Original exam-style questions with full mark schemes. Use the filters to choose level and difficulty, and tick questions to build a printable test.
105questions
871marks
Fullmark schemes
Original questions in the style of each exam boardFull worked mark schemes with method and accuracy marksCreated by Evan Kotronis with AI assistanceUpdated 15 October 2026 · new questions added continuouslyFree access until 30 November 2026How every question is checked →
Make your own test. Tap + Add to test on any question, from any topic, then “Build & print test”. Or let the test builder choose questions for you in seconds.
1IBMI-033-00-01Sampling and descriptive statisticsPaper SLEasy5 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
Marking point
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 SLEasy6 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
Marking point
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 SLEasy4 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
Marking point
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 SLEasy5 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 SLEasy5 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
Marking point
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 SLEasy6 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 SLEasy6 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
Marking point
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 SLEasy7 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
Marking point
Mark
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 SLEasy5 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
Marking point
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 SLEasy7 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
Marking point
Mark
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 SLEasy5 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
Marking point
Mark
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 SLEasy6 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
Marking point
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 SLEasy6 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
Marking point
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 SLEasy5 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
Marking point
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 SLEasy5 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
Marking point
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 SLEasy6 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 SLEasy5 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
Marking point
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 SLEasy5 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
Marking point
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 SLEasy5 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 SLEasy6 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 SLEasy7 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 SLEasy6 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 SLEasy7 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 SLEasy6 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 SLEasy7 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 SLEasy4 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
Marking point
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 SLEasy7 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 SLEasy5 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 SLEasy5 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 SLEasy7 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 SLEasy7 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 SLEasy6 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 SLEasy6 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.
34IBMI-033-00-03Sampling and descriptive statisticsPaper SLMedium8 marks
A consumer group tests the battery life, in hours, of 15 smartphones of the same model under identical conditions. The results are:
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).
36IBMI-033-00-07Sampling and descriptive statisticsPaper SLMedium6 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)
Show mark scheme
Marking point
Mark
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
✓ 1
4.74 cakes A1
✓ 1
Part (c)
Mean remains 37 A1
✓ 1
Standard deviation decreases, because the new value has zero deviation from the mean while the number of values increases R1
✓ 1
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.
37IBMI-033-00-09Sampling and descriptive statisticsPaper SLMedium11 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).
38IBMI-033-00-11Sampling and descriptive statisticsPaper SLMedium10 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.
39IBMI-033-00-12Sampling and descriptive statisticsPaper SLMedium6 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)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Σfx = 58 + 2p and Σf = 30 + p M1
✓ 1
(58 + 2p) ÷ (30 + p) = 1.95 A1
✓ 1
p = 10 A1
✓ 1
Part (b)
2 books A1
✓ 1
Part (c)
Uses p = 10 in GDC, or √(218 ÷ 40 − 1.952) M1
✓ 1
1.28 books A1
✓ 1
Exam tip: When a frequency is unknown, remember it appears in both the numerator (Σfx) and the denominator (Σf) of the mean.
40IBMI-033-07-r2-03Probability and the binomial distributionPaper SLMedium7 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
Marking point
Mark
Notes
Part (a)
0.04 × 0.95 + 0.96 × 0.03 M1
✓ 1
= 0.0668 A1
✓ 1
Part (b)
(0.04 × 0.95) ÷ 0.0668 = 0.038 ÷ 0.0668 M1
✓ 1
= 0.569 A1
✓ 1
Part (c)
Recognises binomial with n = 50 M1
✓ 1
and p = 0.0668, using P(Y ≤ 2) M1
✓ 1
= 0.342 A1
✓ 1
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.
41IBMI-033-07-r2-05Probability and the binomial distributionPaper SLMedium6 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
Marking point
Mark
Notes
Part (a)
1 − 0.98n > 0.5, i.e. 0.98n < 0.5 M1
✓ 1
n > 34.3 A1
✓ 1
Smallest n = 35 A1
✓ 1
Part (b)
X ~ B(40, 0.02) M1
✓ 1
P(X ≥ 2) = 1 − P(X ≤ 1) M1
✓ 1
= 0.190 A1
✓ 1
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.
42IBMI-033-08-r2-03Normal distribution and chi-squared testsPaper SLMedium11 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.
43IBMI-033-08-r2-04Normal distribution and chi-squared testsPaper SLMedium7 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
Marking point
Mark
Notes
Part (a)
H0: customers choose coffee, tea, hot chocolate and juice in the ratio 5 : 2 : 2 : 1
✓ 1
Part (b)
Uses proportions of 160, e.g. \(\frac{5}{10} \times 160\)
✓ 1
Coffee 80, tea 32, hot chocolate 32, juice 16
✓ 1
Part (c)
Correct use of \(\sum \frac{(O-E)^2}{E}\) or GDC with correct observed and expected lists
✓ 1
χ2calc = 2.05
✓ 1
Part (d)
2.05 < 7.815 (or p-value 0.562 > 0.05), so do not reject H0
✓ 1
There is insufficient evidence to doubt the owner's claim; the data are consistent with the 5 : 2 : 2 : 1 ratio
✓ 1
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.
44IBMI-033-05-r1-03Sampling and descriptive statisticsPaper SLMedium12 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.
45IBMI-033-05-r1-04Sampling and descriptive statisticsPaper SLMedium6 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
Marking point
Mark
Notes
Part (a)
M1: 1.8 × 18.4 + 32
✓ 1
A1: 65.1 °F (65.12)
✓ 1
Part (b)
M1: 1.8 × 3.5, with the +32 correctly ignored
✓ 1
A1: 6.30 °F
✓ 1
Part (c)
M1: \(\frac{30\times 65.12+20\times 70.2}{50}\)
✓ 1
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.
46IBMI-033-05-r1-05Sampling and descriptive statisticsPaper SLMedium12 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.
47IBMI-033-03-r2-03Normal distribution and chi-squared testsPaper SLMedium11 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'.
48IBMI-033-03-r2-04Normal distribution and chi-squared testsPaper SLMedium7 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)
Show mark scheme
Marking point
Mark
Notes
Part (a)
H0: complaints are uniformly distributed across the five weekdays (equally likely on each day) (A1)
✓ 1
Part (b)
Expected frequency = 50 (A1)
✓ 1
Degrees of freedom = 5 − 1 = 4 (A1)
✓ 1
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)
✓ 1
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%.
