of age is associated with about 8.74 thousand euros (€8740) more annual maintenance cost
✓ 1
Part (c)
M1 solve 250 = 8.74a + 22.1
✓ 1
A1 26.1 years
✓ 1
R1 unreliable: extrapolation well beyond the data (ages 3 to 15)
✓ 1
Exam tip: Regression of C on a is for predicting cost from age; use it only within the range of ages in the data, and treat any estimate outside that range with caution.
188IBMI3S-4-02r25-02Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 1Medium5 marks
The grain mass, X grams, in an ear of wheat is normally distributed with mean μ and standard deviation σ. It is known that 10% of ears have a grain mass less than 48 g and that 20% have a grain mass greater than 63 g.
(a)
Find the standardised value z corresponding to each of the masses 48 g and 63 g.
(2)
(b)
Hence find the value of μ and the value of σ, each correct to 3 significant figures. Use unrounded values of z.
Exam tip: Use the unrounded z-values from your calculator in the simultaneous equations; rounding them early can shift the third significant figure.
189IBMI3S-4-02r25-07Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 1Medium7 marks
A streaming service records the lengths, l minutes, of 50 songs in a playlist. 2 ≤ l < 3: 6 songs 3 ≤ l < 4: 19 songs 4 ≤ l < 5: 15 songs 5 ≤ l < 6: 7 songs 6 ≤ l < 7: 3 songs
(a)
Write down the modal class.
(1)
(b)
Estimate the mean length of a song.
(2)
(c)
Estimate the population standard deviation of the lengths, correct to 3 significant figures.
(2)
(d)
Assuming the lengths in each class are spread evenly, estimate the percentage of songs longer than 4.5 minutes.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 3 ≤ l < 4
✓ 1
Part (b)
M1 midpoints 2.5, 3.5, 4.5, 5.5, 6.5 used with the frequencies
✓ 1
A1 4.14 minutes
✓ 1
Part (c)
M1 use of the midpoints and frequencies on a calculator (statistics mode)
✓ 1
A1 1.05 minutes
✓ 1
Part (d)
M1 half of the 4 ≤ l < 5 class: 0.5 × 15 = 7.5, plus 7 and 3
✓ 1
A1 17.5 ÷ 50 = 35%
✓ 1
Exam tip: State that you are using class midpoints; the result for grouped data is an estimate, not the exact mean.
190IBMI3S-4-02r25-08Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 1Medium6 marks
An airline sells 52 tickets for a flight with 50 seats. Each ticket holder independently turns up with probability 0.93.
(a)
Let X be the number of ticket holders who turn up. State the distribution of X.
(1)
(b)
Find the probability that all 52 ticket holders turn up.
(2)
(c)
Find the probability that more than 50 ticket holders turn up, so that at least one passenger cannot be seated.
Common error: “More than 50” means X ≥ 51, so use 1 − P(X ≤ 50), not 1 − P(X ≤ 51).
191IBMI3S-4-02r25-09Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 1Medium5 marks
At a music school, 35% of students practise daily. A student who practises daily passes the grade exam with probability 0.9; a student who does not practise daily passes with probability 0.55.
(a)
Find the probability that a randomly chosen student practises daily and passes.
(1)
(b)
Find the probability that a randomly chosen student passes.
(2)
(c)
Given that a student passed, find the probability that the student practises daily. Give your answer correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 0.35 × 0.9 = 0.315
✓ 1
Part (b)
M1 0.315 + 0.65 × 0.55
✓ 1
A1 0.6725
✓ 1
Part (c)
M1 0.315 ÷ 0.6725 (conditional probability P(daily | pass))
✓ 1
A1 0.468
✓ 1
Exam tip: For conditional probability, the denominator is the probability of the event that is given (here, passing), not of the whole sample space.
192IBMI3S-4-02r31-04Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium8 marks
Calculator
A marine biologist measures the daily growth, G mm, of kelp fronds at different sea temperatures, T °C. T: 14, 15, 16, 17, 18, 19, 20, 21 G: 2.1, 2.9, 2.7, 3.6, 3.5, 4.6, 4.4, 5.3
(a)
Find the Pearson's product-moment correlation coefficient, r, and describe the correlation.
(2)
(b)
Find the equation of the regression line of G on T.
(2)
(c)
Interpret the gradient of the regression line in context.
(1)
(d)
Use the unrounded regression line to estimate the daily growth at 18.5 °C. Give your answer to 3 significant figures.
(2)
(e)
The biologist suggests using the line to predict growth at 30 °C. Comment on the reliability of this prediction.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 r = 0.964
✓ 1
R1 very strong positive linear correlation
✓ 1
Part (b)
A1 gradient 0.423
✓ 1
A1 G = 0.423T − 3.76
✓ 1
Part (c)
R1 each 1 °C rise in temperature is associated with an increase of about 0.423 mm in daily growth
✓ 1
Part (d)
M1 substituting T = 18.5
✓ 1
A1 4.06 mm
✓ 1
Part (e)
R1 30 °C is outside the range of the data (extrapolation) so the prediction is unreliable
✓ 1
Common error: Predicting outside the range of the data (extrapolation) assumes the linear trend continues, which may not be true; only interpolation is reliable.
193IBMI3S-4-02r31-05Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium7 marks
Calculator
A courier company finds that 8% of parcels are delivered late, independently of one another. A driver delivers 25 parcels in a day. Let X be the number of late parcels.
(a)
State the distribution of X.
(1)
(b)
Find the probability that exactly 3 parcels are late.
(2)
(c)
Find the probability that at least 2 parcels are late.
(2)
(d)
Find the expected number of late parcels in a day.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 X ~ B(25, 0.08)
✓ 1
Part (b)
M1 using the binomial probability formula or calculator for P(X = 3)
✓ 1
A1 0.188
✓ 1
Part (c)
M1 1 − P(X ≤ 1)
✓ 1
A1 0.605
✓ 1
Part (d)
M1 25 × 0.08
✓ 1
A1 2 parcels
✓ 1
Common error: 'At least 2' means 2 or more, so use 1 − P(X ≤ 1), not 1 − P(X ≤ 2).
194IBMI3S-4-02r31-06Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium7 marks
Calculator
An architect surveys 220 clients about their preferred window style, choosing from sliding, casement and fixed. The sites are either coastal or inland. Coastal sites: sliding 34, casement 46, fixed 20. Inland sites: sliding 41, casement 38, fixed 41. A χ2 test for independence is carried out at the 5% significance level.
(a)
State the null hypothesis and the alternative hypothesis.
(1)
(b)
Write down the number of degrees of freedom.
(1)
(c)
Use your calculator to find the p-value, and state the conclusion of the test, giving a reason.
(3)
(d)
Find the expected frequency for coastal sites with fixed windows, and state why the test conditions are met for this cell.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: window style preference is independent of site type; H1: window style preference is not independent of site type (association exists)
✓ 1
Part (b)
A1 (2 − 1)(3 − 1) = 2
✓ 1
Part (c)
A1 p-value = 0.0320
✓ 1
R1 comparing with 0.05: p < 0.05
✓ 1
A1 reject H0: there is evidence of an association between window style and site type
✓ 1
Part (d)
M1 expected = 100 × 61 ÷ 220
✓ 1
A1 27.7, which is greater than 5, so the test is valid for this cell
✓ 1
Exam tip: Always compare the p-value with the stated significance level and write the conclusion in the context of the question, not just 'reject H₀'.
195IBMI3S-4-02r31-08Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium6 marks
Calculator
The masses of loaded shipping containers at a port are normally distributed with mean 24.5 tonnes and standard deviation 1.2 tonnes.
(a)
Find the probability that a container has a mass greater than 26.5 tonnes.
(2)
(b)
A crane has a safe lifting limit set so that only 2% of containers exceed it. Find this limit, to 3 significant figures.
(2)
(c)
A ship carries 500 containers. Find the expected number with a mass greater than 26.5 tonnes. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to find P(M > 26.5)
✓ 1
A1 0.0478
✓ 1
Part (b)
M1 inverse normal with P(M < k) = 0.98
✓ 1
A1 27.0 tonnes
✓ 1
Part (c)
M1 500 × P(M > 26.5) (ECF from (a))
✓ 1
A1 23.9
✓ 1
Exam tip: 'Greater than' probabilities are found from the upper tail; check that your answer is below 0.5 when the value is above the mean.
196IBMI3S-4-02r31-10Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium7 marks
Calculator
Dolphins photographed by a research team are recognised as already catalogued with probability 0.35, independently of one another. In one survey, 12 dolphins are photographed. Let X be the number that are recognised.
(a)
Find the probability that at most 4 dolphins are recognised.
(2)
(b)
The team wants the probability that at least one dolphin is recognised to exceed 0.94. Find the smallest number of dolphins that must be photographed.
(3)
(c)
Find the probability that exactly 3 of the 12 dolphins are recognised.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to find P(X ≤ 4) using B(12, 0.35)
✓ 1
A1 0.583
✓ 1
Part (b)
M1 1 − 0.65n > 0.94 (or trial of values of n)
✓ 1
A1 n = 6 gives 0.925, which is too small
✓ 1
A1 n = 7 (probability 0.951)
✓ 1
Part (c)
M1 12C3 × 0.353 × 0.659 or calculator
✓ 1
A1 0.195
✓ 1
Exam tip: 'At least one' is best handled with the complement: P(X ≥ 1) = 1 − P(X = 0).
197IBMI3S-4-02r31-13Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium6 marks
Calculator
The floor areas of 50 offices in a building are summarised below. Area, a (m2): 10 ≤ a < 15, 15 ≤ a < 20, 20 ≤ a < 25, 25 ≤ a < 30, 30 ≤ a < 35 Frequency: 6, 14, 18, 9, 3
(a)
Write down the modal class.
(1)
(b)
Estimate the mean floor area.
(2)
(c)
Explain why the answer to (b) is only an estimate.
(1)
(d)
Estimate the population standard deviation of the floor areas.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 20 ≤ a < 25
✓ 1
Part (b)
M1 using midpoints 12.5, 17.5, 22.5, 27.5, 32.5
✓ 1
A1 21.4 m2
✓ 1
Part (c)
R1 the exact values within each class are unknown, so the midpoints are used
✓ 1
Part (d)
M1 using midpoints with the frequencies in a standard deviation calculation
✓ 1
A1 5.32 m2
✓ 1
Exam tip: Check whether the question asks for the population or the sample standard deviation; for the whole group of 50 offices use the population value (σ).
198IBMI3S-4-02r31-14Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium6 marks
Calculator
A logistics company ranks eight depots (1 = best) on delivery speed and on customer satisfaction. Depot A to H, speed rank: 1, 2, 3, 4, 5, 6, 7, 8 Satisfaction rank: 2, 1, 4, 3, 6, 8, 5, 7
(a)
Find the sum of the squares of the differences in rank, Σd2.
Give one reason why Spearman's coefficient is used here rather than Pearson's.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 differences −1, 1, −1, 1, −1, −2, 2, 1
✓ 1
A1 Σd2 = 14
✓ 1
Part (b)
M1 1 − 6 × 14 ÷ (8 × 63) (ECF from (a))
✓ 1
A1 0.833
✓ 1
Part (c)
R1 a strong positive association: depots that deliver faster tend to have higher customer satisfaction
✓ 1
Part (d)
R1 the data are ranks (ordinal), not measured values
✓ 1
Exam tip: Spearman's coefficient measures how well the ranks agree; a value close to 1 means a strong increasing relationship.
199IBMI3S-4-04r33-05Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium6 marks
A robotics team measures how long a battery lasts when a robot carries different loads. The table shows the load, x kg, and the running time, y minutes. x: 0.5, 1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0 y: 116, 113, 101, 99, 93, 80, 83, 70
(a)
Find Pearson's product-moment correlation coefficient, r, and describe the correlation.
(2)
(b)
Find the equation of the regression line of y on x. Give the coefficients to 4 significant figures.
(2)
(c)
Interpret the gradient of the regression line in context.
(1)
(d)
Use the equation from (b) (or the unrounded equation) to estimate the running time with a load of 2.8 kg. Give your answer to 3 significant figures.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(r=-0.981\)
✓ 1
A1 strong negative correlation
✓ 1
Part (b)
M1 use of GDC linear regression
✓ 1
A1 \(y=-12.88x+123.4\)
✓ 1
Part (c)
R1 each extra kilogram of load reduces the running time by about 12.9 minutes
✓ 1
Part (d)
A1 87.3 minutes (the unrounded equation also gives 87.3)
✓ 1
Exam tip: Only use the regression line to estimate within the range of the data (interpolation); 2.8 kg lies between 2.5 kg and 3.0 kg.
200IBMI3S-4-04r33-06Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium7 marks
A fashion retailer asks 240 customers to choose their favourite colour for a new jacket from three options. The results are shown below. Under 30 years (120 customers): black 42, neutral 31, bright 47 30 years or over (120 customers): black 38, neutral 55, bright 27 The retailer carries out a χ2 test for independence at the 5% significance level.
(a)
State the null hypothesis.
(1)
(b)
Show that the expected frequency for under-30 customers choosing black is 40.
(2)
(c)
Write down the number of degrees of freedom.
(1)
(d)
Find the χ2 statistic and the p-value.
(2)
(e)
State the conclusion of the test, giving a reason.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: colour choice and age group are independent
✓ 1
Part (b)
M1 use of \(\frac{\text{row total}\times\text{column total}}{\text{grand total}}=\frac{120\times80}{240}\)
✓ 1
A1 40 (shown)
✓ 1
Part (c)
A1 \((2-1)(3-1)=2\)
✓ 1
Part (d)
A1 χ2 = 12.3
✓ 1
A1 p = 0.0021
✓ 1
Part (e)
R1 p < 0.05, so reject H0: there is evidence that colour choice is associated with age group
✓ 1
Common error: Do not say that the test 'proves' an association; say that there is sufficient evidence at the stated significance level.
201IBMI3S-4-04r33-07Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium6 marks
A robot arm is assembled using joints from two suppliers. Supplier A provides 60% of the joints and 3% of these are defective. Supplier B provides the rest and 5% of these are defective.
(a)
Find the probability that a randomly chosen joint is defective.
(2)
(b)
A joint is found to be defective. Find the probability that it came from supplier A.
(2)
(c)
Two joints are chosen at random and independently of one another. Find the probability that exactly one of them is defective.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(0.6\times0.03+0.4\times0.05\) (tree diagram or equivalent)
✓ 1
A1 0.038
✓ 1
Part (b)
M1 use of conditional probability \(\frac{0.6\times0.03}{0.038}\)
✓ 1
A1 0.474
✓ 1
Part (c)
M1 use of \(2\times0.038\times(1-0.038)\)
✓ 1
A1 0.0731
✓ 1
Exam tip: For conditional probability, the denominator is the probability of the condition given (here, being defective), not 1.
202IBMI3S-4-04r33-09Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium6 marks
A biologist compares the lengths, in cm, of trout from two lakes. Random samples gave the following lengths. Lake A: 31.2, 29.8, 33.5, 30.4, 32.1, 28.9, 31.7, 30.6 Lake B: 27.9, 29.1, 26.8, 30.2, 28.4, 27.5, 29.6, 28.0 Assume that the lengths in both lakes are normally distributed with equal variances. The biologist carries out a two-tailed t-test at the 5% significance level to test whether the mean lengths differ.
(a)
State the null and alternative hypotheses.
(2)
(b)
Use a pooled two-sample t-test to find the p-value.
(2)
(c)
State the conclusion of the test, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: the mean lengths of trout in the two lakes are equal (\(\mu_A=\mu_B\))
✓ 1
A1 H1: the mean lengths are not equal (\(\mu_A\neq\mu_B\))
✓ 1
Part (b)
M1 use of GDC pooled two-sample t-test with the two data sets
✓ 1
A1 p = 0.00129 (accept 0.0013)
✓ 1
Part (c)
M1 compare p-value with 0.05 (0.0013 < 0.05)
✓ 1
R1 reject H0: there is evidence that the mean lengths of trout differ between the lakes
✓ 1
Exam tip: The test needs the data entered as two lists; state the type of test (pooled, two-sample, two-tailed) when quoting the p-value.
203IBMI3S-4-04r33-10Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium5 marks
A robotics club records the time, t seconds, that its robot takes to complete 60 practice runs of an obstacle course. 20 ≤ t < 30: 6 runs 30 ≤ t < 40: 14 runs 40 ≤ t < 50: 23 runs 50 ≤ t < 60: 11 runs 60 ≤ t < 70: 6 runs
(a)
Write down the modal class.
(1)
(b)
Estimate the mean time.
(2)
(c)
Estimate the population standard deviation of the times. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 40 ≤ t < 50
✓ 1
Part (b)
M1 use of class midpoints 25, 35, 45, 55, 65 with the frequencies
✓ 1
A1 44.5 seconds
✓ 1
Part (c)
M1 use of GDC one-variable statistics with midpoints and frequencies (or a correct formula)
✓ 1
A1 11.0 seconds
✓ 1
Exam tip: With grouped data, every answer is an estimate because each value is replaced by its class midpoint.
204IBMI3S-4-04r33-13Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium6 marks
A fashion shop surveys 120 customers. Of these, 65 bought a dress, 48 bought shoes and 20 bought neither item.
(a)
Find the number of customers who bought both a dress and shoes.
(2)
(b)
A customer is chosen at random. Let D be the event 'the customer bought a dress' and S the event 'the customer bought shoes'. Show that D and S are not independent.
(2)
(c)
Three different customers are chosen at random from the 120. Find the probability that all three bought shoes. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(65+48-x+20=120\) (or a Venn diagram with total 120)
✓ 1
A1 13 customers
✓ 1
Part (b)
M1 find \(P(D)\times P(S)=\frac{65}{120}\times\frac{48}{120}=0.2167\) and \(P(D\cap S)=\frac{13}{120}=0.1083\)
✓ 1
R1 \(P(D\cap S)\neq P(D)\times P(S)\), so the events are not independent
✓ 1
Part (c)
M1 use of \(\frac{48}{120}\times\frac{47}{119}\times\frac{46}{118}\) (without replacement)
✓ 1
A1 0.0616
✓ 1
Common error: When customers are chosen without replacement, both the numerator and the denominator decrease after each choice.
205IBMI3S-4-04r33-14Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium6 marks
At a pop-up shop, each customer who plays a game wins a voucher of value €x. The probability distribution of x is shown. x = 0: probability 0.50 x = 4: probability k x = 10: probability 0.12 x = 25: probability 0.04
(a)
Find the value of k.
(2)
(b)
Find the expected value of the voucher, in euros.
(2)
(c)
A customer plays the game three times, independently. Find the probability that the customer wins a voucher of value greater than €0 at least once.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(\sum P(X=x)=1\): \(0.50+k+0.12+0.04=1\)
✓ 1
A1 \(k=0.34\)
✓ 1
Part (b)
M1 use of \(E(X)=\sum xP(X=x)=4\times0.34+10\times0.12+25\times0.04\)
✓ 1
A1 €3.56
✓ 1
Part (c)
M1 use of \(1-0.5^3\) (complement of three games with no voucher)
✓ 1
A1 0.875
✓ 1
Exam tip: The probabilities in a distribution must sum to 1, which is the quickest way to find a missing probability.
206IBMI3S-4-02r34-03Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Probability and the binomial distributionPaper 1Medium6 marks
The durations of 100 customer calls to a telecoms helpline are summarised in the table.
Duration t (seconds)
0 ≤ t < 60
60 ≤ t < 120
120 ≤ t < 180
180 ≤ t < 240
240 ≤ t < 300
300 ≤ t < 360
Frequency
14
32
27
15
8
4
(a)
Write down the modal class.
(1)
(b)
Estimate the mean duration of a call, to 3 significant figures, using the midpoint of each class.
(2)
(c)
Calculate an estimate of the population standard deviation of the durations, to 3 significant figures.
(2)
(d)
Find the class that contains the median, and justify your answer.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(60\le t<120\)
✓ 1
Part (b)
M1 use of midpoints 30, 90, 150, 210, 270, 330 with frequencies
✓ 1
A1 140 seconds (139.8)
✓ 1
Part (c)
M1 use of the GDC or the formula with the midpoints and frequencies
✓ 1
A1 77.8 seconds (77.79…)
✓ 1
Part (d)
R1 cumulative frequencies 14, 46, 73, … so the 50th and 51st values are in \(120\le t<180\)
✓ 1
Common error: Using class boundaries instead of midpoints when estimating the mean from grouped data.
207IBMI3S-4-02r34-04Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Probability and the binomial distributionPaper 1Medium6 marks
Jars of honey are filled by a machine. The mass of honey in a jar is normally distributed with mean 502 g and standard deviation 4.5 g.
(a)
Find the probability that a randomly chosen jar contains less than 495 g. Give your answer to 3 significant figures.
(2)
(b)
Find the probability that a randomly chosen jar contains between 495 g and 510 g. Give your answer to 3 significant figures.
(2)
(c)
The producer wants only 2 % of jars to contain less than the mass printed on the label. Find the mass that should be printed on the label, correct to 1 decimal place.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(P(M<495)\) using the normal distribution function on the GDC
✓ 1
A1 0.0599 (0.05990…)
✓ 1
Part (b)
M1 \(P(495<M<510)\)
✓ 1
A1 0.902 (0.9023…)
✓ 1
Part (c)
M1 inverse normal with probability 0.02 (\(z=-2.054\))
✓ 1
A1 \(502-2.054\times4.5=492.8\) g (492.76…)
✓ 1
Common error: Using the standard deviation as the variance (or the other way round) when entering values into the GDC.
208IBMI3S-4-02r34-08Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Probability and the binomial distributionPaper 1Medium8 marks
A telecoms company logged 180 fault reports over several weeks. The numbers received on each day of the working week were: Monday 41, Tuesday 32, Wednesday 29, Thursday 31, Friday 47. The company tests, at the 5 % significance level, whether faults are equally likely on each of the five days.
(a)
State the null hypothesis and the alternative hypothesis.
(2)
(b)
Write down the expected number of reports on each day if H0 is true.
(1)
(c)
Find the number of degrees of freedom and the χ2 statistic, giving the statistic to 3 significant figures. The p-value is 0.161 to 3 significant figures.
(3)
(d)
State the conclusion of the test, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: faults are equally likely on each day (the data fit a uniform distribution)
✓ 1
A1 H1: faults are not equally likely on each day
✓ 1
Part (b)
A1 \(180\div5=36\)
✓ 1
Part (c)
A1 degrees of freedom \(=5-1=4\)
✓ 1
M1 \(\sum\dfrac{(O-E)^2}{E}\) or χ2 goodness-of-fit test on the GDC
✓ 1
A1 \(\chi^2=6.56\) (6.5556…)
✓ 1
Part (d)
R1 \(0.161>0.05\) (or \(6.56<9.49\), the critical value)
✓ 1
A1 do not reject H0: there is insufficient evidence that faults are not equally likely on each day
✓ 1
Common error: In a goodness-of-fit test the degrees of freedom are the number of categories minus 1, not the number of categories minus 2 or the sample size minus 1.
209IBMI3S-4-02r34-14Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Probability and the binomial distributionPaper 1Medium6 marks
The download speed X Mbps of a mobile network in a valley is normally distributed with mean 52 and standard deviation 10. A speed below 35 Mbps is classed as 'slow'.
(a)
Find the probability that a speed test gives a slow result, to 3 significant figures.
(2)
(b)
Out of 200 speed tests, find the expected number of slow results, to 3 significant figures. Use the unrounded probability from (a).
(2)
(c)
Eight speed tests are carried out, independently. Find the probability that exactly two are slow, to 3 significant figures. Use the unrounded probability from (a).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(P(X<35)\) using the normal distribution function on the GDC
✓ 1
A1 0.0446 (0.04457…)
✓ 1
Part (b)
M1 \(200\times0.04457\ldots\)
✓ 1
A1 8.91 (8.913…)
✓ 1
Part (c)
M1 \(Y\sim B(8,0.04457\ldots)\) and \(P(Y=2)\)
✓ 1
A1 0.0423 (0.04230…)
✓ 1
Common error: Entering the variance in place of the standard deviation when using normal distribution functions.
210IBMI3S-4-02r36-07Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium6 marks
A test is used to check a space-probe component for a fault. 4% of components are faulty. If a component is faulty, the test is positive with probability 0.95. If it is not faulty, the test is positive with probability 0.03.
(a)
Find the probability that a randomly chosen component tests positive.
(3)
(b)
Given that a component tests positive, find the probability that it is faulty. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use two branches (faulty and positive, not faulty and positive)
✓ 1
A1 0.04 × 0.95 + 0.96 × 0.03
✓ 1
A1 0.0668
✓ 1
Part (b)
M1 P(faulty ∩ positive) ÷ P(positive)
✓ 1
A1 0.038 ÷ 0.0668
✓ 1
A1 0.569
✓ 1
Common error: P(faulty | positive) is not the same as P(positive | faulty); the first divides by the chance of testing positive.
211IBMI3S-4-02r36-11Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium6 marks
The time, in minutes, that a bridge sensor takes to transmit a data packet was recorded for 44 packets.
Time t: 0 ≤ t < 10 (4), 10 ≤ t < 20 (11), 20 ≤ t < 30 (17), 30 ≤ t < 40 (9), 40 ≤ t < 50 (3) (frequency in brackets)
(a)
Write down the modal class.
(1)
(b)
Calculate an estimate of the mean time.
(3)
(c)
Calculate an estimate of the standard deviation (use the population standard deviation, σn).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 20 ≤ t < 30
✓ 1
Part (b)
M1 use of midpoints 5, 15, 25, 35, 45
✓ 1
M1 Σfx ÷ Σf = 1060 ÷ 44
✓ 1
A1 24.1 minutes
✓ 1
Part (c)
M1 use of midpoints on a GDC or √(Σfx2/Σf − mean2)
✓ 1
A1 10.4 minutes
✓ 1
Exam tip: For grouped data every value in a class is represented by its midpoint, so state that the results are estimates.
212IBMI3S-4-04r37-03Sampling and descriptive statistics; T-tests, confidence intervals and Markov chains; Probability and the binomial distribution; Normal distribution and chi-squared tests; Correlation and regression; Normal distribution and chi-squared tests; Probability and the binomial distributionPaper 1Medium7 marks
A robotics team tests two designs of gripper, A and B, on the time taken to pick up a component. Ten tests of design A gave a mean time of 14.2 s with sample standard deviation 1.1 s. Twelve tests of design B gave a mean time of 13.1 s with sample standard deviation 1.3 s. Both sets of times may be assumed to come from normal populations with equal variances. The team tests, at the 5% significance level, whether the mean time for design B is less than the mean time for design A.
(a)
State the null hypothesis and the alternative hypothesis.
(2)
(b)
Use your GDC to perform a pooled two-sample t-test and write down the p-value.
(1)
(c)
State the conclusion of the test, giving a reason.
(2)
(d)
Calculate the pooled estimate of the common variance.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(H_0:\mu_A=\mu_B\)
✓ 1
A1 \(H_1:\mu_B<\mu_A\)
✓ 1
Part (b)
A1 \(p=0.0235\)
✓ 1
Part (c)
R1 \(0.0235<0.05\)
✓ 1
A1 reject \(H_0\): there is sufficient evidence that the mean time for design B is less than for design A
Exam tip: The direction of the alternative hypothesis matters: a claim that one mean is smaller is a one-tailed test, so do not double the p-value.
213IBMI3S-4-04r37-06Sampling and descriptive statistics; T-tests, confidence intervals and Markov chains; Probability and the binomial distribution; Normal distribution and chi-squared tests; Correlation and regression; Normal distribution and chi-squared tests; Probability and the binomial distributionPaper 1Medium6 marks
A robotics laboratory tests 120 robots. Of these, 30 have a sensor fault (S), 24 have a motor fault (M) and 12 have both faults. One robot is chosen at random.
(a)
Find the probability that the robot has neither a sensor fault nor a motor fault.
(2)
(b)
Find \(P(S\mid M)\).
(2)
(c)
Determine whether the events S and M are independent. Justify your answer.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 number with at least one fault \(=30+24-12=42\)
✓ 1
A1 \(\frac{78}{120}=0.65\)
✓ 1
Part (b)
M1 \(\frac{P(S\cap M)}{P(M)}=\frac{0.1}{0.2}\) or \(\frac{12}{24}\)
✓ 1
A1 0.5
✓ 1
Part (c)
M1 \(P(S)\times P(M)=0.25\times0.2=0.05\)
✓ 1
R1 this is not equal to \(P(S\cap M)=0.1\), so the events are not independent
✓ 1
Common error: The 12 robots with both faults are counted in both the 30 and the 24; subtract them once when finding the number with at least one fault.
214IBMI3S-4-04r37-07Sampling and descriptive statistics; T-tests, confidence intervals and Markov chains; Probability and the binomial distribution; Normal distribution and chi-squared tests; Correlation and regression; Normal distribution and chi-squared tests; Probability and the binomial distributionPaper 1Medium7 marks
A robotic arm places a component correctly with probability 0.85 on each attempt, independently of other attempts. In one batch the arm makes 12 attempts. Let X be the number of components placed correctly in a batch.
(a)
Find the probability that all 12 components in a batch are placed correctly. Give your answer to 3 significant figures.
(2)
(b)
Find the probability that at least 11 components in a batch are placed correctly. Give your answer to 3 significant figures.
(2)
(c)
Find the expected number of components placed correctly in a batch.
