480 free, high-quality past-paper-style questions on Statistics and probability for IB Maths AA SL, each with a full mark scheme. Covers: Sampling, data and outliers; Presentation of data; Correlation and regression of y on x; and more. For when you have finished the past papers, or want to practise one topic at a time.Topic 4 at Higher Level →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.3,687 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1AASL4-q00014.1 Sampling, data and outliersPaper 1Easy9 marks
A school has 600 students: 240 in Grade 10, 200 in Grade 11 and 160 in Grade 12. The principal wants to survey a sample of 60 students about how they travel to school.
(a)(i)
State whether each of the following variables is discrete or continuous.
The number of siblings a student has.
(1)
(a)(ii)
The time a student takes to travel to school.
(1)
(b)
A stratified sample by grade is chosen. Find the number of students from each grade in the sample.
(2)
(c)
A teacher suggests surveying the first 60 students who arrive at school on Monday. Name this sampling technique and give one reason why it may produce a biased sample.
(2)
(d)
Describe how a systematic sample of 60 students could be chosen from an alphabetical list of all 600 students.
(1)
(e)
State one difference between a quota sample and a stratified sample.
(1)
(f)
One student in the sample leaves the question about travel time blank. Suggest how this missing value should be dealt with when the data are analysed.
(1)
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Notes
(a)(i)
discrete
A1
(a)(ii)
continuous
A1
(b)
\(\dfrac{240}{600}\times60\) (or using the fraction \(\tfrac{60}{600}=\tfrac{1}{10}\) of each grade)
(M1)
Grade 10: 24, Grade 11: 20, Grade 12: 16
A1
All three required.
(c)
convenience sampling
A1
e.g. the earliest arrivals are likely to share a way of travelling (for example they come by car or live nearby), so they are not representative of all students
R1
Accept any reason that refers to the sample not being representative of the population.
(d)
choose a random starting student among the first 10 on the list, then select every 10th student after that
A1
Must see the interval \(600\div60=10\) and a random start.
(e)
in a quota sample the members of each group are not chosen at random (they are chosen until each quota is filled), whereas in a stratified sample they are chosen randomly from each stratum
A1
Accept any valid distinction about random selection within groups.
(f)
leave that student out of the analysis of travel time (use the other 59 responses), rather than recording the value as 0 or guessing it
R1
Accept replacing the student with another randomly chosen student from the same grade.
2AASL4-q00024.1 Sampling, data and outliersPaper 1Easy7 marks
A city council wants to know the opinion of the 24 000 adults who live in the city about a proposed new cycle lane. The council places a poll on the website of a local cycling club and receives 850 responses. A local newspaper later reports that “70% of residents support the cycle lane”, based on these responses.
(a)
State the population the council is interested in.
(1)
(b)
Write down the size of the sample.
(1)
(c)
Explain why this sample is likely to be biased.
(2)
(d)
Describe how the council could select a simple random sample of 400 adults from the city.
(2)
(e)
Suggest one reason why the newspaper’s figure of 70% may be unreliable, other than the method of sampling.
(1)
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Notes
(a)
all (24 000) adults who live in the city
A1
Do not accept “people who use the website” or “cyclists”.
(b)
\(850\)
A1
(c)
the respondents chose themselves from visitors to a cycling-club website (self-selected / convenience sample)
R1
so cyclists, who are more likely to support a cycle lane, are over-represented; adults who do not cycle or do not visit the site are unlikely to be included
R1
Must refer to the context for this mark.
(d)
use a complete list of the adults in the city (e.g. the electoral register) and number them 1 to 24 000
M1
use a random number generator to choose 400 different numbers (ignoring repeats) and survey those adults
A1
Award A1 only if each adult has an equal chance of being chosen.
(e)
e.g. the 70% applies only to those who responded, not to all residents; the newspaper may have its own interest in the result; people could respond more than once online
R1
Accept any reasonable reason relating to the reliability of the source or of the data.
Answers: (a) all adults living in the city · (b) 850
3AASL4-q00034.1 Sampling, data and outliersPaper 1Easy6 marks
A factory makes 2000 light bulbs each day on three machines. Every bulb is given a serial number as it is made. A quality inspector needs to test 40 of the bulbs made on one day.
(a)(i)
Name the sampling technique described in each of the following.
The inspector tests the first 40 bulbs made that morning.
(1)
(a)(ii)
The inspector uses a random number generator to choose 40 different serial numbers from 1 to 2000.
(1)
(a)(iii)
The inspector tests every 50th bulb made, starting with a bulb chosen at random from the first 50.
(1)
(a)(iv)
The inspector chooses bulbs from each machine in proportion to the number of bulbs it makes, selecting them at random.
(1)
(b)
The bulbs made in the afternoon are more likely to be faulty. State, with a reason, which method in part (a) is most likely to underestimate the proportion of faulty bulbs.
(1)
(c)
Machine 1 has 50 moulds that are used in turn, so every 50th bulb comes from the same mould. Explain why the method in part (a)(iii) may give a misleading result.
(1)
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Mark
Notes
(a)(i)
convenience sampling
A1
(a)(ii)
simple random sampling
A1
(a)(iii)
systematic sampling
A1
(a)(iv)
stratified sampling
A1
Do not accept quota: the bulbs are chosen at random within each machine.
(b)
method (i) (convenience): all the bulbs tested were made in the morning, so the faulty afternoon bulbs are not included
R1
(c)
the sample may contain bulbs from only one mould (of machine 1), so a faulty (or perfect) mould would be over-represented and the sample would not represent all the moulds
R1
Answers: (a)(i) convenience (ii) simple random (iii) systematic (iv) stratified · (b) (i)
4AASL4-q00104.2 Presentation of dataPaper 1Easy8 marks
The times, in minutes, taken by 15 students to complete a quiz are
Time (min)
23
17
29
12
26
21
33
18
52
24
20
30
15
27
21
(a)
Find the median, the lower quartile and the upper quartile of the times.
(3)
(b)
Show that the time of 52 minutes is an outlier.
(2)
(c)
There are no other outliers. Draw a box-and-whisker diagram to represent the data on a scale from 10 to 55 minutes.
Answers: (a) median 23, \(Q_{1}=18\), \(Q_{3}=29\) · (c) box 18–29, median 23, whiskers 12 and 33, cross at 52
Model sketch
Model sketch
5AASL4-q00124.2 Presentation of dataPaper 1Easy7 marks
The box-and-whisker diagram shows the departure delays, in minutes, of 120 flights from an airport on one day.
Diagram drawn to scale
(a)
Write down the median delay.
(1)
(b)
Find the interquartile range.
(1)
(c)
Write down the approximate number of flights delayed by between 14 and 27 minutes.
(1)
(d)
The airport aims for at least 75% of its flights to be delayed by less than 30 minutes. Determine, with a reason, whether this aim was met on this day.
(2)
(e)
Explain why the upper whisker ends at 44 rather than at 52.
(1)
(f)
Explain why the delays are unlikely to be normally distributed.
(1)
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Mark
Notes
(a)
\(19\) minutes
A1
(b)
\(27-14=13\) minutes
A1
(c)
\(60\) flights
A1
Half of 120.
(d)
the upper quartile is 27, so (about) 75% of the flights were delayed by at most 27 minutes
R1
\(27\lt 30\), so the aim was met
A1
Do not award A1R0.
(e)
the whisker extends to the largest value that is not an outlier; 52 is more than \(27+1.5\times13=46.5\), so it is an outlier and is shown separately with a cross
R1
(f)
the box and whiskers are not symmetrical: the median is much closer to \(Q_{1}\) than to \(Q_{3}\) and the upper whisker is much longer than the lower whisker
6AASL4-q00294.4 Correlation and regression of y on xPaper 1Easy5 marks
The four scatter diagrams A, B, C and D are drawn on the same scales.
Diagram drawn to scale
(a)
The values of Pearson’s product-moment correlation coefficient, \(r\), for the four data sets are \(0.98\), \(-0.74\), \(-0.04\) and \(-0.99\). Match each diagram with its value of \(r\).
(3)
(b)
Using diagram C, explain why a value of \(r\) close to zero does not mean that there is no relationship between \(x\) and \(y\).
(1)
(c)
Diagram A shows, for 12 summer days, the number of ice creams sold at a beach, \(x\) (hundreds), and the number of people treated for sunburn, \(y\) (tens). A journalist says that eating ice cream causes sunburn. Comment on this claim.
(1)
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Mark
Notes
(a)
A: \(0.98\)
A1
B: \(-0.74\) and D: \(-0.99\)
A1
Both required for this mark.
C: \(-0.04\)
A1
(b)
the points in C follow a clear curve (U-shape); \(r\) only measures the strength of a linear relationship
R1
(c)
correlation does not imply causation; both variables are probably caused by a third factor, such as hot sunny weather
8AASL4-q00404.5 Probability and expected occurrencesPaper 1Easy7 marks
A fair four-sided die, numbered 1 to 4, and a fair six-sided die, numbered 1 to 6, are rolled. The score, \(D\), is the positive difference between the two numbers obtained (so \(D=0\) when the numbers are equal).
(a)
Find \(\mathrm{P}(D=0)\).
(2)
(b)
Find \(\mathrm{P}(D\ge 3)\).
(2)
(c)
The two dice are rolled 120 times. Find the expected number of times that \(D=1\).
(2)
(d)
In an experiment the two dice were rolled 120 times and \(D=1\) occurred 41 times. Write down the relative frequency of \(D=1\) in this experiment.
(1)
Show mark scheme
Working
Mark
Notes
(a)
sample space of 24 equally likely outcomes, e.g. a table of differences
(M1)
\(\dfrac{4}{24}=\dfrac{1}{6}\)
A1
(b)
\(4+2+1=7\) outcomes with \(D=3,\,4,\,5\)
(A1)
\(\dfrac{7}{24}\)
A1
(c)
\(\mathrm{P}(D=1)=\dfrac{7}{24}\), and \(120\times\dfrac{7}{24}\)
9AASL4-q00414.5 Probability and expected occurrencesPaper 1Easy5 marks
A spinner is divided into four sectors coloured red, blue, yellow and green. The angles of the sectors are shown in the diagram, which is drawn to scale. When the spinner is spun, the probability that it stops on a colour is proportional to the angle of that sector.
(a)
Find the probability that the spinner stops on red.
(1)
(b)
Find the probability that the spinner does not stop on yellow.
(2)
(c)
The spinner is spun 240 times. Find the expected number of times it stops on green or blue.
(2)
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Working
Mark
Notes
(a)
\(\dfrac{150}{360}=\dfrac{5}{12}\)
A1
(b)
\(1-\dfrac{75}{360}\)
(M1)
Or \(\dfrac{150+90+45}{360}\).
\(\dfrac{285}{360}=\dfrac{19}{24}\)
A1
(c)
\(\mathrm{P}(\text{green or blue})=\dfrac{135}{360}=\dfrac{3}{8}\), and \(240\times\dfrac{3}{8}\)
10AASL4-q00464.6 Combined events and diagramsPaper 1Easy7 marks
In a group of 50 students, 20 play a musical instrument (\(M\)), 26 play a sport (\(S\)) and 12 do neither. The Venn diagram shows the events \(M\) and \(S\). The values \(p\), \(q\), \(r\) and \(s\) are the numbers of students in each region.
Diagram not to scale
(a)
Find the values of \(p\), \(q\) and \(r\).
(3)
(b)
A student is chosen at random from the group. Find the probability that the student does exactly one of the two activities.
(2)
(c)
A student is chosen at random from those who play a sport. Find the probability that this student also plays a musical instrument.
13AASL4-q00514.5 Probability and expected occurrencesPaper 1Easy6 marks
Spinner P has three equal sectors numbered 1, 2 and 3. Spinner Q has four equal sectors numbered 1, 2, 3 and 4. Both spinners are spun and the score is the product of the two numbers obtained.
(a)
Draw a sample space diagram showing all the possible scores.
(2)
(b)
Find the probability that the score is an even number.
(2)
(c)
Given that at least one of the spinners shows a 3, find the probability that the score is greater than 6.
14AASL4-q00604.7 Discrete random variablesPaper 1Easy7 marks
At a school fair, a player pays $5 to draw one card at random from 20 cards. One card is gold, four cards are silver and the rest are white. A gold card wins $40, a silver card wins $10 and a white card wins nothing. Let \(X\) be the player’s gain, in dollars, from one game (the amount won minus the $5 paid).
(a)
Find the probability distribution of \(X\), giving your answer as a table.
(3)
(b)
Find \(\mathrm{E}(X)\).
(2)
(c)
Explain whether the game is fair.
(1)
(d)
The game is played 200 times during the fair. Find the profit that the fair organisers can expect to make.