49IBMI-033-03-r2-05Normal distribution and chi-squared testsPaper SLMedium8 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
Marking point
Mark
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)
✓ 1
Part (b)
Use of inverse normal with area 0.9 (M1)
✓ 1
t = 46.4 minutes (A1)
✓ 1
Part (c)
Journey times on different days are independent (or the probability of a journey exceeding 45 minutes is the same each day) (R1)
✓ 1
Part (d)
Recognising Y ~ B(20, 0.14723) (M1)
✓ 1
P(Y > 3) = 1 − P(Y ≤ 3) (M1)
✓ 1
= 0.339 (accept 0.338 from p = 0.147) (A1)
✓ 1
Common error: 'More than 3' means 1 − P(Y ≤ 3), not 1 − P(Y ≤ 4) or 1 − P(Y ≤ 2).
50IBMI-033-02-r2-03Probability and the binomial distributionPaper SLMedium7 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
Marking point
Mark
Notes
Part (a)
M1: multiplies along branches, e.g. \(0.03 \times 0.95\) or \(0.97 \times 0.08\)
✓ 1
M1: adds the two positive branches \(0.0285 + 0.0776\)
✓ 1
A1: \(0.106\)
✓ 1
Part (b)
M1: numerator \(0.03 \times 0.95 = 0.0285\)
✓ 1
M1: divides by their answer to (a), \(\frac{0.0285}{0.1061}\)
✓ 1
A1: \(0.269\)
✓ 1
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
✓ 1
Exam tip: Draw a tree diagram first. Conditional probability questions become one division once both positive branches are labelled.
51IBMI-033-02-r1-04Probability and the binomial distributionPaper SLMedium8 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
Marking point
Mark
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)
✓ 1
A1: least \(n=10\)
✓ 1
Part (c)
M1: interval \(10.8\pm1.74\), i.e. \(9.06
✓ 1
A1: identifies \(X=10,\,11,\,12\), i.e. \(P(10\le X\le12)\)
✓ 1
A1: 0.614
✓ 1
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.
52IBMI-033-02-r1-05Probability and the binomial distributionPaper SLMedium11 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.
53IBMI-033-07-r1-04Probability and the binomial distributionPaper SLMedium12 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”.
54IBMI-033-01-r1-03Correlation and regressionPaper SLMedium6 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
Marking point
Mark
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.
55IBMI-033-05-r2-03Sampling and descriptive statisticsPaper SLMedium12 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.
56IBMI-033-05-r2-04Sampling and descriptive statisticsPaper SLMedium6 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
Marking point
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
✓ 1
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
✓ 1
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.
57IBMI-033-06-r1-03Correlation and regressionPaper SLMedium7 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
Marking point
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.
58IBMI-033-06-r1-04Correlation and regressionPaper SLMedium12 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.
59IBMI-033-06-r1-06Correlation and regressionPaper SLMedium6 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
Marking point
Mark
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.
60IBMI-033-04-r2-06T-tests, confidence intervals and Markov chainsPaper SLMedium12 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.
61IBMI-033-03-r1-03Normal distribution and chi-squared testsPaper SLMedium12 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.
62IBMI-033-03-r1-04Normal distribution and chi-squared testsPaper SLMedium6 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.
63IBMI-033-08-r1-03Normal distribution and chi-squared testsPaper SLMedium11 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.
64IBMI-033-01-r1-04Correlation and regressionPaper SLMedium12 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.
65IBMI-033-02-r1-03Probability and the binomial distributionPaper SLMedium7 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.
66IBMI-033-08-r1-06Normal distribution and chi-squared testsPaper SLMedium11 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.
67IBMI-033-06-r2-03Correlation and regressionPaper SLMedium7 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.
68IBMI-033-06-r2-04Correlation and regressionPaper SLMedium10 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.
69IBMI-033-07-r1-05Probability and the binomial distributionPaper SLMedium8 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
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.
71IBMI-033-08-r1-04Normal distribution and chi-squared testsPaper SLMedium7 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.
72IBMI-033-02-r2-04Probability and the binomial distributionPaper SLMedium11 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.
73IBMI-033-01-r2-04Correlation and regressionPaper SLMedium7 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)
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.
75IBMI-033-07-r1-03Probability and the binomial distributionPaper SLMedium7 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.
76IBMI-033-00-05Sampling and descriptive statisticsPaper SLHard7 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.
77IBMI-033-00-10Sampling and descriptive statisticsPaper SLHard11 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.
78IBMI-033-05-r2-05Sampling and descriptive statisticsPaper SLHard12 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.
79IBMI-033-03-r2-02Normal distribution and chi-squared testsPaper HLEasy5 marks
A machine fills bottles of sparkling water. The volume, V ml, in a bottle is normally distributed with mean 500 ml and unknown standard deviation σ ml. Quality checks show that 3% of bottles contain less than 490 ml.
(a)
Find the value of σ.
(3)
(b)
The machine is adjusted so that its mean changes but σ does not. After the adjustment only 1% of bottles contain less than 490 ml. Find the new mean volume.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Inverse normal of 0.03 gives z = −1.8808 (A1)
✓ 1
Setting up (490 − 500)/σ = −1.8808 (M1)
✓ 1
σ = 5.32 ml (5.3169…) (A1)
✓ 1
Part (b)
Using z = −2.3263 with (490 − μ)/5.3169 = −2.3263 (M1)
✓ 1
μ = 502 ml (502.37…) (A1)
✓ 1
Common error: Use the unrounded value of σ from part (a) in part (b); using 5.3 instead of 5.3169 shifts the new mean noticeably.
80IBMI-033-04-r2-03T-tests, confidence intervals and Markov chainsPaper HLEasy6 marks
Each working day Priya travels to work either by bicycle (C) or by bus (B). If she cycles one day, the probability that she cycles the next day is 0.7. If she takes the bus one day, the probability that she cycles the next day is 0.4.
(a)
Write down the transition matrix T for this Markov chain, with columns representing today's state (C, B) and rows representing tomorrow's state (C, B).
(2)
(b)
Priya cycles on Monday. Find the probability that she cycles on Thursday of the same week.
(2)
(c)
Find the long-term proportion of working days on which Priya cycles.
Use of \(\mathbf{T}^3\begin{pmatrix}1\\0\end{pmatrix}\)
✓ 1
0.583
✓ 1
Part (c)
Steady state: solve \(\mathbf{T}\mathbf{s}=\mathbf{s}\), e.g. \(p=0.7p+0.4(1-p)\)
✓ 1
\(p=\frac{4}{7}\) = 0.571
✓ 1
Exam tip: From Monday to Thursday is three transitions, so use T³. Check that each column of your matrix sums to 1.
81IBMI-033-04-r1-02T-tests, confidence intervals and Markov chainsPaper HLEasy5 marks
A machine fills bags of rice. The label states that each bag contains 500 g. The masses of the bags filled are normally distributed with a known standard deviation of 4.8 g. A consumer group weighs a random sample of 40 bags and finds a sample mean of 502.3 g.
(a)
Find a 95% confidence interval for the mean mass of rice in a bag. Give the bounds to 1 decimal place.