(1)
(d)
The arm works on 40 batches. Find the expected number of batches in which all 12 components are placed correctly. Use the unrounded probability from part (a). Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(X\sim B(12,0.85)\) and \(P(X=12)=0.85^{12}\)
✓ 1
A1 0.142
✓ 1
Part (b)
M1 \(P(X=11)+P(X=12)\) or \(P(X\geq11)\) on the GDC
✓ 1
A1 0.443
✓ 1
Part (c)
A1 \(12\times0.85=10.2\)
✓ 1
Part (d)
M1 \(40\times0.85^{12}\)
✓ 1
A1 5.69
✓ 1
Common error: "At least 11" means 11 or 12, so add two probabilities (or use \(1-P(X\leq10)\)) rather than finding \(P(X=11)\) alone.
215IBMI3S-4-04r37-11Sampling and descriptive statistics; T-tests, confidence intervals and Markov chains; Probability and the binomial distribution; Normal distribution and chi-squared tests; Correlation and regression; Normal distribution and chi-squared tests; Probability and the binomial distributionPaper 1Medium7 marks
A clothing retailer asks 270 customers in which setting they prefer to buy clothes. The results are shown below. Under 25: 48 prefer online, 22 prefer boutique. 25–44: 62 prefer online, 58 prefer boutique. 45 and over: 30 prefer online, 50 prefer boutique. A χ2 test for independence is carried out at the 5% significance level.
(a)
State the null hypothesis and the alternative hypothesis.
(1)
(b)
Write down the number of degrees of freedom and, using your GDC, the p-value for the test.
(2)
(c)
State the conclusion of the test, giving a reason.
(2)
(d)
Calculate the expected frequency for customers aged 25–44 who prefer boutique shopping. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(H_0\): age group and preferred setting are independent; \(H_1\): they are not independent (associated)
✓ 1
Part (b)
A1 degrees of freedom \(=(3-1)(2-1)=2\)
✓ 1
A1 \(p=0.000732\)
✓ 1
Part (c)
R1 \(0.000732<0.05\)
✓ 1
A1 reject \(H_0\): there is evidence of an association between age group and preferred setting
✓ 1
Part (d)
M1 \(\frac{120\times130}{270}\)
✓ 1
A1 57.8
✓ 1
Exam tip: Expected frequency = (row total × column total) ÷ grand total. The degrees of freedom depend on the number of rows and columns, not on the grand total.
216IBMI3S-4-04r37-13Sampling and descriptive statistics; T-tests, confidence intervals and Markov chains; Probability and the binomial distribution; Normal distribution and chi-squared tests; Correlation and regression; Normal distribution and chi-squared tests; Probability and the binomial distributionPaper 1Medium7 marks
A marine biologist measures the length, x cm, and mass, y g, of eight fish of the same species. Length x: 28, 31, 33, 35, 38, 40, 42, 45 Mass y: 184, 324, 392, 452, 544, 604, 660, 744
(a)
Use your GDC to find the Pearson product-moment correlation coefficient, r. Give your answer to 3 significant figures.
(1)
(b)
Find the equation of the regression line of y on x.
(2)
(c)
Interpret the gradient of the regression line in this context.
(1)
(d)
Use the regression line to estimate the mass of a fish of length 38 cm.
(2)
(e)
Explain why the regression line should not be used to estimate the mass of a fish of length 60 cm.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 0.996
✓ 1
Part (b)
M1 uses a GDC to find the gradient and the intercept
✓ 1
A1 \(y=32x-680\)
✓ 1
Part (c)
R1 each additional 1 cm of length is associated with an increase of about 32 g in mass
✓ 1
Part (d)
M1 \(32\times38-680\)
✓ 1
A1 536 g
✓ 1
Part (e)
R1 60 cm is outside the range of the data (extrapolation), so the linear relationship may not hold
✓ 1
Exam tip: Interpret a gradient using the units of both variables: say "for each extra cm of length ..." rather than just stating a number.
217IBMI3S-4-02r38-03Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 1Medium8 marks
Calculator
A data centre records the power drawn by a server rack, x kW, and the temperature of the exhaust air, y °C, at seven times of day. x: 2.0, 2.5, 3.1, 3.6, 4.2, 4.8, 5.5 y: 21.5, 23.0, 25.1, 26.2, 28.9, 30.1, 33.4
(a)
Write down Pearson's product-moment correlation coefficient, r, and describe the correlation.
(2)
(b)
Find the equation of the regression line of y on x. Give the coefficients to 3 significant figures.
(2)
(c)
Interpret the gradient of the regression line in context.
(1)
(d)
Estimate the exhaust temperature when the rack draws 3.9 kW. Give your answer to 3 significant figures.
(2)
(e)
The operator uses the line to estimate the exhaust temperature when the rack draws 9 kW. Comment on the reliability of this estimate.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 r = 0.996
✓ 1
A1 very strong positive linear correlation
✓ 1
Part (b)
A1 gradient 3.32
✓ 1
A1 y = 3.32x + 14.7
✓ 1
Part (c)
R1 each extra 1 kW of power drawn is associated with a rise of about 3.32 °C in exhaust temperature
✓ 1
Part (d)
M1 3.32 × 3.9 + 14.7 (or unrounded coefficients)
✓ 1
A1 27.6 °C
✓ 1
Part (e)
R1 9 kW is far outside the range of the data (2.0 to 5.5 kW), so this is extrapolation and the estimate may be unreliable
✓ 1
Exam tip: Always interpret a gradient with its units and the words 'for each additional', and do not claim that a strong correlation proves cause.
218IBMI3S-4-02r38-11Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 1Medium7 marks
Calculator
The response times, t milliseconds, of 100 requests to a data centre are summarised in the table. 0 ≤ t < 10: frequency 14 10 ≤ t < 20: frequency 38 20 ≤ t < 30: frequency 31 30 ≤ t < 40: frequency 12 40 ≤ t < 50: frequency 5
(a)
Use the midpoints of the classes to estimate the mean response time.
(2)
(b)
Use the midpoints to estimate the population standard deviation, to 3 significant figures.
(2)
(c)
Write down the class that contains the median.
(1)
(d)
Estimate the median response time using linear interpolation within this class. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of midpoints 5, 15, 25, 35, 45 with the frequencies
✓ 1
A1 20.6 ms
✓ 1
Part (b)
M1 use of GDC with midpoints and frequencies for σn
✓ 1
A1 10.3 ms
✓ 1
Part (c)
A1 10 ≤ t < 20 (cumulative frequencies 14 and 52 show the 50th value is in this class)
✓ 1
Part (d)
M1 10 + (50 − 14) ÷ 38 × 10
✓ 1
A1 19.5 ms
✓ 1
Common error: Using the class boundaries instead of the class midpoints when estimating the mean from grouped data.
219IBMI3S-4-02r38-12Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 1Medium7 marks
Calculator
A cyclist records her average speed, y km/h, on routes of different steepness, where x is the average gradient of the route in percent. For routes with gradients between 1% and 8%, the regression line of y on x is y = 31.4 − 2.1x, and the correlation coefficient is r = −0.94.
(a)
Describe the correlation between gradient and average speed.
(1)
(b)
Interpret the value of the gradient of the regression line, −2.1, in context.
(1)
(c)
Use the regression line to estimate the average speed on a route with gradient 7.0%.
(2)
(d)
On a route with gradient 6.0%, her average speed was 19.6 km/h. Find the residual for this route (observed value − predicted value).
(2)
(e)
The cyclist uses the line to predict her average speed on a route with gradient 15%. Explain why this prediction is not reliable.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 strong negative correlation (as gradient increases, average speed decreases)
✓ 1
Part (b)
R1 for each extra 1% of average gradient, the average speed decreases by about 2.1 km/h
✓ 1
Part (c)
M1 31.4 − 2.1 × 7.0
✓ 1
A1 16.7 km/h
✓ 1
Part (d)
M1 predicted value 31.4 − 2.1 × 6.0 = 18.8
✓ 1
A1 residual = 0.8 km/h
✓ 1
Part (e)
R1 15% is outside the range of the data (1% to 8%), so this is extrapolation (accept: the line gives about −0.1 km/h, an impossible speed)
✓ 1
Exam tip: A residual is observed minus predicted: a positive residual means the point lies above the regression line.
220IBMI3S-4-02r44-03Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium7 marks
In a Formula 1 team, the time T seconds taken for a pit stop is normally distributed with mean 2.45 and standard deviation 0.18.
(a)
Find the probability that a pit stop takes more than 2.7 seconds.
(2)
(b)
Find the probability that a pit stop takes between 2.3 and 2.6 seconds.
(2)
(c)
The team's fastest 5% of pit stops are faster than t seconds. Find t.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 normal distribution with \(\mu=2.45\), \(\sigma=0.18\) and \(P(T>2.7)\) (for example \(z=1.39\))
✓ 1
A1 0.0824
✓ 1
Part (b)
M1 \(P(2.3
✓ 1
A1 0.595
✓ 1
Part (c)
M1 \(z=-1.645\) for a lower tail of 0.05
✓ 1
M1 \(t=2.45-1.645\times 0.18\)
✓ 1
A1 \(t=2.15\)
✓ 1
Exam tip: Sketch the normal curve and shade the required area before using the calculator; it shows which tail you need.
221IBMI3S-4-02r44-04Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium7 marks
A researcher records the maximum daily temperature, x °C, and the number of bees, y, leaving a hive each minute on eight days.
x
12
14
16
18
20
22
24
26
y
18
25
29
38
41
52
55
64
(a)
Find the Pearson's product-moment correlation coefficient, r.
(2)
(b)
Describe the correlation between x and y.
(1)
(c)
Find the equation of the regression line of y on x, in the form \(y=ax+b\) with a and b to 3 significant figures.
(2)
(d)
Interpret the value of a in this context.
(1)
(e)
Use your equation from part (c), with the rounded values of a and b, to estimate the number of bees leaving per minute when the temperature is 21 °C.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of GDC statistics mode with the data
✓ 1
A1 \(r=0.995\)
✓ 1
Part (b)
A1 very strong positive correlation
✓ 1
Part (c)
A1 \(a=3.24\)
✓ 1
A1 \(b=-21.3\)
✓ 1
Part (d)
R1 each 1 °C rise in maximum temperature is associated with about 3.24 more bees leaving per minute
✓ 1
Part (e)
A1 \(3.24(21)-21.3=46.7\) (ECF from (c)); estimate is within the data range so reliable
✓ 1
Common error: Only use a regression line to predict inside the range of the data; the intercept here would predict a negative number of bees at 0 °C, which is meaningless.
222IBMI3S-4-02r44-05Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium6 marks
At a hospital, 4% of patients screened have a certain condition. The screening test is positive for 95% of patients who have the condition and for 8% of patients who do not. A patient is screened at random.
(a)
Find the probability that the test result is positive.
(3)
(b)
Given that the test is positive, find the probability that the patient has the condition.
(2)
(c)
Explain why your answer to part (b) is much lower than 0.95.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(0.04\times 0.95=0.038\) (true positives)
✓ 1
M1 \(0.96\times 0.08=0.0768\) (false positives)
✓ 1
A1 \(0.1148\)
✓ 1
Part (b)
M1 \(\frac{0.04\times 0.95}{0.1148}\) (ECF from (a))
✓ 1
A1 0.331
✓ 1
Part (c)
R1 most patients do not have the condition, so the false positives outnumber the true positives
✓ 1
Exam tip: A tree diagram or a table of 1000 imagined patients makes conditional probability questions easier to organise.
223IBMI3S-4-02r44-12Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium7 marks
The waiting times, in minutes, of 75 patients in a hospital emergency department are summarised.
Waiting time t (min)
0 ≤ t < 10
10 ≤ t < 20
20 ≤ t < 30
30 ≤ t < 40
40 ≤ t < 50
50 ≤ t < 60
Frequency
6
14
25
18
9
3
(a)
Estimate the mean waiting time.
(2)
(b)
Estimate the population standard deviation.
(2)
(c)
Write down the interval containing the median.
(1)
(d)
Estimate the median waiting time using linear interpolation within this interval.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of mid-interval values 5, 15, 25, 35, 45, 55
✓ 1
A1 27.5 minutes
✓ 1
Part (b)
M1 use of mid-interval values and frequencies on the GDC
✓ 1
A1 12.3 minutes
✓ 1
Part (c)
A1 \(20\le t<30\) (the 37.5th value lies here since the cumulative frequency reaches 20 then 45)
✓ 1
Part (d)
M1 \(20+\frac{37.5-20}{25}\times 10\)
✓ 1
A1 27.0 minutes
✓ 1
Common error: Grouped data give only estimates, because mid-interval values stand in for the unknown individual times.
224IBMI3S-4-02r47-04Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium7 marks
A games coach records the number of hours per week, x, that eight players spend training, and each player's skill rating, y (out of 100). x: 2, 3, 5, 6, 8, 9, 11, 12 y: 41, 48, 52, 63, 66, 74, 79, 90
(a)
Find the Pearson product-moment correlation coefficient, r, and describe the correlation.
(2)
(b)
Find the equation of the regression line of y on x. Give the coefficients to 3 significant figures.
(2)
(c)
Interpret the value of the gradient in this context.
(1)
(d)
A player trains 30 hours per week. Use the regression line to estimate their rating, and comment on the validity of the estimate. Use the unrounded line from your calculator.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 r = 0.986
✓ 1
A1 strong positive linear correlation
✓ 1
Part (b)
M1 attempt at regression line y = ax + b
✓ 1
A1 y = 4.52x + 32.5
✓ 1
Part (c)
R1 each extra hour of weekly training is associated with an increase of about 4.5 rating points
✓ 1
Part (d)
A1 168
✓ 1
R1 not valid: 30 hours is outside the range of the data (extrapolation) and the rating cannot exceed 100
✓ 1
Common error: A high correlation does not show that more training causes higher ratings, and a regression line should only be used to predict within the range of the data.
225IBMI3S-4-02r47-05Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium8 marks
A water company records complaints from two zones of a city. Each of 200 complaints is classified by type. The results are shown below. North: low pressure 48, discoloured water 31, taste or odour 21 South: low pressure 62, discoloured water 24, taste or odour 14 A χ2 test for independence is carried out at the 5% significance level to decide whether the type of complaint depends on the zone.
(a)
State the null and alternative hypotheses.
(1)
(b)
Show that the expected frequency for North with low pressure is 55.
(2)
(c)
Write down the number of degrees of freedom.
(1)
(d)
Find the χ2 statistic and the p-value.
(2)
(e)
State the conclusion of the test, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: complaint type and zone are independent; H1: complaint type and zone are not independent (dependent)
✓ 1
Part (b)
M1 (row total × column total) ÷ grand total = (100 × 110) ÷ 200
A1 do not reject H0: no evidence that complaint type depends on zone
✓ 1
Exam tip: Always compare the p-value with the stated significance level and write the conclusion in the context of the question, not just 'accept' or 'reject'.
226IBMI3S-4-02r47-06Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium7 marks
A game claims that when a player opens a chest, the item received is common with probability 0.50, uncommon 0.30, rare 0.15 and epic 0.05. A player opens 200 chests and receives 78 common, 70 uncommon, 36 rare and 16 epic items. A χ2 goodness-of-fit test is carried out at the 5% significance level.
(a)
State the null hypothesis.
(1)
(b)
Find the expected frequencies for the four item types.
(1)
(c)
Find the number of degrees of freedom, the χ2 statistic and the p-value.
(3)
(d)
State the conclusion of the test, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: the item probabilities are as claimed (0.50, 0.30, 0.15, 0.05)
A1 reject H0: evidence that the item probabilities are not as claimed
✓ 1
Common error: The number of degrees of freedom for goodness of fit is the number of categories minus 1 when no parameters are estimated from the data.
227IBMI3S-4-02r47-07Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium7 marks
The mass of a pharmaceutical tablet, in milligrams, is normally distributed with mean μ and standard deviation σ. 10% of tablets have mass less than 194.6 mg and 2% of tablets have mass greater than 206.2 mg.
(a)
Find the values of μ and σ, each correct to 3 significant figures.
(5)
(b)
Using these values, find the interquartile range of the tablet masses, to 3 significant figures.
M1 solves the two equations simultaneously, for example σ = 11.6 ÷ (2.0537… + 1.2816…)
✓ 1
A1 σ = 3.48
✓ 1
A1 μ = 199
✓ 1
Part (b)
M1 IQR = 2 × 0.6745σ (z = ±0.6745 for the quartiles)
✓ 1
A1 4.69 mg
✓ 1
Exam tip: For 'greater than' probabilities, convert to the area to the left (here 0.98) before using inverse normal.
228IBMI3S-4-02r47-08Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium8 marks
Seventeen patients with high blood pressure are randomly assigned to one of two drugs. The reductions in systolic blood pressure (mmHg) after four weeks are Drug A (8 patients): 12.1, 9.8, 14.3, 11.2, 10.5, 13.0, 12.6, 9.4 Drug B (9 patients): 8.9, 10.2, 7.6, 9.1, 11.0, 8.2, 9.5, 7.9, 10.4 A two-tailed t-test at the 5% significance level is used to decide whether the mean reductions differ.
(a)
State the null and alternative hypotheses.
(2)
(b)
State two assumptions needed for the pooled two-sample t-test.
(2)
(c)
Find the p-value for the test.
(2)
(d)
State the conclusion in context, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: μA = μB
✓ 1
A1 H1: μA ≠ μB
✓ 1
Part (b)
R1 both populations are normally distributed
✓ 1
R1 the populations have equal variances
✓ 1
Part (c)
M1 two-sample pooled t-test (df = 15)
✓ 1
A1 p = 0.00357 (t = 3.45)
✓ 1
Part (d)
R1 p = 0.00357 < 0.05
✓ 1
A1 reject H0: evidence that the mean reductions in blood pressure differ between the drugs
✓ 1
Common error: Choose the pooled option only when the equal-variance assumption is stated; and remember a two-tailed test compares the p-value directly with the significance level.
229IBMI3S-4-02r47-10Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium8 marks
In a video game, each attack by a character is a critical hit with probability 0.15, independently of other attacks. In a battle the character makes 40 attacks. Let X be the number of critical hits.
(a)
Write down the mean and the variance of X.
(2)
(b)
Find the probability that there are at least 8 critical hits.
(2)
(c)
Find the smallest number of attacks, n, needed so that the probability of at least one critical hit is greater than 0.95.
(3)
(d)
Find the probability of at least one critical hit in this number of attacks. Give your answer to 3 significant figures.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 mean = 6
✓ 1
A1 variance = 5.1
✓ 1
Part (b)
M1 P(X ≥ 8) = 1 − P(X ≤ 7)
✓ 1
A1 0.244
✓ 1
Part (c)
M1 sets 1 − 0.85n > 0.95 (or 0.85n < 0.05)
✓ 1
A1 n > 18.4 (n = 18 gives 0.946…)
✓ 1
A1 n = 19
✓ 1
Part (d)
A1 1 − 0.8519 = 0.954
✓ 1
Common error: 'At least one' is solved most easily through the complement: 1 − P(none). Test integer values around your solution to confirm the smallest n.
230IBMI3S-4-02r47-11Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium7 marks
The times, in hours, taken to repair 94 burst water pipes are summarised below. 0 ≤ t < 5: 8 5 ≤ t < 10: 21 10 ≤ t < 15: 34 15 ≤ t < 20: 18 20 ≤ t < 25: 9 25 ≤ t < 35: 4
(a)
Write down the modal class.
(1)
(b)
Estimate the mean repair time. Give your answer to 3 significant figures.
(2)
(c)
Estimate the standard deviation of the repair times, treating the data as a population. Give your answer to 3 significant figures.
(2)
(d)
Use the cumulative frequencies 8, 29, 63, 81, 90, 94 to estimate the median repair time to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 10 ≤ t < 15
✓ 1
Part (b)
M1 uses midpoints 2.5, 7.5, 12.5, 17.5, 22.5, 30
✓ 1
A1 13.2 hours
✓ 1
Part (c)
M1 attempt at standard deviation using midpoints and frequencies
✓ 1
A1 6.40 hours
✓ 1
Part (d)
M1 the 47th value lies in 10 ≤ t < 15 and linear interpolation is used
✓ 1
A1 10 + (47 − 29)/34 × 5 = 12.6 hours
✓ 1
Exam tip: For grouped data the mean and standard deviation are estimates because midpoints replace the actual values; say so if asked to comment.
231IBMI3S-4-04r47-02Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium7 marks
A robotics club times a robot as it completes a maze twelve times. The times, in seconds, are 14.2, 15.1, 15.8, 16.0, 16.4, 16.9, 17.3, 17.8, 18.5, 19.0, 19.6, 27.4.
(a)
Find the mean time, and the population standard deviation, each correct to 3 significant figures.
(2)
(b)
Find the median and the interquartile range of the times.
(3)
(c)
A time is classed as an outlier if it is more than 1.5 × IQR above Q3. Show that 27.4 s is an outlier.
A1 upper boundary 23.025, and 27.4 > 23.025 so it is an outlier (R1 comparison stated)
✓ 1
Common error: Quote the boundary value and compare it with the data point; stating "it is far from the others" does not show an outlier.
232IBMI3S-4-04r47-03Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium6 marks
The length, L cm, of herring caught by a trawler is normally distributed with mean 28.4 cm and standard deviation 2.6 cm.
(a)
Find the probability that a randomly chosen herring is longer than 30 cm.
(2)
(b)
A catch contains 800 herring. Estimate how many are shorter than 25 cm.
(2)
(c)
Only the longest 15% of herring are kept for breeding. Find the minimum length of a herring that is kept, correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 normal distribution calculation, L ~ N(28.4, 2.62)
✓ 1
A1 0.269
✓ 1
Part (b)
M1 P(L < 25) = 0.0955 (0.09548…) multiplied by 800
✓ 1
A1 76 herring
✓ 1
Part (c)
M1 inverse normal with P(L < l) = 0.85, z = 1.0364
✓ 1
A1 l = 28.4 + 2.6 × 1.0364 = 31.1 cm
✓ 1
Exam tip: For "the longest 15%" use the cumulative probability 0.85 in the inverse normal function, not 0.15.
233IBMI3S-4-04r47-04Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium7 marks
A machine cuts strips of fabric for a fashion label. The length of a strip, X cm, is normally distributed with mean 150 cm. It is found that 4% of strips are shorter than 148.6 cm.
(a)
Find the standard deviation of the lengths, correct to 3 significant figures.
(3)
(b)
Using σ = 0.800, find the probability that a strip has a length between 149 cm and 151.5 cm.
(2)
(c)
Strips longer than 151.5 cm are rejected. Using σ = 0.800, estimate the number of rejected strips in a batch of 2000, to the nearest whole number.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 z = −1.7507 for P(Z < z) = 0.04
✓ 1
M1 (148.6 − 150)/σ = −1.7507
✓ 1
A1 σ = 0.800 cm
✓ 1
Part (b)
M1 P(149 < X < 151.5) with N(150, 0.8002)
✓ 1
A1 0.864
✓ 1
Part (c)
M1 P(X > 151.5) = 0.0304 multiplied by 2000
✓ 1
A1 61 strips
✓ 1
Common error: A standardised value for a probability below 0.5 is negative; keep the sign when you substitute into (x − μ)/σ.
234IBMI3S-4-04r47-05Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium8 marks
A fashion retailer asks 200 customers to choose between a casual design and a formal design. The results, by age group, are shown.
Casual
Formal
Under 25
48
22
25 to 44
41
39
45 and over
21
29
A χ2 test for independence is carried out at the 5% significance level.
(a)
State the null and alternative hypotheses.
(1)
(b)
Find the expected frequency for customers under 25 who choose the casual design.
(2)
(c)
Write down the number of degrees of freedom.
(1)
(d)
Find the χ2 statistic and the p-value.
(2)
(e)
State the conclusion of the test, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: choice of design is independent of age group; H1: choice of design is not independent of age group (associated)
✓ 1
Part (b)
M1 70 × 110/200
✓ 1
A1 38.5
✓ 1
Part (c)
A1 (3 − 1)(2 − 1) = 2
✓ 1
Part (d)
A1 χ2 = 9.08
✓ 1
A1 p = 0.0107
✓ 1
Part (e)
R1 p = 0.0107 < 0.05 (or 9.08 > 5.991)
✓ 1
A1 reject H0: there is evidence that choice of design depends on age group
✓ 1
Exam tip: State the comparison (p-value against the significance level) and then the conclusion in the words of the context.
235IBMI3S-4-04r47-08Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium6 marks
A warehouse robot picks up a parcel successfully with probability 0.85 on each attempt, independently of other attempts. The robot makes 12 attempts. Let X be the number of successful attempts.
(a)
Find P(X = 10).
(2)
(b)
Find the smallest number of consecutive attempts for which the probability that every attempt is successful is below 0.2.
Common error: "At least 10" includes 10 itself: use 1 − P(X ≤ 9), not 1 − P(X ≤ 10).
236IBMI3S-4-04r47-10Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium5 marks
Two judges each score eight robots in a design competition. The scores out of 100 are shown.
Robot
P
Q
R
S
T
U
V
W
Judge 1
82
74
91
65
88
70
79
95
Judge 2
78
76
85
60
90
72
70
92
(a)
Find Spearman's rank correlation coefficient for these scores, correct to 3 significant figures.
(3)
(b)
Interpret your value in the context of the competition.
(1)
(c)
Give one reason why Spearman's coefficient, rather than Pearson's, is suitable for comparing the judges.
R1 strong positive association: the two judges tend to rank the robots in a similar order
✓ 1
Part (c)
R1 it compares the rankings, so it needs only an ordering and does not require a linear relationship between the raw scores
✓ 1
Exam tip: Rank both sets from the same end (for example, 1 for the lowest score) before calculating differences.
237IBMI3S-4-04r47-13Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Medium7 marks
A robot is programmed to choose one of four directions (north, east, south, west) at random each time it reaches a junction. In 200 junctions it chose north 62 times, east 44 times, south 51 times and west 43 times. A χ2 goodness-of-fit test is carried out at the 5% significance level to test whether the four directions are chosen with equal probability.
(a)
State the null and alternative hypotheses.
(1)
(b)
Find the expected frequency for each direction.
(1)
(c)
Find the χ2 statistic and the p-value for 3 degrees of freedom.
(3)
(d)
State the conclusion of the test, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: the four directions are equally likely (the data fit a uniform distribution); H1: the directions are not equally likely
✓ 1
Part (b)
A1 200/4 = 50
✓ 1
Part (c)
M1 Σ(O − E)2/E
✓ 1
A1 χ2 = 4.6
✓ 1
A1 p = 0.204
✓ 1
Part (d)
R1 p = 0.204 > 0.05 (or 4.6 < 7.815)
✓ 1
A1 do not reject H0: there is insufficient evidence that the directions are not equally likely
✓ 1
Common error: Failing to reject H₀ does not prove the directions are equally likely; it only means the data give insufficient evidence against it.
238IBMI3S-4-02r48-05Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 1Medium7 marks
Calculator
A game developer states that the loot boxes in a game contain a Common, Rare, Epic or Legendary item with probabilities 0.50, 0.30, 0.15 and 0.05 respectively. A player opens 200 boxes and records the items.
Item
Common
Rare
Epic
Legendary
Frequency
112
63
18
7
A \(\chi^2\) goodness-of-fit test at the 5% significance level is used to test whether the stated probabilities are correct.
(a)
State the null and alternative hypotheses.
(2)
(b)
Write down the number of degrees of freedom.
(1)
(c)
Calculate the \(\chi^2\) statistic.
(2)
(d)
The \(p\)-value for this test is 0.0632. State, with a reason, whether there is sufficient evidence at the 5% level to reject the developer's stated probabilities.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(H_0\): the stated probabilities (0.50, 0.30, 0.15, 0.05) are correct (the data fit the distribution)
✓ 1
A1 \(H_1\): the stated probabilities are not correct
✓ 1
Part (b)
A1 3
✓ 1
Part (c)
M1 expected frequencies 100, 60, 30, 10 and use of \(\sum\frac{(f_o-f_e)^2}{f_e}\)
R1 0.0632 > 0.05 (or \(\chi^2=7.29\) is below the critical value 7.815)
✓ 1
A1 do not reject \(H_0\); no sufficient evidence that the stated probabilities are wrong
✓ 1
Common error: Failing to reject \(H_0\) does not prove the stated probabilities are correct; it means there is not enough evidence against them.
239IBMI3S-4-02r48-06Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 1Medium5 marks
Calculator
The mean daily maximum temperature, \(T\) °C, and the mean daily water use, \(W\) megalitres (ML), of a town are recorded in eight summer weeks.
T
12
15
18
21
24
27
30
33
W
310
324
331
346
352
371
374
392
(a)
Find the equation of the regression line of \(W\) on \(T\), giving the gradient correct to 2 decimal places and the intercept correct to 1 decimal place.
(3)
(b)
Use your equation from part (a) to estimate the mean daily water use in a week with a mean daily maximum temperature of 26 °C. Give your answer to the nearest ML.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of linear regression on the calculator
✓ 1
A1 gradient 3.77
✓ 1
A1 intercept 265.2, so \(W=3.77T+265.2\)
✓ 1
Part (b)
M1 \(W=3.77\times26+265.2\) (ECF from (a))
✓ 1
A1 363 ML
✓ 1
Exam tip: Substitute into the rounded equation you stated, and make sure the units of your answer match the variable on the left-hand side.