(1)
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Working
Mark
Notes
(a)
values of \(X\): \(35,\ 5,\ -5\)
A1
\(\mathrm{P}(X=35)=\dfrac{1}{20}\) and \(\mathrm{P}(X=5)=\dfrac{4}{20}\)
15AASL4-q00644.7 Discrete random variablesPaper 1Easy8 marks
The probability that Kai passes his driving test at any attempt is 0.6, independently of previous attempts. He will take the test until he passes, but he is allowed at most three attempts. Let \(X\) be the number of attempts that Kai takes.
Diagram not to scale
(a)
Show that \(\mathrm{P}(X=3)=0.16\).
(2)
(b)
Find \(\mathrm{E}(X)\).
(3)
(c)
Each attempt costs €45. Find the expected total amount that Kai pays for his attempts.
(1)
(d)
Find the probability that Kai passes his driving test.
(2)
Show mark scheme
Working
Mark
Notes
(a)
he fails the first two attempts: \(0.4\times0.4\)
M1
Award M0 for \(0.4\times0.4\times0.6\).
third attempt is taken whatever its result, so \(\mathrm{P}(X=3)=0.4^{2}=0.16\)
16AASL4-q00814.9 The normal distributionPaper 1Easy6 marks
The random variables \(X\), \(Y\) and \(W\) are normally distributed, with \(X\sim\mathrm{N}(50,\,5^{2})\), \(Y\sim\mathrm{N}(50,\,10^{2})\) and \(W\sim\mathrm{N}(65,\,5^{2})\). Their probability density curves are labelled \(P\), \(Q\) and \(R\) on the diagram.
Graph drawn to scale
(a)
Write down which curve represents each of \(X\), \(Y\) and \(W\).
(2)
(b)(i)
Using the fact that approximately 68% of a normal distribution lies within one standard deviation of the mean and 95% lies within two standard deviations of the mean, estimate
\(\mathrm{P}(X\gt 60)\);
(1)
(b)(ii)
\(\mathrm{P}(Y\gt 60)\);
(2)
(b)(iii)
\(\mathrm{P}(W\gt 60)\).
(1)
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Mark
Notes
(a)
\(X\): curve \(R\) and \(W\): curve \(Q\)
A1
Same spread, different centres.
\(Y\): curve \(P\)
A1
Award A1A1 for all three correct; A1A0 for one correct.
(b)(i)
\(60=\mu+2\sigma\), so \(\mathrm{P}(X\gt 60)\approx\dfrac{1-0.95}{2}=0.025\)
A1
Accept 2.5%.
(b)(ii)
\(60=\mu+\sigma\) for \(Y\), so \(\mathrm{P}(Y\gt 60)=\dfrac{1-0.68}{2}\)
(M1)
\(0.16\)
A1
Accept 16%.
(b)(iii)
\(60=\mu-\sigma\) for \(W\), so \(\mathrm{P}(W\gt 60)\approx0.5+0.34=0.84\)
19AASL4-q01074.11 Conditional probability and independencePaper 1Easy7 marks
An integer \(n\) is chosen at random from the integers 1 to 30 inclusive. Let \(A\) be the event “\(n\) is a multiple of 3”, \(B\) the event “\(n\) is a multiple of 5” and \(C\) the event “\(n\) is a prime number”.
(a)
Write down \(\mathrm{P}(A)\).
(1)
(b)
Find \(\mathrm{P}(A\mid B)\).
(2)
(c)
State, with a reason, whether \(A\) and \(B\) are independent.
(1)
(d)
Find \(\mathrm{P}(A\mid C)\) and hence determine whether \(A\) and \(C\) are independent.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac{10}{30}=\dfrac{1}{3}\)
A1
(b)
\(B=\{5,10,15,20,25,30\}\), of which 15 and 30 are multiples of 3
(M1)
Or \(\dfrac{\mathrm{P}(A\cap B)}{\mathrm{P}(B)}=\dfrac{2/30}{6/30}\).
\(\dfrac{2}{6}=\dfrac{1}{3}\)
A1
(c)
independent, because \(\mathrm{P}(A\mid B)=\dfrac{1}{3}=\mathrm{P}(A)\)
R1
Accept \(\mathrm{P}(A\cap B)=\dfrac{1}{15}=\dfrac{1}{3}\times\dfrac{1}{5}\). FT from (a) and (b).
(d)
10 primes \(\{2,3,5,7,11,13,17,19,23,29\}\), of which only 3 is a multiple of 3
(M1)
\(\mathrm{P}(A\mid C)=\dfrac{1}{10}\)
A1
\(\dfrac{1}{10}\ne\dfrac{1}{3}=\mathrm{P}(A)\), so \(A\) and \(C\) are not independent
20IBMA2S-050-02-024.9 The normal distributionPaper 1Easy5 marks
No calculator
The finishing times, T minutes, of runners in a city marathon are modelled by a normal distribution with mean 240 minutes and standard deviation σ minutes. It is known that P(T > 270) = 0.12.
Diagram not to scale
(a)
Write down P(T < 210), justifying your answer.
(2)
(b)
Find P(210 < T < 270).
(1)
(c)
Find P(240 < T < 270).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 the normal curve is symmetric about the mean 240, and 210 and 270 are both 30 minutes from the mean ; A1 P(T < 210) = 0.12
✓ 2
Part (b)
A1 1 − 2(0.12) = 0.76
✓ 1
Part (c)
M1 use of P(T > 240) = 0.5, e.g. 0.5 − 0.12 ; A1 0.38
✓ 2
Exam tip: In no-calculator normal questions, the mean splits the distribution into two halves of 0.5 each. Most answers come from combining 0.5 with the given tail probability.
21IBMA2S-050-07-024.7 Discrete random variablesPaper 1Easy6 marks
No calculator
In a mobile game, the number of bonus coins, X, awarded after each level is a discrete random variable with probability distribution
\(P(X=x)=\frac{k}{x}\) for \(x\in\{1,2,3,6\}\),
where k is a constant.
(a)
Show that \(k=\frac{1}{2}\).
(2)
(b)
Find E(X).
(2)
(c)
After one level, a player is told that they received at most 3 coins. Find the probability that they received exactly 1 coin.
Common error: In the conditional part, the denominator is P(X ≤ 3) = 11/12, not 1. Restrict to the given event before dividing.
22IBMA2S-050-07-034.7 Discrete random variablesPaper 1Easy6 marks
No calculator
At a school sports-day stall, a player pays $4 to roll a fair eight-sided die numbered 1 to 8. • If the die shows 8, the player receives $20. • If the die shows 6 or 7, the player receives $5. • Otherwise, the player receives nothing.
Let X be the player's profit, in dollars, from one game (the amount received minus the $4 paid).
(a)
Write down the probability distribution of X.
(2)
(b)
Find E(X).
(2)
(c)
The organisers change the top prize (for rolling an 8) to $W, keeping everything else the same, so that the game is fair. Find W.
M1 sets expected profit to zero, e.g. \(\frac{W-4}{8}+\frac{2}{8}-\frac{20}{8}=0\) or \(\frac{W}{8}+\frac{10}{8}=4\)
✓ 1
A1 \(W=22\)
✓ 1
Common error: Profit is the prize minus the stake. Using the prize amounts (20, 5, 0) as the values of X gives expected winnings, not expected profit.
23IBMA2S-050-07-044.6 Combined events and diagramsPaper 1Easy6 marks
No calculator
A box of 7 replacement sensors for a wind turbine contains 2 defective sensors. An engineer selects 2 sensors at random from the box without replacement. Let D be the number of defective sensors selected.
(a)
Find the probability distribution of D.
(4)
(b)
Find E(D).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 multiplies conditional probabilities for selection without replacement
Common error: P(D = 1) needs the factor 2: a defective sensor can be chosen first or second. Check your three probabilities add to 1.
24IBMA2S-050-01-014.9 The normal distributionPaper 1Easy4 marks
No calculator
A drone delivery company finds that the probability that any one delivery arrives late is 0.04, independently of all other deliveries.
(a)
In one week the company makes 350 deliveries. Find the expected number of late deliveries.
(2)
(b)
Two deliveries are chosen at random. Find the probability that both arrive on time.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses expected number = \(np\), i.e. \(350 \times 0.04\)
✓ 1
A1 14
✓ 1
Part (b)
M1 uses P(on time) = 0.96 and multiplies for independent events, \(0.96 \times 0.96\)
✓ 1
A1 0.9216
✓ 1
Exam tip: An expected number does not have to be a whole number, but here 14 happens to be exact. Never round an expected value to an integer unless the question asks.
25IBMA2S-050-01-034.8 The binomial distributionPaper 1Easy5 marks
No calculator
A tennis player's first serve lands in with probability \(\frac{3}{5}\), independently on each serve.
(a)
In a practice session she hits 40 first serves. Find the expected number that land in.
(2)
(b)
She hits 4 first serves. Find the probability that exactly 3 of them land in. Give your answer as a fraction in its simplest form.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(40 \times \frac{3}{5}\)
✓ 1
A1 24
✓ 1
Part (b)
M1 recognises binomial with \(n = 4\), \(p = \frac{3}{5}\)
Common error: Forgetting the factor \(\binom{4}{3} = 4\) gives only the probability of one particular order (in, in, in, out), not of exactly 3 in any order.
26IBMA2S-050-04-014.11 Conditional probability and independencePaper 1Easy6 marks
No calculator
A phone manufacturer records settings on its handsets. Let \(A\) be the event that a randomly chosen handset has battery-saver mode switched on, and \(B\) the event that it has dark mode switched on. It is known that \(P(A)=0.6\), \(P(B)=0.35\) and \(P(B\mid A)=0.25\).
(a)
Find \(P(A\cap B)\).
(2)
(b)
Find \(P(A\cup B)\).
(2)
(c)
Determine whether \(A\) and \(B\) are independent. Justify your answer.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(P(A\cap B)=P(B\mid A)\,P(A)=0.25\times0.6\)
✓ 1
A1 \(P(A\cap B)=0.15\)
✓ 1
Part (b)
M1 use of \(P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.6+0.35-0.15\)
✓ 1
A1 \(P(A\cup B)=0.8\)
✓ 1
Part (c)
M1 calculates \(P(A)P(B)=0.6\times0.35=0.21\) (or compares \(P(B\mid A)=0.25\) with \(P(B)=0.35\))
✓ 1
R1 \(0.21\neq0.15\) (or \(0.25\neq0.35\)), so \(A\) and \(B\) are not independent
✓ 1
Exam tip: For an independence test, state both values you are comparing and then say clearly that they are unequal; a bare 'not independent' earns no reasoning mark.
27IBMA2S-050-04-024.11 Conditional probability and independencePaper 1Easy5 marks
No calculator
A sports clinic follows 200 junior athletes for one season. 60 are swimmers, 80 are cyclists and 60 are runners; each athlete does exactly one of these sports. During the season, 9 of the swimmers, 20 of the cyclists and 21 of the runners suffer an injury. One athlete is chosen at random.
(a)
Find the probability that the athlete was injured.
(1)
(b)
Given that the athlete was injured, find the probability that the athlete is a cyclist.
(2)
(c)
Determine whether the event 'the athlete was injured' is independent of the event 'the athlete is a cyclist'. Justify your answer.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(\frac{50}{200}=\frac{1}{4}\)
✓ 1
Part (b)
M1 restricts to the 50 injured athletes, \(\frac{20}{50}\)
R1 this equals \(P(\text{injured})=\frac{1}{4}\), so the events are independent
✓ 1
Common error: In (b) the denominator is the number injured (50), not the number of cyclists (80). Identify the 'given' group first and use its total.
28IBMA2S-050-04-034.11 Conditional probability and independencePaper 1Easy5 marks
No calculator
Two satellites, X and Y, independently attempt to detect a signal from a ground station. Satellite X detects the signal with probability \(p\) and satellite Y detects it with probability \(0.4\). The probability that at least one satellite detects the signal is \(0.7\).
(a)
Find the value of \(p\).
(3)
(b)
Find the probability that exactly one of the satellites detects the signal.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(P(X\cup Y)=P(X)+P(Y)-P(X)P(Y)\) with independence (or \(1-(1-p)(0.6)=0.7\))
✓ 1
M1 forms equation \(p+0.4-0.4p=0.7\)
✓ 1
A1 \(p=0.5\)
✓ 1
Part (b)
M1 \(0.5\times0.6+0.5\times0.4\)
✓ 1
A1 \(0.5\)
✓ 1
Exam tip: 'At least one' is often quickest as 1 − P(neither): here \(1-(1-p)(0.6)=0.7\) gives \(p\) in one line.
29IBMA2S-050-04-044.6 Combined events and diagramsPaper 1Easy4 marks
No calculator
On a delivery drone's test flight, let \(M\) be the event that a motor fault occurs and \(G\) the event that a GPS fault occurs. \(P(M)=0.3\) and \(P(G)=0.2\).