(3)
(b)
The consumer group suspects that the machine is putting more than 500 g of rice in the bags on average. Comment on this suspicion, referring to your interval.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: uses \(502.3 \pm 1.960 \times \frac{4.8}{\sqrt{40}}\) or a GDC z-interval
✓ 1
A1: lower bound 500.8 g
✓ 1
A1: upper bound 503.8 g
✓ 1
Part (b)
R1: 500 g lies below the whole interval / is not contained in the interval
✓ 1
A1: so the data support the suspicion that the mean mass is greater than 500 g
✓ 1
Common error: Here the population standard deviation is known, so a z-interval is used, not a t-interval. Using 4.8 without dividing by \(\sqrt{40}\) gives an interval far too wide.
82IBMI-033-04-r2-02T-tests, confidence intervals and Markov chainsPaper HLMedium7 marks
A café advertises that a large cup of its filter coffee contains a mean of 110 mg of caffeine. A consumer group buys 10 large cups at random and measures the caffeine content, in mg:
92, 105, 98, 110, 101, 95, 99, 107, 103, 96
Assume the caffeine content is normally distributed.
(a)
Find the sample mean and an unbiased estimate of the population variance.
(2)
(b)
Find a 95% confidence interval for the population mean caffeine content. Give the limits to one decimal place.
(3)
(c)
Comment on the café's advertised value, with a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
\(\bar{x}\) = 100.6 mg
✓ 1
\(s_{n-1}^2\) = 32.3 mg2 (32.2667)
✓ 1
Part (b)
Use of t-distribution with 9 degrees of freedom (critical value 2.262) or GDC t-interval
✓ 1
Lower limit 96.5 mg
✓ 1
Upper limit 104.7 mg
✓ 1
Part (c)
110 mg lies outside (above) the 95% confidence interval
✓ 1
So the advertised mean is not supported; the true mean caffeine content is likely to be lower than 110 mg
✓ 1
Common error: The population standard deviation is unknown, so the interval uses the t-distribution with n − 1 degrees of freedom, not a z-interval.
83IBMI-033-04-r2-04T-tests, confidence intervals and Markov chainsPaper HLMedium8 marks
A running coach believes a six-week training programme reduces 5 km race times. Eight runners record their times, in minutes, before and after the programme.
Reject H0: there is significant evidence that the programme reduces the mean 5 km time
✓ 1
Common error: Using a two-sample test on paired data ignores the pairing and usually gives a much larger p-value. Always test the differences.
84IBMI-033-03-r1-05Normal distribution and chi-squared testsPaper HLMedium10 marks
The masses of adults using a lift are normally distributed with mean 78 kg and standard deviation 12 kg. The masses of children using the lift are normally distributed with mean 32 kg and standard deviation 6 kg. All masses are independent.
(a)
Twelve adults enter the lift. Find the probability that their total mass exceeds 1000 kg.
(4)
(b)
An adult and a child are chosen at random. Find the probability that the adult's mass is more than twice the child's mass.
(3)
(c)
The manufacturer wants the probability that the total mass of twelve adults exceeds the lift's rated load L to be 0.01. Find L, giving your answer to the nearest kg.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: mean of total = 12 × 78 = 936 kg
✓ 1
M1: variance of total = 12 × 122 = 1728
✓ 1
M1: P(T > 1000) where T ~ N(936, 1728)
✓ 1
A1: 0.0618
✓ 1
Part (b)
M1: A − 2C has mean 78 − 64 = 14
✓ 1
M1: variance = 122 + 22 × 62 = 288
✓ 1
A1: P(A − 2C > 0) = 0.795
✓ 1
Part (c)
M1: uses T ~ N(936, 1728)
✓ 1
M1: P(T > L) = 0.01, using inverse normal with upper tail
✓ 1
A1: L = 1033 kg (1032.7…)
✓ 1
Common error: Var(2C) = 4Var(C), whereas Var(C₁ + C₂) = 2Var(C). Twice one child's mass is not the same as the total mass of two children.
85IBMI-033-04-r1-04T-tests, confidence intervals and Markov chainsPaper HLMedium8 marks
A phone manufacturer claims that the mean battery life of its new model is 12.0 hours of continuous video playback. A reviewer suspects the true mean is lower. She tests a random sample of 10 phones and records a sample mean of 11.4 hours and an unbiased estimate of the standard deviation \(s_{n-1} = 0.85\) hours. Battery lives may be assumed to be normally distributed.
She carries out a one-sample t-test at the 5% significance level. The critical value for this test is −1.833.
(a)
State suitable hypotheses for the test.
(2)
(b)
Calculate the value of the test statistic.
(2)
(c)
State, with a reason, the conclusion of the test in context.
(2)
(d)
Using \(t_{9} = 1.833\) for 90% confidence, find a 90% confidence interval for the mean battery life.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1: H0: \(\mu = 12.0\), where \(\mu\) is the population mean battery life in hours
✓ 1
A1: H1: \(\mu < 12.0\)
✓ 1
Part (b)
M1: \(t = \frac{11.4 - 12.0}{0.85/\sqrt{10}}\)
✓ 1
A1: \(t = -2.23\)
✓ 1
Part (c)
R1: \(-2.23 < -1.833\), so the test statistic lies in the critical region
✓ 1
A1: reject H0; there is sufficient evidence at the 5% level that the mean battery life is less than 12.0 hours
Common error: The population standard deviation is unknown, so a t-distribution with 9 degrees of freedom is used. Using the normal value 1.645 instead of 1.833 gives an interval that is too narrow.
86IBMI-033-04-r1-05T-tests, confidence intervals and Markov chainsPaper HLMedium9 marks
A language school claims that a one-week intensive course improves vocabulary test scores. Eight students take a vocabulary test before and after the course. Their scores are:
Model B, because its sum of square residuals is smaller (1.03 < 5.59)
✓ 1
Part (d)
M1: substitutes x = 6 into the chosen model, e.g. Model B: 0.6 × 36 + 0.1 × 6 + 2.4 = 24.6
✓ 1
A1: about 2460 charging points
✓ 1
(Follow through from (c): Model A gives 19.6 hundred = 1960 charging points, which earns M1 A1 FT.)
—
Common error: Forgetting the units of y: the model gives y in hundreds, so 24.6 must be converted to 2460 charging points to answer in context.
88IBMI-033-05-r2-06Sampling and descriptive statisticsPaper HLMedium6 marks
A machine fills bags of coffee labelled 250 g. A quality controller takes every 50th bag from the production line until she has 12 bags. She weighs each one and records the mass x grams. The results give \(\sum x = 3018\) and \(\sum x^2 = 759\,160\).
(a)
Find an unbiased estimate of the mean mass of all bags filled by the machine.