240IBMI3S-4-02r48-07Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 1Medium6 marks
Calculator
The concentration of a drug in the blood of patients, measured in mg per litre, one hour after a dose is normally distributed with mean \(\mu\) and standard deviation \(\sigma\). 10% of patients have a concentration below 20 mg per litre and 20% of patients have a concentration above 30 mg per litre.
(a)
Find the \(z\)-values corresponding to these two probabilities. Give your answers to 4 significant figures.
(2)
(b)
Write down two equations connecting \(\mu\) and \(\sigma\).
(2)
(c)
Find the value of \(\mu\). Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(P(X<20)=0.1\): \(z=-1.282\)
✓ 1
A1 \(P(X>30)=0.2\): \(P(X<30)=0.8\), \(z=0.8416\)
✓ 1
Part (b)
M1 \(\frac{20-\mu}{\sigma}=-1.2816\)
✓ 1
A1 \(\frac{30-\mu}{\sigma}=0.8416\)
✓ 1
Part (c)
M1 subtract or substitute to find \(\sigma=\frac{10}{1.2816+0.8416}=4.71\)
✓ 1
A1 \(\mu=20+1.2816\times4.71=26.0\) mg per litre
✓ 1
Common error: A probability of 0.2 in the upper tail corresponds to an area of 0.8 to the left, so the inverse normal input is 0.8.
241IBMI3S-4-02r48-08Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 1Medium7 marks
Calculator
In a trial, 10 patients take drug X and 10 different patients take drug Y. The reduction in pain score for each patient is recorded. The scores for each drug are assumed to be normally distributed with the same variance.
Mean reduction
Sample standard deviation
Drug X
14.2
2.1
Drug Y
11.6
2.4
A one-tailed pooled \(t\)-test is carried out at the 5% significance level to test whether drug X gives a greater mean reduction in pain score than drug Y.
(a)
State the null and alternative hypotheses.
(2)
(b)
Write down the number of degrees of freedom and calculate the test statistic \(t\). Give \(t\) to 3 significant figures.
(3)
(c)
The \(p\)-value for this test is 0.009. State the conclusion of the test, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(H_0\): \(\mu_X=\mu_Y\)
✓ 1
A1 \(H_1\): \(\mu_X>\mu_Y\)
✓ 1
Part (b)
A1 degrees of freedom \(=10+10-2=18\)
✓ 1
M1 pooled variance \(s_p^2=\frac{9\times2.1^2+9\times2.4^2}{18}=5.085\)
A1 reject \(H_0\); there is sufficient evidence that drug X gives a greater mean reduction in pain score
✓ 1
Exam tip: For a one-tailed test compare the one-tailed \(p\)-value with the significance level; the alternative hypothesis fixes the direction.
242IBMI3S-4-02r48-10Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 1Medium6 marks
Calculator
In an online game, each attack independently scores a critical hit with probability 0.15.
(a)
Find the probability that at least one critical hit is scored in 10 attacks. Give your answer to 3 significant figures.
(2)
(b)
Find the smallest number of attacks, \(n\), needed so that the probability of at least one critical hit is greater than 0.9.
(2)
(c)
Using the value of \(n\) from part (b), find the probability of exactly 2 critical hits in \(n\) attacks. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(1-0.85^{10}\)
✓ 1
A1 0.803
✓ 1
Part (b)
M1 solve \(1-0.85^n>0.9\), e.g. table, graph or logarithms (\(n>14.2\))
✓ 1
A1 \(n=15\)
✓ 1
Part (c)
M1 \(\binom{15}{2}(0.15)^2(0.85)^{13}\) or binomial pdf with \(n=15\)
✓ 1
A1 0.286 (ECF from (b))
✓ 1
Exam tip: 'At least one' means \(1-P(X=0)\); for the smallest \(n\) check one value either side of the boundary.
243IBMI3S-4-02r48-14Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 1Medium7 marks
Calculator
120 gamers are surveyed. 70 play game A, 55 play game B and 20 play neither game. A gamer is chosen at random. Let \(A\) be the event that the gamer plays game A and \(B\) the event that the gamer plays game B.
(a)
Find the number of gamers who play both games.
(2)
(b)
Find \(P(A\cap B')\).
(1)
(c)
Determine whether \(A\) and \(B\) are independent events. Justify your answer.
(2)
(d)
Find \(P(A\mid B)\). Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(70+55-(120-20)\) or equivalent Venn diagram equation
✓ 1
A1 25
✓ 1
Part (b)
A1 \(\frac{45}{120}=0.375\)
✓ 1
Part (c)
M1 \(P(A)\times P(B)=\frac{70}{120}\times\frac{55}{120}=0.267\) and \(P(A\cap B)=\frac{25}{120}=0.208\)
✓ 1
R1 these are not equal, so \(A\) and \(B\) are not independent
Exam tip: If \(P(A\mid B)\ne P(A)\) the events are not independent: here \(0.455\ne0.583\).
244IBMI-033-00-05Sampling and descriptive statisticsPaper 1Hard7 marks
Two running clubs take part in a 10 km race. Club A has 25 runners, whose finishing times have mean 42.0 minutes and standard deviation 4.0 minutes. Club B has 15 runners, whose finishing times have mean 46.8 minutes and standard deviation 5.0 minutes.
(a)
Find the mean finishing time of all 40 runners.
(2)
(b)
Using σ2 = Σx2/n − (mean)2, show that Σx2 for the Club A runners is 44 500.
(2)
(c)
Find the standard deviation of the finishing times of all 40 runners.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
(25 × 42.0 + 15 × 46.8) ÷ 40 = 1752 ÷ 40 M1
✓ 1
43.8 minutes A1
✓ 1
Part (b)
Σx2 = 25(4.02 + 42.02) M1
✓ 1
= 25 × 1780 = 44 500 A1
✓ 1
Part (c)
Club B: Σx2 = 15(5.02 + 46.82) = 33 228.6, so combined Σx2 = 77 728.6 M1
✓ 1
Variance = 77 728.6 ÷ 40 − 43.82 = 24.775 M1
✓ 1
Standard deviation = 4.98 minutes A1
✓ 1
Common error: You cannot average the two standard deviations. The extra spread caused by the difference between the group means must be included, which is why the combined value is close to the larger one.
245IBMI3S-4-01r14-08Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 1Hard5 marks
The lifetime L years of a solar inverter is normally distributed with mean μ and standard deviation σ. The manufacturer finds that 10% of inverters last less than 8 years and that 5% last more than 14 years.
(a)
Find the standardised values corresponding to these two probabilities, and hence write down two equations in μ and σ.
(2)
(b)
Solve the equations to find σ and μ, each to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(P(L<8)=0.10\) gives \(z=-1.2816\) and \(P(L>14)=0.05\) gives \(z=1.6449\)
✓ 1
A1 \(\frac{8-\mu}{\sigma}=-1.2816\) and \(\frac{14-\mu}{\sigma}=1.6449\)
Common error: The 10% point lies below the mean, so its z-value is negative; using +1.2816 produces wrong values for both μ and σ.
246IBMI3S-4-01r14-09Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 1Hard7 marks
In an outbreak, each close contact of an infected person becomes infected independently with probability 0.08. Let X be the number of infected people among n close contacts of one infected person.
(a)
For n = 15, find the probability that at least one contact becomes infected.
(2)
(b)
Find the smallest value of n for which the probability that at least one contact becomes infected exceeds 0.95.
(3)
(c)
For this value of n, find the probability that at least one contact becomes infected, to 4 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(1-0.92^{15}\) (or \(1-P(X=0)\))
✓ 1
A1 0.714
✓ 1
Part (b)
M1 \(1-0.92^{n}>0.95\) or \(0.92^n<0.05\)
✓ 1
M1 attempt to solve by logarithms or by trying values (\(n=35\) gives 0.9460, which is too small)
✓ 1
A1 n = 36
✓ 1
Part (c)
M1 substitution into \(1-0.92^n\)
✓ 1
A1 0.9503
✓ 1
Exam tip: "At least one" is best handled with the complement: 1 − P(X = 0). Solving the inequality with logarithms reverses the sign when you divide by ln 0.92, which is negative.
247IBMI3S-4-02r20-12Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Hard7 marks
A housing association installs smart thermostats in some homes. The daily energy use, in kWh, is recorded for a random sample of homes with a smart thermostat (sample A) and a random sample of homes without one (sample B).
It may be assumed that both populations are normally distributed with equal variances. A one-tailed two-sample t-test, using the pooled variance, is carried out at the 5% significance level to test whether smart thermostats reduce mean daily energy use.
(a)
Write down the null and alternative hypotheses.
(2)
(b)
Find the p-value for the test.
(2)
(c)
State the conclusion of the test, giving a reason.
(2)
(d)
State why the pooled version of the test can be used here.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: μA = μB (the mean energy use is the same)
✓ 1
A1 H1: μA < μB (the mean is lower with a smart thermostat)
✓ 1
Part (b)
M1 use of a pooled two-sample t-test on the GDC (t = −3.06, df = 17)
✓ 1
A1 p = 0.00353
✓ 1
Part (c)
M1 comparison of p = 0.00353 with 0.05
✓ 1
R1 reject H0: there is evidence that smart thermostats reduce mean daily energy use
✓ 1
Part (d)
R1 the populations are assumed to have equal variances (the sample standard deviations 0.845 and 0.876 are similar)
✓ 1
Exam tip: For a one-tailed test, make sure the GDC alternative hypothesis matches H₁, otherwise the p-value will be double what it should be.
248IBMI3S-4-04r33-08Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Hard7 marks
The length, X metres, of fabric on a roll is normally distributed with mean μ and standard deviation σ. It is known that 10% of rolls are shorter than 48.5 m and 5% of rolls are longer than 51.2 m.
(a)
Write down the z-values that correspond to 48.5 m and to 51.2 m. Give each to 3 significant figures.
(2)
(b)
Find μ and σ. Give each to 3 significant figures.
(3)
(c)
Find the probability that a roll is between 49 m and 50 m long. Use the unrounded values of μ and σ.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 48.5 m: \(z=-1.28\)
✓ 1
A1 51.2 m: \(z=1.64\)
✓ 1
Part (b)
M1 set up \(48.5=\mu-1.2816\sigma\) and \(51.2=\mu+1.6449\sigma\) (accept values from (a))
✓ 1
A1 σ = 0.923 m
✓ 1
A1 μ = 49.7 m
✓ 1
Part (c)
M1 use of \(P(49
✓ 1
A1 0.405
✓ 1
Exam tip: Two probabilities given for one normal distribution lead to two equations in μ and σ; write each as \(x=\mu+z\sigma\) and solve simultaneously.
249IBMI3S-4-02r34-05Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Probability and the binomial distributionPaper 1Hard8 marks
The time, X milliseconds, for a mobile network to connect a call is normally distributed with mean μ and standard deviation σ. It is known that 10 % of calls connect in less than 20 ms and 20 % of calls take longer than 45 ms.
(a)
Find the value of μ and the value of σ, each to 3 significant figures.
(6)
(b)
Find the connection time that is exceeded by 5 % of calls, to 3 significant figures. Use your answers from (a), or the unrounded values.
M1 \(P(X>45)=0.2\) so \(P(X<45)=0.8\) gives \(z=0.8416\)
✓ 1
A1 \(20=\mu-1.282\sigma\) and \(45=\mu+0.8416\sigma\)
✓ 1
M1 solve the two equations simultaneously
✓ 1
A1 \(\sigma=11.8\) (11.77…)
✓ 1
A1 \(\mu=35.1\) (35.09…)
✓ 1
Part (b)
M1 \(z=1.645\) for \(P(Z<z)=0.95\), then \(\mu+1.645\sigma\)
✓ 1
A1 54.5 ms (54.46… unrounded; 54.5 if 35.1 and 11.8 are used)
✓ 1
Exam tip: Write each condition as a probability about X, convert it to a z-value using inverse normal, then form one equation per condition.
250IBMI3S-4-02r36-09Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Hard8 marks
A space probe carries n independent dust sensors. Each sensor detects a passing dust particle with probability 0.06, independently of the others.
(a)
For n = 20, find the probability that at least one sensor detects the particle.
(2)
(b)
Find the smallest value of n for which the probability that at least one sensor detects the particle is greater than 0.99.
(4)
(c)
Find the probability that at least one sensor detects the particle when n = 75. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 1 − 0.9420
✓ 1
A1 0.710
✓ 1
Part (b)
M1 1 − 0.94n > 0.99, so 0.94n < 0.01
✓ 1
M1 attempt to solve, by logarithms or by table/GDC
✓ 1
A1 n > 74.4
✓ 1
R1 n = 75 (74 sensors gives 0.9897, below 0.99)
✓ 1
Part (c)
M1 1 − 0.9475
✓ 1
A1 0.990
✓ 1
Common error: With an inequality for n, give the smallest whole number that works and check the value just below it.
251IBMI3S-4-02r44-09Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 1Hard7 marks
The mass of honey X kg harvested from a hive in a region is normally distributed with mean μ and standard deviation σ. Ten percent of hives yield less than 20 kg, and five percent of hives yield more than 40 kg.
(a)
Write down two equations connecting μ and σ.
(3)
(b)
Find the values of μ and σ.
(2)
(c)
The top 20% of hives are classed as high-yielding. Using the unrounded values of μ and σ, find the least yield for a hive to be high-yielding.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(z=-1.282\) for \(P(X<20)=0.1\) and \(z=1.645\) for \(P(X>40)=0.05\)
M1 \(z=0.8416\) for the upper 20% and \(\mu+0.8416\sigma\)
✓ 1
A1 34.5 kg
✓ 1
Exam tip: For unknown mean and standard deviation, convert each probability into a z-value using the inverse normal function, then solve the pair of linear equations.
252IBMI-033-00-06Sampling and descriptive statisticsPaper 2Medium11 marks
Two coffee shops, P and Q, record the time, in minutes, that customers wait to be served during one morning. The five-number summaries are:
Shop P: minimum 1.2, lower quartile 2.8, median 3.9, upper quartile 5.1, maximum 9.6 Shop Q: minimum 2.0, lower quartile 3.3, median 4.2, upper quartile 4.9, maximum 6.4
An outlier is a value more than 1.5 × IQR above the upper quartile or below the lower quartile.
(a)
Find the range and the interquartile range of the waiting times at Shop P.
(2)
(b)
Show that Shop P has at least one outlier.
(3)
(c)
Determine whether Shop Q has any outliers.
(2)
(d)
Compare the waiting times at the two shops.
(2)
(e)
Shop Q served 160 customers that morning. Estimate the number of customers at Shop Q who waited between 3.3 and 4.9 minutes.
Median at P (3.9) is lower than at Q (4.2), so a typical customer waits less at P R1
✓ 1
IQR/range at P is larger, so waiting times at P are less consistent R1
✓ 1
Part (e)
50% of customers lie between the quartiles M1
✓ 1
80 customers A1
✓ 1
Exam tip: When comparing distributions, make one comment about a measure of centre and one about a measure of spread, and link each to the context (waiting times, not just numbers).
253IBMI-033-00-09Sampling and descriptive statisticsPaper 2Medium11 marks
An orchard weighs a random sample of 200 apples. The masses, m grams, are recorded in the table.
100 < m ≤ 120 : 14 120 < m ≤ 140 : 38 140 < m ≤ 160 : 70 160 < m ≤ 180 : 52 180 < m ≤ 200 : 26
(a)
Write down the cumulative frequencies for m ≤ 120, 140, 160, 180 and 200.
(2)
(b)
Use linear interpolation within the appropriate class to estimate the median mass.
(3)
(c)
Find an estimate of the mean mass.
(2)
(d)
Find an estimate of the standard deviation of the masses.
(2)
(e)
Assuming masses are evenly spread within each class, estimate the percentage of apples with mass greater than 170 g.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
At least three correct A1
✓ 1
All correct: 14, 52, 122, 174, 200 A1
✓ 1
Part (b)
Median is the 100th value, in class 140 < m ≤ 160 M1
Midpoints and frequencies in GDC, or √(4 828 000 ÷ 200 − 153.82) M1
✓ 1
22.0 g A1
✓ 1
Part (e)
½ × 52 + 26 = 52 apples M1
✓ 1
26.0% A1
✓ 1
Common error: In the interpolation, subtract the cumulative frequency at the start of the median class (52), not the frequency of the class before it (38).
254IBMI-033-00-11Sampling and descriptive statisticsPaper 2Medium10 marks
A traffic officer records the number of people, x, in each of 100 cars passing a checkpoint.
x = 1 : 48 cars x = 2 : 31 cars x = 3 : 12 cars x = 4 : 6 cars x = 5 : 3 cars
(a)
State whether x is discrete or continuous.
(1)
(b)
Find the mean number of people per car.
(2)
(c)
Find the standard deviation of the number of people per car.
(2)
(d)
Find the median and the interquartile range.
(3)
(e)
A local newspaper claims that 'on average, cars carry more than 2 people'. Comment on this claim.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Discrete A1
✓ 1
Part (b)
Σfx ÷ 100 = 185 ÷ 100 M1
✓ 1
1.85 A1
✓ 1
Part (c)
Values and frequencies entered in GDC, or √(451 ÷ 100 − 1.852) M1
✓ 1
1.04 (1.0428…) A1
✓ 1
Part (d)
Median = 2 (50th and 51st values are both 2) A1
✓ 1
Q1 = 1 and Q3 = 2 A1
✓ 1
IQR = 1 A1
✓ 1
Part (e)
Correct comparison of at least one average with 2, e.g. mean 1.85 < 2, or median = 2 (not more than 2), or mode = 1 < 2 R1
✓ 1
Conclusion: the claim is not supported, since no measure of average exceeds 2 R1
✓ 1
Common error: With a frequency table, the median is the value of the 50th/51st car (found from cumulative frequencies), not the middle value of the list 1, 2, 3, 4, 5.
255IBMI-033-08-r2-03Normal distribution and chi-squared testsPaper 2Medium11 marks
A mill fills bags labelled as containing 1 kg of flour. The mass, M grams, of flour in a bag is normally distributed with mean 1010 g and standard deviation 6 g. A bag is classed as underweight if it contains less than 1000 g.
(a)
Find the probability that a randomly chosen bag is underweight.
(2)
(b)
Find P(1005 < M < 1020).
(2)
(c)
A bag is chosen at random and found to contain more than 1005 g. Find the probability that it contains more than 1020 g.
(3)
(d)
The bags are packed in boxes of 12. Assuming the masses of bags are independent, find the probability that a box contains at least two underweight bags.
Correct values substituted, e.g. \(\frac{0.04779}{0.79767}\)
✓ 1
0.0599
✓ 1
Part (d)
Recognises binomial distribution, X ~ B(12, 0.04779…)
✓ 1
Uses P(X ≥ 2) = 1 − P(X ≤ 1)
✓ 1
P(X ≤ 1) = 0.890 (0.89028…)
✓ 1
0.110
✓ 1
Exam tip: Store the probability from part (a) in your calculator and use it in part (d). Using the rounded value 0.0478 can be enough to move the final answer off the correct 3 s.f. value.
256IBMI-033-05-r1-03Sampling and descriptive statisticsPaper 2Medium12 marks
A consumer magazine tests the battery life, in hours, of 15 smartphones of the same model under identical conditions. The results, in order, are: 8.2, 9.1, 9.4, 9.8, 10.0, 10.3, 10.5, 10.7, 11.0, 11.2, 11.4, 11.9, 12.3, 12.8, 17.5
(a)
Write down the median, the lower quartile and the upper quartile of the battery lives.
(3)
(b)
An outlier is a value more than 1.5 × IQR beyond the nearer quartile. Show that the phone with battery life 17.5 hours is an outlier.
(3)
(c)
Find the mean and the standard deviation of all 15 battery lives.
(2)
(d)
The magazine discovers that the 17.5-hour phone had its screen switched off by mistake during the test, so its result is removed. Find the mean battery life of the remaining 14 phones.
(2)
(e)
By considering how the mean and median change when this value is removed, explain which measure of central tendency was more appropriate for the original data.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1: Median = 10.7 h
✓ 1
A1: Q1 = 9.8 h
✓ 1
A1: Q3 = 11.9 h
✓ 1
Part (b)
A1: IQR = 11.9 − 9.8 = 2.1
✓ 1
M1: Upper fence = 11.9 + 1.5 × 2.1 = 15.05
✓ 1
R1: 17.5 > 15.05, so 17.5 h is an outlier
✓ 1
Part (c)
A1: Mean = 11.1 h (11.07…)
✓ 1
A1: Standard deviation = 2.09 h
✓ 1
Part (d)
M1: \(\frac{166.1-17.5}{14}=\frac{148.6}{14}\)
✓ 1
A1: 10.6 h (10.61…)
✓ 1
Part (e)
A1: Median of remaining 14 values is \(\frac{10.5+10.7}{2}=10.6\) h, changing by only 0.1 h, while the mean falls by about 0.46 h
✓ 1
R1: The median is more appropriate because it is not strongly affected by the extreme value, whereas the mean was pulled upwards by the outlier
✓ 1
Exam tip: Show the fence calculation explicitly when asked to 'show' an outlier; simply stating that 17.5 looks large does not earn the reasoning mark.
257IBMI-033-05-r1-05Sampling and descriptive statisticsPaper 2Medium12 marks
An orchard weighs a sample of 200 apples. No apple weighed 80 g or less. The cumulative frequencies of mass, m grams, are: m ≤ 100: 12 m ≤ 120: 46 m ≤ 140: 108 m ≤ 160: 164 m ≤ 180: 190 m ≤ 200: 200 Assume that masses are evenly spread within each class.
(a)
Find the number of apples with 140 < m ≤ 160.
(1)
(b)
Estimate the median mass.
(2)
(c)
Estimate the interquartile range of the masses.
(3)
(d)
Apples heavier than 170 g are sold as premium. Estimate the number of premium apples in the sample.
(2)
(e)
Using mid-interval values, find an estimate of the mean mass of the apples.
(2)
(f)
The 200 apples were all taken from the top layer of crates on the packing line. Explain why this sample may not be representative of the orchard's harvest.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1: 164 − 108 = 56
✓ 1
Part (b)
M1: 100th apple lies in 120 < m ≤ 140; \(120+\frac{100-46}{62}\times 20\)
✓ 1
A1: 137 g (137.4…)
✓ 1
Part (c)
M1: Q1 from 50th apple: \(120+\frac{4}{62}\times 20=121.3\) g
✓ 1
A1: Q3 from 150th apple: \(140+\frac{42}{56}\times 20=155\) g
✓ 1
A1: IQR = 33.7 g
✓ 1
Part (d)
M1: Cumulative frequency at 170 g ≈ 164 + 0.5 × 26 = 177
R1: The sample is not random / every apple did not have an equal chance of selection
✓ 1
R1: Valid context reason, e.g. packers may place the largest or best-looking apples on top, so the estimates of mass are likely to be biased upwards
✓ 1
Exam tip: Convert cumulative frequencies to class frequencies before estimating the mean; using the cumulative values as frequencies gives a meaningless answer.
258IBMI-033-03-r2-03Normal distribution and chi-squared testsPaper 2Medium11 marks
A city council surveys 300 commuters to investigate whether the main way they travel to work is associated with their age group. The results are shown below.
Under 30: Car 42, Bus 50, Bicycle 38 (total 130) 30 and over: Car 88, Bus 52, Bicycle 30 (total 170) Column totals: Car 130, Bus 102, Bicycle 68 (grand total 300)
A χ2 test for independence is carried out at the 5% significance level.
(a)
State the null and alternative hypotheses.
(2)
(b)
Show that the expected frequency of commuters under 30 who travel by bicycle is 29.5, correct to 3 significant figures.
(2)
(c)
Write down the number of degrees of freedom for this test.
(1)
(d)
Find the χ2 statistic and the p-value.
(2)
(e)
State the conclusion of the test, giving a reason.
(2)
(f)
By comparing observed and expected frequencies, describe how the travel choices of commuters under 30 differ from what would be expected under H0.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
H0: the main way of travelling to work is independent of age group (A1)
✓ 1
H1: the main way of travelling to work is not independent of age group (A1)
✓ 1
Part (b)
(130 × 68)/300 (M1)
✓ 1
= 29.466… = 29.5 (A1)
✓ 1
Part (c)
(2 − 1)(3 − 1) = 2 (A1)
✓ 1
Part (d)
χ2 = 12.1 (12.139…) (A1)
✓ 1
p = 0.00231 (A1)
✓ 1
Part (e)
0.00231 < 0.05 so reject H0 (R1)
✓ 1
There is sufficient evidence that the way of travelling to work is associated with (not independent of) age group (A1)
✓ 1
Do not award A1 without R1.
—
Part (f)
Under-30s use bus (50 vs 44.2) and bicycle (38 vs 29.5) more than expected (A1)
✓ 1
Under-30s use the car less than expected (42 vs 56.3) (A1)
✓ 1
Exam tip: Write the conclusion in context: say what is associated with what, not just 'reject H₀'.
259IBMI-033-02-r1-05Probability and the binomial distributionPaper 2Medium11 marks
A company records how its 240 employees travel to work and whether they arrived late on a particular Monday.
Bus: 18 late, 102 on time Car: 12 late, 68 on time Bicycle: 4 late, 36 on time
(a)
Find the probability that a randomly chosen employee was late.
(1)
(b)
Given that an employee was late, find the probability that they travelled by bus.
(2)
(c)
Determine whether the events "travels by bus" and "was late" are independent. Justify your answer.
(3)
(d)
Two different employees are chosen at random. Find the probability that both were late.
(2)
(e)
On each of 5 days, one employee is chosen at random from all 240, independently each day, and their Monday record is checked. Find the probability that exactly 2 of the 5 chosen employees were late.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1: \(\frac{34}{240}=0.142\)
✓ 1
Part (b)
M1: uses the 'late' total as denominator, \(\frac{18}{34}\)
✓ 1
A1: 0.529
✓ 1
Part (c)
M1: compares \(P(B\cap L)\) with \(P(B)\times P(L)\)
✓ 1
A1: \(P(B\cap L)=\frac{18}{240}=0.075\) and \(P(B)P(L)=0.5\times0.14167=0.0708\)
✓ 1
R1: these are not equal, so the events are not independent
✓ 1
Part (d)
M1: without replacement, \(\frac{34}{240}\times\frac{33}{239}\)
Exam tip: A statement that events are 'not independent' earns the reasoning mark only when it is backed by numerical values of both \(P(A\cap B)\) and \(P(A)P(B)\), compared explicitly.
260IBMI-033-07-r1-04Probability and the binomial distributionPaper 2Medium12 marks
A football player scores from a penalty kick with probability 0.82. The outcomes of her penalty kicks are assumed to be independent. During a training session she takes 10 penalty kicks. Let S be the number she scores.
(a)
Find the probability that she scores exactly 8 penalties.
(2)
(b)
Find the probability that she scores more than 8 penalties.
(2)
(c)
In a longer session she takes n penalties. Find the least value of n for which the probability that she misses at least one penalty is greater than 0.99.
(3)
(d)
In the 10-penalty session, her coach notes that she scored at least 8 penalties. Find the probability that she scored all 10.
(3)
(e)
Suggest, with a reason, why the binomial model may not be appropriate for her penalties in a real match.
Probability of scoring may not be constant / outcomes may not be independent R1
✓ 1
Valid contextual reason, e.g. pressure or confidence after a miss affects the next kick, or different goalkeepers R1
✓ 1
Common error: In the conditional part, the numerator is P(S = 10), not P(S = 10) × P(S ≥ 8): scoring all 10 already lies inside the event “at least 8”.
261IBMI-033-05-r2-03Sampling and descriptive statisticsPaper 2Medium12 marks
A phone company records the monthly mobile data usage, in gigabytes (GB), of 15 randomly chosen customers. The values, in ascending order, are:
Write down the median, the lower quartile and the upper quartile.
(3)
(b)
Find the interquartile range.
(1)
(c)
Show that 18.2 GB is an outlier.
(3)
(d)
Find the mean data usage of the 15 customers.
(1)
(e)
The customer who used 18.2 GB is removed from the data. Find the mean of the remaining 14 customers.
(2)
(f)
The company wants to report a typical monthly usage for the original 15 customers. State, with a reason, whether the mean or the median is more appropriate.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Median = 6.0 GB A1
✓ 1
Q1 = 4.0 GB A1
✓ 1
Q3 = 8.1 GB A1
✓ 1
Part (b)
8.1 − 4.0 = 4.1 GB A1
✓ 1
Part (c)
Upper fence = Q3 + 1.5 × IQR M1
✓ 1
= 8.1 + 6.15 = 14.25 A1
✓ 1
18.2 > 14.25, so 18.2 is an outlier R1
✓ 1
Part (d)
\(\frac{99.8}{15}\) = 6.65 GB A1
✓ 1
Part (e)
\(\frac{99.8-18.2}{14}\) or \(\frac{81.6}{14}\) M1
✓ 1
5.83 GB A1
✓ 1
Part (f)
The median A1
✓ 1
Because it is not affected by the extreme value of 18.2 GB, whereas the mean is pulled upwards by it R1
✓ 1
Exam tip: To show a value is an outlier, write the fence calculation in full and then compare the value with the fence in a clear sentence. The final comparison earns its own mark.
262IBMI-033-06-r1-04Correlation and regressionPaper 2Medium12 marks
A car dealer records the age, x years, and the resale value, y thousand dollars, of eight cars of the same model.
Find the value of Pearson's product-moment correlation coefficient, r.
(2)
(b)
Describe the correlation between age and resale value.
(1)
(c)
Write down the equation of the regression line of y on x.
(2)
(d)
Interpret, in context, the gradient and the y-intercept of your regression line.