(a)
Find \(P(M\cup G)\) if \(M\) and \(G\) are mutually exclusive.
(1)
(b)
Find \(P(M\cup G)\) if instead \(M\) and \(G\) are independent.
(2)
(c)
Assuming \(M\) and \(G\) are independent, write down \(P(M\mid G)\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(0.3+0.2=0.5\)
✓ 1
Part (b)
M1 \(P(M\cap G)=0.3\times0.2=0.06\) used in \(0.3+0.2-0.06\)
✓ 1
A1 \(0.44\)
✓ 1
Part (c)
A1 \(P(M\mid G)=P(M)=0.3\)
✓ 1
Common error: Mutually exclusive and independent are different: mutually exclusive means \(P(M\cap G)=0\), independent means \(P(M\cap G)=P(M)P(G)\).
30IBMA2S-050-05-014.2 Presentation of dataPaper 1Easy6 marks
No calculator
A technology reviewer tested the battery life, in hours, of a sample of smartphones under identical conditions. The five-number summary of the results is: minimum = 9, lower quartile = 14, median = 17, upper quartile = 20, maximum = 31. An outlier is defined as a value more than 1.5 × IQR above the upper quartile or more than 1.5 × IQR below the lower quartile.
Diagram not to scale
(a)
Write down the interquartile range.
(1)
(b)
Determine whether the maximum value of 31 hours is an outlier.
(3)
(c)
Find the lower boundary for outliers and hence state whether the minimum value is an outlier.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 IQR = 20 − 14 = 6 hours
✓ 1
Part (b)
M1 upper boundary = Q3 + 1.5 × IQR = 20 + 1.5 × 6 A1 = 29 hours R1 31 > 29, so 31 hours is an outlier
✓ 3
Part (c)
A1 lower boundary = Q1 − 1.5 × IQR = 14 − 1.5 × 6 = 5 hours R1 9 > 5, so the minimum is not an outlier
✓ 2
Exam tip: Always compare the data value with the boundary and state the conclusion in words. A correct boundary with no conclusion does not earn the reasoning mark.
31IBMA2S-050-05-034.2 Presentation of dataPaper 1Easy4 marks
No calculator
A space agency recovered 35 meteorite fragments and recorded their masses, m grams. The data are shown in a histogram with equal class widths of 10 g, where bar heights represent frequencies. The bar heights are: 0 ≤ m < 10: 3 10 ≤ m < 20: 7 20 ≤ m < 30: k 30 ≤ m < 40: 9 40 ≤ m < 50: 4
(a)
Find the value of k.
(2)
(b)
Write down the modal class.
(1)
(c)
Find the number of fragments with a mass of at least 30 g.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 3 + 7 + k + 9 + 4 = 35 A1 k = 12
✓ 2
Part (b)
A1 20 ≤ m < 30
✓ 1
Part (c)
A1 9 + 4 = 13 fragments
✓ 1
Exam tip: With equal class widths, bar height is proportional to frequency, so the bar heights must add to the total number of items.
32IBMA2S-050-00-014.11 Conditional probability and independencePaper 1Easy6 marks
No calculator
A phone manufacturer surveys owners of its latest model. For a randomly chosen owner, let \(W\) be the event that they use wireless charging and \(F\) be the event that they use face unlock. It is known that \(P(W)=0.45\), \(P(F)=0.6\) and \(P(W\cup F)=0.8\).
(a)
Find \(P(W\cap F)\).
(2)
(b)
Determine whether the events \(W\) and \(F\) are independent. Justify your answer.
(2)
(c)
Find the probability that a randomly chosen owner uses wireless charging, given that they use face unlock.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(P(W\cap F)=P(W)+P(F)-P(W\cup F)\), i.e. 0.45 + 0.6 − 0.8
✓ 1
A1 \(P(W\cap F)=0.25\)
✓ 1
Part (b)
M1 \(P(W)\times P(F)=0.45\times0.6=0.27\) (or equivalent: \(P(W\mid F)=\frac{0.25}{0.6}=\frac{5}{12}\) compared with \(P(W)=0.45\))
✓ 1
R1 since \(0.27\neq0.25\) (or \(\frac{5}{12}\neq0.45\)), \(W\) and \(F\) are not independent
Exam tip: To test independence, compare \(P(A)P(B)\) with \(P(A\cap B)\), or compare \(P(A\mid B)\) with \(P(A)\). Then state the conclusion explicitly: the comparison alone does not earn the reasoning mark.
33IBMA2S-050-00-044.11 Conditional probability and independencePaper 1Easy5 marks
No calculator
An email provider estimates that 30% of incoming emails are spam. Its filter flags 95% of spam emails and also wrongly flags 5% of emails that are not spam. An incoming email is chosen at random.
Diagram not to scale
(a)
Find the probability that the email is flagged.
(3)
(b)
Given that the email is flagged, find the probability that it is spam.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 one correct branch product, e.g. \(0.3\times0.95=0.285\) or \(0.7\times0.05=0.035\)
✓ 1
M1 adding the two branch products \(0.285+0.035\)
✓ 1
A1 \(0.32\)
✓ 1
Part (b)
M1 \(\frac{0.285}{0.32}\) (follow through from their answer to (a))
✓ 1
A1 \(\frac{57}{64}\) (accept 0.890625 or 0.891 to 3 s.f.)
✓ 1
Exam tip: Sketch a tree diagram first: the conditional probability in (b) is one branch product divided by the total from (a).
34IBMA2S-050-06-024.3 Measures of central tendency and dispersionPaper 1Easy4 marks
No calculator
Six prototype rocket engines are each fired on a test stand and their burn times, in seconds, are recorded. Five of the burn times are 40, 45, 38, 44, 41 The sixth burn time is t seconds. The mean burn time of all six engines is 42 seconds.
(a)
Find the value of t.
(2)
(b)
Find the median burn time of the six engines.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 total of all six = 6 × 42 = 252
✓ 1
A1 t = 252 − 208 = 44
✓ 1
Part (b)
M1 ordered list 38, 40, 41, 44, 44, 45 and averages 3rd and 4th values
✓ 1
A1 median = 42.5 s
✓ 1
Common error: Forgetting to include the new value 44 when ordering the data, or taking the median of the original five values, gives 41 instead of 42.5.
35IBMA2S-050-06-034.3 Measures of central tendency and dispersionPaper 1Easy5 marks
No calculator
The temperature, T °C, of a component on a satellite is recorded every hour for a week. The readings have mean 12 °C and standard deviation 5 °C.
(a)
Each reading is converted to an absolute temperature K using K = T + 273. Write down the mean and the standard deviation of the values of K.
(2)
(b)
Each reading is instead converted to degrees Fahrenheit using F = 1.8T + 32. Find the mean and the variance of the values of F.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 mean = 285
✓ 1
A1 standard deviation = 5 (unchanged)
✓ 1
Part (b)
A1 mean = 1.8 × 12 + 32 = 53.6
✓ 1
M1 standard deviation = 1.8 × 5 = 9
✓ 1
A1 variance = 92 = 81
✓ 1
Common error: Adding a constant shifts the mean but not the spread; multiplying by 1.8 multiplies the standard deviation by 1.8 but the variance by 1.8². The +32 never affects the spread.
36IBMA2S-050-03-014.4 Correlation and regression of y on xPaper 1Easy5 marks
No calculator
A student records, over 30 days, the daily screen-on time of her phone, x hours, and the percentage of battery used that day, y%. The screen-on times ranged from 1 hour to 7 hours. The mean screen-on time was 4 hours and the mean battery used was 47%. The regression line of y on x has equation \(y = 11x + c\).
(a)
Find the value of c.
(2)
(b)
Interpret the value 11 in the context of the question.
(1)
(c)
Use the regression line to estimate the percentage of battery used on a day with 6 hours of screen-on time.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitutes the mean point (4, 47) into the line: \(47 = 11(4) + c\)
✓ 1
A1 \(c = 3\)
✓ 1
Part (b)
A1 each additional hour of screen-on time is associated with, on average, about 11 percentage points more battery used (11% of the full battery). Accept equivalent wording that refers to an average or predicted increase per hour
✓ 1
Part (c)
M1 substitutes \(x = 6\): \(y = 11(6) + 3\)
✓ 1
A1 \(y = 69\)%
✓ 1
Exam tip: The regression line of y on x always passes through the mean point \((\bar{x}, \bar{y})\). If you are given the means and one coefficient, substitute the mean point to find the other coefficient.
37IBMA2S-050-03-064.4 Correlation and regression of y on xPaper 1Easy5 marks
No calculator
Three data sets are collected and Pearson's product-moment correlation coefficient, r, is calculated for each.
A: running speed (km/h) on a treadmill and heart rate measured by a smartwatch, \(r = 0.92\) B: age of a smartphone (months) and maximum battery capacity (%), \(r = -0.85\) C: shoe size and reaction time in a video game, \(r = 0.04\)
(a)
Describe the correlation shown in data set B.
(2)
(b)
State, with a reason, whether it would be useful to use a regression line to predict reaction time from shoe size.
(1)
(c)
The running speeds in data set A are converted from km/h to m/s. State the value of r for the converted data, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 strong
✓ 1
A1 negative linear correlation: older phones tend to have a lower maximum battery capacity
✓ 1
Part (b)
R1 not useful, because \(r = 0.04\) shows almost no linear correlation
✓ 1
Part (c)
A1 \(r = 0.92\) (unchanged)
✓ 1
R1 converting units multiplies every x-value by the same positive constant, which does not change the strength or direction of the linear correlation
✓ 1
Exam tip: A full description of correlation states both the strength (strong, moderate, weak) and the direction (positive or negative), ideally in the context of the variables.
38IBMA2S-050-03-124.10 Regression of x on yPaper 1Easy5 marks
No calculator
A sports physiologist studies adults aged 20 to 60 years. For each person she records age, x years, and maximum heart rate, y beats per minute (bpm). The regression line of y on x is \(y = 208 - 0.7x\) and \(r = -0.89\).
(a)
Interpret the value −0.7 in context.
(1)
(b)
Estimate the maximum heart rate of a 45-year-old.
(2)
(c)
A parent uses the line to estimate the maximum heart rate of an 8-year-old child. Comment on whether this estimate is reliable.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 maximum heart rate decreases by about 0.7 bpm for each additional year of age
✓ 1
Part (b)
M1 substitutes \(x = 45\): \(208 - 0.7(45)\)
✓ 1
A1 176.5 bpm
✓ 1
Part (c)
R1 age 8 is outside the range of the data (20 to 60 years), so this is extrapolation
✓ 1
R1 the relationship may be different for children, so the estimate is unreliable
✓ 1
Exam tip: A strong value of r does not make a prediction reliable outside the data range. Always check whether the given x-value is interpolation or extrapolation.
39IBMA2S-050-09-014.1 Sampling, data and outliersPaper 1Easy5 marks
No calculator
An engineering consultancy employs 225 mechanical engineers, 150 electrical engineers and 125 software engineers. A survey of a sample of the engineers records two variables for each person: the diameter, in mm, of a test bolt they machine during a skills assessment, and the number of patents they filed last year.
(a)
Classify each of the two variables as discrete or continuous.
(2)
(b)
The sample is stratified by department and contains 18 mechanical engineers. Find the total size of the sample.
(2)
(c)
State one advantage of using a stratified sample rather than a convenience sample in this survey.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 bolt diameter is continuous
✓ 1
A1 number of patents is discrete
✓ 1
Part (b)
M1 \(\frac{225}{500}\times n=18\) or equivalent
✓ 1
A1 \(n=40\)
✓ 1
Part (c)
R1 each department is represented in proportion to its size, so the sample reflects the structure of the population; a convenience sample might over-represent one department and so be biased
✓ 1
Exam tip: In stratified sampling the fraction of the sample taken from each stratum equals that stratum's fraction of the population. Here 225 out of 500 is 45%, so the 18 mechanical engineers must be 45% of the sample.
40IBMA2S-050-09-024.2 Presentation of dataPaper 1Easy5 marks
No calculator
The reaction times of the sprinters at an athletics meeting were recorded by the starting blocks. For this data, the lower quartile is 142 ms and the upper quartile is 171 ms. An outlier is a value more than 1.5 × IQR below the lower quartile or more than 1.5 × IQR above the upper quartile.
(a)
Find the interquartile range.
(1)
(b)
Find the values below and above which a reaction time is an outlier.
(2)
(c)
Two of the recorded times were 95 ms and 210 ms. Determine, giving a reason, whether each is an outlier.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 IQR = 171 − 142 = 29 ms
✓ 1
Part (b)
M1 142 − 1.5 × 29 or 171 + 1.5 × 29
✓ 1
A1 98.5 ms and 214.5 ms
✓ 1
Part (c)
R1 95 < 98.5, so 95 ms is an outlier
✓ 1
R1 98.5 < 210 < 214.5, so 210 ms is not an outlier
✓ 1
Common error: Do not add 1.5 × IQR to the median or to the maximum. The fences are measured out from the quartiles: Q₁ − 1.5 × IQR and Q₃ + 1.5 × IQR.