(1)
(b)
Find an unbiased estimate of the variance of the mass of all bags filled by the machine.
(3)
(c)
Name the sampling method used, and give one reason why it could produce a biased sample.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
\(\bar{x}=\frac{3018}{12}\) = 251.5 g A1
✓ 1
Part (b)
Correct use of \(\sum x^2\) and \(\bar{x}\), e.g. \(759\,160-12(251.5)^2=133\) or \(\frac{759\,160}{12}-251.5^2=11.08...\) M1
✓ 1
Correct use of divisor \(n-1\), e.g. \(\frac{133}{11}\) or \(11.08...\times\frac{12}{11}\) M1
✓ 1
\(s_{n-1}^2\) = 12.1 g2 A1
✓ 1
Part (c)
Systematic sampling A1
✓ 1
If a fault or fill pattern in the machine repeats at an interval that matches the sampling interval (e.g. every 25 or 50 bags), the sample could keep selecting the faulty bags or always miss them R1
✓ 1
Common error: Σx²/n − mean² gives the variance with divisor n. For an unbiased estimate of the population variance, multiply by n/(n − 1), or divide the sum of squared deviations by n − 1.
89IBMI-033-04-r1-03T-tests, confidence intervals and Markov chainsPaper HLMedium8 marks
Priya travels to work each weekday either by bicycle (C) or by bus (B). If she cycles on one day, the probability that she cycles the next day is 0.7. If she takes the bus on one day, the probability that she takes the bus the next day is 0.6. Assume her choice depends only on the previous day's choice.
(a)
Write down the transition matrix T, with columns representing the current day's mode of travel (C then B) and rows the next day's.
(2)
(b)
Priya cycles on Monday. Find the probability that she cycles on Thursday of the same week.
(2)
(c)
Find the long-term proportion of days on which Priya cycles.
(3)
(d)
Priya works 210 days a year. Estimate the number of days in a year on which she cycles to work.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1: one column fully correct, e.g. first column \(\begin{pmatrix}0.7\\0.3\end{pmatrix}\)
✓ 1
A1: \(\mathbf{T}=\begin{pmatrix}0.7 & 0.4\\ 0.3 & 0.6\end{pmatrix}\) fully correct, in the stated orientation
M1: forms \(\mathbf{T}\mathbf{s}=\mathbf{s}\) with \(\mathbf{s}=\begin{pmatrix}p\\q\end{pmatrix}\), e.g. \(0.7p+0.4q=p\), or finds an eigenvector of T for eigenvalue 1
✓ 1
M1: uses \(p+q=1\) (or normalises the eigenvector) to get one equation in one unknown, e.g. \(0.3p=0.4(1-p)\)
✓ 1
A1: \(p=\frac{4}{7}\approx 0.571\)
✓ 1
Part (d)
A1: \(210 \times \frac{4}{7} = 120\) days
✓ 1
Exam tip: Count transitions carefully: Monday to Thursday is three steps, so use \(\mathbf{T}^3\), not \(\mathbf{T}^4\).
90IBMI-033-01-r2-05Correlation and regressionPaper HLMedium8 marks
The number of downloads, N, of a new mobile game was recorded at the end of each of its first 9 weeks after launch, where t is the number of weeks. A developer suspects exponential growth, so she calculates ln N for each week. The regression line of ln N on t is \(\ln N = 0.231t + 5.12\) with correlation coefficient r = 0.994.
(a)
Show that the model can be written as \(N = Ae^{kt}\), and find the values of A and k.
(3)
(b)
Use the model to estimate the number of downloads at the end of week 10.
(2)
(c)
Find the value of t at which the model predicts that downloads first reach 5000.
(2)
(d)
Give one reason why this model is unlikely to be valid in the long term.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Exponentiates both sides: \(N = e^{5.12}e^{0.231t}\) (M1)
✓ 1
A = \(e^{5.12}\) ≈ 167 (A1)
✓ 1
k = 0.231 (A1)
✓ 1
Part (b)
Substitutes t = 10, e.g. \(N = e^{7.43}\) or \(167e^{2.31}\) (M1)
✓ 1
N ≈ 1690 downloads (accept 1680 from rounded A) (A1)
Any one valid reason, e.g. exponential growth is unbounded but downloads are limited by the number of potential users or by falling interest; OR long-term predictions are extrapolation far beyond the 9 weeks of data (R1)
✓ 1
Common error: The constant A is \(e^{5.12}\), not 5.12. The intercept of the linearised model is ln A.
91IBMI-033-08-r2-05Normal distribution and chi-squared testsPaper HLMedium10 marks
The times, T seconds, that a swimmer takes to complete a 100 m race can be modelled by a normal distribution with mean μ and standard deviation σ. Records show that 12% of her times are under 62.0 s and 30% of her times are over 66.5 s.
(a)
Find the value of μ and the value of σ.
(5)
(b)
To qualify for a national event, the swimmer must record a time under 64.0 s. Find the probability that she qualifies in a single race.
(2)
(c)
During a season she swims 15 independent races. Find the probability that she records a qualifying time in at least 4 of these races.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
z = −1.175 (−1.17498…) for the lower 12%
✓ 1
z = 0.5244 (0.52440…) for the upper 30%
✓ 1
Forms two equations, e.g. μ − 1.175σ = 62.0 and μ + 0.5244σ = 66.5
✓ 1
μ = 65.1 s (65.111…)
✓ 1
σ = 2.65 s (2.6480…)
✓ 1
Part (b)
Finds P(T < 64.0) using the values of μ and σ found in the previous part
✓ 1
0.337 (0.33735…)
✓ 1
Part (c)
Recognises X ~ B(15, 0.337…)
✓ 1
Uses 1 − P(X ≤ 3)
✓ 1
0.800
✓ 1
Exam tip: Carry the unrounded values of μ and σ into the qualifying-time probability. Using the rounded values 65.1 and 2.65 gives 0.339, which is not accurate enough.
92IBMI-033-07-r2-06Probability and the binomial distributionPaper HLHard13 marks
An airline states that 8% of its flights are delayed by more than 30 minutes. A consumer group believes the true proportion is higher. It selects a random sample of 60 of the airline's flights and finds that 9 are delayed by more than 30 minutes. The group carries out a hypothesis test at the 5% significance level. Let X be the number of delayed flights in a sample of 60.
(a)
State the null and alternative hypotheses.
(2)
(b)
Find the p-value for this test.
(2)
(c)
State the conclusion of the test in context, giving a reason.
(2)
(d)
Find the critical region for this test.
(3)
(e)
Write down the probability of a Type I error for this test.