(2)
(e)
Estimate the resale value of a car of this model that is 6.5 years old.
(2)
(f)
A car of this model is valued at $20 000. By finding a suitable regression line, estimate its age.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Correct use of GDC
✓ 1
r = −0.998
✓ 1
Part (b)
Strong negative (linear) correlation
✓ 1
Part (c)
Gradient −2.17
✓ 1
y = −2.17x + 26.3
✓ 1
Part (d)
Gradient: the value falls by about $2170 for each extra year of age
✓ 1
Intercept: the estimated value of a new car (age 0) is about $26 300
✓ 1
Part (e)
Substitutes x = 6.5: y = −2.1702… × 6.5 + 26.3285…
✓ 1
y = 12.2, i.e. about $12 200
✓ 1
Part (f)
Recognises that the regression line of x on y is needed
✓ 1
x = −0.459y + 12.1
✓ 1
Substitutes y = 20: x = 2.92 years
✓ 1
Common error: To estimate x from a given y, use the regression line of x on y. Rearranging the y on x line gives a different estimate unless |r| = 1.
263IBMI-033-04-r2-06T-tests, confidence intervals and Markov chainsPaper 2Medium12 marks
A food delivery company compares delivery times, in minutes, from two of its kitchens. A random sample of 10 deliveries from Kitchen P has mean 52.3 and standard deviation \(s_{n-1}\) = 6.1. A random sample of 12 deliveries from Kitchen Q has mean 47.8 and \(s_{n-1}\) = 5.4. A pooled two-sample t-test is carried out at the 5% significance level to determine whether there is a difference in mean delivery times.
(a)
State the null and alternative hypotheses.
(2)
(b)
Find the value of the test statistic and the p-value.
(3)
(c)
State the conclusion of the test in context.
(2)
(d)
State one assumption about the populations needed for a pooled test and comment on whether the sample data support it.
(2)
(e)
Assuming the same sample sizes and standard deviations, find the smallest difference in sample means that would have led to rejecting H0 at the 5% level.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
H0: \(\mu_P=\mu_Q\) (population mean delivery times are equal)
✓ 1
H1: \(\mu_P\neq\mu_Q\)
✓ 1
Part (b)
Pooled two-sample t-test with summary statistics, two-tailed, 20 degrees of freedom
✓ 1
t = 1.84
✓ 1
p = 0.0813
✓ 1
Part (c)
0.0813 > 0.05
✓ 1
Do not reject H0: insufficient evidence of a difference in mean delivery times between the two kitchens
✓ 1
Part (d)
Population variances are equal (both normally distributed)
✓ 1
Sample standard deviations 6.1 and 5.4 are similar, so the assumption appears reasonable
✓ 1
Part (e)
Critical value \(t_{20}\) = 2.086
✓ 1
Difference = 2.086 × \(s_p\sqrt{\frac{1}{10}+\frac{1}{12}}\) with \(s_p\) = 5.73
✓ 1
5.11 minutes
✓ 1
Common error: The question asks about 'a difference', so the test is two-tailed: compare the two-tailed p-value with 5%, not 2.5% or a one-tailed value.
264IBMI-033-03-r1-03Normal distribution and chi-squared testsPaper 2Medium12 marks
The time, T minutes, for a commuter's train journey is normally distributed with mean μ and standard deviation σ. Records show that 15% of journeys take more than 48 minutes and 5% of journeys take less than 30 minutes.
(a)
Find the value of μ and the value of σ.
(4)
(b)
Find the probability that a journey takes between 35 and 45 minutes.
(2)
(c)
The commuter makes 10 journeys, which may be assumed independent. Find the probability that at least 3 of them take more than 48 minutes.
(3)
(d)
On one day the journey has already lasted 40 minutes. Find the probability that it will last more than 48 minutes in total.
Exam tip: Keep μ and σ to full GDC accuracy when you use them in later parts. Rounding to 41.0 and 6.71 too early can change the third significant figure of your answers.
265IBMI-033-08-r1-03Normal distribution and chi-squared testsPaper 2Medium11 marks
A machine produces bolts whose lengths, L mm, are normally distributed. It is known that 8% of bolts are shorter than 49.5 mm and 15% of bolts are longer than 51.2 mm. A bolt is accepted if its length is between 49.5 mm and 51.2 mm.
(a)
Find the mean and the standard deviation of L.
(5)
(b)
Given that a bolt is accepted, find the probability that its length is greater than 50.0 mm.
(3)
(c)
Bolts are packed in boxes of 20. Find the probability that at least 18 bolts in a randomly chosen box are accepted.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Use of inverse normal with P(Z < z) = 0.08 and P(Z < z) = 0.85
✓ 1
z = −1.405 and z = 1.036
✓ 1
Forming μ − 1.405σ = 49.5 and μ + 1.036σ = 51.2 and solving simultaneously
Common error: When finding σ, remember that 15% above 51.2 mm means 85% lies below it, so the z-value is positive (1.036). Using invNorm(0.15) gives a negative z-value and leads to a negative or meaningless σ.
266IBMI-033-01-r1-04Correlation and regressionPaper 2Medium12 marks
A teacher records the number of hours, x, that eight students spent revising and their score, y, on a test marked out of 100.
Find Pearson's product-moment correlation coefficient, r, for these data.
(2)
(b)
Describe the correlation between revision time and test score.
(2)
(c)
Write down the equation of the regression line of y on x, in the form y = ax + b.
(2)
(d)
Interpret, in context, the value of b.
(1)
(e)
A student plans to revise for 20 hours. Use the regression line to calculate her predicted score, and give two reasons why this prediction should not be relied upon.
(3)
(f)
Use the regression line to estimate the score of a student who revises for 9 hours.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: correct use of GDC with both lists entered, or a correct substitution into the formula for r
✓ 1
A1: r = 0.998 (0.99777…)
✓ 1
Part (b)
Strong
✓ 1
positive
✓ 1
Part (c)
a = 3.70 (3.69898…)
✓ 1
b = 34.1 (34.1326…)
✓ 1
Part (d)
A student who does no revision would be expected to score about 34 marks
✓ 1
Part (e)
M1: y = 3.69898 × 20 + 34.1326 ≈ 108
✓ 1
R1: 20 hours is outside the range of the data (2 to 13 hours), so this is extrapolation
✓ 1
R1: the prediction exceeds the maximum possible score of 100, so it is impossible
✓ 1
Part (f)
M1: substitutes x = 9 into their line
✓ 1
A1: 67.4
✓ 1
Exam tip: Use the unrounded coefficients stored in your GDC when making predictions, then round only the final answer to 3 significant figures.
267IBMI-033-08-r1-06Normal distribution and chi-squared testsPaper 2Medium11 marks
A psychologist claims that adult reaction times, T ms, follow the distribution N(250, 302). To test the claim, the reaction times of 200 adults are recorded and grouped:
t < 220: 24 220 ≤ t < 250: 61 250 ≤ t < 280: 80 t ≥ 280: 35
A \(\chi^2\) goodness of fit test is carried out at the 5% significance level.
(a)
Find the expected frequency for each group under the claimed distribution.
(3)
(b)
State the null hypothesis.
(1)
(c)
Write down the degrees of freedom, and find the \(\chi^2\) statistic and the p-value.
(3)
(d)
State the conclusion of the test, giving a reason.
(2)
(e)
The psychologist wants to split the last group into 280 ≤ t < 310 and t ≥ 310. Show that this would not be appropriate for a \(\chi^2\) test.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
P(T < 220) = P(T ≥ 280) = 0.1587 and P(220 ≤ T < 250) = P(250 ≤ T < 280) = 0.3413
✓ 1
Expected frequency for t < 220 and for t ≥ 280: 200 × 0.1587 = 31.7
✓ 1
Expected frequency for each of the two middle groups: 200 × 0.3413 = 68.3
✓ 1
Part (b)
H0: reaction times follow the distribution N(250, 302)
✓ 1
Part (c)
Degrees of freedom = 4 − 1 = 3
✓ 1
$\chi^2_{calc}$ = 5.01
✓ 1
p = 0.171
✓ 1
Part (d)
0.171 > 0.05 (or 5.01 < critical value 7.815), so do not reject H0
✓ 1
There is insufficient evidence to reject the claim that reaction times follow N(250, 302)
✓ 1
Part (e)
Expected frequency for t ≥ 310 = 200 × P(T ≥ 310) = 200 × 0.02275 = 4.55
✓ 1
4.55 < 5, so this expected frequency is too small and the groups would need to be combined
✓ 1
Exam tip: The mean and standard deviation are given here, not estimated from the data, so the degrees of freedom are simply (number of groups − 1) = 3. Check that every expected frequency is at least 5 before carrying out the test.
268IBMI-033-06-r2-04Correlation and regressionPaper 2Medium10 marks
A dealer records the age, x years, and the selling price, y thousand dollars, of eight used cars of the same model.
A1: residual = 16.9 − 16.617 = 0.283 thousand, i.e. $283 (accept $250 to $300 from rounded coefficients)
✓ 1
Common error: A residual is observed value minus predicted value, not the other way round. A positive residual means the car sold for more than the model predicts. Use unrounded GDC values in later calculations to avoid rounding errors.
269IBMI-033-06-r2-06Correlation and regressionPaper 2Medium10 marks
A researcher surveys eight teenagers about their average daily phone use, x hours, and their average nightly sleep, y hours.
Find Pearson's product-moment correlation coefficient, r, for all eight teenagers.
(2)
(b)
The researcher discovers that the sleep value for the teenager with 2.5 hours of phone use was recorded incorrectly, and removes this data pair. Find the value of r for the remaining seven teenagers.
(2)
(c)
Comment on the effect of the incorrect data pair on the correlation coefficient.
(1)
(d)
For the seven remaining teenagers, find the equation of the regression line of y on x.
(2)
(e)
Use your equation to estimate the average nightly sleep of a teenager who uses their phone for 4 hours per day.
(2)
(f)
A newspaper claims that this study shows that phone use causes teenagers to sleep less. Comment on this claim.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: correct use of GDC with all eight data pairs
✓ 1
A1: r = −0.256
✓ 1
Part (b)
M1: removes (2.5, 4.0) and recalculates
✓ 1
A1: r = −0.974
✓ 1
Part (c)
R1: the single incorrect value (an outlier) made the correlation much weaker. With it, r ≈ −0.26 (weak negative). Without it, r ≈ −0.97 (strong negative). The outlier hid the strong linear relationship
✓ 1
Part (d)
A1: gradient −0.472
✓ 1
A1: y = −0.472x + 9.21
✓ 1
Part (e)
M1: substitutes x = 4 into their equation
✓ 1
A1: 7.32 hours (accept 7.3)
✓ 1
Part (f)
R1: the claim is not justified. Award for any one valid reason: correlation does not imply causation; other factors (e.g. homework load or late evening activities) could affect both variables; the sample of seven is very small
✓ 1
Exam tip: Before calculating r, check the data for values that look out of place. One outlier can change r dramatically, so mention it when you interpret the result.
270IBMI-033-02-r2-04Probability and the binomial distributionPaper 2Medium11 marks
At a customer support centre, the probability that a call is resolved on first contact is 0.7. An operator handles 20 calls in a shift. Let \(X\) be the number of calls in a shift that are resolved on first contact.
(a)
State two assumptions needed to model \(X\) with a binomial distribution, in the context of the question.
(2)
(b)
Find the probability that exactly 15 calls are resolved on first contact.
(2)
(c)
Find the probability that at least 12 calls are resolved on first contact.
(2)
(d)
The operator earns a bonus for a shift if at least 16 calls are resolved on first contact. Find the probability that the operator earns a bonus in a given shift.
(2)
(e)
The operator works 5 shifts in a week. Shifts are independent. Find the probability that the operator earns a bonus in at least 2 of the 5 shifts.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1: the outcome of each call is independent of the other calls
✓ 1
A1: the probability of resolving a call on first contact is constant at 0.7 for every call
✓ 1
Part (b)
M1: \(X \sim B(20, 0.7)\), attempts \(P(X=15)\)
✓ 1
A1: \(0.179\) (0.17886…)
✓ 1
Part (c)
M1: \(1 - P(X \le 11)\)
✓ 1
A1: \(0.887\) (0.8867…)
✓ 1
Part (d)
M1: \(P(X \ge 16) = 1 - P(X \le 15)\)
✓ 1
A1: \(0.238\) (0.2375…)
✓ 1
Part (e)
M1: recognises \(W \sim B(5, p)\) with \(p\) from the previous part
✓ 1
M1: \(1 - P(W \le 1)\) or \(1 - P(W=0) - P(W=1)\)
✓ 1
A1: \(0.341\) using unrounded \(p = 0.2375…\) (accept \(0.342\) from \(p = 0.238\))
✓ 1
Exam tip: Carry the unrounded probability 0.2375… into the final part. Rounding it to 0.238 or 0.24 before the last calculation can change the third significant figure of the answer.
271IBMI-033-01-r2-03Correlation and regressionPaper 2Medium10 marks
A household records the mean daily outdoor temperature, x °C, and the electricity used for heating, y kWh, on seven winter days.
Find the value of the Pearson product-moment correlation coefficient, r.
(2)
(b)
Describe the correlation between x and y in context.
(2)
(c)
Write down the equation of the regression line of y on x, giving the values in the form y = ax + b.
(2)
(d)
Interpret the gradient of the regression line in context.
(1)
(e)
Use your regression line to estimate the electricity used on a day with mean temperature 8 °C.
(2)
(f)
Explain why the regression line should not be used to estimate the electricity used on a day with mean temperature 30 °C.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Negative value, with the data correctly entered (r = −0.99864...) (A1)
✓ 1
r = −0.999 (3 s.f.) (A1)
✓ 1
Part (b)
Very strong negative linear correlation (A1)
✓ 1
As the outdoor temperature increases, the electricity used for heating decreases (R1)
✓ 1
Part (c)
a = −2.06 (−2.0584...) (A1)
✓ 1
b = 52.4 (52.433...) (A1)
✓ 1
Part (d)
For each 1 °C rise in mean daily temperature, heating electricity use is predicted to fall by about 2.06 kWh (R1)
✓ 1
Part (e)
Substitutes x = 8 into the regression line, e.g. −2.06(8) + 52.4 (M1)
✓ 1
y = 36.0 kWh using full-accuracy coefficients (35.966...); accept 35.9 kWh from the rounded line (A1)
✓ 1
Part (f)
30 °C is outside the data range (2 °C to 18 °C), so this is extrapolation and unreliable; the line would predict negative electricity use (about −9.3 kWh), which is impossible (R1)
✓ 1
Exam tip: Keep the full-accuracy regression coefficients in your GDC when making a prediction. Using rounded values can change the third significant figure.
272IBMI3S-4-07-07Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 2Medium12 marks
A council surveys 250 adults about their main way of travelling to work. The results are shown below.
Cycle
Bus
Car
Under 30
44
30
26
30 or over
56
70
24
A χ2 test for independence is carried out at the 5% significance level.
(a)
Find the probability that a randomly chosen respondent travels by bus.
(1)
(b)
Given that a respondent is aged 30 or over, find the probability that they cycle.
(2)
(c)
State the null hypothesis.
(1)
(d)
Show that the expected frequency for respondents under 30 who travel by car is 20.
(2)
(e)
Write down the number of degrees of freedom.
(1)
(f)
Calculate the χ2 statistic.
(2)
(g)
Find the p-value and state the conclusion of the test.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(\frac{100}{250}=0.4\)
✓ 1
Part (b)
M1 conditional probability with denominator 150
✓ 1
A1 \(\frac{56}{150}=0.373\)
✓ 1
Part (c)
A1 H0: age group and method of travel are independent
✓ 1
Part (d)
M1 \(\frac{100\times50}{250}\)
✓ 1
A1 20 (shown)
✓ 1
Part (e)
A1 \((2-1)(3-1)=2\)
✓ 1
Part (f)
M1 attempt to use \(\sum\frac{(O-E)^2}{E}\) with all six expected frequencies (40, 40, 20, 60, 60, 30)
✓ 1
A1 χ2 = 7.83
✓ 1
Part (g)
A1 p-value = 0.0199
✓ 1
R1 0.0199 < 0.05
✓ 1
A1 reject H0: evidence that travel method depends on age group
✓ 1
Common error: Do not conclude that one variable causes the other; the test only shows whether the variables are associated.
273IBMI3S-4-07-08Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 2Medium13 marks
In a region, 2% of people have a particular condition. A screening test gives a positive result for 95% of people who have the condition and for 6% of people who do not have it.
(a)
Find the probability that a randomly chosen person tests positive.
(3)
(b)
Find the probability that a person who tests positive has the condition.
(2)
(c)
Explain, in context, why this probability is much lower than 95%.
(1)
(d)
Eight people who tested positive are chosen. State the distribution of the number, X, who have the condition, and write down one assumption you made.
(2)
(e)
Find \(P(X\ge2)\).
(2)
(f)
Given that at least one of the eight has the condition, find the probability that at least two do.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(P(D\cap +)=0.02\times0.95\)
✓ 1
M1 \(P(D'\cap +)=0.98\times0.06\)
✓ 1
A1 \(0.019+0.0588=0.0778\)
✓ 1
Part (b)
M1 \(\frac{0.019}{0.0778}\)
✓ 1
A1 0.244
✓ 1
Part (c)
R1 most people do not have the condition, so the false positives from the large healthy group outnumber the true positives
✓ 1
Part (d)
A1 \(X\sim B(8,0.244)\)
✓ 1
A1 assumption: the people are independent (e.g. chosen at random from all positive results)
✓ 1
Part (e)
M1 \(1-P(X\le1)\)
✓ 1
A1 0.618
✓ 1
Part (f)
M1 \(P(X\ge1)=1-(1-0.244)^8\)
✓ 1
M1 \(\frac{P(X\ge2)}{P(X\ge1)}\)
✓ 1
A1 0.692
✓ 1
Common error: Keep the unrounded value of \(P(D\mid +)\) for later parts; using 0.24 changes the third significant figure.
274IBMI3S-4-07b-06Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 2Medium9 marks
A sports medicine clinic records the type of injury for 240 footballers: 110 who play on natural grass and 130 who play on artificial turf. The results are shown below.
Natural grass: 34 muscle, 52 ligament, 24 bone Artificial turf: 26 muscle, 63 ligament, 41 bone
A χ2 test for independence is carried out at the 5% significance level.
(a)
Write down the null and alternative hypotheses.
(2)
(b)
Show that the expected frequency for bone injuries on artificial turf is 35.2, correct to 3 significant figures.
(2)
(c)
State the number of degrees of freedom.
(1)
(d)
Find the χ2 statistic and the p-value.
(2)
(e)
State, with a reason, the conclusion of the test.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: injury type is independent of playing surface
✓ 1
A1 H1: injury type and playing surface are not independent
✓ 1
Part (b)
M1 (130 × 65) / 240
✓ 1
A1 35.208… = 35.2
✓ 1
Part (c)
A1 (2 − 1)(3 − 1) = 2
✓ 1
Part (d)
A1 χ2 = 4.93
✓ 1
A1 p-value = 0.0849
✓ 1
Part (e)
M1 compares p = 0.0849 with 0.05
✓ 1
R1 0.0849 > 0.05, so do not reject H0: no significant evidence that injury type depends on surface
✓ 1
Exam tip: Always compare the p-value with the stated significance level and give the conclusion in the words of the context.
275IBMI3S-4-07b-07Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 2Medium12 marks
An astronomer tests how the signal-to-noise ratio, y, of a star image depends on the exposure time, x minutes. The results are:
Exposure time x: 2, 4, 6, 8, 10, 12, 15 Signal-to-noise ratio y: 11.8, 14.9, 17.6, 19.1, 21.7, 22.4, 25.3
(a)
Find the product-moment correlation coefficient r, and the equation of the regression line of y on x.
(3)
(b)
Interpret the gradient of the regression line.
(2)
(c)
Estimate the signal-to-noise ratio for an exposure of 9 minutes.
(2)
(d)
The astronomer wants a signal-to-noise ratio of 60. Explain why the regression line should not be used to find the exposure time needed.
(2)
(e)
Calculate the residual for the observation at x = 10.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 r = 0.989
✓ 1
A1 gradient 1.005
✓ 1
A1 y = 1.005x + 10.785
✓ 1
Part (b)
A1 an extra minute of exposure is associated with
✓ 1
R1 an increase of about 1.01 in the signal-to-noise ratio
✓ 1
Part (c)
M1 substitute x = 9
✓ 1
A1 19.8
✓ 1
Part (d)
R1 a ratio of 60 is far outside the range of the data (extrapolation)
✓ 1
R1 the relationship may not remain linear for long exposures
✓ 1
Part (e)
M1 predicted value 1.0054(10) + 10.785
✓ 1
A1 20.839
✓ 1
A1 residual = 21.7 − 20.839 = 0.861
✓ 1
Exam tip: A residual is observed value minus predicted value. A positive residual means the point lies above the line.
276IBMI3S-4-01r7-11Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Medium11 marks
In a theatre, the audience stands to applaud at the end of some shows. In 200 randomly chosen groups of four consecutive shows, the number of shows, X, that received a standing ovation was recorded.
X
0
1
2
3
4
Frequency
9
52
71
54
14
The manager proposes the model X ~ B(4, 0.5). A χ2 goodness-of-fit test at the 5% significance level is used to test whether this model is suitable.
(a)
Show that the expected frequency for X = 2 under the model is 75.
(2)
(b)
Write down the expected frequencies for X = 0, 1, 3 and 4.
(1)
(c)
State the null hypothesis and the number of degrees of freedom.
(2)
(d)
Calculate the χ2 statistic, correct to 3 significant figures.
(3)
(e)
The critical value at the 5% level for this test is 9.488. State the conclusion, with a reason.
(2)
(f)
State one condition on the expected frequencies that must be met for the χ2 test to be valid.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(X = 2) = C(4,2) × 0.54 = 0.375
✓ 1
A1 200 × 0.375 = 75 (shown)
✓ 1
Part (b)
A1 12.5, 50, 50, 12.5
✓ 1
Part (c)
A1 H0: the binomial model B(4, 0.5) fits the data
✓ 1
A1 degrees of freedom = 4
✓ 1
Part (d)
M1 Σ(O − E)2/E with the expected frequencies
✓ 1
A1 values 0.98, 0.08, 0.2133, 0.32, 0.18 (any three correct)
✓ 1
A1 χ2 = 1.77
✓ 1
Part (e)
M1 1.77 < 9.488
✓ 1
R1 do not reject H0; the model B(4, 0.5) is a suitable fit
✓ 1
Part (f)
A1 every expected frequency is at least 5 (or no more than 20% below 5)
✓ 1
Exam tip: The expected frequencies must sum to the total of the observed frequencies, which gives a quick check.
277IBMI3S-4-01r7-13Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Medium12 marks
Ash thickness was measured at 100 sites around a volcano after an eruption.
Thickness t (mm)
0 ≤ t < 10
10 ≤ t < 20
20 ≤ t < 30
30 ≤ t < 40
40 ≤ t < 50
50 ≤ t < 60
Number of sites
6
18
31
25
14
6
(a)
Use the midpoints of the classes to estimate the mean thickness.
(2)
(b)
Explain why this value is only an estimate.
(1)
(c)
Estimate the standard deviation of the thickness, correct to 3 significant figures.
(2)
(d)
Write down the modal class.
(1)
(e)
Estimate the percentage of sites with thickness of at least 20 mm and less than 40 mm.
(1)
(f)
The thickness is now modelled by a normal distribution with mean 29.1 mm and standard deviation 12.7 mm. Using this model, estimate the number of the 100 sites with thickness greater than 50 mm.
(2)
(g)
Compare your answer to the previous part with the data in the table.
(1)
(h)
Assuming the sites are evenly spread within each class, estimate the number of sites with thickness less than 25 mm.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sum of (midpoint × frequency)/100 = 2910/100
✓ 1
A1 29.1 mm
✓ 1
Part (b)
R1 the exact thickness at each site is not known; midpoints are used for each class
✓ 1
Part (c)
M1 use of the midpoints and frequencies in the calculator’s statistics function (or the standard deviation formula)
✓ 1
A1 12.7 mm
✓ 1
Part (d)
A1 20 ≤ t < 30
✓ 1
Part (e)
A1 (31 + 25)/100 = 56%
✓ 1
Part (f)
M1 P(T > 50) using N(29.1, 12.72)
✓ 1
A1 about 5 sites
✓ 1
Part (g)
R1 the data show 6 sites in the class 50 to 60 mm, so the model is consistent (close) with the data
✓ 1
Part (h)
M1 24 sites below 20 mm, plus half of the 31 sites in the class 20 to 30 mm
✓ 1
A1 24 + 15.5 = 39.5 sites
✓ 1
Exam tip: Cumulative counts at class boundaries are exact; values between boundaries need an assumption such as even spread.
278IBMI3S-4-01r14-10Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 2Medium12 marks
A council surveys 400 households about rooftop solar panels. The households are classified by the age of the main householder and whether they have panels installed. | Panels installed | No panels Under 35 | 18 | 62 35 to 59 | 52 | 108 60 or over | 40 | 120
(a)
A household from the survey is chosen at random. Find the probability that it has panels installed, and the probability that the main householder is under 35 given that the household has panels installed.
(3)
(b)
The council uses a χ2 test for independence at the 5% significance level. State the null and alternative hypotheses.
(2)
(c)
Write down the number of degrees of freedom.
(1)
(d)
Show that the expected frequency for the under-35 households with panels installed is 22.
(2)
(e)
Find the χ2 statistic.
(2)
(f)
The p-value for this test is 0.173. State the conclusion, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(\frac{110}{400}=0.275\)
✓ 1
M1 \(\frac{18}{110}\) (conditional probability using the panels-installed total)
✓ 1
A1 0.164
✓ 1
Part (b)
A1 H0: age of main householder and having panels installed are independent
✓ 1
A1 H1: age of main householder and having panels installed are not independent (associated)
✓ 1
Part (c)
A1 \((3-1)(2-1)=2\)
✓ 1
Part (d)
M1 \(\frac{80\times110}{400}\) or \(400\times0.2\times0.275\)
✓ 1
A1 22 (shown, with the working visible)
✓ 1
Part (e)
M1 attempt at \(\sum\frac{(O-E)^2}{E}\) with the expected frequencies 22, 58, 44, 116, 44, 116
✓ 1
A1 χ2 = 3.51
✓ 1
Part (f)
R1 0.173 > 0.05 (or 3.51 < critical value 5.991)
✓ 1
A1 do not reject H0: no significant evidence of an association between age group and having panels installed
✓ 1
Common error: Expected frequencies come from (row total × column total) ÷ grand total. Using the observed frequency or an overall average for every cell gives the wrong statistic.
279IBMI3S-4-01r14-11Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 2Medium13 marks
A manufacturer compares the efficiency (%) of solar cells from two production lines. Samples of cells were tested. Line A (8 cells): 17.8, 18.4, 17.1, 18.9, 17.6, 18.2, 17.3, 18.0 Line B (9 cells): 18.6, 19.1, 18.2, 19.5, 18.8, 17.9, 19.3, 18.4, 19.0 Assume both samples come from normal populations with equal variances.
(a)
Find the mean and the sample standard deviation (using n − 1) for each line, to 3 significant figures.
(4)
(b)
The manufacturer tests, at the 5% significance level, whether the mean efficiency differs between the two lines. State the hypotheses.
(2)
(c)
State why a pooled two-sample t-test is appropriate, referring to the information in the question.
(1)
(d)
Perform the test and write down the p-value.
(2)
(e)
State the conclusion in context.
(2)
(f)
The manufacturer now claims that Line B is better than Line A. Write down the p-value for the one-tailed test and state whether the claim is supported at the 5% level.
(1)
(g)
Find the difference between the sample means (Line B − Line A), in percentage points, to 3 significant figures.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 Line A mean = 17.9
✓ 1
A1 Line A standard deviation = 0.591
✓ 1
A1 Line B mean = 18.8
✓ 1
A1 Line B standard deviation = 0.527
✓ 1
Part (b)
A1 H0: μA = μB
✓ 1
A1 H1: μA ≠ μB (two-tailed)
✓ 1
Part (c)
R1 the samples are independent, the populations are normal and the variances are assumed equal (any two of these in context)
✓ 1
Part (d)
M1 pooled two-sample t-test on the data (two-tailed)
✓ 1
A1 p = 0.0072 (accept 0.00718)
✓ 1
Part (e)
R1 0.007 < 0.05
✓ 1
A1 reject H0: there is significant evidence that mean efficiency differs between Line A and Line B
✓ 1
Part (f)
A1 p = 0.0036 (half the two-tailed value) < 0.05, so the claim is supported
✓ 1
Part (g)
A1 0.843
✓ 1
Exam tip: Decide on one-tailed or two-tailed from the wording of the claim before reading the p-value: "differs" is two-tailed, "better" is one-tailed.
280IBMI3S-4-02r18-03Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared testsPaper 2Medium9 marks
A textile company compares two dyeing processes, A and B. For each process, 8 fabric strips are tested and the breaking force, in newtons (N), is recorded. Process A: 212, 218, 225, 209, 221, 215, 228, 219 Process B: 224, 231, 219, 236, 227, 233, 222, 229 The company carries out a two-tailed t-test at the 5 % significance level, to test whether the two processes give different mean breaking forces.
(a)
State the null and alternative hypotheses.
(2)
(b)
State one assumption that must be made about the populations for a pooled t-test to be valid.