41IBMA2S-050-09-034.1 Sampling, data and outliersPaper 1Easy5 marks
No calculator
A hospital outpatient clinic wants to survey its patients about their waiting times. Three methods of choosing the patients are proposed. Method I: patients are listed alphabetically, a random starting point is chosen from the first 12 names, and then every 12th patient on the list is selected. Method II: the first 30 patients who arrive on Monday morning are selected. Method III: the receptionist chooses patients until she has 15 from each of three age bands.
(a)
Name the sampling technique used in each method.
(3)
(b)
Identify the method most likely to give a biased view of waiting times across the whole week. Justify your answer.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 Method I: systematic sampling
✓ 1
A1 Method II: convenience sampling
✓ 1
A1 Method III: quota sampling
✓ 1
Part (b)
A1 Method II
✓ 1
R1 only patients arriving on Monday morning are included, so other days and times are excluded. Early arrivals may also wait less than patients who come later, so the sample is not representative of the whole week
✓ 1
Note: the A1 may be awarded only if the R1 is awarded. A comment that Method III is also non-random is acceptable extra detail, but it must not replace the reason given for Method II.
—
Exam tip: Quota and stratified sampling both use groups. In quota sampling the people within each group are not chosen at random, which is the difference to state.
42IBMA2S-050-09-044.1 Sampling, data and outliersPaper 1Easy5 marks
No calculator
A factory produces 3600 smartphone batteries in a shift, numbered 1 to 3600 in the order they are made. A systematic sample of 45 batteries is taken for safety testing by selecting batteries at a fixed interval.
(a)
Find the sampling interval.
(1)
(b)
The first battery selected is number 37. Write down the number of the third battery selected.
(1)
(c)
Find the number of the last battery selected.
(2)
(d)
State one possible disadvantage of systematic sampling in this context.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 3600 ÷ 45 = 80
✓ 1
Part (b)
A1 37 + 2 × 80 = 197
✓ 1
Part (c)
M1 37 + 44 × 80
✓ 1
A1 3557
✓ 1
Part (d)
R1 any one valid disadvantage , for example:
✓ 1
• a periodic fault (e.g. one machine cycle repeating every 80 batteries) could be missed every time or picked every time, giving a biased sample;
—
• not every possible sample of 45 batteries can be chosen, so it is not a simple random sample.
—
Common error: The 45th battery is 44 intervals after the first, not 45. Writing 37 + 45 × 80 = 3637 gives a number outside the production run.
43IBMA2S-050-08-014.10 Regression of x on yPaper 1Easy5 marks
No calculator
A phone manufacturer records, for a sample of users over a 4-hour period, the screen-on time x minutes and the battery drain y%. The regression line of y on x is \(y = 0.5x + 20\). The regression line of x on y is \(x = 1.6y - 26\).
(a)
Find the mean screen-on time \(\bar{x}\) and the mean battery drain \(\bar{y}\).
(3)
(b)
A user's battery drain over the period was 50%. Estimate this user's screen-on time.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 recognises both lines pass through \((\bar{x},\bar{y})\) and attempts to solve simultaneously, e.g. \(x = 1.6(0.5x+20) - 26\)
✓ 1
A1 \(\bar{x} = 30\) minutes
✓ 1
A1 \(\bar{y} = 35\)%
✓ 1
Part (b)
M1 uses the x on y line: \(x = 1.6(50) - 26\)
✓ 1
A1 54 minutes
✓ 1
Exam tip: Both regression lines always pass through the mean point, so solving the two equations simultaneously gives \((\bar{x},\bar{y})\) directly.
44IBMA2S-050-08-054.10 Regression of x on yPaper 1Easy5 marks
No calculator
A forestry team estimates the age of trees in a plantation from their trunk circumference. For a sample of trees of known age, the regression line of age x years on trunk circumference y cm is \(x = 0.8y + 5\).
(a)
Estimate the age of a tree with trunk circumference 60 cm.
(2)
(b)
Interpret the value 0.8 in the context of the question.
(1)
(c)
The mean trunk circumference of the sample was 45 cm. Find the mean age of the trees in the sample.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitutes \(y = 60\): \(x = 0.8(60) + 5\)
✓ 1
A1 53 years
✓ 1
Part (b)
R1 for each extra 1 cm of circumference, the estimated age increases by 0.8 years
✓ 1
Part (c)
M1 recognises the line passes through \((\bar{x},\bar{y})\): \(\bar{x} = 0.8(45) + 5\)
✓ 1
A1 41 years
✓ 1
Exam tip: When interpreting a gradient, give the change in the subject variable per one unit of the other variable, with both units stated.
45IBMA3S-4-16r5-014.11 Conditional probability and independencePaper 1Easy4 marks
No calculator
On a regional airline, the event L is that a flight departs late and the event H is that a flight meets a headwind. It is known that P(L) = 0.60, P(H) = 0.45 and P(L ∪ H) = 0.85.
(a)
Find P(L ∩ H).
(2)
(b)
Determine whether L and H are independent. Justify your answer.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of P(L ∪ H) = P(L) + P(H) − P(L ∩ H)
✓ 1
A1 P(L ∩ H) = 0.2
✓ 1
Part (b)
M1 P(L) × P(H) = 0.60 × 0.45 = 0.27
✓ 1
R1 0.27 ≠ 0.2 so the events are not independent
✓ 1
Common error: To test independence compare P(A ∩ B) with P(A) × P(B). Showing only that the events overlap says nothing about independence.
46IBMA3S-4-16r5-024.7 Discrete random variablesPaper 1Easy5 marks
No calculator
The number of lambs X born to a ewe in a flock has probability distribution P(X = x) = k(x + 2) for x = 0, 1, 2, 3, where k is a constant.
(a)
Find the value of k.
(2)
(b)
Find E(X).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sum of probabilities = 1: 2k + 3k + 4k + 5k = 1
✓ 1
A1 k = 1/14
✓ 1
Part (b)
M1 attempt to use E(X) = ΣxP(X = x)
✓ 1
A1 (0 × 2 + 1 × 3 + 2 × 4 + 3 × 5)/14
✓ 1
A1 E(X) = 26/14 = 13/7
✓ 1
Exam tip: Write out each probability in a table before using E(X) = Σ x P(X = x). Check that your probabilities add to 1.
47IBMA3S-4-16r5-134.6 Combined events and diagramsPaper 1Easy7 marks
No calculator
Events A and B are independent, with P(A) = 0.4 and P(A ∪ B) = 0.7. In an aviation safety study, A is a bird strike on take-off and B is a radio fault.
(a)
Find P(B).
(3)
(b)
Event C is mutually exclusive with A, and P(A ∪ C) = 0.7. Find P(C).
(2)
(c)
Explain why A and B are not mutually exclusive.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of independence: P(A ∩ B) = 0.4P(B)
✓ 1
M1 0.7 = 0.4 + P(B) − 0.4P(B)
✓ 1
A1 P(B) = 0.5
✓ 1
Part (b)
M1 for mutually exclusive events P(A ∪ C) = P(A) + P(C)
✓ 1
A1 P(C) = 0.7 − 0.4 = 0.3
✓ 1
Part (c)
R1 P(A ∩ B) = 0.4 × 0.5 = 0.2
✓ 1
R1 this is not zero, so A and B can occur together
✓ 1
Common error: Independent events (with non-zero probabilities) always overlap. Mutually exclusive events have P(A ∩ B) = 0. These are different ideas.
48IBMA3S-4-16r6-034.2 Presentation of dataPaper 1Easy5 marks
No calculator
A data centre records the number of cooling faults, x, found in each server rack during a week. The results for the racks inspected are shown below, where k is a positive integer.
x
0
1
2
3
4
Frequency
3
k
9
5
2
The mean number of faults per rack is 2.
(a)
Find the value of k.
(3)
(b)
Find the median number of faults.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 total frequency = 19 + k and ∑fx = 41 + k
✓ 1
M1 \(\frac{41+k}{19+k}=2\)
✓ 1
A1 k = 3
✓ 1
Part (b)
M1 22 racks, so median lies between the 11th and 12th values (cumulative frequencies 3, 6, 15)
✓ 1
A1 median = 2
✓ 1
Exam tip: With a frequency table, work out the cumulative frequencies before looking for the middle value.
49IBMA3S-4-16r8-014.6 Combined events and diagramsPaper 1Easy5 marks
No calculator
On summit day of a mountaineering expedition the weather is either clear (C) or poor (C'). P(C) = 0.7. When the weather is clear, the probability that a climber reaches the summit (S) is 0.8. When it is poor, this probability is 0.2.
Tree diagram
(a)
Find P(S).
(3)
(b)
Given that a climber reaches the summit, find the probability that the weather was clear. Give your answer as a fraction.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use P(C ∩ S) = 0.7 × 0.8 = 0.56
✓ 1
M1 attempt to use P(C' ∩ S) = 0.3 × 0.2 = 0.06
✓ 1
A1 P(S) = 0.62
✓ 1
Part (b)
M1 P(C | S) = P(C ∩ S) / P(S) = 0.56 / 0.62
✓ 1
A1 28/31
✓ 1
Common error: P(C | S) is not the same as P(S | C). Put the given event (the one after the bar) in the denominator.
50IBMA3S-4-16r9-014.2 Presentation of dataPaper 1Easy7 marks
No calculator
A fishing co-operative records the mass of fish landed (in tonnes) on eight days: 12, 15, 9, 18, 15, 11, 14, 16.
(a)
Find the mean daily landing.
(2)
(b)
Find the median.
(2)
(c)
Find the interquartile range, taking each quartile as the median of the lower or upper four values.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to sum the data and divide by 8
✓ 1
A1 13.75 tonnes
✓ 1
Part (b)
M1 ordered list 9, 11, 12, 14, 15, 15, 16, 18 (or middle two values identified)
✓ 1
A1 14.5 tonnes
✓ 1
Part (c)
A1 lower quartile 11.5
✓ 1
A1 upper quartile 15.5
✓ 1
A1 IQR = 15.5 − 11.5 = 4
✓ 1
Exam tip: Write the data in order before finding any median or quartile; most lost marks come from using the unordered list.
51IBMA3S-4-16r9-024.2 Presentation of dataPaper 1Easy7 marks
No calculator
In a boutique, D is the event that a customer buys a dress and J is the event that the customer buys a jacket. It is known that P(D) = 0.55, P(J) = 0.40 and P(D ∪ J) = 0.80.
(a)
Find P(D ∩ J).
(2)
(b)
Determine whether D and J are independent events. Justify your answer.
(2)
(c)
Find P(D | J).
(2)
(d)
On a busy day 600 customers enter the shop. Estimate how many buy neither a dress nor a jacket.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of P(D ∪ J) = P(D) + P(J) − P(D ∩ J)
✓ 1
A1 0.15
✓ 1
Part (b)
M1 P(D) × P(J) = 0.55 × 0.40 = 0.22
✓ 1
R1 0.22 ≠ 0.15 so the events are not independent
✓ 1
Part (c)
M1 P(D ∩ J) / P(J) = 0.15 / 0.40
✓ 1
A1 3/8 (0.375)
✓ 1
Part (d)
A1 P(neither) = 0.20, so 0.20 × 600 = 120
✓ 1
Common error: Do not test independence by checking whether P(A ∩ B) is small; compare it with P(A) × P(B).
52IBMA3S-4-16r9-144.2 Presentation of dataPaper 1Easy6 marks
No calculator
A robot completes a maze 20 times. The cumulative frequency table shows the times t seconds. t < 10: 2 t < 20: 7 t < 30: 15 t < 40: 19 t < 50: 20 Use linear interpolation within classes where needed.
(a)
Write down the number of times for which 20 ≤ t < 30.
(1)
(b)
Estimate the median time.
(2)
(c)
Estimate the interquartile range.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 15 − 7 = 8
✓ 1
Part (b)
M1 10th value lies in 20 ≤ t < 30 with 7 below: 20 + (3/8) × 10
Exam tip: Quartiles are read at n/4 and 3n/4 on the cumulative frequency scale, then matched to the class that contains that position.
53IBMA3S-4-16r21-034.12 Standardization and inverse normalPaper 1Easy5 marks
No calculator
The concentration C mg/L of chlorine in a town's water supply is normally distributed with mean μ and standard deviation σ. A reading of 0.8 mg/L has a z-score of −2 and a reading of 1.4 mg/L has a z-score of 1.
Exam tip: Write each z-score as (value − mean) ÷ standard deviation and treat the pair as simultaneous equations.