(1)
(f)
The true proportion of delayed flights is 0.12. Find the probability of a Type II error.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
H0: p = 0.08 A1
✓ 1
H1: p > 0.08 A1
✓ 1
Part (b)
Under H0, X ~ B(60, 0.08); p-value = P(X ≥ 9) = 1 − P(X ≤ 8) M1
✓ 1
= 0.0482 A1
✓ 1
Part (c)
0.0482 < 0.05 R1
✓ 1
Reject H0: there is sufficient evidence that more than 8% of the airline's flights are delayed by over 30 minutes A1
✓ 1
Part (d)
Considers upper-tail probabilities, e.g. P(X ≥ 8) = 0.104 M1
✓ 1
and P(X ≥ 9) = 0.0482 A1
✓ 1
Critical region: X ≥ 9 A1
✓ 1
Part (e)
P(X ≥ 9 | p = 0.08) = 0.0482 A1
✓ 1
Part (f)
Type II error: accepting H0, i.e. X ≤ 8 M1
✓ 1
using X ~ B(60, 0.12) M1
✓ 1
P(X ≤ 8) = 0.710 A1
✓ 1
Exam tip: For a discrete test the Type I error probability is the actual probability of the critical region (0.0482), not the nominal 5%. A Type II error uses the non-rejection region with the true value of p.
93IBMI-033-05-r1-06Sampling and descriptive statisticsPaper HLHard10 marks
A sports-science company develops an app that measures reaction time, in milliseconds (ms). To assess the app, 8 volunteers each take the test twice, one week apart, under the same conditions. The results are: Volunteer: A, B, C, D, E, F, G, H First test (x ms): 210, 245, 230, 198, 260, 225, 240, 215 Second test (y ms): 215, 240, 236, 205, 255, 220, 248, 212
(a)
State what is meant by the reliability of a test, and name the type of reliability test being carried out here.
(2)
(b)
Calculate the Pearson product-moment correlation coefficient, r, between the first and second test results.
(2)
(c)
Interpret your value of r in terms of the reliability of the app.
(2)
(d)
Explain why a high value of r does not show that the app is valid, and describe how criterion-related validity could be tested.
(2)
(e)
The company later groups reaction times into categories for a χ2 goodness-of-fit test with a larger sample. State one consideration when choosing the categories, and why it matters.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1: Reliability is the extent to which the test produces consistent results when repeated under the same conditions
✓ 1
A1: Test-retest reliability
✓ 1
Part (b)
M1: Correct use of GDC with both lists entered as paired data
✓ 1
A1: r = 0.957
✓ 1
Part (c)
A1: Very strong positive linear correlation between first and second results
✓ 1
R1: Volunteers obtain similar times on both occasions, so the app appears to be reliable
✓ 1
Part (d)
R1: Validity concerns whether the app measures true reaction time; it could be consistently wrong (e.g. adding a fixed screen delay) and still give a high r
✓ 1
A1: Compare app results with those from an established, accurate laboratory reaction-time device on the same people and check for strong correlation/agreement
✓ 1
Part (e)
A1: Categories should be chosen so that each expected frequency is at least 5 (combine sparse classes if necessary)
✓ 1
R1: Otherwise the χ2 distribution is a poor approximation to the test statistic, making the conclusion unreliable
✓ 1
Common error: Reliability and validity are different: a strong test-retest correlation shows consistency only, not that the measurement is accurate.
94IBMI-033-02-r2-06Probability and the binomial distributionPaper HLHard11 marks
Faults in a type of woven fabric occur randomly at a mean rate of 0.6 faults per metre. The fabric is sold in rolls of length 5 metres. Let \(Y\) be the number of faults in a roll.
(a)
Find the probability that a roll contains at most 2 faults.
(2)
(b)
A shop receives a batch of 12 rolls. Find the probability that at least 8 of these rolls contain at most 2 faults.
(3)
(c)
The manufacturer also sells short offcuts of length \(L\) metres. Find the greatest value of \(L\) for which the probability that an offcut has no faults is at least 0.2.
(3)
(d)
Given that a 5 metre roll contains at least one fault, find the probability that it contains exactly 3 faults.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: \(Y \sim Po(3)\)
✓ 1
A1: \(P(Y \le 2) = 0.423\)
✓ 1
Part (b)
M1: \(R \sim B(12, 0.42319…)\)
✓ 1
M1: \(P(R \ge 8) = 1 - P(R \le 7)\)
✓ 1
A1: \(0.0794\)
✓ 1
Part (c)
M1: \(e^{-0.6L} \ge 0.2\)
✓ 1
M1: \(L \le \frac{\ln 5}{0.6}\) or solves graphically
Exam tip: Scale the Poisson mean to the interval in the question: 0.6 per metre becomes 3 per 5 m roll and 0.6L for an offcut of length L.
95IBMI-033-07-r1-06Probability and the binomial distributionPaper HLHard14 marks
The number of calls received by a customer helpline can be modelled by a Poisson distribution with mean 3.2 calls per 10-minute interval. The numbers of calls in non-overlapping intervals are independent.
(a)
Find the probability that exactly 5 calls are received in a 10-minute interval.
(2)
(b)
Each operator can handle at most one call per 10-minute interval. Find the least number of operators needed so that the probability that all calls in a 10-minute interval can be handled is at least 0.95.
(2)
(c)
Find the probability that more than 6 calls are received in a 20-minute period.
(3)
(d)
Given that exactly 6 calls are received in a 20-minute period, find the probability that exactly 3 of them arrive in the first 10 minutes.
(3)
(e)
An hour is divided into six 10-minute intervals. An interval is classed as “busy” if more than 4 calls are received in it. Find the probability that exactly 2 of the six intervals in an hour are busy.
She believes the data can be modelled by \(N = ae^{kt}\), where a and k are positive constants.
(a)
Show that, if the model is correct, the graph of ln N against t is a straight line.
(2)
(b)
Find the equation of the regression line of ln N on t.
(2)
(c)
Find the value of Pearson's correlation coefficient for ln N and t, and state what it indicates about the model.
(2)
(d)
Hence find the values of a and k.
(2)
(e)
Use the model to estimate the number of bacteria after 7 hours.
(2)
(f)
Find the time at which the model predicts that the number of bacteria first reaches 5000.