(1)
(c)
Use your calculator to find the p-value of the test, to 3 significant figures.
(2)
(d)
State the conclusion of the test, giving a reason.
(2)
(e)
Calculate the difference between the sample mean breaking forces of process B and process A.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: the mean breaking forces are equal (μA = μB)
✓ 1
A1 H1: the mean breaking forces are not equal (μA ≠ μB)
✓ 1
Part (b)
R1 Both populations are normally distributed with equal variances (accept either)
✓ 1
Part (c)
M1 attempt to use a two-sample pooled t-test
✓ 1
A1 p = 0.00868
✓ 1
Part (d)
R1 comparison of the p-value with 0.05: 0.00868 < 0.05
✓ 1
A1 reject H0: there is sufficient evidence of a difference in mean breaking force between the processes
✓ 1
Part (e)
M1 mean of B − mean of A = 227.625 − 218.375
✓ 1
A1 9.25 N
✓ 1
Exam tip: A conclusion must refer to the context: say what the evidence shows about the dyeing processes, not just "reject H₀".
281IBMI3S-4-02r18-09Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared testsPaper 2Medium10 marks
A marine biologist claims that the fish caught in a net are cod, haddock, plaice and other species in the ratio 4 : 3 : 2 : 1. A random sample of 300 fish from the net contains 108 cod, 97 haddock, 62 plaice and 33 other fish. A χ2 goodness-of-fit test is carried out at the 5 % significance level.
(a)
State the null hypothesis and the alternative hypothesis.
(2)
(b)
Show that the expected frequencies are 120, 90, 60 and 30.
(2)
(c)
Write down the number of degrees of freedom and the critical value of χ2 for this test.
(2)
(d)
Calculate the χ2 test statistic to 3 significant figures and state the conclusion of the test.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: the species are in the ratio 4 : 3 : 2 : 1 (the data fit the claimed distribution)
✓ 1
A1 H1: the species are not in the ratio 4 : 3 : 2 : 1
A1 do not reject H0: the data are consistent with the claimed ratio
✓ 1
Exam tip: Compare the test statistic with the critical value (or the p-value with 0.05) in a sentence, and then give a conclusion in context.
282IBMI3S-4-02r18-10Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared testsPaper 2Medium9 marks
A fashion retailer asks 360 customers whether they usually buy online or in a shop. The results are shown in the table.
Age group
Online
In shop
Total
Under 25
70
50
120
25 to 44
62
58
120
45 and over
48
72
120
Total
180
180
360
A χ2 test for independence is carried out at the 5 % significance level.
(a)
State the null hypothesis.
(1)
(b)
Write down the number of degrees of freedom.
(1)
(c)
Use your calculator to find the χ2 statistic and the p-value, each to 3 significant figures.
(3)
(d)
State the conclusion of the test, giving a reason.
(2)
(e)
Show that the expected frequency for customers under 25 who buy online is 60.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: age group and buying method (online or in shop) are independent
✓ 1
Part (b)
A1 (3 − 1)(2 − 1) = 2
✓ 1
Part (c)
M1 attempt to use a χ2 test on the 3 × 2 table of observed values
✓ 1
A1 χ2 = 8.27
✓ 1
A1 p = 0.016
✓ 1
Part (d)
R1 0.016 < 0.05
✓ 1
A1 reject H0: there is evidence that buying method is associated with age group
✓ 1
Part (e)
M1 (120 × 180) ÷ 360
✓ 1
A1 60 (AG)
✓ 1
Exam tip: Expected frequency for a cell = (row total × column total) ÷ grand total.
283IBMI3S-4-01r22-05Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regressionPaper 2Medium12 marks
A theatre manager records the seating area chosen for 200 tickets sold, classified by performance time. The results are shown in the table.
Stalls
Circle
Gallery
Total
Matinee
34
52
24
110
Evening
41
30
19
90
Total
75
82
43
200
The manager carries out a χ2 test for independence at the 5% significance level.
(a)
State the null and alternative hypotheses.
(2)
(b)
Show that the expected frequency for an evening ticket in the gallery is 19.35.
(2)
(c)
The χ2 test is valid only if a condition on the expected frequencies is met. State this condition and verify that it holds here, given that the smallest expected frequency is 19.35.
(2)
(d)
Write down the number of degrees of freedom and calculate the χ2 statistic, to 3 significant figures.
(3)
(e)
The critical value of χ2 at the 5% level with this number of degrees of freedom is 5.99. State the conclusion of the test, giving a reason.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: seating area is independent of performance time
✓ 1
A1 H1: seating area is not independent of (is associated with) performance time
✓ 1
Part (b)
M1 90 × 43 / 200 (row total × column total ÷ grand total)
✓ 1
A1 19.35 (must be seen as 3870/200 or equivalent)
✓ 1
Part (c)
R1 all expected frequencies must be at least 5
✓ 1
A1 smallest expected frequency 19.35 ≥ 5, so the condition is met
✓ 1
Part (d)
A1 degrees of freedom = (2 − 1)(3 − 1) = 2
✓ 1
M1 use of Σ(O − E)2/E with expected frequencies
✓ 1
A1 χ2 = 5.19
✓ 1
Part (e)
M1 compare the χ2 statistic 5.19 with the critical value 5.99
✓ 1
A1 5.19 < 5.99, so do not reject H0
✓ 1
R1 there is insufficient evidence at the 5% level of an association between seating area and performance time
✓ 1
Exam tip: Check the expected-frequency condition before trusting a chi-squared result, and use the degrees of freedom (rows - 1)(columns - 1) for contingency tables.
284IBMI3S-4-02r25-12Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 2Medium12 marks
An airline records the payload, x tonnes, and the take-off distance, y metres, for eight flights of the same aircraft type from the same runway in similar conditions. x: 12, 15, 18, 21, 24, 27, 30, 33 y: 1020, 1105, 1250, 1320, 1475, 1530, 1690, 1760
(a)
Find the product-moment correlation coefficient, r, correct to 3 significant figures.
(1)
(b)
Describe the correlation between payload and take-off distance.
(1)
(c)
Find the equation of the regression line of y on x, with coefficients correct to 3 significant figures.
(2)
(d)
Interpret, in context, the gradient and the constant term of the regression line.
(2)
(e)
Use the regression line to estimate the take-off distance for a payload of 20 tonnes. Give your answer correct to 3 significant figures.
(2)
(f)
A planner proposes using the regression line to predict the take-off distance for a payload of 55 tonnes. Comment on the reliability of this prediction.
(2)
(g)
For the flight with payload 24 tonnes, find the residual (actual distance − predicted distance). Use the unrounded regression coefficients and give your answer to the nearest metre.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 r = 0.997
✓ 1
Part (b)
A1 strong positive linear correlation
✓ 1
Part (c)
A1 gradient 36.1
✓ 1
A1 y = 36.1x + 581
✓ 1
Part (d)
A1 each extra tonne of payload increases the take-off distance by about 36.1 m
✓ 1
A1 with no payload the model predicts a take-off distance of about 581 m
✓ 1
Part (e)
M1 substitution of x = 20 into the regression equation
✓ 1
A1 1300 m
✓ 1
Part (f)
R1 55 is outside the range of the data (12 to 33 tonnes), so this is extrapolation
✓ 1
R1 the linear relationship may not continue (for example, heavier loads may need a disproportionately longer run), so the prediction may be unreliable
✓ 1
Part (g)
M1 1475 − (581.25 + 36.1111 × 24) = 1475 − 1447.9
✓ 1
A1 residual = 27 m (the flight needed 27 m more than predicted)
✓ 1
Common error: Using rounded coefficients in later calculations can change the final digits; keep the unrounded regression values in your calculator.
285IBMI3S-4-02r34-07Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Probability and the binomial distributionPaper 2Medium12 marks
A food company tests three new recipes (B, C, D) of a snack bar. Each of 200 volunteers states which recipe they prefer. The results, by age group, are shown in the table.
Recipe B
Recipe C
Recipe D
Total
Under 30
34
46
20
100
30 and over
28
35
37
100
Total
62
81
57
200
A χ2 test for independence is carried out at the 5 % significance level.
(a)
Write down the null hypothesis and the alternative hypothesis.
(2)
(b)
Show that the expected frequency for volunteers aged 30 and over who prefer recipe C is 40.5.
(2)
(c)
Find the number of degrees of freedom, the χ2 statistic and the p-value. Give the χ2 statistic to 3 significant figures and the p-value to 3 significant figures.
(4)
(d)
State the conclusion of the test, giving a reason.
(2)
(e)
Recipe D has the largest contribution to the χ2 statistic for the under-30 volunteers. Find that contribution, to 3 significant figures, and state whether under-30 volunteers chose recipe D more or less often than expected under independence.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: recipe preference is independent of age group
✓ 1
A1 H1: recipe preference is not independent of age group (there is an association)
✓ 1
Part (b)
M1 \(\dfrac{100\times81}{200}\)
✓ 1
A1 \(=40.5\)
✓ 1
Part (c)
A1 degrees of freedom \(=(2-1)(3-1)=2\)
✓ 1
M1 χ2 test on the GDC with the observed table
✓ 1
A1 \(\chi^2=7.14\) (7.1446…)
✓ 1
A1 \(p=0.0281\) (0.02809…)
✓ 1
Part (d)
R1 \(0.0281<0.05\) (or \(7.14>5.99\), the critical value)
✓ 1
A1 reject H0: there is evidence of an association between age group and preferred recipe
✓ 1
Part (e)
M1 expected frequency \(\dfrac{100\times57}{200}=28.5\) and contribution \(\dfrac{(20-28.5)^2}{28.5}\)
✓ 1
A1 2.54 (2.535…); under-30 volunteers chose recipe D less often than expected (20 observed against 28.5 expected)
✓ 1
Exam tip: A χ² test supports 'association' or 'no evidence of association', never 'proves'. Always compare the p-value with the significance level and state the comparison.
286IBMI3S-4-02r34-11Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Probability and the binomial distributionPaper 2Medium10 marks
A bakery tests two preservatives, P and Q, on loaves. Eight loaves were treated with each preservative and the number of days before mould appeared was recorded. Preservative P: 5.1, 5.8, 6.2, 5.5, 6.0, 5.7, 6.3, 5.4 Preservative Q: 6.1, 6.5, 5.9, 6.8, 6.2, 7.0, 6.4, 6.6 The bakery uses a two-sample t-test with pooled variance at the 5 % significance level to test whether the mean shelf life is different for the two preservatives.
(a)
State the null hypothesis and the alternative hypothesis.
(2)
(b)
State two assumptions that must hold for this test to be valid.
(2)
(c)
Carry out the test. Give the t-statistic, the degrees of freedom and the p-value, each to 3 significant figures.
(3)
(d)
State the conclusion of the test, giving a reason.
(2)
(e)
State which preservative appears to give the longer shelf life, using the sample means.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: the mean shelf life is the same for both preservatives, \(\mu_P=\mu_Q\)
✓ 1
A1 H1: the mean shelf lives are different, \(\mu_P\ne\mu_Q\)
✓ 1
Part (b)
A1 the shelf lives for each preservative are normally distributed (or come from normally distributed populations)
✓ 1
A1 the two populations have equal variances (and the samples are independent random samples)
✓ 1
Part (c)
M1 use of the pooled two-sample t-test on the GDC
✓ 1
A1 \(t=-3.53\) (or 3.53 if the order of the samples is reversed) and degrees of freedom \(=14\)
✓ 1
A1 \(p=0.00330\) (0.003301…)
✓ 1
Part (d)
R1 \(0.00330<0.05\)
✓ 1
A1 reject H0: there is evidence that the mean shelf lives differ
✓ 1
Part (e)
A1 Q, with sample mean 6.44 days compared with 5.75 days for P
✓ 1
Common error: Saying 'accept H₀' or 'prove': a test either rejects H₀ or finds insufficient evidence to reject it.
287IBMI3S-4-02r36-02Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Medium10 marks
A roaster records the roasting time x (minutes) and the percentage mass loss y of six batches of beans.
Find Pearson's product-moment correlation coefficient r and describe the correlation.
(2)
(b)
Find the equation of the regression line of y on x.
(2)
(c)
Interpret the gradient of the regression line in this context.
(2)
(d)
Use the regression line to estimate the mass loss for a roasting time of 10.5 minutes.
(2)
(e)
The regression line is used to estimate the mass loss for a 20-minute roast. Find this estimate and comment on its reliability.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 r = 0.993
✓ 1
A1 strong positive correlation
✓ 1
Part (b)
A1 gradient 0.929
✓ 1
A1 y = 0.929x + 3.83
✓ 1
Part (c)
A1 each extra minute of roasting is associated with about 0.929 more percentage points of mass loss
✓ 1
R1 correct reference to context: percentage mass loss, per minute
✓ 1
Part (d)
M1 substitute x = 10.5
✓ 1
A1 13.6%
✓ 1
Part (e)
A1 22.4%
✓ 1
R1 20 minutes is far outside the data range 8 to 13 minutes (extrapolation), so the estimate is unreliable
✓ 1
Common error: Do not claim causation from a high correlation coefficient, and avoid using a regression line far outside the range of the data.
288IBMI3S-4-02r36-03Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Medium9 marks
The breaking strength of steel cables used in a footbridge is normally distributed with mean 480 kN and standard deviation 12 kN.
(a)
Find the probability that a randomly chosen cable has a breaking strength below 465 kN.
(2)
(b)
95% of cables have a breaking strength greater than k kN. Find k.
(2)
(c)
A delivery contains 200 cables. Estimate the number with breaking strength below 465 kN, giving your answer to the nearest whole number.
(2)
(d)
Three cables are chosen independently. Find the probability that all three have breaking strength of at least 465 kN.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of normal distribution with μ = 480, σ = 12
✓ 1
A1 0.106
✓ 1
Part (b)
M1 inverse normal with probability 0.05 (or 0.95)
✓ 1
A1 460 kN (460.26…)
✓ 1
Part (c)
M1 200 × 0.10565 (or ECF from (a))
✓ 1
A1 21
✓ 1
Part (d)
A1 P(at least 465 kN) = 1 − 0.10565 = 0.89435
✓ 1
M1 (0.89435)3
✓ 1
A1 0.715
✓ 1
Exam tip: Sketch the normal curve and shade the required region before using the calculator; it prevents mixing up tails.
289IBMI3S-4-02r36-04Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Medium9 marks
A café asks 230 customers whether they usually order a hot or an iced drink, and which roast they prefer. The results are shown below.
Hot drink: light 48, medium 62, dark 40 Iced drink: light 32, medium 18, dark 30
A χ2 test for independence is carried out at the 5% significance level.
(a)
Write down the null and alternative hypotheses.
(2)
(b)
Write down the number of degrees of freedom.
(1)
(c)
Show that the expected frequency for customers who order a hot drink and prefer light roast is 52.2, correct to 3 significant figures.
(2)
(d)
Find the χ2 test statistic.
(2)
(e)
The p-value for this test is 0.0158. State the conclusion of the test, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: type of drink and roast preference are independent
✓ 1
A1 H1: type of drink and roast preference are not independent
✓ 1
Part (b)
A1 (2 − 1)(3 − 1) = 2
✓ 1
Part (c)
M1 150 × 80 ÷ 230
✓ 1
A1 52.17… = 52.2 (AG)
✓ 1
Part (d)
M1 attempt to use Σ(O − E)2/E or GDC
✓ 1
A1 8.29
✓ 1
Part (e)
R1 0.0158 < 0.05
✓ 1
A1 reject H0: there is evidence that drink type and roast preference are associated
✓ 1
Common error: A significant result shows an association, not which roast causes which choice; do not claim cause and effect.
290IBMI3S-4-02r36-05Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Medium9 marks
A supplier claims that the beans it sells are graded as follows: 10% small, 40% medium, 35% large and 15% extra large. A café takes a random sample of 200 beans and finds 28 small, 70 medium, 72 large and 30 extra large. A χ2 goodness-of-fit test is carried out at the 5% significance level.
(a)
State the null and alternative hypotheses.
(2)
(b)
Find the expected frequency for each grade.
(2)
(c)
State the number of degrees of freedom.
(1)
(d)
Find the χ2 test statistic.
(2)
(e)
The critical value at the 5% level with this number of degrees of freedom is 7.81. State the conclusion in context.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: the grading proportions are as the supplier claims
✓ 1
A1 H1: the grading proportions are not as the supplier claims
✓ 1
Part (b)
M1 200 × the stated proportions
✓ 1
A1 20, 80, 70, 30
✓ 1
Part (c)
A1 4 − 1 = 3
✓ 1
Part (d)
M1 Σ(O − E)2/E = 3.2 + 1.25 + 0.0571 + 0
✓ 1
A1 4.51
✓ 1
Part (e)
R1 4.51 < 7.81
✓ 1
A1 do not reject H0: no evidence against the supplier's claim
✓ 1
Exam tip: State the degrees of freedom as (number of categories − 1) when no parameters are estimated from the data.
291IBMI3S-4-02r36-12Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Medium9 marks
The number of attempts X needed for a probe to dock successfully has probability distribution P(X = x) = k(6 − x) for x = 1, 2, 3, 4, 5, where k is a constant.
(a)
Show that k = 1/15.
(2)
(b)
Find P(X ≥ 3).
(2)
(c)
Find E(X).
(3)
(d)
Each attempt costs £120. Find the expected total cost, using the exact value of E(X).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sum of probabilities 5k + 4k + 3k + 2k + k = 1
✓ 1
A1 15k = 1, so k = 1/15 (AG)
✓ 1
Part (b)
M1 (3 + 2 + 1) ÷ 15
✓ 1
A1 0.4 (or 2/5)
✓ 1
Part (c)
M1 ΣxP(X = x)
✓ 1
A1 (5 + 8 + 9 + 8 + 5) ÷ 15
✓ 1
A1 7/3 = 2.33
✓ 1
Part (d)
M1 120 × 7/3
✓ 1
A1 £280
✓ 1
Common error: For E(X) multiply each value of x by its probability; do not just average the probabilities.
292IBMI3S-4-04r37-08Sampling and descriptive statistics; T-tests, confidence intervals and Markov chains; Probability and the binomial distribution; Normal distribution and chi-squared tests; Correlation and regression; Normal distribution and chi-squared tests; Probability and the binomial distributionPaper 2Medium12 marks
In a trawl catch, 7% of the fish are undersized. A sample of 25 fish is taken from a trawl. Let X be the number of undersized fish in the sample.
(a)
State the distribution of X, including its parameters.
(1)
(b)
State one assumption needed for this distribution to be a good model, and explain why it might not hold for fish in a trawl.
(2)
(c)
Find the probability that none of the 25 fish is undersized. Give your answer to 3 significant figures.
(2)
(d)
Find the probability that at least three of the 25 fish are undersized. Give your answer to 3 significant figures.
(2)
(e)
Find the smallest sample size n for which the probability of at least one undersized fish in the sample exceeds 0.95.
(3)
(f)
Find the mean and the standard deviation of X. Give the standard deviation to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(X\sim B(25,0.07)\)
✓ 1
Part (b)
R1 the fish are independent of each other (or the probability of being undersized is constant)
✓ 1
R1 fish in a shoal tend to be similar in size, so one fish being undersized makes others more likely to be undersized
✓ 1
Part (c)
M1 \(0.93^{25}\)
✓ 1
A1 0.163
✓ 1
Part (d)
M1 \(1-P(X\leq2)\)
✓ 1
A1 0.253
✓ 1
Part (e)
M1 \(1-0.93^n>0.95\), so \(0.93^n<0.05\)
✓ 1
M1 solves, for example \(n>\frac{\ln0.05}{\ln0.93}=41.3\), or tests \(n=41\) (0.949) and \(n=42\) (0.953)
✓ 1
A1 \(n=42\)
✓ 1
Part (f)
A1 mean \(=25\times0.07=1.75\)
✓ 1
A1 standard deviation \(=\sqrt{25\times0.07\times0.93}=1.28\)
✓ 1
Exam tip: For "smallest n such that" questions, set up the inequality, solve with logarithms or a table on your GDC, and always round up to the next whole number.
293IBMI3S-4-04r37-12Sampling and descriptive statistics; T-tests, confidence intervals and Markov chains; Probability and the binomial distribution; Normal distribution and chi-squared tests; Correlation and regression; Normal distribution and chi-squared tests; Probability and the binomial distributionPaper 2Medium12 marks
A robot's vision system sorts 200 items into four categories, A, B, C and D. The manufacturer claims that items fall into these categories in the ratio 40% : 30% : 20% : 10%. In a test, the numbers of items sorted into the categories were: A: 70, B: 74, C: 38, D: 18. A χ2 goodness-of-fit test is carried out at the 5% significance level.
(a)
State the null hypothesis and the alternative hypothesis.
(2)
(b)
Find the expected frequencies for the four categories.
(2)
(c)
Write down the number of degrees of freedom.
(1)
(d)
Use your GDC to find the χ2 statistic. Give your answer to 3 significant figures.
(1)
(e)
Find the p-value.
(1)
(f)
State the conclusion of the test, giving a reason.
(2)
(g)
Write down the critical value at the 5% significance level and show that it leads to the same conclusion.
(1)
(h)
Show that the contribution of category B to the χ2 statistic is 3.27, to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(H_0\): the manufacturer's proportions (40%, 30%, 20%, 10%) are correct, that is, the data fit the stated distribution
✓ 1
A1 \(H_1\): the data do not fit the stated distribution
✓ 1
Part (b)
M1 \(200\times\) each proportion
✓ 1
A1 A: 80, B: 60, C: 40, D: 20
✓ 1
Part (c)
A1 \(4-1=3\)
✓ 1
Part (d)
A1 4.82
✓ 1
Part (e)
A1 0.186
✓ 1
Part (f)
R1 \(0.186>0.05\)
✓ 1
A1 do not reject \(H_0\): there is insufficient evidence that the claimed proportions are wrong
✓ 1
Part (g)
R1 critical value 7.81, and \(4.82<7.81\), so \(H_0\) is not rejected
✓ 1
Part (h)
M1 \(\frac{(74-60)^2}{60}\)
✓ 1
A1 3.27 (AG)
✓ 1
Exam tip: A goodness-of-fit test compares observed frequencies with expected frequencies worked out from the claimed proportions, and the degrees of freedom are the number of categories minus one.
294IBMI3S-4-02r38-06Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 2Medium9 marks
Calculator
A cycling club asks 200 of its members which type of bike they use most often. The results are Under 40 (100 members): road bike 36, hybrid 41, e-bike 23 Aged 40 or over (100 members): road bike 20, hybrid 38, e-bike 42 A χ2 test for independence is carried out at the 5% significance level.
(a)
Write down the null hypothesis and the alternative hypothesis.
(2)
(b)
Show that the expected number of e-bike users aged 40 or over, if H0 is true, is 32.5.
(2)
(c)
Write down the number of degrees of freedom.
(1)
(d)
Use your GDC to find the χ2 statistic and the p-value.
(2)
(e)
State the conclusion of the test, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: type of bike is independent of age group
✓ 1
A1 H1: type of bike is not independent of (is associated with) age group
✓ 1
Part (b)
M1 100 × 65 ÷ 200
✓ 1
A1 32.5 (AG)
✓ 1
Part (c)
A1 (2 − 1)(3 − 1) = 2
✓ 1
Part (d)
A1 p-value = 0.00598
✓ 1
A1 χ2 = 10.2
✓ 1
Part (e)
R1 0.00598 < 0.05 (or 10.2 > 5.99, the critical value for 2 degrees of freedom)
✓ 1
A1 reject H0: there is evidence that bike type is associated with age group
✓ 1
Exam tip: Always compare the p-value with the significance level in words ('since 0.00598 < 0.05') before stating the conclusion in context.
295IBMI3S-4-02r38-07Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 2Medium9 marks
Calculator
Glass beads recovered from an earlier excavation were 40% blue, 30% green, 20% amber and 10% clear. At a new site, 200 beads are found: 88 blue, 64 green, 36 amber and 12 clear. A χ2 goodness-of-fit test at the 5% significance level is used to decide whether the new beads follow the earlier proportions.
(a)
State the null hypothesis and the alternative hypothesis.
(2)
(b)
Write down the expected frequency for each colour if H0 is true.
(1)
(c)
Write down the number of degrees of freedom.
(1)
(d)
Find the χ2 statistic and the p-value. Give both to 3 significant figures.
(3)
(e)
State the conclusion of the test, with a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: the colours of beads at the new site follow the proportions 40% : 30% : 20% : 10%
✓ 1
A1 H1: the colours do not follow these proportions
✓ 1
Part (b)
A1 blue 80, green 60, amber 40, clear 20
✓ 1
Part (c)
A1 4 − 1 = 3
✓ 1
Part (d)
A1 p-value = 0.198
✓ 1
M1 Σ(O − E)2 ÷ E = 82/80 + 42/60 + 42/40 + 82/20
✓ 1
A1 χ2 = 4.67
✓ 1
Part (e)
R1 0.198 > 0.05
✓ 1
A1 do not reject H0: no evidence that the colour proportions differ from the earlier site
✓ 1
Common error: The degrees of freedom for a goodness-of-fit test are the number of categories minus 1 (when no parameters are estimated), not the sample size minus 1.
296IBMI3S-4-02r44-06Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Medium12 marks
A Formula 1 analyst records the pit-stop strategy used by 130 drivers in a season and the result they achieved.
Podium
Points, no podium
No points
Total
One stop
14
18
8
40
Two stops
9
22
19
50
Three stops
4
11
25
40
Total
27
51
52
130
She carries out a χ2 test for independence at the 5% significance level.
(a)
State the null and alternative hypotheses.
(2)
(b)
Show that the expected frequency for one stop and podium is 8.31, to 3 significant figures.
(2)
(c)
Write down the number of degrees of freedom.
(2)
(d)
Find the χ2 test statistic.
(2)
(e)
Find the p-value.
(1)
(f)
State, with a reason, the conclusion of the test.
(2)
(g)
State why the χ2 test is valid for this table.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(H_0\): strategy and result are independent
✓ 1
A1 \(H_1\): strategy and result are not independent (associated)
✓ 1
Part (b)
M1 \(\frac{40\times 27}{130}\)
✓ 1
A1 \(8.3077\ldots=8.31\)
✓ 1
Part (c)
M1 \((3-1)(3-1)\)
✓ 1
A1 4
✓ 1
Part (d)
M1 use of GDC χ2 test for independence with the observed table
✓ 1
A1 17.5
✓ 1
Part (e)
A1 0.00157
✓ 1
Part (f)
R1 \(p=0.00157<0.05\)
✓ 1
A1 reject \(H_0\): there is evidence of an association between pit-stop strategy and result
✓ 1
Part (g)
R1 all expected frequencies are at least 5 (the smallest is 8.31)
✓ 1
Common error: Do not include the row and column totals when entering the observed matrix into the calculator.
297IBMI3S-4-02r44-10Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Medium12 marks
A Formula 1 team compares lap times, in seconds, on two tyre compounds at the same circuit. Eight laps were recorded on each compound.
Soft
87.4
87.9
88.3
87.1
88.8
87.6
88.1
87.7
Medium
88.9
89.4
88.6
89.9
89.1
88.7
90.2
89.3
The team assumes that both sets of lap times are normally distributed with equal population variances, and carries out a one-tailed pooled two-sample t-test at the 5% significance level to see whether the soft tyre gives a lower mean lap time.
(a)
Find the mean and the sample standard deviation for each compound. Give means to 4 significant figures and standard deviations to 3 significant figures.
(4)
(b)
State the null and alternative hypotheses.
(2)
(c)
State one reason why a pooled test is appropriate.
(1)
(d)
Find the t-statistic and the p-value.
(3)
(e)
State the conclusion of the test, with a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 soft mean 87.86
✓ 1
A1 medium mean 89.26
✓ 1
A1 soft standard deviation 0.537
✓ 1
A1 medium standard deviation 0.563
✓ 1
Part (b)
A1 \(H_0\): \(\mu_S=\mu_M\)
✓ 1
A1 \(H_1\): \(\mu_S<\mu_M\)
✓ 1
Part (c)
R1 the sample standard deviations are similar, consistent with equal population variances
✓ 1
Part (d)
M1 use of GDC pooled two-sample t-test with the one-tailed alternative
✓ 1
A1 \(t=-5.09\)
✓ 1
A1 \(p=8.24\times 10^{-5}\)
✓ 1
Part (e)
R1 \(p<0.05\)
✓ 1
A1 reject \(H_0\): there is evidence that the soft tyre gives a lower mean lap time
✓ 1
Common error: The alternative hypothesis sets the tail: because the claim is 'lower', use the one-tailed option and compare the p-value directly with 0.05.
298IBMI-033-00-10Sampling and descriptive statisticsPaper 2Hard11 marks
A psychologist measures the reaction times, in seconds, of 30 participants. She calculates the mean as 0.312 s and the population standard deviation as 0.045 s. She later discovers that one reaction time was recorded as 0.512 s when it should have been 0.215 s.
(a)
Find the sum of the 30 recorded reaction times.
(2)
(b)
Find the correct mean reaction time.
(2)
(c)
Show that, for the recorded data, Σx2 = 2.98 correct to 3 significant figures.
(2)
(d)
Find the correct population standard deviation of the reaction times.
(3)
(e)
Explain, in context, why correcting the error reduces the standard deviation.
The incorrect value 0.512 s was far above the mean (about 0.2 s), so it added a large squared deviation R1
✓ 1
The correct value 0.215 s is closer to the mean, so the participants' reaction times are more consistent (less spread out) than first calculated R1
✓ 1
Exam tip: Keep full unrounded values (2.98107, 0.3021) in your calculator until the end. Rounding Σx² to 2.98 first gives about 0.0295 s instead of 0.0301 s.