54IBMA3S-4-16r21-084.1 Sampling, data and outliersPaper 1Easy4 marks
No calculator
A water company has 12 000 customer households in four districts: North 4800, East 3600, South 2400 and West 1200. It plans to survey 60 households using a stratified sample, with the number from each district proportional to the number of households in that district.
(a)
Find the number of households to be sampled from each district. [3]
(3)
(b)
State one advantage of a stratified sample over a simple random sample of 60 households from all 12 000. [1]
R1 every district is represented in proportion to its size (a simple random sample might under-represent the small district)
✓ 1
Exam tip: For a stratified sample, sample size in a stratum = (stratum size ÷ total) × total sample.
55IBMA3S-4-16r22-024.3 Measures of central tendency and dispersionPaper 1Easy5 marks
No calculator
During an outbreak, a health authority records the number of new cases each day. For the daily counts, the lower quartile is 42 and the upper quartile is 78. A value is classed as an outlier if it is more than 1.5 × IQR above the upper quartile or below the lower quartile.
(a)
Find the interquartile range.
(1)
(b)
Find the value above which a daily count is classed as an outlier.
(3)
(c)
Two daily counts are 128 and 140. State which of them is an outlier, giving a reason.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 IQR = 78 − 42 = 36
✓ 1
Part (b)
M1 1.5 × 36 = 54
✓ 1
M1 78 + 54
✓ 1
A1 132
✓ 1
Part (c)
R1 140, because it is greater than 132 (and 128 is not)
✓ 1
Exam tip: Without a calculator, find the IQR first, then multiply it by 1.5, and only then add to Q3 or subtract from Q1.
56IBMA3S-4-16r24-014.11 Conditional probability and independencePaper 1Easy6 marks
No calculator
In an apiary, let A be the event that a hive has varroa mites and B the event that a hive needs extra feeding. P(A) = 0.4, P(B) = 0.3 and P(A ∪ B) = 0.58.
Exam tip: To test independence, compare P(A∩B) with P(A)P(B); showing they are equal is the reasoning mark.
57IBMA3S-4-16r24-044.8 The binomial distributionPaper 1Easy6 marks
No calculator
The probability that a Formula 1 driver starts a race from pole position is 0.4. If he starts from pole, the probability that he wins is 0.5. If he does not start from pole, the probability that he wins is 0.2.
Tree diagram
(a)
Find the probability that he wins the race.
(3)
(b)
Given that he wins the race, find the probability that he started from pole position. Give your answer as a fraction in its simplest form.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(pole ∩ win) = 0.4 × 0.5 = 0.2
✓ 1
M1 P(not pole ∩ win) = 0.6 × 0.2 = 0.12
✓ 1
A1 P(win) = 0.2 + 0.12 = 0.32
✓ 1
Part (b)
M1 P(pole | win) = P(pole ∩ win) ÷ P(win)
✓ 1
A1 0.2 ÷ 0.32 (ECF from (a))
✓ 1
A1 5/8
✓ 1
Common error: The conditional probability P(pole | win) divides by P(win), not by P(pole).
58IBMA3S-4-18r36-014.2 Presentation of dataPaper 1Easy6 marks
No calculator
A marine biologist measures the lengths, in cm, of a sample of juvenile fish. The lower quartile is 24 and the upper quartile is 36. The shortest fish measured 3 cm and the longest measured 59 cm. A value is an outlier if it is more than 1.5 × IQR below the lower quartile or more than 1.5 × IQR above the upper quartile.
(a)
Find the interquartile range.
(2)
(b)
Determine whether either the shortest or the longest fish is an outlier.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 Q3 − Q1 = 36 − 24
✓ 1
A1 12
✓ 1
Part (b)
M1 attempt to use 1.5 × IQR (= 18) with a quartile
✓ 1
A1 lower fence 24 − 18 = 6
✓ 1
A1 upper fence 36 + 18 = 54
✓ 1
R1 3 < 6 and 59 > 54, so both are outliers
✓ 1
Exam tip: State both fences and compare each extreme value with them; a conclusion without the comparison does not earn the reasoning mark.
59IBMA3S-4-18r37-014.11 Conditional probability and independencePaper 1Easy6 marks
No calculator
An airline operates a morning flight from Lisbon. Let \(F\) be the event that the morning is foggy and \(D\) the event that the flight is delayed. It is known that \(P(F)=0.2\), \(P(D\mid F)=0.5\) and \(P(D\mid F')=0.1\).
Tree diagram
(a)
Find \(P(D\cap F)\).
(2)
(b)
Find \(P(D)\).
(2)
(c)
Given that the flight is delayed, find the probability that the morning was foggy.
Exam tip: Write the conditional probability formula \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\) before substituting.
60IBMA3S-4-18r37-024.11 Conditional probability and independencePaper 1Easy4 marks
No calculator
At a concert, let \(A\) be the event that the band plays its slow ballad and \(B\) the event that the band plays an encore. \(P(A)=0.4\), \(P(B)=0.3\) and \(P(A\cup B)=0.6\).
(a)
Find \(P(A\cap B)\).
(2)
(b)
Determine whether \(A\) and \(B\) are independent. Justify your answer.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), so \(0.6=0.4+0.3-P(A\cap B)\)
✓ 1
A1 0.1
✓ 1
Part (b)
M1 \(P(A)\times P(B)=0.4\times0.3=0.12\)
✓ 1
R1 \(0.12\neq0.1=P(A\cap B)\), so \(A\) and \(B\) are not independent
✓ 1
Common error: Independence requires \(P(A\cap B)=P(A)P(B)\); showing only that the events can both happen does not test it.
61IBMA3S-4-18r37-054.8 The binomial distributionPaper 1Easy4 marks
No calculator
A farmer plants 400 sunflower seeds. Each seed has probability 0.85 of germinating.
(a)
Find the expected number of seeds that germinate.
(2)
(b)
Each plant that germinates has probability 0.9 of surviving until harvest. Find the expected number of the 400 seeds that give a plant that survives until harvest.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(400\times0.85\)
✓ 1
A1 340
✓ 1
Part (b)
M1 \(340\times0.9\) or \(400\times0.85\times0.9\)
✓ 1
A1 306
✓ 1
Exam tip: Expected number of occurrences is the number of trials multiplied by the probability of one occurrence; multiply probabilities when events happen in sequence.
62IBMA3S-4-18r37-084.2 Presentation of dataPaper 1Easy5 marks
No calculator
A violinist records her practice time, in minutes, on eight days: 20, 30, 35, 35, 40, 45, 55, 60.
(a)
Find the mean practice time.
(2)
(b)
Find the median practice time.
(1)
(c)
On a ninth day she practises for \(x\) minutes and the mean of the nine days is 42 minutes. Find \(x\).
Common error: With an even number of values the median is the mean of the two middle values, not the lower of them.
63IBMA3S-4-17r38-024.6 Combined events and diagramsPaper 1Easy7 marks
No calculator
A space probe carries two cameras, C1 and C2, that fail independently. At the end of the mission, the probability that C1 is still working is 0.8 and the probability that C2 is still working is 0.7.
(a)
Find the probability that both cameras are still working.
(2)
(b)
Find the probability that at least one camera is still working.
(2)
(c)
Given that at least one camera is still working, find the probability that C1 is working. Give your answer as a fraction in its simplest form.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.8 × 0.7 (independence)
✓ 1
A1 0.56
✓ 1
Part (b)
M1 0.8 + 0.7 − 0.56 (or 1 − 0.2×0.3)
✓ 1
A1 0.94
✓ 1
Part (c)
M1 conditional probability formula P(C1 | at least one) = P(C1)/P(at least one)
✓ 1
A1 0.8/0.94
✓ 1
A1 40/47
✓ 1
Exam tip: For "at least one", use the complement or the addition rule rather than adding the separate probabilities.
64IBMA3S-4-17r38-054.4 Correlation and regression of y on xPaper 1Easy6 marks
No calculator
For a set of roasting runs, the regression line of y on x is y = 2.5x + c, where x is the roasting time in minutes and y is the final bean temperature in °C. The mean roasting time is 14 minutes and the mean final temperature is 62 °C.
(a)
Find the value of c.
(2)
(b)
Estimate the final temperature for a run of 18 minutes.
(2)
(c)
A student uses the line to estimate the roasting time needed for a final temperature of 100 °C. Explain why this is not appropriate.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 the regression line passes through (14, 62): 62 = 2.5×14 + c
✓ 1
A1 c = 27
✓ 1
Part (b)
M1 substitute x = 18
✓ 1
A1 72 °C (ECF from (a))
✓ 1
Part (c)
R1 the line of y on x is for predicting y from x
✓ 1
R1 predicting x needs the regression line of x on y
✓ 1
Exam tip: The regression line of y on x always passes through the mean point (x̄, ȳ).
65IBMA3S-4-18r39-014.7 Discrete random variablesPaper 1Easy6 marks
No calculator
The discrete random variable \(X\) is the number of pieces of checked baggage carried by a randomly chosen passenger on a regional flight. Its probability distribution is \(P(X=x)=k(x+1)\) for \(x\in\{0,1,2,3\}\), where \(k\) is a constant.
(a)
Find the value of \(k\).
(2)
(b)
Find \(E(X)\).
(2)
(c)
Find \(P(X\ge2\mid X\ge1)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(k(1+2+3+4)=1\)
✓ 1
A1 \(k=0.1\)
✓ 1
Part (b)
M1 \(0(0.1)+1(0.2)+2(0.3)+3(0.4)\)
✓ 1
A1 \(E(X)=2\)
✓ 1
Part (c)
M1 \(\frac{P(X\ge2)}{P(X\ge1)}=\frac{0.7}{0.9}\)
✓ 1
A1 \(\frac79\) (= 0.7778 to 4 d.p.)
✓ 1
Exam tip: In a conditional probability, the denominator is the probability of the event that is given, not of the whole sample space.
66IBMA3S-4-18r39-034.3 Measures of central tendency and dispersionPaper 1Easy6 marks
No calculator
A music teacher records the number of hours, \(x\), that each of 20 students practised last week.
A 21st student's practice time is added, and the mean of all 21 values is then exactly 4 hours. Find this student's practice time.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 the 10th and 11th values are needed (cumulative frequencies 3, 8, 14)
✓ 1
A1 median \(=4\)
✓ 1
Part (b)
M1 \(\sum fx=6+15+24+20+12=77\)
✓ 1
A1 mean \(=\frac{77}{20}=3.85\)
✓ 1
Part (c)
M1 \(\frac{77+x}{21}=4\)
✓ 1
A1 \(x=7\) hours
✓ 1
Exam tip: When a value is added to a data set, work with the total: new total \(=\) new mean \(\times\) new number of values.
67IBMA3S-4-18r39-074.6 Combined events and diagramsPaper 1Easy7 marks
No calculator
In a youth orchestra, \(A\) is the event that a randomly chosen member plays the piano, \(B\) is the event that the member sings in the choir and \(C\) is the event that the member plays the drums. It is given that \(P(A)=0.4\), that \(A\) and \(B\) are independent, and that \(P(A\cup B)=0.7\).
(a)
Find \(P(B)\).
(3)
(b)
Write down \(P(A\mid B')\), giving a reason.
(2)
(c)
The events \(A\) and \(C\) are mutually exclusive and \(P(C)=0.25\). Find the probability that a member plays neither the piano nor the drums.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(P(A\cap B)=P(A)P(B)\)
✓ 1
M1 \(0.7=0.4+P(B)-0.4P(B)\)
✓ 1
A1 \(P(B)=0.5\)
✓ 1
Part (b)
R1 \(A\) and \(B\) independent implies \(A\) and \(B'\) independent (or \(\frac{0.2}{0.5}\))
✓ 1
A1 \(0.4\)
✓ 1
Part (c)
M1 \(P(A\cup C)=0.4+0.25=0.65\)
✓ 1
A1 \(1-0.65=0.35\)
✓ 1
Common error: Independent and mutually exclusive are different ideas: independence gives \(P(A\cap B)=P(A)P(B)\), whereas mutually exclusive events have \(P(A\cap C)=0\).
68IBMA3S-4-19r41-014.2 Presentation of dataPaper 1Easy6 marks
No calculator
The times, in minutes, taken for a pain-relief tablet to take effect in 11 patients are 12, 15, 15, 17, 18, 20, 21, 23, 24, 26, 41.
(a)
Write down the median time.
(1)
(b)
Find the lower quartile and the upper quartile.
(2)
(c)
A value is classified as an outlier if it is greater than Q3 + 1.5 × (Q3 − Q1). Determine whether the time of 41 minutes is an outlier.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 median = 20 minutes
✓ 1
Part (b)
A1 lower quartile = 15
✓ 1
A1 upper quartile = 24
✓ 1
Part (c)
M1 IQR = 24 − 15 = 9 (ECF from (b))
✓ 1
M1 Q3 + 1.5 × IQR = 24 + 13.5 = 37.5
✓ 1
R1 41 > 37.5, so 41 minutes is an outlier
✓ 1
Exam tip: Show the boundary value you compare against, not just the word 'outlier'; the comparison is what earns the final mark.