(3)
(g)
Comment on the reliability of your answer to the previous part.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: takes natural logs of both sides, ln N = ln a + ln(ekt)
✓ 1
R1: ln N = kt + ln a, which is of the form Y = mt + c with gradient k and intercept ln a, so linear
✓ 1
Part (b)
Gradient 0.435 (0.434604…)
✓ 1
Intercept 4.79 (4.78977…), giving ln N = 0.435t + 4.79
✓ 1
Part (c)
r = 1.00 (0.99995…)
✓ 1
Almost perfect positive linear relationship between ln N and t, so the exponential model is highly appropriate
✓ 1
Part (d)
M1: k = gradient = 0.435 and a = e4.78977…
✓ 1
A1: a = 120 (120.27…)
✓ 1
Part (e)
M1: N = 120.27e0.434604 × 7
✓ 1
A1: N ≈ 2520
✓ 1
Part (f)
M1: sets up equation 120.27e0.434604t = 5000
✓ 1
M1: solves, e.g. t = ln(5000 ÷ 120.27) ÷ 0.434604 or by GDC
✓ 1
A1: t = 8.58 hours (accept 8.57 from rounded values)
✓ 1
Part (g)
Unreliable: t = 8.58 is outside the data range (0–5 hours), and growth is likely to slow as nutrients or space run out
✓ 1
Exam tip: Keep the full GDC values of the gradient and intercept stored in memory. Rounding a to 120 and k to 0.435 before solving can shift the final answer in the third significant figure.
97IBMI-033-04-r2-05T-tests, confidence intervals and Markov chainsPaper HLHard16 marks
In a city, every household subscribes to exactly one of two streaming services, X or Y. Each month, 85% of X subscribers stay with X and the rest switch to Y. Of Y subscribers, 75% stay with Y and the rest switch to X. Initially, 40% of households subscribe to X and 60% to Y.
(a)
Write down the transition matrix T, with columns representing the current service (X, Y).
(2)
(b)
Find the proportion of households subscribing to X after 3 months.
(2)
(c)
Find the eigenvalues of T and a corresponding eigenvector for each.
(4)
(d)
Hence, by writing \(\mathbf{T}=\mathbf{PDP}^{-1}\), show that the proportion subscribing to X after n months is \(0.625-0.225(0.6)^n\).
(3)
(e)
Find the least number of months after which more than 62% of households subscribe to X.
(3)
(f)
State the long-term proportion of households subscribing to X, justifying your answer.
Exam tip: Eigenvalues of a transition matrix always include 1; the other eigenvalue (here 0.6) controls how quickly the chain approaches its steady state.
98IBMI-033-06-r2-05Correlation and regressionPaper HLHard11 marks
A cup of coffee is left to cool in a room kept at 20 °C. The excess temperature, D °C, is the coffee's temperature minus the room temperature. It is recorded every 5 minutes.
t (minutes): 0, 5, 10, 15, 20, 25 D (°C): 62.0, 47.5, 37.0, 28.6, 22.0, 17.1
It is suggested that the data can be modelled by \(D = ab^t\).
(a)
By considering ratios of consecutive values of D, justify the use of an exponential model.
(2)
(b)
Use exponential regression to find the values of a and b.
(3)
(c)
Interpret the value of b in context.
(1)
(d)
A linear model for D against t has coefficient of determination R2 = 0.965. Despite this high value, explain why the linear model is not suitable for predicting the coffee's temperature over a longer period.
(2)
(e)
Use your exponential model to estimate the time at which the temperature of the coffee is 45 °C.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: calculates ratios such as 47.5/62.0 = 0.766, 37.0/47.5 = 0.779, 28.6/37.0 = 0.773, 22.0/28.6 = 0.769, 17.1/22.0 = 0.777
✓ 1
R1: the ratios are approximately constant (about 0.77) over equal time intervals, which is the property of an exponential model
✓ 1
Part (b)
M1: exponential regression on the GDC, or a linear regression of ln D on t
✓ 1
A1: a = 61.8
✓ 1
A1: b = 0.950
✓ 1
Part (c)
Each minute the excess temperature is multiplied by 0.950, i.e. it falls by about 5% per minute
✓ 1
Part (d)
R1: the linear model predicts that D becomes zero and then negative (at about 33 minutes), so the coffee would cool below room temperature, which is unrealistic
✓ 1
R1: the data show a curved, decreasing rate of cooling. The exponential model approaches D = 0 asymptotically and fits this behaviour, so a high R2 alone does not make a model appropriate
✓ 1
Part (e)
M1: recognises that the excess temperature is D = 45 − 20 = 25
✓ 1
M1: solves 61.8(0.950)t = 25, e.g. graphically or with logarithms
✓ 1
A1: t = 17.6 minutes
✓ 1
Common error: The model is for the excess temperature, not the actual temperature. Subtract the room temperature before substituting into the model.
99IBMI-033-03-r1-06Normal distribution and chi-squared testsPaper HLHard12 marks
A marine biologist measures the lengths, L cm, of 200 fish of one species. From the raw data, the mean is estimated as 30.0 cm and the standard deviation as 5.0 cm. She tests, at the 5% significance level, whether the lengths can be modelled by a normal distribution. The observed frequencies are:
L < 22: 15 22 ≤ L < 26: 25 26 ≤ L < 30: 62 30 ≤ L < 34: 50 34 ≤ L < 38: 36 L ≥ 38: 12
Under the model N(30, 52), the expected frequencies for 22 ≤ L < 26, 30 ≤ L < 34, 34 ≤ L < 38 and L ≥ 38 are 31.41, 57.63, 31.41 and 10.96 respectively.
(a)
Show that the expected frequency for 26 ≤ L < 30 is 57.6 to 3 significant figures, and find the expected frequency for L < 22.
(3)
(b)
State the null hypothesis and explain why the test has 3 degrees of freedom.
(3)
(c)
Find the χ2 statistic.
(2)
(d)
The critical value is 7.815. State the conclusion of the test in context.
(2)
(e)
A colleague suggests using twelve classes, each 2 cm wide, to make the test more detailed. Explain one problem with this suggestion.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: P(26 ≤ L < 30) = 0.28814…
✓ 1
A1: 200 × 0.28814… = 57.63 ≈ 57.6
✓ 1
A1: 200 × P(L < 22) = 200 × 0.054799 = 10.96
✓ 1
Part (b)
A1: H0: the fish lengths can be modelled by a normal distribution
✓ 1
R1: two parameters (the mean and the standard deviation) were estimated from the data, and the expected frequencies must total 200
✓ 1
A1: 6 − 1 − 2 = 3
✓ 1
Part (c)
M1: \(\sum \frac{(O-E)^2}{E}\) over all six classes
✓ 1
A1: χ2 = 4.91 (4.908…)
✓ 1
Part (d)
R1: 4.91 < 7.815
✓ 1
A1: do not reject H0; there is insufficient evidence that the lengths are not normally distributed, so the normal model may be used
✓ 1
Part (e)
R1: the expected frequencies in the tail classes would fall below 5
✓ 1
R1: those classes would have to be combined, which reduces the degrees of freedom and removes the extra detail gained
✓ 1
Exam tip: Subtract one degree of freedom for every parameter estimated from the sample, as well as one for the fixed total. For a normal model with both μ and σ estimated, that is 3 in total.