299IBMI-033-05-r2-05Sampling and descriptive statisticsPaper 2Hard12 marks
A fitness company has two branches. Over one week it records the length of every workout session, in minutes, at each branch.
Branch A: 40 sessions, mean 52 minutes, standard deviation 8 minutes Branch B: 60 sessions, mean 47 minutes, standard deviation 11 minutes
Treat each branch's data as a complete set, so the standard deviations are calculated with divisor n.
(a)
Find the mean length of all 100 sessions.
(2)
(b)
The manager will select 20 of the 100 recorded sessions for a follow-up survey, using a sample stratified by branch. Find the number of sessions to be selected from Branch B.
(2)
(c)
Show that, for Branch A, the sum of the squares of the session lengths is 110 720.
(2)
(d)
Find the standard deviation of the lengths of all 100 sessions.
(4)
(e)
Compare the consistency of session lengths at the two branches.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
\(\frac{40(52)+60(47)}{100}\) M1
✓ 1
49 minutes A1
✓ 1
Part (b)
\(\frac{60}{100}\times 20\) M1
✓ 1
12 A1
✓ 1
Part (c)
Use of \(\sigma^2=\frac{\sum x^2}{n}-\bar{x}^2\), i.e. \(64=\frac{\sum x^2}{40}-52^2\) M1
Branch B has the larger standard deviation (11 compared with 8) A1
✓ 1
So session lengths at Branch B are less consistent (more varied) than at Branch A R1
✓ 1
Exam tip: You cannot combine two standard deviations by averaging them. Rebuild each group's sum of squares from n(σ² + mean²), add the sums, then apply the variance formula to the whole group.
300IBMI3S-4-07-13Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 2Hard12 marks
A machine fills bottles with juice. The volume, V ml, is normally distributed with standard deviation 4 ml. The mean is set so that 1% of bottles contain less than 495 ml.
(a)
Find the mean volume.
(3)
(b)
A bottle is rejected if it contains less than 495 ml or more than 512 ml. Find the probability that a bottle is rejected.
(3)
(c)
Given that a bottle is rejected, find the probability that it contains less than 495 ml.
(2)
(d)
A random sample of 30 bottles is taken. Find the probability that at least two are rejected.
(3)
(e)
State one assumption required for the model in part (d).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(P(V<495)=0.01\), standardise with \(z=-2.326\)
✓ 1
M1 \(\frac{495-\mu}{4}=-2.326\)
✓ 1
A1 μ = 504 ml (504.3)
✓ 1
Part (b)
M1 \(P(V>512)\) using the unrounded mean
✓ 1
A1 0.0272
✓ 1
A1 \(0.01+0.0272=0.0372\)
✓ 1
Part (c)
M1 \(\frac{0.01}{0.0372}\)
✓ 1
A1 0.269
✓ 1
Part (d)
A1 \(X\sim B(30,0.0372)\)
✓ 1
M1 \(1-P(X\le1)\)
✓ 1
A1 0.308
✓ 1
Part (e)
R1 bottles are filled independently of each other (the probability of rejection is constant)
✓ 1
Exam tip: Keep unrounded values from (a) and (b) in your calculator memory so the later parts are not affected by rounding.
301IBMI3S-4-07b-11Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 2Hard12 marks
The distance, in thousands of km, that a type of lorry tyre lasts before replacement is modelled by a normal distribution with mean μ and standard deviation σ. Data show that 10% of tyres last less than 57 thousand km and 20% last more than 70 thousand km.
(a)
Find the standardised values z corresponding to P(Z < z) = 0.10 and to P(Z < z) = 0.80.
(2)
(b)
Hence find μ and σ.
(4)
(c)
Find the probability that a tyre lasts between 60 and 68 thousand km.
(3)
(d)
A fleet fits 500 of these tyres. Estimate how many will last between 60 and 68 thousand km.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 z = −1.2816
✓ 1
A1 z = 0.8416
✓ 1
Part (b)
M1 (57 − μ)/σ = −1.2816
✓ 1
M1 (70 − μ)/σ = 0.8416
✓ 1
A1 σ = 13 / (0.8416 + 1.2816) = 6.12
✓ 1
A1 μ = 64.8
✓ 1
Part (c)
M1 normal cdf with μ and σ from (b)
✓ 1
M1 limits 60 and 68 used
✓ 1
A1 0.482
✓ 1
Part (d)
M1 500 × 0.4824
✓ 1
A1 241.2
✓ 1
R1 round to 241 since a count must be a whole number
✓ 1
Exam tip: Two unknown parameters need two equations. Write each as (value − μ)/σ = z and solve simultaneously.
302IBMI3S-4-07b-12Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 2Hard12 marks
A laboratory inspects sets of three fitness trackers. In 200 sets, the number of faulty trackers per set was recorded.
Faulty trackers: 0 | 1 | 2 | 3 Number of sets: 150 | 41 | 8 | 1
The manufacturer claims that each tracker is faulty with probability 0.1, independently. The laboratory carries out a χ2 goodness-of-fit test, at the 5% significance level, for the model X ~ B(3, 0.1).
(a)
State one assumption of the binomial model that the laboratory should check.
(1)
(b)
Write down the null and alternative hypotheses.
(2)
(c)
Calculate the expected frequencies. The categories '2' and '3' are combined; explain why this is necessary.
(3)
(d)
State the number of degrees of freedom and the critical value.
(2)
(e)
Calculate the χ2 test statistic.
(2)
(f)
State the conclusion of the test, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 trackers are faulty independently of one another, with constant probability
✓ 1
Part (b)
A1 H0: the number of faulty trackers per set follows B(3, 0.1)
✓ 1
A1 H1: it does not follow B(3, 0.1)
✓ 1
Part (c)
A1 expected: 145.8, 48.6
✓ 1
A1 combined category: 5.4 + 0.2 = 5.6
✓ 1
R1 the expected frequency for '3' is only 0.2, below the minimum of 5 required for the test
✓ 1
Part (d)
A1 3 categories − 1 = 2 degrees of freedom
✓ 1
A1 critical value 5.99
✓ 1
Part (e)
M1 sum of (O − E)2/E over three categories
✓ 1
A1 χ2 = 3.37
✓ 1
Part (f)
M1 compares 3.37 with the critical value 5.99
✓ 1
R1 3.37 < 5.99, so do not reject H0: the data are consistent with B(3, 0.1)
✓ 1
Common error: Expected frequencies below 5 must be merged with a neighbouring category, and the degrees of freedom counted after merging.
303IBMI3S-4-07b-13Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 2Hard12 marks
A bike-share company inspects its e-scooters. Each scooter has a faulty battery with probability 0.08, independently. A depot receives a delivery of 20 scooters. Let X be the number with a faulty battery.
(a)
Find P(X = 0).
(2)
(b)
Find the expected value and the standard deviation of X.
(2)
(c)
A larger delivery has n scooters. Find the smallest n for which the probability of at least one faulty battery exceeds 0.95.
(3)
(d)
Find P(X = 2).
(2)
(e)
Given that at least one scooter in the delivery has a faulty battery, find the probability that exactly two do.
Common error: In a conditional probability, divide by the probability of the condition (here P(X ≥ 1)), not by 1.
304IBMI3S-4-01r7-10Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Hard15 marks
The time, T minutes, taken to load a ship at a port is normally distributed with mean 46 and standard deviation 7.5.
(a)
Find the probability that a ship takes less than 40 minutes to load, correct to 3 significant figures.
(2)
(b)
Find the probability that a ship takes more than one hour to load, correct to 3 significant figures.
(2)
(c)
Find the probability that a ship takes more than 55 minutes to load, correct to 4 decimal places.
(2)
(d)
In one day, 12 ships are loaded independently. Find the probability that at least 3 of them take more than 55 minutes. Use the unrounded value of the probability from the previous part, and give your answer to 3 significant figures.
(3)
(e)
Given that a ship takes longer than the mean time to load, find the probability that it takes more than 55 minutes. Give your answer to 3 significant figures.
(3)
(f)
The longest 5% of loading times exceed k minutes. Find k, correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(T < 40) using N(46, 7.52)
✓ 1
A1 0.212
✓ 1
Part (b)
M1 P(T > 60)
✓ 1
A1 0.031
✓ 1
Part (c)
M1 P(T > 55)
✓ 1
A1 0.1151
✓ 1
Part (d)
A1 X ~ B(12, 0.1151…) where X is the number taking more than 55 minutes
✓ 1
M1 P(X ≥ 3) = 1 − P(X ≤ 2)
✓ 1
A1 0.152
✓ 1
Part (e)
M1 P(T > 55 | T > 46) = P(T > 55)/P(T > 46)
✓ 1
A1 0.1151/0.5 (P(T > 46) = 0.5)
✓ 1
A1 0.230
✓ 1
Part (f)
M1 use of inverse normal, z = 1.6449 for the upper 5%
✓ 1
M1 k = 46 + 1.6449 × 7.5
✓ 1
A1 k = 58.3 minutes
✓ 1
Exam tip: In a conditional normal probability, the given condition becomes the denominator; here it is exactly 0.5 by symmetry.
305IBMI3S-4-01r7-12Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Hard13 marks
A shipping company records the average speed, x knots, and the fuel used, y tonnes per day, for eight voyages of the same cargo ship.
Find the equation of the regression line of y on x.
(2)
(c)
Interpret the gradient of the regression line in context.
(2)
(d)
The company notes that the y-intercept is negative. Comment on whether this is sensible in context.
(2)
(e)
Describe the correlation between x and y.
(1)
(f)
The company wants to use the line to estimate fuel use at 40 knots. Comment on the reliability of this estimate.
(2)
(g)
Use the regression line to estimate the fuel used per day at an average speed of 15 knots.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of the statistics function of a calculator with the eight data pairs
✓ 1
A1 r = 0.999
✓ 1
Part (b)
M1 use of regression on a calculator
✓ 1
A1 y = 2.6x − 8
✓ 1
Part (c)
M1 each extra knot of average speed
✓ 1
A1 is associated with an extra 2.6 tonnes of fuel used per day
✓ 1
Part (d)
M1 the intercept −8 would be a negative fuel use at a speed of 0 knots
✓ 1
R1 not meaningful; the model is only valid for speeds within the range of the data (10 to 24 knots)
✓ 1
Part (e)
A1 strong positive linear correlation
✓ 1
Part (f)
M1 40 knots is outside the range of the data (extrapolation)
✓ 1
R1 the relationship may not remain linear, so the estimate is unreliable
✓ 1
Part (g)
M1 y = 2.6 × 15 − 8
✓ 1
A1 31 tonnes per day
✓ 1
Common error: A value of r close to 1 shows a strong linear association but does not by itself make predictions outside the data range reliable.
306IBMI3S-4-01r7-14Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Hard12 marks
At a theatre, 1200 seats are numbered 1 to 1200. A researcher compares the amount spent on refreshments (in £) by 8 randomly chosen customers at a matinee performance with that of 9 randomly chosen customers at an evening performance. Matinee: 4.2, 5.1, 3.8, 6.0, 4.9, 5.5, 4.4, 5.2 Evening: 5.6, 6.3, 5.9, 7.1, 6.0, 5.4, 6.8, 6.5, 7.2 The researcher carries out a two-tailed t-test at the 5% significance level, assuming equal population variances, to test whether the mean spend differs between the two performances.
(a)
State the null and alternative hypotheses.
(2)
(b)
State two assumptions required for this test to be valid.
(2)
(c)
Use your calculator to find the p-value, correct to 3 significant figures.
(2)
(d)
State the conclusion of the test, with a reason.
(2)
(e)
In a different study, the researcher selects 80 of the 1200 seats using systematic sampling. The first seat chosen is seat 7. Find the number of the 10th seat chosen.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: the mean spend is the same at the matinee and evening performances
✓ 1
A1 H1: the mean spends are different
✓ 1
Part (b)
A1 both populations are normally distributed
✓ 1
A1 the population variances are equal (and samples are independent)
✓ 1
Part (c)
M1 use of a two-sample pooled t-test
✓ 1
A1 p = 6.22 × 10−4
✓ 1
Part (d)
M1 comparison of the p-value with 0.05
✓ 1
R1 reject H0; there is evidence that the mean spend differs between performances
✓ 1
Part (e)
M1 sampling interval = 1200 ÷ 80
✓ 1
A1 15
✓ 1
M1 7 + 9 × 15
✓ 1
A1 seat 142
✓ 1
Exam tip: Systematic sampling uses a fixed interval after a random start; the nth item is the start plus (n − 1) intervals.
307IBMI3S-4-01r14-14Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared testsPaper 2Hard16 marks
A virus has an incubation period I days that is normally distributed with mean 5.2 days and standard deviation 1.4 days. In a town, 70% of people are vaccinated. When exposed to the virus, a vaccinated person becomes infected with probability 0.06 and an unvaccinated person with probability 0.35.
(a)
Find the probability that an infected person has an incubation period of more than 7 days.
(2)
(b)
Find the incubation period that is exceeded by only 10% of infected people, to 3 significant figures.
(2)
(c)
A person from the town is exposed to the virus. Find the probability that they become infected.
(2)
(d)
Given that an exposed person becomes infected, find the probability that they were vaccinated.
(2)
(e)
Twenty exposed people are chosen at random. Let X be the number who become infected. State the distribution of X, and find the probability that at least 4 become infected.
(3)
(f)
Find the smallest integer k such that P(X ≤ k) ≥ 0.95.
(2)
(g)
Give one reason why the binomial model may not be suitable for people exposed during an outbreak.
(1)
(h)
Each infection costs the health service €850. Using 0.147 as the probability of infection, find the expected total cost when 2000 people are exposed.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 normal cdf with mean 5.2 and standard deviation 1.4 (upper tail)
✓ 1
A1 0.099
✓ 1
Part (b)
M1 inverse normal with probability 0.9 (lower tail), or \(z=1.2816\)
✓ 1
A1 6.99 days
✓ 1
Part (c)
M1 \(0.7\times0.06+0.3\times0.35\) (tree diagram or law of total probability)
✓ 1
A1 0.147
✓ 1
Part (d)
M1 \(\frac{0.7\times0.06}{0.147}\)
✓ 1
A1 0.286
✓ 1
Part (e)
A1 X ~ B(20, 0.147)
✓ 1
M1 \(1-P(X\le3)\)
✓ 1
A1 0.338
✓ 1
Part (f)
M1 comparing cumulative probabilities (P(X ≤ 5) = 0.9380, which is below 0.95)
✓ 1
A1 k = 6 (P(X ≤ 6) = 0.9802)
✓ 1
Part (g)
R1 e.g. infections are not independent (people in the same household or workplace are linked) / the probability of infection may vary between people
✓ 1
Part (h)
M1 \(2000\times0.147\times850\)
✓ 1
A1 €249900
✓ 1
Exam tip: When a question gives two stages of probabilities (vaccinated, then infected), draw a tree diagram first; every later part can be read from it.
308IBMI3S-4-01r15-06Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Hard12 marks
A clothing retailer surveys 300 customers about their preferred style of clothing and records the age group of each. The results are shown in the table.
Casual
Smart
Sportswear
Under 25
48
30
42
25 to 44
36
38
26
45 and over
22
40
18
The retailer carries out a χ2 test for independence at the 5% significance level to decide whether preferred style is associated with age group. Use the χ2 test on your calculator to obtain the test statistic and the p-value.
(a)
Write down the null hypothesis and the alternative hypothesis.
(2)
(b)
Write down the number of degrees of freedom, and show that the expected frequency for customers under 25 who prefer casual clothing is 42.4.
(3)
(c)
Find the χ2 test statistic and the p-value.
(2)
(d)
State, with a reason, the conclusion of the test.
(3)
(e)
Find which cell makes the greatest contribution to the χ2 statistic, and calculate that contribution.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: preferred style and age group are independent
✓ 1
A1 H1: preferred style and age group are not independent
✓ 1
Part (b)
A1 degrees of freedom = (3 − 1)(3 − 1) = 4
✓ 1
M1 expected = (row total × column total) ÷ grand total = 120 × 106 ÷ 300
✓ 1
A1 expected = 42.4 (AG)
✓ 1
Part (c)
A1 χ2 = 13.63
✓ 1
A1 p-value = 0.00858
✓ 1
Part (d)
R1 compares p-value with the significance level: 0.00858 < 0.05
✓ 1
A1 reject H0
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R1 there is sufficient evidence that preferred style of clothing is associated with age group
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Part (e)
A1 the cell is 45 and over, smart (observed 40, expected 28.80)
✓ 1
A1 contribution = (40 − 28.80)2 ÷ 28.80 = 4.36
✓ 1
Exam tip: The expected frequency of every cell should be at least 5 before the χ² test is valid; check this from the expected-value matrix.
309IBMI3S-4-01r15-07Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Hard13 marks
A warehouse uses robots that stop with a fault from time to time. Over one working week (Monday to Friday) the warehouse logged 100 faults: 33 on Monday, 17 on Tuesday, 15 on Wednesday, 16 on Thursday and 19 on Friday. The manager suspects that faults are not equally likely on each day and carries out a χ2 goodness of fit test at the 5% significance level.
(a)
Write down the null hypothesis and the alternative hypothesis.
(2)
(b)
Write down the expected frequency for each day, and the number of degrees of freedom.
(2)
(c)
Calculate the χ2 test statistic.
(2)
(d)
Find the p-value. State the conclusion of the test at the 5% level and say whether this conclusion would change at the 1% level.
(3)
(e)
The manager believes that the unusually high number of faults on Monday is the only reason for the result. Carry out a second χ2 goodness of fit test at the 5% level on the data for Tuesday to Friday only, assuming equal likelihood. Give the expected frequency, the χ2 statistic, the p-value and the conclusion.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: faults are equally likely on each day of the week (uniform distribution)
✓ 1
A1 H1: faults are not equally likely on each day
✓ 1
Part (b)
A1 expected frequency = 100 ÷ 5 = 20
✓ 1
A1 degrees of freedom = 5 − 1 = 4
✓ 1
Part (c)
M1 uses Σ(O − E)2 ÷ E
✓ 1
A1 χ2 = 11.0
✓ 1
Part (d)
A1 p-value = 0.0266
✓ 1
R1 0.0266 < 0.05, so reject H0 at the 5% level
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R1 0.0266 > 0.01, so H0 would not be rejected at the 1% level
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Part (e)
A1 expected frequency = 67 ÷ 4 = 16.75
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A1 χ2 = 0.522
✓ 1
A1 p-value = 0.914 (df = 3)
✓ 1
R1 0.914 > 0.05, so do not reject H0: no evidence that faults are unequally likely on Tuesday to Friday, which supports the manager
✓ 1
Common error: The degrees of freedom for a goodness of fit test with equal expected frequencies are the number of categories minus 1, not the number of categories.
310IBMI3S-4-01r15-10Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Hard12 marks
A machine cuts steel rods for a robot arm. The length of a rod, L mm, is normally distributed with mean μ and standard deviation σ. It is found that 10% of rods are shorter than 49.2 mm and 5% of rods are longer than 50.9 mm.
(a)
Find the standardised values (z-values) corresponding to 49.2 mm and to 50.9 mm.
(2)
(b)
Hence find the values of μ and σ, giving each to 3 significant figures.
(4)
(c)
A rod is accepted if its length is between 49.0 mm and 51.0 mm. Use the unrounded values of μ and σ to find the probability that a randomly chosen rod is accepted.
(2)
(d)
A batch contains 15 rods chosen at random. Using the probability in part (c) to 3 significant figures, find the expected number of accepted rods in the batch.
(1)
(e)
Find the probability that at least 13 of the 15 rods in the batch are accepted. Use the probability in part (c) to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 z for 49.2 is -1.282
✓ 1
A1 z for 50.9 is 1.645
✓ 1
Part (b)
M1 forms (49.2 − μ)/σ = -1.282 and (50.9 − μ)/σ = 1.645
Exam tip: Write each condition as an equation in the form (value − μ)/σ = z and solve the pair simultaneously; keep full calculator values until the final answer.
311IBMI3S-4-01r16-04Sampling and descriptive statistics; T-tests, confidence intervals and Markov chains; Probability and the binomial distribution; Normal distribution and chi-squared testsPaper 2Hard12 marks
A fashion retailer compares the amounts spent per customer, in £, at two of its stores. In all parts, standard deviations are population standard deviations.
Store A: 40 customers, mean £62.50, standard deviation £14.20. Store B: 60 customers, mean £48.00, standard deviation £11.00.
(a)
Find the mean amount spent per customer for all 100 customers.
(2)
(b)
Find the standard deviation of the amounts spent by all 100 customers, giving your answer to 3 significant figures.
(5)
(c)
The retailer then gives every customer a 15% discount on their bill and adds a £1.50 charge for a bag, so each new amount is y = 0.85x + 1.5. Using the unrounded standard deviation from (b), find the mean and standard deviation of the new amounts, each to 2 decimal places.
(4)
(d)
Explain why the £1.50 charge does not change the standard deviation.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (40 × 62.5 + 60 × 48) ÷ 100
✓ 1
A1 mean = £53.80
✓ 1
Part (b)
M1 use of Σx2 = n(σ2 + mean2) for a store
✓ 1
A1 Store A: Σx2 = 164 316 (164 315.6)
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A1 Store B: Σx2 = 145 500
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M1 variance = (Σx2 total ÷ 100) − 53.82 = 203.7…
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A1 standard deviation = £14.3
✓ 1
Part (c)
M1 meannew = 0.85 × 53.8 + 1.5
✓ 1
A1 mean = £47.23
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M1 standard deviationnew = 0.85 × (standard deviation from (b))
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A1 standard deviation = £12.13
✓ 1
Part (d)
R1 adding a constant shifts every value equally, so the spread about the mean is unchanged
✓ 1
Common error: When combining groups, you cannot average the standard deviations. Combine the sums of squares, then subtract the square of the combined mean.
312IBMI3S-4-01r16-11Sampling and descriptive statistics; T-tests, confidence intervals and Markov chains; Probability and the binomial distribution; Normal distribution and chi-squared testsPaper 2Hard13 marks
A robot manufacturer tests each robot at the end of the production line. Each robot independently fails the test with probability 0.08. Robots are tested in batches of 20. A batch is accepted if at most one robot in it fails. Let X be the number of robots in a batch of 20 that fail.
(a)
State two assumptions needed for X to follow a binomial distribution.
(2)
(b)
Find the probability that no robot in a batch fails.
(2)
(c)
Find the probability that a batch is accepted.
(2)
(d)
Given that a batch is accepted, find the probability that none of its robots failed.
(2)
(e)
Eight batches are tested. Using the unrounded probability from (c), find the probability that at least 6 batches are accepted.
(3)
(f)
After a process improvement, the failure probability is reduced to f. Find the greatest value of f, to 2 significant figures, for which the probability that no robot in a batch of 20 fails is greater than 0.5.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 each robot fails independently of the others
✓ 1
A1 the probability of failing is constant (0.08) for every robot
✓ 1
Part (b)
M1 X ~ B(20, 0.08), P(X = 0) = 0.9220
✓ 1
A1 0.189
✓ 1
Part (c)
M1 P(X = 0) + P(X = 1) (or P(X ≤ 1))
✓ 1
A1 0.517
✓ 1
Part (d)
M1 P(X = 0) ÷ P(X ≤ 1)
✓ 1
A1 0.365
✓ 1
Part (e)
M1 Y ~ B(8, 0.5169…) for the number of accepted batches
✓ 1
M1 P(Y ≥ 6) = 1 − P(Y ≤ 5)
✓ 1
A1 0.168
✓ 1
Part (f)
M1 (1 − f)20 > 0.5
✓ 1
A1 f < 0.0341, so greatest value 0.034
✓ 1
Exam tip: When a later part depends on an earlier probability, keep the unrounded value in your calculator to avoid a rounding error in the final answer. In conditional probability, identify the "given" event first: it forms the denominator.
313IBMI3S-4-01r16-14Sampling and descriptive statistics; T-tests, confidence intervals and Markov chains; Probability and the binomial distribution; Normal distribution and chi-squared testsPaper 2Hard12 marks
A fashion retailer surveys 250 customers about their preferred style (Casual, Smart or Sports) and records their age group. A χ2 test for independence is carried out at the 5% significance level. Age group | Casual | Smart | Sports Under 25 | 32 | 28 | 20 25 to 44 | 26 | 34 | 30 45 and over | 18 | 25 | 37
(a)
State the null hypothesis.
(1)
(b)
Show that the expected frequency for "Under 25" and "Casual" is 24.32.
(2)
(c)
Write down the number of degrees of freedom.
(1)
(d)
Find the χ2 statistic and the p-value for the test.
(3)
(e)
State the conclusion of the test at the 5% level, and at the 1% level. The critical values for 4 degrees of freedom are 9.488 (5%) and 13.28 (1%).
(2)
(f)
State why it is valid to use this test with these data.
(1)
(g)
The largest single contribution to χ2 comes from the "45 and over", "Sports" cell, for which the expected frequency is 27.84. Calculate this contribution to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: age group and preferred style are independent
✓ 1
Part (b)
M1 80 × 76 ÷ 250 (row total × column total ÷ grand total)
✓ 1
A1 24.32 (shown)
✓ 1
Part (c)
A1 (3 − 1)(3 − 1) = 4
✓ 1
Part (d)
M1 use of the χ2 test for independence on the calculator
✓ 1
A1 χ2 = 9.93
✓ 1
A1 p-value = 0.0416
✓ 1
Part (e)
A1 5%: 9.93 > 9.488 (or 0.0416 < 0.05), reject H0: evidence of an association
✓ 1
A1 1%: 9.93 < 13.28 (or 0.0416 > 0.01), do not reject H0
✓ 1
Part (f)
R1 every expected frequency is greater than 5 (the smallest is 24.32)
✓ 1
Part (g)
M1 (37 − 27.84)2 ÷ 27.84
✓ 1
A1 3.01
✓ 1
Common error: Do not choose the number of degrees of freedom as the number of cells minus 1. For a table with r rows and c columns it is (r − 1)(c − 1).
314IBMI3S-4-02r16-03Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regressionPaper 2Hard12 marks
A telecoms company measures the download speed, S Mbps, at test points across a town. S is normally distributed with mean 48 Mbps and standard deviation 9 Mbps. A test point is classed as good if S > 40.
(a)
Find the probability that a randomly chosen test point records a speed below 30 Mbps.
(2)
(b)
Find the probability that a randomly chosen test point is classed as good.
(2)
(c)
Five test points are chosen at random. Using the unrounded probability from (b), find the probability that at least 4 are classed as good.
(3)
(d)
Given that a test point is classed as good, find the probability that its speed exceeds 60 Mbps. Give your answer to 3 significant figures.
(3)
(e)
The company advertises a minimum speed v that is exceeded at 95% of test points. Find v to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(S < 30) with S ~ N(48, 92)
✓ 1
A1 0.0228
✓ 1
Part (b)
M1 P(S > 40)
✓ 1
A1 0.813
✓ 1
Part (c)
M1 X ~ B(5, 0.8129…) for the number of good points
M1 48 + (−1.645) × 9, or inverse normal with area 0.05
✓ 1
A1 v = 33.2 Mbps
✓ 1
Exam tip: In a conditional probability for a continuous variable, the numerator is the probability of both events together. Here the event "S > 60" lies entirely inside "S > 40", so the numerator is just P(S > 60).
315IBMI3S-4-02r16-08Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regressionPaper 2Hard12 marks
The times, t hours, taken by 80 climbers to complete a summit day are grouped as follows. Assume the times in each class are spread evenly across that class. Time (hours) | 6 ≤ t < 8 | 8 ≤ t < 10 | 10 ≤ t < 12 | 12 ≤ t < 14 | 14 ≤ t < 16 Frequency | 6 | 18 | 30 | 20 | 6
(a)
Write down the cumulative frequencies for the five classes.
(1)
(b)
Estimate the median time.
(2)
(c)
Estimate the interquartile range, using unrounded values of the quartiles.
(3)
(d)
The slowest 10% of climbers are given extra safety support. Estimate the time above which a climber is in this group.
(2)
(e)
The race organiser says that "more than half of the climbers took at least 11.5 hours". Use your results to comment on this statement.
(2)
(f)
Estimate the mean time using mid-interval values.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 6, 24, 54, 74, 80 (all correct)
✓ 1
Part (b)
M1 the 40th value lies in the class 10 ≤ t < 12; 10 + (40 − 24) ÷ 30 × 2
✓ 1
A1 11.1 hours
✓ 1
Part (c)
M1 Q1 from the 20th value: 8 + (20 − 6) ÷ 18 × 2 = 9.56
✓ 1
M1 Q3 from the 60th value: 12 + (60 − 54) ÷ 20 × 2 = 12.6
✓ 1
A1 IQR = 3.04 hours
✓ 1
Part (d)
M1 90th percentile is the 72nd value, in the class 12 ≤ t < 14: 12 + (72 − 54) ÷ 20 × 2
✓ 1
A1 13.8 hours
✓ 1
Part (e)
M1 compare 11.5 with the estimated median 11.1 (or the 40th value)
✓ 1
R1 11.5 is above the median, so fewer than half took at least 11.5 hours; the statement is not supported (ECF from (b))
Exam tip: To estimate a percentile from grouped data, find which class holds the required position, then add the fraction of the class width reached, assuming the values are evenly spread within the class.