69IBMA3S-4-18r45-014.11 Conditional probability and independencePaper 1Easy6 marks
No calculator
A group of mountaineers is planning an expedition. Let A be the event that a randomly chosen mountaineer carries a supplementary oxygen supply and B the event that the mountaineer carries a satellite phone. It is known that P(A) = 0.6, P(B) = 0.5 and P(A ∩ B) = 0.24.
(a)
Find P(A ∪ B).
(2)
(b)
Find P(B | A).
(2)
(c)
Determine whether A and B are independent. Justify your answer.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.6 + 0.5 − 0.24
✓ 1
A1 0.86
✓ 1
Part (b)
M1 P(B | A) = P(A ∩ B) ÷ P(A) = 0.24 ÷ 0.6
✓ 1
A1 0.4
✓ 1
Part (c)
M1 P(A) × P(B) = 0.6 × 0.5 = 0.30
✓ 1
R1 0.30 ≠ 0.24 = P(A ∩ B), so A and B are not independent
✓ 1
Exam tip: To test independence, compare P(A ∩ B) with P(A) × P(B) and state the conclusion in words.
70IBMA3S-4-19r45-034.6 Combined events and diagramsPaper 1Easy6 marks
No calculator
At a music school, A is the event that a randomly chosen student plays the piano and B is the event that the student sings in the choir. P(A) = 0.4, P(B) = 0.35 and P(A ∩ B) = 0.14.
(a)
Show that A and B are independent.
(2)
(b)
Find P(A ∪ B).
(2)
(c)
Find P(A | B′).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(A) × P(B) = 0.4 × 0.35 = 0.14
✓ 1
R1 this equals P(A ∩ B), so the events are independent
✓ 1
Part (b)
M1 use of P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
✓ 1
A1 0.61
✓ 1
Part (c)
M1 P(A ∩ B′)/P(B′) = (0.4 − 0.14)/0.65
✓ 1
A1 0.4
✓ 1
Common error: Independence is checked by comparing P(A ∩ B) with P(A)P(B); do not confuse it with the events being mutually exclusive.
71IBMA3S-4-18r46-024.8 The binomial distributionPaper 1Easy6 marks
No calculator
Each of four stagehands independently arrives late for a performance with probability \(\frac13\). Let \(X\) be the number of stagehands who arrive late.
(a)
Find \(P(X=0)\), giving your answer as a fraction.
(2)
(b)
Find \(P(X=2)\), giving your answer as a fraction in its simplest form.
(2)
(c)
Find the probability that at least one stagehand arrives late.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\left(\frac23\right)^4\)
✓ 1
A1 \(\frac{16}{81}\)
✓ 1
Part (b)
M1 \(\binom42\left(\frac13\right)^2\left(\frac23\right)^2\) or \(6\times\frac19\times\frac49\)
✓ 1
A1 \(\frac{8}{27}\)
✓ 1
Part (c)
M1 \(1-P(X=0)\)
✓ 1
A1 \(\frac{65}{81}\) (ECF from (a))
✓ 1
Common error: In an exact binomial calculation do not forget the coefficient \(\binom{n}{r}\): \(\left(\frac13\right)^2\left(\frac23\right)^2\) alone is only the probability of one particular order.
72IBMA3S-4-18r46-144.1 Sampling, data and outliersPaper 1Easy6 marks
No calculator
At a performance there are \(50\) patrons. Of the \(30\) adults, \(18\) bought a programme and \(12\) did not. Of the \(20\) children, \(4\) bought a programme and \(16\) did not.
(a)
A patron is chosen at random. Find the probability that the patron bought a programme.
(1)
(b)
Given that a randomly chosen patron bought a programme, find the probability that the patron is an adult.
(2)
(c)
Two different patrons are chosen at random. Find the probability that both bought a programme, giving your answer as a fraction in its simplest form.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(\frac{22}{50}=\frac{11}{25}\)
✓ 1
Part (b)
M1 \(\frac{18}{22}\)
✓ 1
A1 \(\frac{9}{11}\)
✓ 1
Part (c)
M1 \(\frac{22}{50}\times\frac{21}{49}\)
✓ 1
A1 \(\frac{462}{2450}\)
✓ 1
A1 \(\frac{33}{175}\)
✓ 1
Exam tip: When two patrons are chosen without replacement, the second probability has one fewer in both the numerator and the denominator.
73IBMA3S-4-18r48-034.8 The binomial distributionPaper 1Easy4 marks
No calculator
A hospital charity sells raffle tickets for €2 each. Each ticket has a 0.01 probability of winning a €100 prize, a 0.05 probability of winning a €10 prize, and otherwise wins nothing. A ticket cannot win more than one prize.
(a)
Find the probability that a ticket wins nothing.
(1)
(b)
Find the expected value of the prize on one ticket, in €.
(2)
(c)
Find the charity's expected profit per ticket sold, in €.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(1 - 0.01 - 0.05 = 0.94\)
✓ 1
Part (b)
M1 attempt to use \(\sum xP(X=x) = 100(0.01) + 10(0.05)\)
✓ 1
A1 €1.50
✓ 1
Part (c)
A1 $2 - 1.50 = $ €0.50 (ECF from (b))
✓ 1
Common error: Expected profit uses the ticket price minus the expected prize, not minus the largest prize.
74IBMA3S-4-18r48-074.1 Sampling, data and outliersPaper 1Easy4 marks
No calculator
A hospital has 800 inpatients: 480 in general wards, 200 in surgical wards and 120 in maternity wards. A manager wants a sample of 40 patients for a satisfaction survey.
(a)
Find how many patients should be sampled from each type of ward if a stratified sample is used.
(3)
(b)
The manager instead suggests asking the first 40 patients who leave the general ward on Monday. State one reason why this sample may be biased.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use \(\frac{\text{stratum size}}{800} \times 40\)
✓ 1
A1 general 24 and surgical 10
✓ 1
A1 maternity 6
✓ 1
Part (b)
R1 any valid reason, e.g. it excludes surgical and maternity patients (not representative), or patients leaving on one day may be those who recovered quickly
✓ 1
Exam tip: A stratified sample keeps the sample proportions the same as the population proportions.
75IBMA3S-4-18r48-144.1 Sampling, data and outliersPaper 1Easy5 marks
No calculator
At a Grand Prix, 1200 grandstand seats are numbered 1 to 1200. The organiser surveys 80 fans using a systematic sample: the sampling interval is chosen, a starting seat is chosen at random from the first interval, and then every seat at that interval is selected.
(a)
Find the sampling interval.
(1)
(b)
The starting seat is 11. Write down the next two seats selected, and find the number of the 80th seat selected.
(3)
(c)
State one possible source of bias in this method if the seat numbers follow a regular pattern.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(\frac{1200}{80} = 15\)
✓ 1
Part (b)
A1 26 and 41
✓ 1
M1 attempt to use \(11 + 79 \times 15\)
✓ 1
A1 seat 1196
✓ 1
Part (c)
R1 any valid reason, e.g. every 15th seat may always be an aisle (or end) seat, so the sample would not represent all fans
✓ 1
Exam tip: In a systematic sample, only the starting point is random; the later selections follow a fixed pattern.
76AASL4-q00094.2 Presentation of dataPaper 1Medium8 marks
The box-and-whisker diagrams show the scores of the students in two classes, A and B, in the same test. An outlier is indicated with a cross.
Diagram drawn to scale
(a)
Write down the median score for Class B.
(1)
(b)
Find the interquartile range for each class.
(2)
(c)
Show that the score of 18 in Class B is an outlier.
(2)
(d)
Compare the distributions of the scores in the two classes.
(2)
(e)
State, with a reason, which class has scores that are more likely to be normally distributed.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(58\)
A1
(b)
Class A: \(63-41=22\)
A1
Class B: \(62-48=14\)
A1
(c)
\(Q_{1}-1.5\times\text{IQR}=48-1.5\times14\)
M1
FT from their IQR for Class B.
\(=27\) and \(18\lt 27\)
A1
Must see 27 and a comparison.
(d)
Class B has the higher median (58 compared with 52), so Class B scored higher on average
R1
Must refer to the median (or a measure of central tendency) with values or a clear comparison.
Class B has the smaller IQR (14 compared with 22), so its scores are less spread out (more consistent)
R1
Accept range (A: 62; B: 52) as the measure of spread.
(e)
Class A: its box and whiskers are approximately symmetrical about the median
R1
Award R1 only with a reason referring to symmetry. Class B is not symmetrical: the median is much closer to \(Q_{3}\).
Answers: (a) 58 · (b) A: 22, B: 14 · (e) Class A
77AASL4-q00114.2 Presentation of dataPaper 1Medium6 marks
Each of the frequency histograms P, Q and R shows a set of whole-number data with values from 1 to 9. The box-and-whisker diagrams 1, 2 and 3 represent the same three data sets, in a different order.
Diagram drawn to scale
(a)
Write down which box-and-whisker diagram represents each histogram.
(2)
(b)
Using histogram P, show that the median of this data set is 5.
(2)
(c)
State, with a reason, which data set could be modelled by a normal distribution.
(1)
(d)
All three box-and-whisker diagrams have the same whiskers. Explain what this shows about using only the range to compare the data sets.
(1)
Show mark scheme
Working
Mark
Notes
(a)
P and 2
A1
Q and 3; R and 1
A1
(b)
\(n=1+2+4+7+9+7+4+2+1=37\), so the median is the 19th value
M1
cumulative frequencies \(1,3,7,14,23\): the 15th to 23rd values are 5, so the 19th value is 5
A1
median \(=5\)
AG
(c)
P (box plot 2), because its histogram and box plot are symmetrical about the median, with frequencies highest in the middle
R1
Award R1 only for P with a reason referring to symmetry.
(d)
the range is the same (8) for all three sets, although the sets have very different centres and shapes; the range alone does not distinguish them
R1
Answers: (a) P–2, Q–3, R–1 · (c) P
78AASL4-q00194.3 Measures of central tendency and dispersionPaper 1Medium8 marks
The table shows the number of books, \(x\), borrowed from a library last month by each student in a group.
Number of books, \(x\)
0
1
2
3
4
Frequency
2
5
\(k\)
3
1
The mean number of books borrowed is 1.8.
(a)
Find the value of \(k\).
(3)
(b)
Find the median number of books borrowed.
(2)
(c)
In a reading scheme, each student receives 3 points for every book borrowed plus 1 bonus point. The variance of the number of books borrowed is 0.96. Find the mean and the variance of the number of points received.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac{0(2)+1(5)+2k+3(3)+4(1)}{11+k}=1.8\)
M1
\(18+2k=1.8(11+k)\), i.e. \(18+2k=19.8+1.8k\)
A1
\(k=9\)
A1
(b)
20 students, so the median is the mean of the 10th and 11th values; cumulative frequencies 2, 7, 16
(M1)
FT from their \(k\).
median \(=2\)
A1
(c)
mean \(=3(1.8)+1=6.4\)
A1
variance multiplied by \(3^{2}\); adding 1 does not change it
(M1)
Award M0 for \(3\times0.96\) or \(3\times0.96+1\).
79AASL4-q00204.3 Measures of central tendency and dispersionPaper 1Medium6 marks
A list of five positive integers has mode 6, median 8, mean 11 and range 15.
(a)
Find the five integers.
(4)
(b)
A sixth positive integer, \(k\), is added to the list. The median of the six integers is 9. Find the value of \(k\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
sum of the integers \(=5\times11=55\)
M1
the mode 6 must occur at least twice, and the 3rd (middle) value is 8, so the two smallest values are 6 and 6
R1
largest value \(=6+15=21\)
M1
fourth value \(=55-6-6-8-21=14\); the integers are \(6,\,6,\,8,\,14,\,21\)
A1
(b)
the median is the mean of the 3rd and 4th values, which must be 8 and \(k\) (\(8\le k\le14\)): \(\dfrac{8+k}{2}=9\)
M1
If \(k\ge14\) the median is 11; if \(k\le8\) the median is at most 8.
\(k=10\)
A1
Answers: (a) 6, 6, 8, 14, 21 · (b) \(k=10\)
80AASL4-q00214.3 Measures of central tendency and dispersionPaper 1Medium7 marks
A test is marked out of 120. In class A, 12 students took the test and their mean mark was 64. In class B, 18 students took the test and their mean mark was 74.
(a)
Find the total of the marks for class A.
(1)
(b)
Find the mean mark of all 30 students.
(2)
(c)
One more student takes the test later. The mean mark of all 31 students is 71. Find this student’s mark.