100IBMI-033-04-r1-06T-tests, confidence intervals and Markov chainsPaper HLHard14 marks
A car-rental company has depots at the Airport (A), the Beach (B) and the City (C). Each car is rented once a day and returned to one of the depots. For a car rented from A, the probabilities of return to A, B and C are 0.6, 0.3 and 0.1 respectively. For a car rented from B they are 0.2, 0.7 and 0.1. For a car rented from C they are 0.2, 0.2 and 0.6. Assume the return depot depends only on the depot of rental.
(a)
Write down the transition matrix T, with columns representing the depot of rental (A, B, C) and rows the depot of return.
(2)
(b)
A car starts at A. Find the probability that it is at C after two rentals.
(2)
(c)
A car starts at A. Find the probability that it is not returned to C at any time during its next three rentals, justifying your method.
(3)
(d)
Find the steady-state probability vector for the location of a car.
(4)
(e)
The company has 150 cars. The City depot has 25 parking spaces. Find the expected number of cars at C in the long term and comment on whether the depot is large enough.
R1: 30 > 25, so in the long term the City depot is expected to be too small
✓ 1
Exam tip: Before multiplying out matrices, look for structure: equal probabilities of moving to C from both A and B make part (c) a one-line calculation.
101IBMI-033-08-r1-05Normal distribution and chi-squared testsPaper HLHard14 marks
At an airport, the masses of adult passengers, M kg, are modelled by N(78, 122), and the masses of their cabin luggage, B kg, are modelled by N(15, 42). All masses are independent.
(a)
Find the probability that a randomly chosen passenger is more than 70 kg heavier than a randomly chosen piece of luggage.
(3)
(b)
A shuttle bus has a maximum safe load of 1100 kg. Eleven passengers, each with one piece of luggage, board the bus. Find the probability that the total mass exceeds the safe load.
(3)
(c)
A student models the total mass of the eleven passengers as 11M. Explain why this model is not appropriate.
(2)
(d)
The operator requires the probability that the total mass of passengers and luggage exceeds 1100 kg to be less than 0.01. Find the largest number of passengers, each with luggage, who may board.
(3)
(e)
A random sample of 25 passengers is weighed. Find the value of k such that the probability that the sample mean mass exceeds k kg is 0.05.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
E(M − B) = 78 − 15 = 63
✓ 1
Var(M − B) = 144 + 16 = 160
✓ 1
P(M − B > 70) = 0.290
✓ 1
Part (b)
Mean of total = 11 × (78 + 15) = 1023
✓ 1
Variance of total = 11 × 160 = 1760
✓ 1
P(T > 1100) = 0.0332
✓ 1
Part (c)
11M assumes all eleven passengers have the same mass, but they are different, independent people, so the total is M1 + … + M11
✓ 1
Var(11M) = 121 × 144 = 17424, whereas Var(M1 + … + M11) = 11 × 144 = 1584, so the model greatly overstates the spread
✓ 1
Part (d)
For n passengers, total ~ N(93n, 160n)
✓ 1
Testing values or solving 1100 − 93n ≥ 2.326√(160n): n = 11 gives 0.0332 > 0.01, while n = 10 gives about 1.1×10−5 < 0.01
✓ 1
n = 10
✓ 1
Part (e)
\(\bar{M}\) ~ N(78, \(\frac{144}{25}\)), so the standard deviation is 2.4
✓ 1
Use of inverse normal: k = 78 + 1.645 × 2.4
✓ 1
k = 81.9 kg
✓ 1
Common error: For a difference of independent variables, the variances add: Var(M − B) = 144 + 16, not 144 − 16. Also distinguish the sum of n independent copies (variance nσ²) from a multiple nM (variance n²σ²).
102IBMI-033-06-r1-05Correlation and regressionPaper HLHard10 marks
A pharmacologist measures the concentration, C mg L−1, of a drug in a patient's blood t hours after it was given.
It is believed that the data can be modelled by an equation of the form C = aekt, where a and k are constants.
(a)
Find the equation of the regression line of ln C on t.
(3)
(b)
Hence find the values of a and k.
(3)
(c)
Interpret the value of a in context.
(1)
(d)
The drug is no longer effective when its concentration falls below 5 mg L−1. Use the model to estimate the time after which the drug is no longer effective.
(2)
(e)
Comment on the reliability of your answer to the previous part.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Calculates ln C for each value, e.g. 4.086, 3.784, 3.490, 3.182, 2.890, 2.580
✓ 1
Gradient = −0.301 (−0.30053…)
✓ 1
ln C = −0.301t + 4.39 (intercept 4.3874…)
✓ 1
Part (b)
Uses a = eintercept with the unrounded intercept, e4.3874…
✓ 1
a = 80.4
✓ 1
k = −0.301
✓ 1
Part (c)
The model estimates the concentration immediately after the drug is given (t = 0) as about 80.4 mg L−1
✓ 1
Part (d)
Sets up 80.43e−0.3005t = 5 and solves it by graph or by logarithms
✓ 1
t = 9.24 hours
✓ 1
Part (e)
9.24 hours is beyond the last measurement at 6 hours, so it is extrapolation and may be unreliable
✓ 1
Exam tip: Keep the unrounded intercept from the ln C regression when you calculate a = eiⁿtercept. Rounding to 4.39 first gives a = 80.6, which loses accuracy.
103IBMI-033-01-r2-06Correlation and regressionPaper HLHard12 marks
An oceanographer measures the speed, v m s−1, of waves travelling over water of depth d m at five coastal sites.
d: 1, 2, 4, 8, 16 v: 3.0, 4.3, 6.1, 8.4, 12.2
She proposes a model of the form \(v = ad^{\,b}\).
(a)
By considering ln v and ln d, find the equation of the regression line of ln v on ln d.
(3)
(b)
Hence find the values of a and b.
(2)
(c)
A colleague fits a linear model, v = md + c, to the original data. Find the coefficient of determination, R2, for this linear model.
(2)
(d)
The regression of ln v on ln d has R2 = 0.9996. Give two reasons why the power model is more appropriate than the linear model.
(2)
(e)
Use the power model to estimate the wave speed where the water depth is 12 m.
(2)
(f)
Comment on the reliability of using the model to estimate the wave speed where the depth is 100 m.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Takes natural logs of both variables and uses the GDC (M1)
✓ 1
gradient = 0.501 (A1)
✓ 1
intercept = 1.10, so ln v = 0.501 ln d + 1.10 (A1)
✓ 1
Part (b)
a = e1.10… = 3.02 (A1)
✓ 1
b = 0.501 (A1)
✓ 1
Part (c)
Uses the GDC for linear regression of v on d, giving v = 0.589d + 3.15 (M1)
✓ 1
R² = 0.975 (A1)
✓ 1
Part (d)
The residuals from the linear model show a systematic pattern. At d = 1, 2, 4, 8 and 16 they are about −0.74, −0.03, +0.60, +0.54 and −0.37: negative, then positive, then negative. This indicates curvature that the power model captures, since it fits the transformed data almost perfectly (R² = 0.9996) (R1)
✓ 1
The linear model has intercept c ≈ 3.15. It predicts a wave speed of about 3.15 m s−1 at zero depth, which is unrealistic, whereas the power model gives v = 0 when d = 0 (R1)
✓ 1
[Accept other valid reasons, e.g. the power model's predictions are much closer to every observed value.]