316IBMI3S-4-02r18-08Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared testsPaper 2Hard10 marks
A robot cuts metal rods whose lengths, in centimetres, are normally distributed with mean μ and standard deviation σ. It is found that 10 % of rods are shorter than 74.0 cm and 5 % of rods are longer than 76.0 cm.
(a)
Find the values of μ and σ. Give each to 3 significant figures.
(5)
(b)
A rod is acceptable if its length is between 74.5 cm and 75.5 cm. Using unrounded values from part (a), find the probability that a randomly chosen rod is acceptable. Give your answer to 3 significant figures.
(2)
(c)
The robot is adjusted so that the mean length is exactly 75.0 cm, while σ changes to a new value s. After the adjustment, 2 % of rods are shorter than 74.0 cm. Find s, to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 inverse normal gives z = −1.2816 for P(Z < z) = 0.10
✓ 1
A1 z = 1.6449 for P(Z > z) = 0.05
✓ 1
M1 μ − 1.2816σ = 74.0 and μ + 1.6449σ = 76.0
✓ 1
A1 σ = 0.683 cm
✓ 1
A1 μ = 74.9 cm
✓ 1
Part (b)
M1 P(74.5 < X < 75.5)
✓ 1
A1 0.528
✓ 1
Part (c)
M1 (74.0 − 75.0) ÷ s = z where P(Z < z) = 0.02
✓ 1
A1 z = −2.0537
✓ 1
A1 s = 0.487 cm
✓ 1
Common error: Both conditions must be converted to z-values with the correct sign: a value below the mean has a negative z-value.
317IBMI3S-4-02r18-13Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared testsPaper 2Hard13 marks
A researcher measures the lengths, in centimetres, of random samples of fish of the same species from two lakes. Lake A (10 fish): 24.1, 26.3, 22.8, 25.0, 27.2, 23.5, 25.8, 24.6, 26.0, 23.9 Lake B (12 fish): 26.8, 28.1, 25.4, 27.5, 29.0, 26.2, 27.9, 25.9, 28.6, 27.1, 26.5, 28.3 The researcher believes that fish in lake B have a greater mean length than fish in lake A, and carries out a one-tailed t-test at the 5 % significance level.
(a)
Calculate the mean and the sample standard deviation (using n − 1) of the lengths for each lake. Give each answer to 3 significant figures.
(3)
(b)
State the null and alternative hypotheses.
(2)
(c)
State two assumptions required for a pooled t-test to be valid.
(2)
(d)
Use your calculator to find the p-value for the test.
(2)
(e)
State the conclusion of the test, giving a reason.
(2)
(f)
Use the unrounded sample standard deviations to calculate the pooled estimate of variance, sp2, to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 lake A mean = 24.9 cm and lake B mean = 27.3 cm
✓ 1
A1 lake A standard deviation = 1.39 cm
✓ 1
A1 lake B standard deviation = 1.14 cm
✓ 1
Part (b)
A1 H0: μA = μB (mean lengths are equal)
✓ 1
A1 H1: μB > μA
✓ 1
Part (c)
R1 the lengths in each lake are normally distributed
✓ 1
R1 the populations have equal variances (and the samples are random and independent)
✓ 1
Part (d)
M1 attempt to use a two-sample pooled t-test (one-tailed)
✓ 1
A1 p = 0.000149
✓ 1
Part (e)
R1 0.000149 < 0.05
✓ 1
A1 reject H0: there is sufficient evidence that fish in lake B have a greater mean length
✓ 1
Part (f)
M1 sp2 = (9 × sA2 + 11 × sB2) ÷ 20
✓ 1
A1 1.58 cm2
✓ 1
Common error: A one-tailed alternative hypothesis must point in the direction of the researcher's belief (here μB > μA), and the p-value is compared with 0.05 in that one tail.
318IBMI3S-4-02r18-14Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared testsPaper 2Hard12 marks
A robot manufacturer buys motors from two suppliers. Supplier X provides 60 % of the motors and supplier Y provides the other 40 %. A motor from supplier X is defective with probability 0.03, and a motor from supplier Y is defective with probability 0.08.
(a)
Show that the probability that a randomly chosen motor is defective is 0.05.
(2)
(b)
Given that a randomly chosen motor is defective, find the probability that it came from supplier Y.
(2)
(c)
A box contains 12 motors chosen at random. Assume each motor is independently defective with probability 0.05. Find the probability that at least one motor in the box is defective. Give your answer to 3 significant figures.
(2)
(d)
Given that a box contains at least one defective motor, find the probability that it contains exactly two defective motors. Give your answer to 3 significant figures.
(4)
(e)
A motor is chosen at random. Find the expected number of motors in a box of 12 that are both defective and from supplier Y.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(defective) = 0.6 × 0.03 + 0.4 × 0.08
✓ 1
A1 0.018 + 0.032 = 0.05 (AG)
✓ 1
Part (b)
M1 P(Y | defective) = (0.4 × 0.08) ÷ 0.05
✓ 1
A1 0.64
✓ 1
Part (c)
M1 1 − 0.9512
✓ 1
A1 0.460
✓ 1
Part (d)
M1 P(exactly 2) = 66 × 0.052 × 0.9510 = 0.0988
✓ 1
A1 conditional probability set up as P(exactly 2) ÷ P(at least 1)
✓ 1
M1 0.0988 ÷ 0.4596 (using unrounded values)
✓ 1
A1 0.215
✓ 1
Part (e)
M1 probability a motor is defective and from Y = 0.4 × 0.08 = 0.032
✓ 1
A1 12 × 0.032 = 0.384
✓ 1
Common error: P(defective | from Y) = 0.08 is not the same as P(from Y | defective); the second needs the total probability of a defective motor in the denominator.
319IBMI3S-4-02r20-13Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Hard12 marks
The efficiency, E %, of solar cells made by a factory is normally distributed with mean 21.4 and standard deviation 0.8. A cell is classed as Grade A if its efficiency is greater than 22.0%, and is rejected if its efficiency is less than 20.0%.
(a)
Find the probability that a randomly chosen cell is Grade A.
(2)
(b)
Find the probability that a randomly chosen cell is rejected.
(2)
(c)
A batch contains 25 randomly chosen cells. Using the unrounded probability from part (a), find the probability that at least 8 cells in the batch are Grade A.
(3)
(d)
Find the expected number of Grade A cells in a batch of 25.
(2)
(e)
A customer takes a random sample of n cells. Find the smallest value of n for which the probability that the sample contains at least one Grade A cell is greater than 0.99. Use the unrounded probability from part (a).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(E > 22.0) with E ~ N(21.4, 0.82)
✓ 1
A1 0.227 (0.2266...)
✓ 1
Part (b)
M1 P(E < 20.0)
✓ 1
A1 0.0401
✓ 1
Part (c)
M1 Y ~ B(25, 0.2266...)
✓ 1
M1 1 − P(Y ≤ 7)
✓ 1
A1 0.187
✓ 1
Part (d)
M1 25 × 0.2266...
✓ 1
A1 5.67
✓ 1
Part (e)
M1 1 − (1 − 0.2266...)n > 0.99
✓ 1
M1 (0.7733...)n < 0.01, so n > ln(0.01)/ln(0.7733...) = 17.9
✓ 1
A1 n = 18
✓ 1
Exam tip: When finding the least n from an inequality, remember that dividing by a negative logarithm reverses the inequality sign.
320IBMI3S-4-02r20-14Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Hard14 marks
A council surveys 300 residents about a proposed wind farm. Each resident is asked to choose Support, Neutral or Oppose.
Support
Neutral
Oppose
Under 40
62
28
30
40 and over
58
45
77
A χ2 test for independence at the 5% significance level is carried out to investigate whether opinion depends on age group.
(a)
State the null and alternative hypotheses.
(2)
(b)
Show that the expected frequency for residents under 40 who oppose is 42.8.
(2)
(c)
Write down the number of degrees of freedom.
(1)
(d)
Find the χ2 statistic and the p-value.
(2)
(e)
The critical value at the 5% level is 5.991. State the conclusion, giving a reason.
(2)
(f)
A resident who opposes the wind farm is chosen at random from those surveyed. Find the probability that this resident is aged 40 or over.
(2)
(g)
The cell "under 40, Support" has the largest contribution to the χ2 statistic. Show that this contribution is 4.08, to 3 significant figures, and state what it suggests.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: opinion on the wind farm and age group are independent
✓ 1
A1 H1: opinion and age group are not independent
✓ 1
Part (b)
M1 107 × 120/300
✓ 1
A1 42.8 (must be shown)
✓ 1
Part (c)
A1 2
✓ 1
Part (d)
A1 χ2 = 13.3 (13.27...)
✓ 1
A1 p = 0.00132
✓ 1
Part (e)
M1 comparison 13.27 > 5.991 (or p < 0.05)
✓ 1
R1 reject H0: there is evidence that opinion depends on age group
✓ 1
Part (f)
M1 77/(30 + 77)
✓ 1
A1 0.720 (0.7196...)
✓ 1
Part (g)
M1 expected frequency 120 × 120/300 = 48 and (62 − 48)2/48
✓ 1
A1 4.08 (4.083...)
✓ 1
R1 more residents under 40 support the wind farm than expected if opinion were independent of age
✓ 1
Exam tip: A χ² test shows that an association exists; the cell contributions show where it comes from.
321IBMI3S-4-01r22-03Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regressionPaper 2Hard14 marks
A shipping company wants to know whether a fuel additive reduces fuel consumption on its container vessels. Fuel consumption, in tonnes per 1000 km, was recorded for 8 voyages without the additive and for 9 voyages with the additive.
The voyages may be treated as independent random samples. A one-tailed two-sample t-test is carried out at the 5% significance level.
(a)
State the null and alternative hypotheses, defining any symbols used.
(2)
(b)
State one assumption that must be made about the populations for the pooled t-test to be valid.
(1)
(c)
Write down the mean and the sample standard deviation of the fuel consumption for each group, to 3 significant figures.
(4)
(d)
Carry out the test, stating the degrees of freedom, the value of the test statistic and the p-value.
(3)
(e)
State the conclusion of the test at the 5% significance level, giving a reason.
(2)
(f)
The company's board will accept the additive only if the result is significant at the 0.1% level. State the conclusion at this level, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: μwithout = μwith (the population mean fuel consumption is the same with and without the additive)
✓ 1
A1 H1: μwithout > μwith (the additive reduces mean fuel consumption)
✓ 1
Part (b)
A1 the populations are normally distributed, or the population variances are equal (accept either)
✓ 1
Part (c)
A1 mean without additive 41.5
✓ 1
A1 sample standard deviation without additive 1.83
✓ 1
A1 mean with additive 38.7
✓ 1
A1 sample standard deviation with additive 1.34
✓ 1
Part (d)
M1 use of the pooled two-sample t-test with a one-tailed alternative
✓ 1
A1 t = 3.61 with 15 degrees of freedom
✓ 1
A1 p-value = 0.00128
✓ 1
Part (e)
R1 0.00128 < 0.05
✓ 1
A1 reject H0: there is sufficient evidence that the additive reduces mean fuel consumption
✓ 1
Part (f)
R1 0.00128 > 0.001
✓ 1
A1 do not reject H0: there is insufficient evidence at the 0.1% level
✓ 1
Common error: Reading a small p-value as proof: the conclusion depends on the significance level chosen before the test, and should be worded as evidence for or against H0.
322IBMI3S-4-01r22-11Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regressionPaper 2Hard13 marks
A theatre has 112 seats. For each performance it sells 120 tickets, because some ticket holders do not attend. Each ticket holder independently attends with probability 0.92. Let Y be the number of ticket holders who attend a performance for which 120 tickets were sold.
(a)
State the distribution of Y and find E(Y).
(2)
(b)
Find the probability that more people attend than there are seats.
(3)
(c)
Find the probability that exactly 112 people attend.
(2)
(d)
Given that at least 109 people attend, find the probability that more than 112 people attend. Use unrounded values from your calculator and give your answer to 3 significant figures.
(3)
(e)
The theatre will sell N tickets for each performance, where each holder still attends with probability 0.92. Find the largest value of N for which the probability that more than 112 people attend is less than 0.05.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 Y ~ B(120, 0.92)
✓ 1
A1 E(Y) = 120 × 0.92 = 110.4
✓ 1
Part (b)
M1 recognises the required probability as P(Y > 112) = 1 − P(Y ≤ 112)
✓ 1
M1 use of the binomial cumulative distribution with n = 120, p = 0.92
✓ 1
A1 0.247
✓ 1
Part (c)
M1 C(120,112)(0.92)112(0.08)8
✓ 1
A1 0.124
✓ 1
Part (d)
M1 P(Y ≥ 113 | Y ≥ 109) = P(Y ≥ 113) / P(Y ≥ 109)
✓ 1
A1 P(Y ≥ 109) = 0.748
✓ 1
A1 0.331
✓ 1
Part (e)
M1 evaluates P(Y > 112) for Y ~ B(N, 0.92) for different N
✓ 1
A1 N = 117 gives 0.0381 (< 0.05)
✓ 1
A1 N = 118 gives 0.0825 (> 0.05), so the largest value is N = 117
✓ 1
Exam tip: 'More than 112' means 113 or more; write P(Y > 112) = 1 - P(Y ≤ 112) before using the calculator.
323IBMI3S-4-01r22-12Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regressionPaper 2Hard12 marks
The mass, M tonnes, of a loaded container passing through a port is normally distributed with mean μ and standard deviation σ. It is known that P(M < 18) = 0.10 and P(M > 26) = 0.20.
(a)
Find the values of z such that P(Z < z) = 0.10 and P(Z < z) = 0.80 for the standard normal variable Z. Give each to 3 significant figures.
(2)
(b)
Write down two equations connecting μ and σ.
(2)
(c)
Hence find σ and μ, each to 3 significant figures.
(3)
(d)
Using the unrounded values of μ and σ, find P(20 < M < 25).
(2)
(e)
Eight loaded containers are chosen at random and independently. Find the probability that at least two of them have a mass greater than 26 tonnes. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 −1.28
✓ 1
A1 0.842
✓ 1
Part (b)
M1 (18 − μ)/σ = −1.2816 (ECF from (a))
✓ 1
M1 (26 − μ)/σ = 0.8416 (ECF from (a))
✓ 1
Part (c)
M1 solves the simultaneous equations (e.g. subtract to get 8 = 2.1232σ)
✓ 1
A1 σ = 3.77
✓ 1
A1 μ = 22.8
✓ 1
Part (d)
M1 normal probability between 20 and 25
✓ 1
A1 0.491
✓ 1
Part (e)
M1 X ~ B(8, 0.2), since P(M > 26) = 0.20
✓ 1
M1 1 − P(X ≤ 1) = 1 − 0.88 − 8 × 0.2 × 0.87
✓ 1
A1 0.497
✓ 1
Exam tip: When both the mean and standard deviation are unknown, convert each probability to a z-value, write two equations, then solve them simultaneously.
324IBMI3S-4-02r25-03Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 2Hard12 marks
A music school records, for 280 students, the family of instrument studied and the time spent practising each week. Strings: under 3 hours 34; 3 to 6 hours 41; over 6 hours 25 Woodwind: under 3 hours 22; 3 to 6 hours 38; over 6 hours 40 Brass and percussion: under 3 hours 14; 3 to 6 hours 21; over 6 hours 45 A χ2 test for independence is carried out at the 5% significance level.
(a)
State the null and alternative hypotheses.
(2)
(b)
Show that, if H0 is true, the expected number of brass and percussion students who practise for 3 to 6 hours is 28.6, correct to 3 significant figures.
(2)
(c)
Write down the number of degrees of freedom.
(1)
(d)
Find the χ2 statistic, correct to 3 significant figures.
(2)
(e)
Find the p-value.
(1)
(f)
State the conclusion of the test, giving a reason.
(2)
(g)
The cell with the largest contribution to χ2 is brass and percussion, over 6 hours. Find this contribution and describe what it shows.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: instrument family and weekly practice time are independent
✓ 1
A1 H1: instrument family and weekly practice time are not independent (associated)
✓ 1
Part (b)
M1 row total 80, column total 100, grand total 280: 80 × 100 ÷ 280
✓ 1
A1 28.571… = 28.6
✓ 1
Part (c)
A1 (3 − 1)(3 − 1) = 4
✓ 1
Part (d)
M1 Σ(O − E)2 ÷ E (or use of the calculator test on the observed matrix)
✓ 1
A1 χ2 = 19.4
✓ 1
Part (e)
A1 p = 0.000655 (0.000655)
✓ 1
Part (f)
R1 p = 0.000655 < 0.05 (or 19.4 > critical value 9.488)
✓ 1
A1 reject H0: there is evidence of an association between instrument family and practice time
✓ 1
Part (g)
A1 (45 − 31.43)2 ÷ 31.43 = 5.86
✓ 1
R1 observed (45) is well above expected (31.4): brass and percussion students are more likely than independence predicts to practise for over 6 hours
✓ 1
Exam tip: Show the expected-frequency calculation (row total × column total ÷ grand total) for one cell; it earns the method mark even if the calculator output is mis-copied.
325IBMI3S-4-02r25-11Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 2Hard12 marks
An airline compares the fuel used per flight on two routes with the same aircraft type. For Route 1, 10 flights used a mean of 3.42 tonnes with sample standard deviation 0.21 tonnes. For Route 2, 12 flights used a mean of 3.61 tonnes with sample standard deviation 0.24 tonnes. Assume that fuel use on each route is normally distributed with the same variance, and use the 5% significance level.
(a)
The airline wants to test whether the mean fuel use differs between the routes. State the null and alternative hypotheses.
(2)
(b)
Write down the number of degrees of freedom for a pooled two-sample t-test.
(1)
(c)
Carry out the test and find the test statistic and the p-value.
(3)
(d)
State the conclusion of the two-tailed test, giving a reason.
(2)
(e)
A manager now claims that Route 2 uses more fuel than Route 1. State the new alternative hypothesis, and state the conclusion of the one-tailed test, giving a reason.
(3)
(f)
Explain why the conclusions in parts (d) and (e) differ even though the data are the same.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: μ1 = μ2
✓ 1
A1 H1: μ1 ≠ μ2
✓ 1
Part (b)
A1 10 + 12 − 2 = 20
✓ 1
Part (c)
M1 two-sample pooled t-test from the summary statistics (two-tailed)
✓ 1
A1 t = -1.95
✓ 1
A1 p = 0.065
✓ 1
Part (d)
R1 p = 0.065 > 0.05
✓ 1
A1 do not reject H0: insufficient evidence of a difference in mean fuel use
✓ 1
Part (e)
A1 H1: μ2 > μ1 (equivalently μ1 < μ2)
✓ 1
R1 one-tailed p = 0.032 < 0.05
✓ 1
A1 reject H0: sufficient evidence that Route 2 uses more fuel
✓ 1
Part (f)
R1 the one-tailed test puts the whole 5% in one tail (the p-value is halved), and the data were tested against a more specific claim
✓ 1
Common error: Do not choose the tail after seeing the data; the hypotheses must be set from the question being asked before the test is run.
326IBMI3S-4-02r31-07Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Hard14 marks
Calculator
The mass, M kg, of a certain species of tuna caught in a marine reserve is normally distributed with mean μ and standard deviation σ. It is known that 10% of the tuna have a mass less than 1.2 kg and 5% have a mass greater than 2.0 kg.
(a)
Find the standardised values z corresponding to these two probabilities. Give your answers to 4 decimal places.
(2)
(b)
Write down two equations connecting μ and σ.
(2)
(c)
Hence find μ and σ, each to 3 significant figures.
(3)
(d)
Using the unrounded values of μ and σ, find the probability that a randomly chosen tuna has a mass between 1.4 kg and 1.8 kg.
(2)
(e)
Six tuna are chosen at random. Using the unrounded values of μ and σ, find the probability that exactly 2 of them have a mass greater than 1.85 kg.
(3)
(f)
A fishing boat lands 500 tuna. Estimate how many have a mass greater than 1.85 kg. Give your answer to the nearest whole number.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 P(Z < z) = 0.1 gives z = −1.2816
✓ 1
A1 P(Z > z) = 0.05 gives z = 1.6449
✓ 1
Part (b)
M1 (1.2 − μ)/σ = −1.2816
✓ 1
A1 (2.0 − μ)/σ = 1.6449 (ECF from (a))
✓ 1
Part (c)
M1 solving the equations simultaneously
✓ 1
A1 σ = 0.273
✓ 1
A1 μ = 1.55
✓ 1
Part (d)
M1 attempt to find P(1.4 < M < 1.8)
✓ 1
A1 0.528
✓ 1
Part (e)
M1 finding p = P(M > 1.85) = 0.1365
✓ 1
M1 binomial with n = 6 and this p: P(X = 2)
✓ 1
A1 0.155
✓ 1
Part (f)
M1 500 × P(M > 1.85) (ECF from (e))
✓ 1
A1 68 tuna
✓ 1
Common error: When using an inverse normal result in later parts, keep unrounded values on your calculator; using rounded values of μ and σ can change the third significant figure.
327IBMI3S-4-02r31-11Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Hard12 marks
Calculator
A logistics manager claims that parcels arrive at a depot equally often on each weekday. Over one week, 820 parcels arrived: Monday 168, Tuesday 152, Wednesday 141, Thursday 175, Friday 184. A χ2 goodness-of-fit test is carried out at the 5% significance level.
(a)
State the null hypothesis and the alternative hypothesis.
(2)
(b)
Find the expected frequency for each day, and state why the test conditions are satisfied.
(2)
(c)
Write down the number of degrees of freedom.
(1)
(d)
Find the value of the χ2 test statistic.
(2)
(e)
Find the p-value.
(1)
(f)
State the conclusion of the test, with a reason.
(2)
(g)
Find the contribution of Friday to the χ2 statistic. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: parcels arrive equally often on each weekday (the data fit a uniform distribution)
✓ 1
A1 H1: parcels do not arrive equally often on each weekday
✓ 1
Part (b)
A1 820 ÷ 5 = 164
✓ 1
R1 all expected frequencies are at least 5
✓ 1
Part (c)
A1 5 − 1 = 4
✓ 1
Part (d)
M1 sum of (O − E)2/E over the five days
✓ 1
A1 7.38 (1210/164)
✓ 1
Part (e)
A1 0.117
✓ 1
Part (f)
R1 0.117 > 0.05
✓ 1
A1 do not reject H0: there is insufficient evidence that parcels are not equally spread across the weekdays
✓ 1
Part (g)
M1 (184 − 164)2/164
✓ 1
A1 2.44
✓ 1
Common error: Not rejecting H₀ does not prove the claim is true; say there is insufficient evidence against it.
328IBMI3S-4-02r31-12Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Hard11 marks
Calculator
A marine biologist compares the weekly growth, in cm, of seaweed grown in standard seawater (sample A) and in nutrient-enriched seawater (sample B). Eight plants were used in each sample, with different plants in each. Sample A: 4.1, 3.8, 4.6, 4.0, 4.4, 3.9, 4.3, 4.1 Sample B: 4.9, 4.5, 4.2, 5.1, 4.7, 4.4, 5.0, 4.7 The biologist carries out a one-tailed two-sample t-test at the 1% significance level to see whether enrichment increases mean growth. Assume that both populations are normally distributed with equal variances.
(a)
Find the mean and sample standard deviation of each sample.
(2)
(b)
State the null hypothesis and the alternative hypothesis.
(2)
(c)
Explain why a paired t-test is not appropriate here.
(1)
(d)
Use your calculator to find the p-value of the test.
(2)
(e)
Find the value of the test statistic t, to 3 significant figures.
(2)
(f)
State the conclusion of the test, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 means: A = 4.15 cm, B = 4.69 cm
✓ 1
A1 sample standard deviations: A = 0.267 cm, B = 0.309 cm
✓ 1
Part (b)
A1 H0: μA = μB (the mean growth is the same)
✓ 1
A1 H1: μB > μA
✓ 1
Part (c)
R1 the plants in the two samples are different and independent, so the data are not paired
✓ 1
Part (d)
M1 using a pooled two-sample t-test with a one-tailed alternative
✓ 1
A1 0.00114
✓ 1
Part (e)
M1 pooled estimate of variance used with the means in (a)
✓ 1
A1 t = 3.72 with 14 degrees of freedom
✓ 1
Part (f)
R1 0.00114 < 0.01
✓ 1
A1 reject H0: there is sufficient evidence that enriched seawater increases mean growth
✓ 1
Exam tip: Check whether the alternative hypothesis is one-tailed or two-tailed before reading the p-value; a two-tailed p-value must be halved for a one-tailed test.
329IBMI3S-4-04r33-11Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Hard13 marks
A salmon farm sells fish of 'market grade', which must have mass greater than 3.5 kg. The mass, W kg, of a salmon is normally distributed with mean 4.2 kg and standard deviation 0.6 kg.
(a)
Find the probability that a randomly chosen salmon is of market grade. Give your answer to 3 significant figures.
(2)
(b)
A random sample of 20 salmon is taken. Find the probability that at least 18 are of market grade. Use the unrounded value of the probability from (a).
(3)
(c)
The farm inspects n randomly chosen salmon. Find the least value of n such that the probability that at least one of them is not of market grade is greater than 0.99.
(4)
(d)
A salmon is chosen at random from those of market grade. Find the probability that its mass is greater than 5.0 kg. Give your answer to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(W\sim N(4.2,0.6^2)\) and \(P(W>3.5)\)
✓ 1
A1 0.878
✓ 1
Part (b)
M1 recognise a binomial distribution \(X\sim B(20,p)\) with \(p=0.878\ldots\)
✓ 1
M1 use of \(P(X\ge18)=1-P(X\le17)\)
✓ 1
A1 0.554
✓ 1
Part (c)
M1 \(P(\text{at least one not market grade})=1-p^{n}\)
✓ 1
M1 set up \(1-p^{n}>0.99\) (so \(p^{n}<0.01\))
✓ 1
M1 solve using logarithms, a table or a solver (\(n>35.5\))
✓ 1
A1 \(n=36\) (since \(p^{35}=0.01067>0.01\) and \(p^{36}=0.00937<0.01\))
✓ 1
Part (d)
M1 recognise conditional probability \(P(W>5.0\mid W>3.5)\)
✓ 1
A1 \(P(W>5.0)=0.0912\)
✓ 1
M1 use of \(\frac{P(W>5.0)}{P(W>3.5)}\) (since \(W>5.0\) implies \(W>3.5\))
✓ 1
A1 0.104
✓ 1
Common error: In 'at least one' questions, work with the complement: the probability that none has the property is \(p^n\), so the required probability is \(1-p^n\).
330IBMI3S-4-04r33-12Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Hard13 marks
A fashion boutique believes that it sells the same number of a particular scarf on each day, Monday to Friday. In one week it sells 300 of these scarves, with the numbers sold on Monday to Friday being 52, 61, 58, 55 and 74. The manager carries out a χ2 goodness-of-fit test at the 5% significance level.
(a)
State the null and alternative hypotheses.
(2)
(b)
Write down the expected number of scarves sold on each day if H0 is true.
(1)
(c)
Write down the number of degrees of freedom.
(1)
(d)
Find the χ2 statistic and the p-value for this week.
(2)
(e)
State the conclusion for this week, giving a reason.
(1)
(f)
After a Friday promotion, a second week has 300 scarf sales, with 40, 44, 41, 50 and 125 sold on Monday to Friday. Find the χ2 statistic and the p-value for the second week. Give the p-value in standard form to 3 significant figures.
(2)
(g)
State the conclusion for the second week, in context.
(2)
(h)
Find the contribution of Friday's sales to the χ2 statistic for the second week. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: the scarves are sold equally often on each of the five days (uniform distribution)
✓ 1
A1 H1: the scarves are not sold equally often on each day
✓ 1
Part (b)
A1 \(300\div5=60\)
✓ 1
Part (c)
A1 \(5-1=4\)
✓ 1
Part (d)
A1 χ2 = 4.83
✓ 1
A1 p = 0.305
✓ 1
Part (e)
R1 p > 0.05, so do not reject H0: there is insufficient evidence that sales differ between the days
✓ 1
Part (f)
A1 χ2 = 89.0
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A1 p = 2.11 × 10−18 (to 3 significant figures; accept a value that is extremely small, below 10−17)
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Part (g)
M1 p < 0.05 (comparison with the significance level)
✓ 1
R1 reject H0: there is strong evidence that the sales are not spread equally over the days
✓ 1
Part (h)
M1 use of \(\frac{(O-E)^2}{E}\) with \(O=125\) and \(E=60\)
✓ 1
A1 70.4
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Exam tip: Each expected frequency should be at least 5 for a χ² test to be valid; here every expected value is 60.
331IBMI3S-4-02r34-10Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Probability and the binomial distributionPaper 2Hard13 marks
A biscuit factory has three ovens. Oven A bakes 50 % of the biscuits, oven B bakes 30 % and oven C bakes 20 %. A biscuit is defective (over-baked) with probability 0.02 if it came from A, 0.03 if it came from B and 0.05 if it came from C.
(a)
A biscuit is chosen at random. Find the probability that it is defective.
(3)
(b)
A defective biscuit is found. Find the probability that it came from oven A, to 3 significant figures.
(3)
(c)
An inspector takes a random sample of 12 biscuits from a very large batch. Assume that each biscuit is independently defective with probability 0.029. Find the probability that at least 2 biscuits in the sample are defective, to 3 significant figures.
(3)
(d)
Find the smallest sample size n for which the probability of finding at least one defective biscuit is greater than 0.95.