(2)
(d)
It is later found that one mark in class B was recorded as 47 instead of 74. Find the correct mean mark for class B.
every time at the lower class boundary: \(\dfrac{0(4)+10(7)+20(10)+30(6)+40(3)}{30}=\dfrac{570}{30}\)
M1
\(19\) minutes
A1
(d)
every time is less than the upper boundary of its class, so the total is less than \(10(4)+20(7)+30(10)+40(6)+50(3)=870\) and the mean is less than \(\dfrac{870}{30}=29\)
R1
Answers: (a) 11 · (b) 24 min · (c) 19 min
82AASL4-q00314.4 Correlation and regression of y on xPaper 1Medium8 marks
For 40 apartments in a city, the floor area, \(x\) m2, and the monthly rent, \(y\) euros, were recorded. The floor areas ranged from 35 m2 to 110 m2. The regression line of \(y\) on \(x\) is \(y=9.5x+120\), and \(r=0.83\).
(a)
Interpret the value 9.5 in the context of the question.
(1)
(b)
Estimate the monthly rent of an apartment with a floor area of 80 m2.
(2)
(c)
Explain why the value 120 should not be interpreted as the rent of an apartment with a floor area of 0 m2.
(1)
(d)
Explain why the equation should not be used to estimate the floor area of an apartment with a monthly rent of €1000.
(1)
(e)(i)
The mean floor area of the 40 apartments is 68 m2.
Find the mean monthly rent of the 40 apartments.
(2)
(e)(ii)
Describe the correlation between floor area and monthly rent.
(1)
Show mark scheme
Working
Mark
Notes
(a)
for each extra square metre of floor area, the monthly rent increases by about €9.50
R1
(b)
\(9.5(80)+120\)
(M1)
€880
A1
(c)
\(x=0\) is far outside the range of the data (35 to 110), and an apartment with no floor area is not meaningful
R1
(d)
this is a regression line of \(y\) on \(x\); it should not be used to predict \(x\) from \(y\) (the regression line of \(x\) on \(y\) is needed)
R1
(e)(i)
the regression line passes through the mean point \((\bar{x},\,\bar{y})\): \(\bar{y}=9.5(68)+120\)
M1
€766
A1
(e)(ii)
(fairly) strong positive (linear) correlation
A1
Answers: (b) €880 · (e)(i) €766 (ii) strong positive
83AASL4-q00424.5 Probability and expected occurrencesPaper 1Medium8 marks
Two fair six-sided dice, each numbered 1 to 6, are rolled. The score on the first die is \(a\) and the score on the second die is \(b\). These values are used to form the equation \(x^{2}+ax+b=0\).
(a)
Show that the equation has real roots if and only if \(a^{2}\ge 4b\).
(1)
(b)
Find the probability that the equation has equal roots.
(2)
(c)
Find the probability that the equation has real roots.
(3)
(d)
The two dice are rolled 72 times and a new equation is formed each time. Find the expected number of equations that have no real roots.
(2)
Show mark scheme
Working
Mark
Notes
(a)
real roots \(\Leftrightarrow\) discriminant \(a^{2}-4b\ge 0\)
R1
\(a^{2}\ge 4b\)
AG
(b)
\(a^{2}=4b\): \((a,b)=(2,1)\) or \((4,4)\)
(M1)
\(\dfrac{2}{36}=\dfrac{1}{18}\)
A1
(c)
systematic listing of outcomes by value of \(a\), or a table of 36 outcomes
84AASL4-q00434.5 Probability and expected occurrencesPaper 1Medium8 marks
A bag contains \(r\) red counters and \(b\) blue counters only. A counter is chosen at random and the probability that it is red is \(\dfrac{2}{5}\).
(a)
Show that \(3r=2b\).
(2)
(b)
Six more red counters are added to the bag. The probability that a counter chosen at random is red is now \(\dfrac{1}{2}\). Find the value of \(r\) and the value of \(b\).
(4)
(c)
After the extra red counters have been added, a counter is chosen at random, its colour noted and it is replaced. This is done 50 times. Find the expected number of times a blue counter is chosen.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac{r}{r+b}=\dfrac{2}{5}\)
M1
\(5r=2r+2b\)
A1
\(3r=2b\)
AG
(b)
\(\dfrac{r+6}{r+b+6}=\dfrac{1}{2}\)
M1
\(2r+12=r+b+6\), so \(b=r+6\)
A1
substituting into \(3r=2b\): \(3r=2(r+6)\)
M1
\(r=12,\ b=18\)
A1
Both values required.
(c)
\(\mathrm{P}(\text{blue})=\dfrac{18}{36}\), and \(50\times\dfrac{1}{2}\)
(M1)
FT from (b).
\(25\)
A1
Answers: (b) \(r=12,\ b=18\) · (c) 25
85AASL4-q00474.6 Combined events and diagramsPaper 1Medium7 marks
Events \(A\) and \(B\) are such that \(\mathrm{P}(A)=0.3\) and \(\mathrm{P}(A\cup B)=0.65\).
(a)
Find \(\mathrm{P}(B)\) if \(A\) and \(B\) are mutually exclusive.
mutually exclusive means \(\mathrm{P}(C\cap D)=0\)
R1
independent means \(\mathrm{P}(C\cap D)=\mathrm{P}(C)\,\mathrm{P}(D)\), which is greater than 0, so both cannot hold
R1
Answers: (a) 0.35 · (b) \(\tfrac{7}{15}\)
86AASL4-q00504.6 Combined events and diagramsPaper 1Medium9 marks
In a club of 60 members, 30 play football (\(F\)), 24 play basketball (\(B\)) and 18 play tennis (\(T\)). Of these, 12 play both football and basketball, 9 play both football and tennis, 8 play both basketball and tennis, and 5 play all three sports. The Venn diagram shows the events, with the number of members in each region labelled \(a\) to \(h\).
Diagram not to scale
(a)
Find the values of \(g\), \(d\), \(e\) and \(f\).
(2)
(b)
Find the values of \(a\), \(b\), \(c\) and \(h\).
(3)
(c)
A member is chosen at random. Find the probability that the member plays exactly two of the three sports.
(2)
(d)
A member who plays at least two of the sports is chosen at random. Find the probability that this member plays tennis.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(g=5\) and subtracting \(g\) from a pairwise total, e.g. \(d=12-5\)
91AASL4-q00624.7 Discrete random variablesPaper 1Medium7 marks
The probability distribution of a discrete random variable \(X\) is shown in the table, where \(k\) is a constant.
\(x\)
1
2
3
4
\(\mathrm{P}(X=x)\)
0.15
\(k\)
\(k^{2}\)
0.1
(a)
Show that \(k^{2}+k-0.75=0\).
(1)
(b)
Find the value of \(k\), justifying your answer.
(3)
(c)
Find \(\mathrm{E}(X)\).
(2)
(d)
Find the probability that \(X\) is an even number.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(0.15+k+k^{2}+0.1=1\)
M1
\(k^{2}+k-0.75=0\)
AG
(b)
\((k+1.5)(k-0.5)=0\), or \(4k^{2}+4k-3=0\Rightarrow(2k+3)(2k-1)=0\)
(M1)
\(k=0.5\) or \(k=-1.5\)
A1
\(k=0.5\) only, since a probability cannot be negative
R1
Award R1 only if \(k=-1.5\) is explicitly rejected with a reason.
(c)
\(1(0.15)+2(0.5)+3(0.25)+4(0.1)\)
(M1)
FT from their positive k.
\(2.3\)
A1
(d)
\(0.5+0.1=0.6\)
A1
FT from their k.
Answers: (b) \(k=0.5\) · (c) 2.3 · (d) 0.6
92AASL4-q00634.7 Discrete random variablesPaper 1Medium7 marks
A company sells one-year insurance for mobile phones. In a year, the probability that a policy holder makes a claim for a repair, costing the company €300, is 0.08. The probability that a policy holder makes a claim for a replacement, costing the company €900, is 0.03. Otherwise no claim is made. Assume that each policy holder makes at most one claim in a year.
(a)
Let \(C\) be the amount, in euros, paid out by the company on one policy in a year. Write down the probability distribution of \(C\).
(2)
(b)
Find \(\mathrm{E}(C)\).
(2)
(c)
The company wants an expected profit of €15 per policy. Find the price that it should charge for a policy.
(2)
(d)
The company sells 2000 policies at this price. Find the expected total profit.
(1)
Show mark scheme
Working
Mark
Notes
(a)
values \(0,\ 300,\ 900\)
A1
probabilities \(0.89,\ 0.08,\ 0.03\)
A1
Must see 0.89.
(b)
\(0(0.89)+300(0.08)+900(0.03)\)
(M1)
€51
A1
(c)
\(\text{price}-51=15\)
(M1)
FT from (b).
€66
A1
(d)
\(2000\times15=\)€30 000
A1
Answers: (b) €51 · (c) €66 · (d) €30 000
93AASL4-q00684.8 The binomial distributionPaper 1Medium7 marks
A random variable \(X\) follows a binomial distribution with mean 6 and variance 4.2.
(a)
Find the probability of success, \(p\), and the number of trials, \(n\).
(4)
(b)
Write down an expression for \(\mathrm{P}(X=0)\).
(1)
(c)
Explain why no binomial random variable can have a mean of 5 and a variance of 6.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(np=6\) and \(np(1-p)=4.2\)
M1
dividing: \(1-p=\dfrac{4.2}{6}\)
M1
\(p=0.3\)
A1
\(n=20\)
A1
(b)
\(0.7^{20}\)
A1
FT from (a).
(c)
\(\dfrac{\text{variance}}{\text{mean}}=1-p\)
M1
\(1-p\lt 1\) since \(p\gt 0\), so the variance must be less than the mean, but \(6\gt 5\)
R1
Or solving gives \(1-p=1.2\Rightarrow p=-0.2\), which is impossible.
Answers: (a) \(p=0.3,\ n=20\) · (b) \(0.7^{20}\)
94AASL4-q00694.8 The binomial distributionPaper 1Medium9 marks
A fair six-sided die is rolled four times. Let \(X\) be the number of rolls that show a 5 or a 6.
(a)
Write down the distribution of \(X\), giving its parameters.
(1)
(b)
Find \(\mathrm{P}(X=0)\) and \(\mathrm{P}(X=4)\).
(2)
(c)
Write down \(\mathrm{E}(X)\).
(1)
(d)
By considering the possible orders of the outcomes, show that \(\mathrm{P}(X=1)=\dfrac{32}{81}\).
95AASL4-q00804.9 The normal distributionPaper 1Medium5 marks
A random variable \(X\) is normally distributed with mean \(\mu\) and standard deviation \(\sigma\). It is known that \(\mathrm{P}(X\lt 42)=0.2\) and \(\mathrm{P}(X\gt 58)=0.2\). This information is shown on the diagram, where each shaded region has area 0.2.
Diagram not to scale
(a)
Write down the value of \(\mu\).
(1)
(b)
Write down \(\mathrm{P}(42\lt X\lt 58)\).
(1)
(c)
By considering the percentage of the distribution that lies within one standard deviation of the mean, deduce whether \(\sigma\) is greater than or less than 8.
(2)
(d)
It is now given that \(\sigma=9.5\). Write down an estimate of \(\mathrm{P}(31\lt X\lt 69)\).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(\mu=50\)
A1
By symmetry, the mean is halfway between 42 and 58.
(b)
\(0.6\)
A1
(c)
about 68% of the distribution lies between \(\mu-\sigma\) and \(\mu+\sigma\), but only 60% lies between \(50-8\) and \(50+8\)
R1
so \(50\pm8\) lies within one standard deviation of the mean, hence \(\sigma\gt 8\)
A1
Award A1 only with the R1.
(d)
\(0.95\)
A1
\(31\) and \(69\) are \(\mu\pm2\sigma\). Accept 95%.
96AASL4-q00824.9 The normal distributionPaper 1Medium6 marks
A random variable \(X\) is normally distributed with mean 30. The diagram shows the probability density curve of \(X\). The shaded region, to the left of 24, has area 0.15.
Diagram not to scale
(a)
Write down \(\mathrm{P}(X\gt 36)\).
(1)
(b)
Find \(\mathrm{P}(24\lt X\lt 36)\).
(2)
(c)
Find \(\mathrm{P}(X\lt 36\mid X\gt 24)\).
(2)
(d)
In a sample of 200 independent observations of \(X\), find the expected number that lie between 24 and 36.
98AASL4-q01034.11 Conditional probability and independencePaper 1Medium8 marks
A fair spinner with five equal sectors numbered 1 to 5 is spun and a fair six-sided die, numbered 1 to 6, is rolled. Let \(A\) be the event “the product of the two scores is even”, \(B\) the event “the sum of the two scores is 7” and \(C\) the event “the sum of the two scores is at least 9”.
(a)
Find \(\mathrm{P}(A)\).
(2)
(b)
Write down \(\mathrm{P}(A\mid B)\).
(1)
(c)
Hence determine whether \(A\) and \(B\) are independent.