—
Part (e)
Substitutes d = 12: v = 3.02 × 120.501 (M1)
✓ 1
v = 10.5 m s−1 (A1)
✓ 1
Part (f)
Unreliable: 100 m is far outside the data range of 1 m to 16 m (extrapolation), and the relationship may change in deeper water (R1)
✓ 1
Exam tip: For a power model \(y = ax^b\), the gradient of the ln y against ln x line is b and the intercept is ln a. R² for ln y against ln x is not on the same scale as R² for y against x, so support a choice of model with residuals or context as well.
104IBMI-033-03-r2-06Normal distribution and chi-squared testsPaper HLHard16 marks
A seed company claims that the heights, H cm, of its tomato seedlings three weeks after planting are normally distributed, with 15% of seedlings shorter than 8 cm and 20% taller than 14 cm.
(a)
Find the mean μ and standard deviation σ of H implied by the company's claim.
(5)
(b)
Using the claim and the unrounded values of μ and σ from (a), find P(8 ≤ H < 11) and P(11 ≤ H < 14).
(3)
(c)
A gardener measures 200 of the company's seedlings. The results are: H < 8: 22; 8 ≤ H < 11: 70; 11 ≤ H < 14: 61; H ≥ 14: 47. She tests the company's claim at the 5% significance level. State the null hypothesis.
(1)
(d)
Find the expected frequencies for the four height groups.
(2)
(e)
Write down the number of degrees of freedom and find the χ2 statistic.
(3)
(f)
The critical value for this test is 7.815. State the gardener's conclusion, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Use of inverse normal for 0.15 and 0.80 (M1)
✓ 1
z = −1.0364 and z = 0.8416 (A1)
✓ 1
μ − 1.0364σ = 8 and μ + 0.8416σ = 14 (M1)
✓ 1
σ = 3.19 cm (3.1948…) (A1)
✓ 1
μ = 11.3 cm (11.311…) (A1)
✓ 1
Part (b)
Use of normal cdf with the unrounded μ and σ from (a) (M1)
✓ 1
P(8 ≤ H < 11) = 0.311 (0.3112…) (A1)
✓ 1
P(11 ≤ H < 14) = 0.339 (0.3388…) (A1)
✓ 1
(Accept 0.312 and 0.338 from μ = 11.3, σ = 3.19.)
—
Part (c)
H0: the heights follow the distribution claimed by the company, N(11.3, 3.192) (A1)
✓ 1
Part (d)
H < 8: 30 and H ≥ 14: 40 (A1)
✓ 1
8 ≤ H < 11: 62.2 and 11 ≤ H < 14: 67.8 (A1)
✓ 1
(Accept 62.5 and 67.5 following through from 0.312 and 0.338.)
—
Part (e)
Degrees of freedom = 4 − 1 = 3, because the parameters come from the claim and are not estimated from the data (A1)
✓ 1
Correct substitution into Σ(O − E)2/E (M1)
✓ 1
χ2 = 5.00 (accept 4.88 following through from 62.5 and 67.5) (A1)
✓ 1
Part (f)
5.00 < 7.815, so do not reject H0 (R1)
✓ 1
There is insufficient evidence to doubt the company's claim about the seedling heights (A1)
✓ 1
Do not award A1 without R1.
—
Exam tip: Degrees of freedom are reduced only for parameters estimated from the sample data. Here μ and σ come from the company's claim, so df = 4 − 1 = 3. Keep unrounded values of μ and σ in your calculator for later parts.
105IBMI-033-08-r2-06Normal distribution and chi-squared testsPaper HLHard13 marks
A biologist measures the lengths, in cm, of 200 trout caught in a lake. From the data, she estimates the mean length to be 32.0 cm and the standard deviation to be 4.0 cm. The lengths are grouped as follows.
Length L < 26: 18 26 ≤ L < 30: 41 30 ≤ L < 34: 72 34 ≤ L < 38: 52 L ≥ 38: 17
She wants to test, at the 5% significance level, whether the lengths can be modelled by a normal distribution.
(a)
State suitable hypotheses for this test.
(1)
(b)
Show that the expected frequency for the class 30 ≤ L < 34 is 76.6, correct to 3 significant figures.
(3)
(c)
Find the expected frequencies for the remaining four classes.
(2)
(d)
Determine the number of degrees of freedom for this test, giving a reason.
(2)
(e)
Find the χ2 test statistic and the p-value.
(3)
(f)
State the conclusion of the test in context.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
H0: trout lengths follow a normal distribution (accept N(32.0, 4.02)); H1: trout lengths do not follow a normal distribution
✓ 1
Part (b)
Uses L ~ N(32.0, 4.02) with P(30 ≤ L < 34)
✓ 1
Probability = 0.38292…
✓ 1
200 × 0.38292… = 76.58… ≈ 76.6
✓ 1
Part (c)
Probabilities 0.0668 and 0.2417 used (or symmetry noted)
✓ 1
L < 26: 13.4; 26–30: 48.3; 34–38: 48.3; L ≥ 38: 13.4
✓ 1
Part (d)
5 − 1 − 2 = 2 degrees of freedom
✓ 1
Reason: two parameters (mean and standard deviation) were estimated from the sample, in addition to the total being fixed
✓ 1
Part (e)
Correct use of \(\sum \frac{(O-E)^2}{E}\) with the five expected frequencies found above
✓ 1
χ2calc = 4.27 (4.268…)
✓ 1
p-value = 0.118 (with 2 degrees of freedom)
✓ 1
Part (f)
0.118 > 0.05, so do not reject H0
✓ 1
There is insufficient evidence to reject a normal model for the trout lengths; N(32.0, 4.02) is a reasonable model
✓ 1
Common error: Calculator goodness of fit programs often default to (number of classes − 1) degrees of freedom. Here that gives 4 instead of 2, so enter the degrees of freedom manually or the p-value will be wrong.
We use essential cookies to make this site work. With your permission we'd also use analytics and advertising cookies to understand which resources help students most and how families find us. Nothing optional is set unless you accept. Privacy policy
EssentialNeeded for the site and your cookie choice to work.
Always on
AnalyticsAnonymous visit statistics that help improve the free resources.
MarketingMeasures whether ads reach the right families. Off unless you turn it on.