(3)
(e)
Give one reason why the independence assumption in (c) might not hold if the 12 biscuits are taken from the same tray.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(0.5\times0.02\), \(0.3\times0.03\) and \(0.2\times0.05\) (probabilities 0.01, 0.009, 0.01)
M1 solve using a table, graph or solver on the GDC: \(n>101.8\)
✓ 1
A1 \(n=102\) (n = 101 gives 0.949, which is not above 0.95)
✓ 1
Part (e)
R1 biscuits on one tray were baked together, so if the oven setting or tray position caused over-baking then the defects are linked (the chance of a defect depends on the other biscuits)
✓ 1
Exam tip: 'At least one' problems usually use the complement: \(P(\text{at least one})=1-P(\text{none})\).
332IBMI3S-4-02r34-12Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Probability and the binomial distributionPaper 2Hard9 marks
At a high-altitude camp, the blood oxygen saturation (%) of 25 climbers is measured. The readings x have \(\sum x=2280\) and \(\sum x^2=208\,300\).
(a)
Calculate the mean and the population standard deviation of the 25 readings. Give the standard deviation to 3 significant figures.
(3)
(b)
One reading of 78 was recorded by a faulty sensor and is removed. Find the mean of the remaining 24 readings.
(2)
(c)
Find the population standard deviation of the remaining 24 readings, to 3 significant figures.
(3)
(d)
State why the standard deviation decreases when the reading of 78 is removed.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 mean \(=2280\div25=91.2\)
✓ 1
M1 \(\sqrt{\dfrac{208\,300}{25}-91.2^2}\)
✓ 1
A1 3.82 (3.8157…)
✓ 1
Part (b)
M1 \(\dfrac{2280-78}{24}\)
✓ 1
A1 91.75
✓ 1
Part (c)
M1 new \(\sum x^2=208\,300-78^2=202\,216\)
✓ 1
M1 \(\sqrt{\dfrac{202\,216}{24}-91.75^2}\)
✓ 1
A1 2.76 (2.7576…)
✓ 1
Part (d)
R1 78 is far from the mean (13.2 below 91.2), so it contributes a large squared deviation; removing it reduces the spread
✓ 1
Exam tip: To update a standard deviation after changing a value, work with \(\sum x\) and \(\sum x^2\) rather than trying to adjust the old standard deviation.
333IBMI3S-4-04r37-10Sampling and descriptive statistics; T-tests, confidence intervals and Markov chains; Probability and the binomial distribution; Normal distribution and chi-squared tests; Correlation and regression; Normal distribution and chi-squared tests; Probability and the binomial distributionPaper 2Hard14 marks
The length, L metres, of a roll of fabric produced by a textile machine is normally distributed with mean μ and standard deviation σ. It is known that 10% of rolls are shorter than 49.2 m and 15% of rolls are longer than 52.1 m.
(a)
Find the z-values for which \(P(Z
(2)
(b)
Write down two equations in μ and σ and hence find the values of μ and σ. Give each to 3 significant figures.
(4)
(c)
Find the probability that a roll is between 50 m and 53 m long. Use the unrounded values of μ and σ. Give your answer to 3 significant figures.
(2)
(d)
Given that a roll is longer than 50 m, find the probability that it is longer than 52 m. Use the unrounded values of μ and σ. Give your answer to 3 significant figures.
(3)
(e)
Five rolls are chosen at random. Find the probability that exactly two of them are longer than 52.1 m. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(z=-1.28\)
✓ 1
A1 \(z=1.04\)
✓ 1
Part (b)
M1 \(49.2=\mu-1.28\sigma\)
✓ 1
M1 \(52.1=\mu+1.04\sigma\)
✓ 1
A1 \(\sigma=1.25\)
✓ 1
A1 \(\mu=50.8\)
✓ 1
Part (c)
M1 \(P(50
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A1 0.700
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Part (d)
M1 \(P(L>52\mid L>50)=\frac{P(L>52)}{P(L>50)}\)
✓ 1
A1 \(\frac{0.1694}{0.7396}\)
✓ 1
A1 0.229
✓ 1
Part (e)
M1 \(X\sim B(5,0.15)\)
✓ 1
M1 \(P(X=2)=\binom{5}{2}(0.15)^2(0.85)^3\)
✓ 1
A1 0.138
✓ 1
Exam tip: When both the mean and the standard deviation are unknown, turn each probability into a z-value and solve the two linear equations simultaneously.
334IBMI3S-4-04r37-14Sampling and descriptive statistics; T-tests, confidence intervals and Markov chains; Probability and the binomial distribution; Normal distribution and chi-squared tests; Correlation and regression; Normal distribution and chi-squared tests; Probability and the binomial distributionPaper 2Hard12 marks
A fashion store records its monthly advertising spend, x (in thousands of euros), and the monthly sales, y (in thousands of euros), for nine months. Spend x: 1.5, 2, 3, 3.5, 4.5, 5, 6, 7, 8 Sales y: 34, 41, 47, 58, 60, 72, 69, 85, 88
(a)
Use your GDC to find the Pearson product-moment correlation coefficient, r. Give your answer to 3 significant figures.
(1)
(b)
Describe the correlation between advertising spend and sales.
(1)
(c)
Find the equation of the regression line of y on x, giving the coefficients to 3 significant figures.
(2)
(d)
Interpret the gradient and the y-intercept of the regression line in this context.
(2)
(e)
Calculate Spearman's rank correlation coefficient. Give your answer to 3 significant figures.
(2)
(f)
Use your regression line to estimate the sales in a month when the advertising spend is €5500. Give your answer in thousands of euros to 3 significant figures.
(2)
(g)
In the month in which the spend was €3500 the sales were €58 000. Find the residual for this month. Use the regression line from part (c). Give your answer to 2 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 0.979
✓ 1
Part (b)
A1 strong positive (linear) correlation
✓ 1
Part (c)
M1 uses a GDC to find the gradient and the intercept
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A1 \(y=8.28x+24.3\)
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Part (d)
R1 each extra €1000 of advertising is associated with about €8280 more sales
✓ 1
R1 with no advertising the model predicts monthly sales of about €24 300
✓ 1
Part (e)
M1 ranks both variables (the ranks differ only for the 69 and 72 values) and uses a GDC or the formula
✓ 1
A1 0.983
✓ 1
Part (f)
M1 \(8.28\times5.5+24.3\)
✓ 1
A1 69.8
✓ 1
Part (g)
M1 predicted sales \(8.28\times3.5+24.3=53.28\)
✓ 1
A1 residual \(=58-53.28=4.7\)
✓ 1
Exam tip: A residual is observed value minus predicted value. A positive residual means the point lies above the regression line.
335IBMI3S-4-02r38-09Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 2Hard12 marks
Calculator
A cyclist's lap times on a track are normally distributed with mean 74 seconds and standard deviation 3.2 seconds, independently from lap to lap. In parts (a) to (d) use the unrounded answer to part (a) in later calculations.
(a)
Find the probability that a lap takes less than 70 seconds. Give your answer to 3 significant figures.
(2)
(b)
In a session of 12 laps, find the probability that exactly 3 laps take less than 70 seconds. Give your answer to 3 significant figures.
(2)
(c)
Find the probability that at least 2 of the 12 laps take less than 70 seconds. Give your answer to 3 significant figures.
(2)
(d)
Find the smallest number of laps, n, for which the probability that at least one lap takes less than 70 seconds is greater than 0.9.
(3)
(e)
The cyclist's slowest 5% of laps are slower than t seconds. Find t, correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(T < 70) with normal cdf, μ = 74, σ = 3.2
✓ 1
A1 0.106
✓ 1
Part (b)
M1 binomial model B(12, 0.10565…)
✓ 1
A1 0.0950
✓ 1
Part (c)
M1 1 − P(X ≤ 1)
✓ 1
A1 0.367
✓ 1
Part (d)
M1 1 − (1 − 0.10565…)n > 0.9
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M1 correct use of logarithms or GDC table (n = 20 gives 0.893, n = 21 gives 0.904)
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A1 n = 21
✓ 1
Part (e)
M1 P(T < t) = 0.95, or equivalent with the upper tail
✓ 1
A1 z = 1.6449 (inverse normal, accept 1.645)
✓ 1
A1 t = 79.3 seconds
✓ 1
Exam tip: 'At least one' problems are almost always easiest as 1 − P(none), which gives an inequality in n that can be solved with logarithms.
336IBMI3S-4-02r38-10Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 2Hard10 marks
Calculator
Archaeologists measure the rim diameters, in millimetres, of clay pots from two settlements. Settlement A (8 pots): 212, 225, 208, 231, 219, 226, 214, 222 Settlement B (9 pots): 203, 216, 209, 198, 211, 205, 214, 207, 200 They carry out a two-sample t-test, at the 5% significance level, to decide whether the mean rim diameter differs between the settlements. Assume that both populations are normally distributed with equal variances, so that a pooled test may be used.
(a)
State the null hypothesis and the alternative hypothesis.
(2)
(b)
Write down the number of degrees of freedom for this test.
(1)
(c)
Use your GDC to find the t-statistic and the p-value.
(3)
(d)
State the conclusion of the test, giving a reason.
(2)
(e)
State why a pooled test is only appropriate if the two sample variances are similar, and write down the two sample means.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: μA = μB (the population mean rim diameters are equal)
✓ 1
A1 H1: μA ≠ μB
✓ 1
Part (b)
A1 8 + 9 − 2 = 15
✓ 1
Part (c)
M1 use of the pooled two-sample t-test on the GDC
✓ 1
A1 p-value = 0.00203
✓ 1
A1 t = 3.72
✓ 1
Part (d)
R1 0.00203 < 0.05
✓ 1
A1 reject H0: there is evidence that the mean rim diameters of pots from the two settlements differ
✓ 1
Part (e)
A1 the pooled test assumes a common population variance, so the samples should have similar spread (sample standard deviations 7.84 and 6.12)
✓ 1
A1 mean A = 219.625 (accept 220) and mean B = 207
✓ 1
Exam tip: State the conclusion in the context of the question (rim diameters of pots), not just 'reject H₀'.
337IBMI3S-4-02r38-14Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 2Hard10 marks
Calculator
The time, T minutes, that a cyclist takes to commute to work is normally distributed with mean μ and standard deviation σ. It is known that P(T < 40) = 0.2 and P(T > 70) = 0.1.
(a)
Write down two equations for μ and σ, using standardisation. Use the values of z from the inverse normal distribution to 4 decimal places.
(2)
(b)
Solve your equations to find σ and μ. Give both answers to 3 significant figures.
(3)
(c)
Find the probability that a commute takes between 45 and 60 minutes. Use the unrounded values of μ and σ. Give your answer to 3 significant figures.
(2)
(d)
The cyclist plans to leave early enough that she is late on only 1% of days. Find the commute time that is exceeded on 1% of days. Use the unrounded values of μ and σ. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 (40 − μ) ÷ σ = −0.8416
✓ 1
A1 (70 − μ) ÷ σ = 1.2816
✓ 1
Part (b)
M1 attempt to eliminate one unknown, e.g. subtracting: 30 = 2.1232σ
✓ 1
A1 σ = 14.1
✓ 1
A1 μ = 51.9
✓ 1
Part (c)
M1 P(45 < T < 60) using normal cdf with μ = 51.89…, σ = 14.13…
✓ 1
A1 0.404
✓ 1
Part (d)
M1 P(T > t) = 0.01 so z = 2.3263
✓ 1
A1 μ + 2.3263σ = 51.89… + 2.3263 × 14.13…
✓ 1
A1 84.8 minutes
✓ 1
Common error: Rounding μ and σ before using them in later parts, which can change the third significant figure of the final answers.
338IBMI3S-4-02r44-11Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Hard12 marks
A beekeeper has 25 colonies. Each colony independently survives the winter with probability 0.88. Let X be the number of colonies that survive.
(a)
Write down the expected value of X.
(1)
(b)
Find \(P(X=22)\).
(2)
(c)
Find \(P(X\ge 24)\).
(2)
(d)
Find the probability that at least 24 colonies survive, given that at least 22 survive.
(3)
(e)
At another apiary with 10 colonies, the probability that all colonies survive is 0.6. Find the survival probability p of each colony there, to 3 significant figures.
(2)
(f)
For the apiary in part (e), using the unrounded value of p, find the probability that exactly 9 colonies survive.
Exam tip: 'Given that' signals conditional probability: the denominator is the probability of the condition, and when one event contains the other the numerator is the smaller event's probability.
339IBMI3S-4-02r44-13Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Hard13 marks
In a Formula 1 season, a team's pit-stop time T seconds is normally distributed with mean 2.6 and standard deviation 0.35. A pit stop is classed as slow if it takes more than 3.0 seconds. Over a season the team makes 22 pit stops, and the times are independent.
(a)
Show that the probability that a pit stop is slow is 0.127, to 3 significant figures.
(2)
(b)
Let X be the number of slow pit stops in the season. State the distribution of X, using the unrounded probability \(q=0.1265\) from part (a).
(1)
(c)
Find the probability that the team has no slow pit stops in the season.
(2)
(d)
Find the probability that the team has at least 3 slow pit stops.
(2)
(e)
Find the expected number and the standard deviation of the number of slow pit stops.
(2)
(f)
In a different race, the team wants the probability of at least one slow stop to exceed 0.75. Find the smallest number of stops, n, for which this is true.
(3)
(g)
Find the probability of at least one slow pit stop in n = 11 stops.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(z=\frac{3.0-2.6}{0.35}=1.14\) and use of the normal distribution for \(P(T>3.0)\)
✓ 1
A1 \(0.1265\ldots=0.127\)
✓ 1
Part (b)
A1 \(X\sim B(22,\,0.1265)\)
✓ 1
Part (c)
M1 \((1-0.1265)^{22}\)
✓ 1
A1 0.0510
✓ 1
Part (d)
M1 \(1-P(X\le 2)\)
✓ 1
A1 0.539
✓ 1
Part (e)
A1 mean \(22\times 0.1265=2.78\)
✓ 1
A1 standard deviation \(\sqrt{22\times 0.1265\times 0.8735}=1.56\)
✓ 1
Part (f)
M1 \(1-(1-0.1265)^n>0.75\)
✓ 1
A1 \(n>10.25\)
✓ 1
A1 \(n=11\)
✓ 1
Part (g)
A1 \(1-(0.8735)^{11}=0.774\)
✓ 1
Exam tip: 'At least one' is best found as 1 minus the probability of none, which also lets you form an inequality in n directly.
340IBMI3S-4-02r47-12Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Hard13 marks
A bottling plant fills bottles with a nominal 500 mL of water. The volume in a bottle, V mL, is normally distributed with mean 500 and standard deviation 6.5. A bottle is rejected if it contains less than 490 mL, and the plant also records any bottle with more than 515 mL as overfilled.
(a)
Find the probability that a bottle is rejected.
(2)
(b)
Find the probability that a bottle is overfilled.
(2)
(c)
A random sample of 30 bottles is taken. Find the probability that at least 2 bottles are rejected. Use the unrounded probability from part (a).
(3)
(d)
The plant manager changes the mean volume to μ′, keeping σ = 6.5, so that only 1% of bottles are rejected. Find μ′ correct to 1 decimal place.
(3)
(e)
With 1% of bottles rejected, find the probability that none of the 30 bottles in a random sample is rejected. Give your answer to 3 significant figures.
(2)
(f)
Suggest one disadvantage to the plant of raising the mean volume rather than reducing the standard deviation.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(V < 490) with μ = 500, σ = 6.5
✓ 1
A1 0.0620
✓ 1
Part (b)
M1 P(V > 515)
✓ 1
A1 0.0105
✓ 1
Part (c)
M1 identifies the binomial distribution B(30, 0.0620…)
✓ 1
M1 uses P(X ≥ 2) = 1 − P(X ≤ 1)
✓ 1
A1 0.562
✓ 1
Part (d)
M1 sets P(V < 490) = 0.01 and uses z = −2.3263…
✓ 1
M1 (490 − μ′)/6.5 = −2.3263…
✓ 1
A1 μ′ = 505.1
✓ 1
Part (e)
M1 0.9930
✓ 1
A1 0.740
✓ 1
Part (f)
R1 any valid reason, for example more water (cost) is used in every bottle, or more bottles are overfilled
✓ 1
Exam tip: In multi-step questions state which probability you carry forward and use the unrounded value from your calculator, so that later answers are not affected by rounding.
341IBMI3S-4-02r47-13Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Hard14 marks
In a pharmacology study the dose of a drug, x mg, and the percentage of patients showing a response, y, are recorded. x: 5, 10, 20, 40, 80, 160 y: 12, 31, 52, 66, 72, 74
(a)
Find the Pearson product-moment correlation coefficient, r.
(2)
(b)
Find Spearman's rank correlation coefficient, rs, showing the ranks used.
(3)
(c)
Explain why rs is larger than r for these data.
(2)
(d)
Find the equation of the regression line of y on x. Give the coefficients to 3 significant figures.
(2)
(e)
Use the unrounded regression line to estimate the percentage responding at a dose of 100 mg.
(2)
(f)
Use the regression line to estimate the percentage responding at a dose of 300 mg, and explain why this is not a sensible estimate.
R1 the relationship is monotonic (always increasing) but not linear
✓ 1
R1 Spearman's coefficient measures monotonic association, whereas r measures linear association, and the response levels off
✓ 1
Part (d)
M1 attempt at regression line
✓ 1
A1 y = 0.315x + 34.7
✓ 1
Part (e)
M1 substitutes x = 100
✓ 1
A1 66.1
✓ 1
Part (f)
A1 129 (using 0.315 × 300 + 34.7 = 129.2 gives the same value to 3 significant figures)
✓ 1
R1 a percentage cannot exceed 100
✓ 1
R1 300 mg is outside the range of the data, and a straight line does not fit data that level off
✓ 1
Common error: Spearman's coefficient of 1 means perfectly monotonic, not perfectly linear. Do not use a linear regression line when the scatter shows a curve.
342IBMI3S-4-02r47-14Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Hard11 marks
A water company has a historical record that burst pipes are caused by corrosion (40%), ground movement (25%), joint failure (15%), third-party damage (10%), freezing (7%) or manufacturing defect (3%). In a recent year 150 bursts are recorded, with observed frequencies 52, 41, 19, 18, 13 and 7 respectively. The company carries out a χ2 goodness-of-fit test at the 5% significance level to see whether the historical distribution still applies.
(a)
State the null and alternative hypotheses.
(2)
(b)
Find the expected frequencies for all six causes.
(2)
(c)
Explain why the last two categories must be combined before the test, and write down the new observed and expected frequencies for the combined category.
(2)
(d)
Find the number of degrees of freedom, the χ2 statistic and the p-value for the test with the combined category.
(3)
(e)
State the conclusion of the test in context, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 H0: the historical distribution still applies
✓ 1
A1 H1: the historical distribution no longer applies
✓ 1
Part (b)
M1 150 × each proportion
✓ 1
A1 60, 37.5, 22.5, 15, 10.5, 4.5
✓ 1
Part (c)
R1 one expected frequency (4.5) is less than 5, which is not allowed in the test
A1 do not reject H0: no evidence that the distribution of causes has changed
✓ 1
Exam tip: Each expected frequency must be at least 5; combine neighbouring categories if not, and use the new number of categories to find the degrees of freedom.
343IBMI3S-4-04r47-06Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Hard13 marks
A robotics team compares the battery life of two designs of robot. Battery life, in hours, was measured on a random sample of each design.
Design A (8 robots)
6.8, 7.2, 7.5, 6.9, 7.8, 7.1, 7.4, 7.3
Design B (9 robots)
7.6, 8.1, 7.9, 8.3, 7.7, 8.0, 7.5, 8.2, 7.8
The team wishes to test, at the 5% significance level, whether Design B has a longer mean battery life than Design A. Battery lives for both designs may be assumed to be normally distributed with equal variances.
(a)
Find the mean and the sample standard deviation for each design, correct to 3 significant figures.
(4)
(b)
State the null and alternative hypotheses.
(2)
(c)
State why a pooled two-sample t-test is suitable here.
(1)
(d)
Carry out the test: find the degrees of freedom, the t-value (3 s.f.) and the p-value (3 s.f.).
(3)
(e)
State your conclusion in context.
(2)
(f)
The p-value for a two-tailed test would be double the value in (d). State whether the conclusion would change at the 5% level, with a reason.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 Design A mean 7.25
✓ 1
A1 Design A standard deviation 0.325
✓ 1
A1 Design B mean 7.90
✓ 1
A1 Design B standard deviation 0.274
✓ 1
Part (b)
A1 H0: μB = μA
✓ 1
A1 H1: μB > μA
✓ 1
Part (c)
R1 the samples are independent, both populations are normal, and equal variances are assumed
✓ 1
Part (d)
A1 degrees of freedom = 8 + 9 − 2 = 15
✓ 1
A1 t = 4.48
✓ 1
A1 p = 0.000222 (one-tailed)
✓ 1
Part (e)
R1 p = 0.000222 < 0.05
✓ 1
A1 reject H0: there is evidence that Design B has a longer mean battery life than Design A
✓ 1
Part (f)
R1 no, 2 × 0.000222 = 0.000444, still less than 0.05
✓ 1
Common error: The alternative hypothesis must match the question: "longer" gives a one-tailed test, so use the one-tailed p-value.
344IBMI3S-4-04r47-11Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Hard12 marks
A fisheries scientist measures the length, x cm, and the mass, y g, of eight sea bass.
x (cm)
22
25
28
30
33
36
39
42
y (g)
118
160
215
250
320
395
470
560
(a)
Find the Pearson correlation coefficient, r, correct to 3 significant figures.
(1)
(b)
Find the equation of the regression line of y on x, with coefficients to 3 significant figures.
(2)
(c)
Use the equation from (b), with the coefficients as rounded there, to estimate the mass of a sea bass of length 31 cm. Give your answer to the nearest gram.
(2)
(d)
The scientist uses the same equation to estimate the mass of a sea bass of length 50 cm. Comment on the reliability of this estimate compared with the estimate in (c).
(2)
(e)
Interpret the gradient of the regression line in context.
(1)
(f)
Using the equation from (b), find the residual (observed mass − predicted mass) for the sea bass of length 28 cm.
(2)
(g)
The scientist wants to estimate the length of a sea bass of mass 300 g and suggests rearranging the equation from (b). Explain why this is not appropriate.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 r = 0.994
✓ 1
Part (b)
A1 gradient 22.2
✓ 1
A1 y = 22.2x − 397
✓ 1
Part (c)
M1 22.2 × 31 − 397
✓ 1
A1 291 g
✓ 1
Part (d)
R1 50 cm is outside the range of the data (extrapolation)
✓ 1
R1 so the estimate is less reliable than the interpolated estimate for 31 cm
✓ 1
Part (e)
R1 on average each extra centimetre of length is associated with about 22.2 g more mass
✓ 1
Part (f)
M1 predicted mass 22.2 × 28 − 397 = 224.6
✓ 1
A1 residual 215 − 224.6 = −9.6 g
✓ 1
Part (g)
R1 the line of y on x minimises the vertical (mass) distances and is for predicting y from x
✓ 1
R1 to estimate length from mass the regression line of x on y should be used
✓ 1
Common error: A regression line of y on x may only be used to predict y; do not rearrange it to predict x.
345IBMI3S-4-04r47-12Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Probability and the binomial distribution; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regression; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; correlation and regression; probability and the binomial distribution; normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chainsPaper 2Hard13 marks
A mill produces fabric whose thickness, T mm, is normally distributed with mean 1.00 mm and standard deviation 0.12 mm. A piece passes inspection if its thickness is between 0.80 mm and 1.20 mm.
(a)
Show that the probability that a piece passes inspection is 0.904, correct to 3 significant figures.
(2)
(b)
A batch contains 20 pieces. Let X be the number that pass inspection. Using 0.904 as the probability that a piece passes, find P(X = 18).
(2)
(c)
Find P(X ≥ 18).
(2)
(d)
Find the mean and the variance of X.
(2)
(e)
The mill adjusts the machine so that the standard deviation changes to σ, with the mean still 1.00 mm. It is required that 97% of pieces pass inspection. Find σ, correct to 3 significant figures.
(3)
(f)
With 97% of pieces passing, find the probability that at least 19 pieces in a batch of 20 pass. Use 0.97 as the probability that a piece passes.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(0.80 < T < 1.20) using N(1.00, 0.122)
✓ 1
A1 0.9044… = 0.904 (AG)
✓ 1
Part (b)
M1 X ~ B(20, 0.904) and a binomial calculation
✓ 1
A1 0.285
✓ 1
Part (c)
M1 1 − P(X ≤ 17) (or sum of P(X = 18, 19, 20))
✓ 1
A1 0.700
✓ 1
Part (d)
A1 mean 20 × 0.904 = 18.1
✓ 1
A1 variance 20 × 0.904 × 0.096 = 1.74
✓ 1
Part (e)
M1 P(Z < z) = 0.985 by symmetry
✓ 1
A1 z = 2.170
✓ 1
A1 σ = 0.20/2.170 = 0.0922 mm
✓ 1
Part (f)
M1 X ~ B(20, 0.97) and P(X ≥ 19) = P(X = 19) + P(X = 20)
✓ 1
A1 0.880
✓ 1
Exam tip: For a symmetric interval about the mean containing probability p, the cumulative probability to use in the inverse normal is (1 + p)/2.
346IBMI3S-4-02r48-11Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 2Hard12 marks
Calculator
A clinic compares two treatments for migraine. Each of 100 patients on treatment P and 100 patients on treatment Q is classed as showing a good, moderate or poor outcome after one month.
Good
Moderate
Poor
Treatment P
48
30
22
Treatment Q
30
34
36
A \(\chi^2\) test for independence is carried out at the 5% significance level.
(a)
State the null and alternative hypotheses.
(2)
(b)
Show that the expected frequency for treatment P and a good outcome is 39.
(2)
(c)
Write down the number of degrees of freedom.
(1)
(d)
Use your calculator to find the \(\chi^2\) statistic and the \(p\)-value. Give both to 3 significant figures.
(3)
(e)
State the conclusion of the test, giving a reason.
(2)
(f)
A patient is chosen at random from those given treatment Q. Find the probability that the patient had a good outcome.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(H_0\): outcome and treatment are independent
✓ 1
A1 \(H_1\): outcome and treatment are not independent (associated)
✓ 1
Part (b)
M1 \(\frac{100\times78}{200}\) (row total × column total ÷ grand total)
✓ 1
A1 39
✓ 1
Part (c)
A1 \((2-1)(3-1)=2\)
✓ 1
Part (d)
M1 enter the observed matrix and use the \(\chi^2\) independence test
✓ 1
A1 \(\chi^2=7.78\)
✓ 1
A1 \(p\)-value \(=0.020\)
✓ 1
Part (e)
R1 0.020 < 0.05
✓ 1
A1 reject \(H_0\); there is sufficient evidence of an association between treatment and outcome
✓ 1
Part (f)
M1 \(\frac{30}{100}\) (relevant row total is 100)
✓ 1
A1 0.300
✓ 1
Common error: Expected frequencies are calculated from the totals, never from the observed cell values; check that every expected frequency is at least 5 before using the test.
347IBMI3S-4-02r48-12Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; T-tests, confidence intervals and Markov chains; Normal distribution and chi-squared tests; Probability and the binomial distribution; Normal distribution and chi-squared tests; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Sampling and descriptive statistics; Correlation and regression; Normal distribution and chi-squared tests; Sampling and descriptive statistics; Probability and the binomial distribution; T-tests, confidence intervals and Markov chains; Correlation and regressionPaper 2Hard12 marks
Calculator
A water company fills storage tanks. The volume of water, \(V\) m3, delivered to a tank is normally distributed with mean 1.50 and standard deviation 0.04. A tank passes inspection if its volume is between 1.43 m3 and 1.57 m3 inclusive.
(a)
Find the probability that a tank contains less than 1.45 m3. Give your answer to 3 significant figures.
(2)
(b)
Show that the probability that a tank passes inspection is 0.9199, correct to 4 decimal places.
(2)
(c)
In a batch of 20 randomly chosen tanks, let \(Y\) be the number that pass inspection. Using the value 0.9199, find the probability that at least 18 pass.
(3)
(d)
Using the value 0.9199, find the expected number of tanks that fail inspection in a delivery of 8000 tanks. Give your answer to the nearest whole number.
(2)
(e)
The company changes the filling machine so that the standard deviation becomes \(\sigma\) while the mean stays at 1.50 m3. It wants only 1% of tanks to hold more than 1.57 m3. Find \(\sigma\), to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of normal cdf for \(P(V<1.45)\)
✓ 1
A1 0.1056, so 0.106
✓ 1
Part (b)
M1 \(P(1.43\le V\le1.57)\) with correct limits entered into normal cdf
✓ 1
A1 0.9199
✓ 1
Part (c)
M1 \(Y\sim B(20,0.9199)\)
✓ 1
M1 \(P(Y\ge18)=1-P(Y\le17)\)
✓ 1
A1 0.787
✓ 1
Part (d)
M1 \(8000\times(1-0.9199)\)
✓ 1
A1 641 tanks
✓ 1
Part (e)
M1 \(P(V>1.57)=0.01\), \(z=2.326\)
✓ 1
M1 \(\frac{1.57-1.50}{\sigma}=2.3263\)
✓ 1
A1 \(\sigma=0.0301\), so 0.0301
✓ 1
Exam tip: When a question gives you a value to use in later parts, use that value rather than your own unrounded answer, so that the numbers match the mark scheme.
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