(1)
(d)
Find \(\mathrm{P}(A\mid C)\).
(2)
(e)
Find \(\mathrm{P}(C\mid A)\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
sample space of 30 equally likely outcomes, or \(\mathrm{P}(\text{product odd})=\dfrac{3}{5}\times\dfrac{3}{6}=\dfrac{3}{10}\)
(M1)
\(\dfrac{21}{30}=\dfrac{7}{10}\)
A1
(b)
\(1\)
A1
Every pair with sum 7 has one odd and one even score, so the product is even.
(c)
\(\mathrm{P}(A\mid B)=1\ne\dfrac{7}{10}=\mathrm{P}(A)\), so not independent
R1
FT from (a) and (b).
(d)
\(C=\{(3,6),(4,5),(4,6),(5,4),(5,5),(5,6)\}\), 6 outcomes, of which 5 have an even product
99AASL4-q01044.11 Conditional probability and independencePaper 1Medium8 marks
In a medical study of a large population, 30% of adults smoke. The probability that a smoker develops a certain disease is three times the probability that a non-smoker develops it. Overall, 4.8% of adults develop the disease.
(a)
Find the probability that a non-smoker develops the disease and the probability that a smoker develops the disease.
(3)
(b)
Given that an adult develops the disease, find the probability that the adult smokes.
(2)
(c)
Given that an adult does not develop the disease, find the probability that the adult smokes.
(2)
(d)
State, with a reason, whether smoking and developing the disease are independent events.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(0.3(3k)+0.7k=0.048\), where \(k=\mathrm{P}(D\mid S')\)
101AASL4-q01124.12 Standardization and inverse normalPaper 1Medium7 marks
The marks in a history test are normally distributed with mean 61 and standard deviation 9. The marks in a geography test are normally distributed with mean 55 and standard deviation 7.5.
(a)
Find the standardized value, \(z\), of a mark of 70 in the history test.
(1)
(b)
Hence estimate the percentage of students who scored more than 70 in history.
(1)
(c)
Find the history mark that has a standardized value of \(-1.5\).
(2)
(d)
Lena’s history mark and her geography mark have the same standardized value. Her geography mark is 5 marks lower than her history mark. Find this standardized value and Lena’s two marks.
Exam tip: Give the conditional probability as the exact fraction 4/7. A rounded decimal is not needed when no calculator is allowed.
103IBMA2S-050-07-064.5 Probability and expected occurrencesPaper 1Medium7 marks
No calculator
In an escape-room puzzle, two fair spinners are spun independently. Spinner A is numbered 1, 2, 3 and spinner B is numbered 1, 2, 3, 4. Let M be the larger of the two numbers obtained (if they are equal, M is that number).
(a)
Find the probability distribution of M.
(3)
(b)
Find E(M).
(2)
(c)
Given that at least one spinner shows a 3, find the probability that M = 3.
M1 identifies 6 outcomes with at least one 3, of which 5 have \(M=3\) (only A = 3, B = 4 excluded)
✓ 1
A1 \(\frac{5}{6}\)
✓ 1
Exam tip: For a 'largest value' variable, P(M ≤ m) is easy to count; P(M = m) = P(M ≤ m) − P(M ≤ m − 1) is a quick check on your table.
104IBMA2S-050-07-084.7 Discrete random variablesPaper 1Medium6 marks
No calculator
A sensor on a space probe counts the number of micrometeoroid impacts, X, in a day. Its distribution is modelled by
\(P(X=x)=kx(6-x)\) for \(x\in\{1,2,3,4,5\}\),
where k is a constant. The numbers of impacts on different days are independent.
(a)
Show that \(k=\frac{1}{35}\).
(2)
(b)
Without calculating a sum, write down E(X), justifying your answer.
(2)
(c)
Find the probability that the total number of impacts over two days is exactly 3.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(k(5+8+9+8+5)=1\)
✓ 1
A1 \(35k=1\) so \(k=\frac1{35}\)
✓ 1
Part (b)
R1 the distribution is symmetric about \(x=3\) since \(P(X=3-d)=P(X=3+d)\)
✓ 1
A1 \(E(X)=3\)
✓ 1
Part (c)
M1 total 3 requires (1, 2) or (2, 1): \(2\times\frac{5}{35}\times\frac{8}{35}\)
✓ 1
A1 \(\frac{80}{1225}=\frac{16}{245}\)
✓ 1
Common error: X cannot be 0 here, so a two-day total of 3 can only arise as 1 + 2 or 2 + 1. Including (0, 3) or (3, 0) uses values outside the support.
105IBMA2S-050-01-064.8 The binomial distributionPaper 1Medium7 marks
No calculator
In each training session a basketball player takes the same number, \(n\), of free throws. Each throw is successful with probability \(p\), independently. The number of successful throws in a session, \(X\), has mean 12 and variance 3.
(a)
Find the value of \(p\) and the value of \(n\).
(4)
(b)
Find the value of \(\dfrac{P(X = 15)}{P(X = 16)}\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writes \(np = 12\) and \(np(1-p) = 3\)
✓ 1
M1 divides to obtain \(1 - p = \frac{3}{12}\)
✓ 1
A1 \(p = \frac{3}{4}\)
✓ 1
A1 \(n = 16\)
✓ 1
Part (b)
M1 \(P(X=15) = 16\left(\frac{3}{4}\right)^{15}\left(\frac{1}{4}\right)\) and \(P(X=16) = \left(\frac{3}{4}\right)^{16}\)
✓ 1
M1 simplifies ratio to \(\frac{16 \times \frac{1}{4}}{\frac{3}{4}}\)
✓ 1
A1 \(\frac{16}{3}\)
✓ 1
Exam tip: Dividing the variance by the mean gives \(1-p\) immediately; this is the quickest route whenever both are given.
106IBMA2S-050-01-084.5 Probability and expected occurrencesPaper 1Medium6 marks
No calculator
A commuter cycles through two sets of traffic lights, A and B, on each trip. The probability that light A is green when she arrives is \(\frac{2}{5}\) and the probability that light B is green is \(\frac{3}{4}\). The two lights operate independently.
(a)
Find the probability that exactly one of the two lights is green on a trip.
(2)
(b)
She makes 200 trips in a year. Find the expected number of trips on which neither light is green.
(2)
(c)
On a particular trip at least one light is green. Find the probability that both lights are green.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(A green, B not) + P(A not, B green) \(= \frac{2}{5}\times\frac{1}{4} + \frac{3}{5}\times\frac{3}{4}\)
✓ 1
A1 \(\frac{11}{20}\)
✓ 1
Part (b)
M1 P(neither) \(= \frac{3}{5}\times\frac{1}{4} = \frac{3}{20}\) and multiplies by 200
✓ 1
A1 30
✓ 1
Part (c)
M1 \(\frac{P(\text{both})}{P(\text{at least one})} = \frac{3/10}{17/20}\)
✓ 1
A1 \(\frac{6}{17}\)
✓ 1
Common error: In the last part the answer is not simply \(\frac{3}{10}\). The condition 'at least one is green' reduces the sample space, so divide by \(1 - \frac{3}{20} = \frac{17}{20}\).
107IBMA2S-050-04-064.6 Combined events and diagramsPaper 1Medium8 marks
No calculator
A gym has 120 members. 50 members use the rowing machine (\(R\)), 70 use the treadmill (\(T\)) and 20 use neither. A member is chosen at random.
Diagram not to scale
(a)
Find the number of members who use both the rowing machine and the treadmill.
(2)
(b)
Find \(P(T\mid R)\).
(2)
(c)
Show that \(R\) and \(T\) are not independent.
(2)
(d)
Find \(P(R\mid T')\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(50+70-x=120-20\)
✓ 1
A1 \(x=20\)
✓ 1
Part (b)
M1 \(\frac{20}{50}\)
✓ 1
A1 \(\frac{2}{5}\)
✓ 1
Part (c)
M1 \(P(R)P(T)=\frac{5}{12}\times\frac{7}{12}=\frac{35}{144}\) and \(P(R\cap T)=\frac{20}{120}=\frac{24}{144}\)
✓ 1
R1 \(\frac{35}{144}\neq\frac{24}{144}\), so not independent
Common error: In (d), \(T'\) includes the 20 members who use neither machine, so \(n(T')=50\), not 30.
108IBMA2S-050-04-084.11 Conditional probability and independencePaper 1Medium5 marks
No calculator
On a given day on Mars, a dust storm occurs (\(A\)) with probability \(\frac{1}{3}\). If there is a dust storm, a rover's solar panel output drops below a critical level (\(B\)) with probability \(\frac{1}{2}\). If there is no dust storm, this happens with probability \(\frac{1}{4}\).
Diagram not to scale
(a)
Find \(P(B)\).
(2)
(b)
Find \(P(A\mid B)\).
(2)
(c)
State, with a reason, whether \(A\) and \(B\) are independent.
R1 not independent since \(P(A\mid B)=\frac{1}{2}\neq P(A)=\frac{1}{3}\)
✓ 1
Common error: \(P(A)=P(B)=\frac{1}{3}\) here does not imply independence; independence requires \(P(A\cap B)=P(A)P(B)\), i.e. \(\frac{1}{6}=\frac{1}{9}\), which is false.
109IBMA2S-050-04-094.8 The binomial distributionPaper 1Medium6 marks
No calculator
A box contains 15 rechargeable batteries, of which 4 are faulty. Two batteries are taken from the box at random, without replacement.
(a)
Find the probability that both batteries are faulty.
(2)
(b)
Find the probability that at least one battery is faulty.
(2)
(c)
Given that at least one battery is faulty, find the probability that both are faulty.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{4}{15}\times\frac{3}{14}\)
✓ 1
A1 \(\frac{2}{35}\)
✓ 1
Part (b)
M1 \(1-\frac{11}{15}\times\frac{10}{14}\)
✓ 1
A1 \(\frac{10}{21}\)
✓ 1
Part (c)
M1 \(\frac{2/35}{10/21}\)
✓ 1
A1 \(\frac{3}{25}\)
✓ 1
Exam tip: In (c), 'both faulty' is contained in 'at least one faulty', so the intersection is just \(P(\text{both faulty})\).
110IBMA2S-050-04-104.11 Conditional probability and independencePaper 1Medium5 marks
No calculator
In a survey of 400 households, 160 own a smart thermostat (\(S\)) and 100 have a heat pump (\(H\)). The number of households with both is \(k\).
(a)
Given that \(S\) and \(H\) are independent for a randomly chosen household, find \(k\).
(3)
(b)
Using your value of \(k\), find the probability that a household has a heat pump, given that it does not own a smart thermostat.
M1 \(\frac{100-40}{400-160}=\frac{60}{240}\), OR uses the independence of \(S'\) and \(H\) to write \(P(H\mid S')=P(H)=\frac{100}{400}\)
✓ 1
A1 \(\frac{1}{4}\)
✓ 1
Exam tip: A neat check: if \(S\) and \(H\) are independent then so are \(S'\) and \(H\), so \(P(H\mid S')\) must equal \(P(H)=\frac{1}{4}\).
111IBMA2S-050-05-054.1 Sampling, data and outliersPaper 1Medium6 marks
No calculator
The resting heart rates, in beats per minute (bpm), of seven patients are listed in ascending order: 52, a, 61, 65, 70, c, 88 In a box-and-whisker diagram of these data, the median lies exactly halfway between the lower and upper quartiles, and the interquartile range is 16 bpm.
(a)
Find the value of a and the value of c.
(3)
(b)
Determine whether the reading of 88 bpm is an outlier.
(2)
(c)
The reading of 88 bpm is replaced by a larger integer value r. Find the smallest value of r for which r is an outlier.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 recognising Q1 = a, Q3 = c and median 65, so a = 65 − 8 A1 a = 57 A1 c = 73
✓ 3
Part (b)
M1 upper boundary = 73 + 1.5 × 16 = 97 bpm R1 88 < 97, so 88 is not an outlier
✓ 2
Part (c)
A1 quartiles unchanged, so need r > 97, giving r = 98 bpm
✓ 1
Exam tip: For an odd number of values, the median is excluded from both halves, so here Q₁ is the 2nd value and Q₃ is the 6th value.
112IBMA2S-050-05-084.2 Presentation of dataPaper 1Medium5 marks
No calculator
A hospital trialled two treatments for a sports injury. The recovery times, in days, are summarised: Treatment A: minimum 4, Q1 7, median 9, Q3 12, maximum 18 Treatment B: minimum 5, Q1 8, median 12, Q3 14, maximum 16
Diagram not to scale
(a)
Find the interquartile range for each treatment.
(2)
(b)
State, with a reason based on the interquartile range, which treatment gives more consistent recovery times.
(1)
(c)
Estimate the percentage of patients on Treatment A who took longer to recover than the median recovery time for Treatment B.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 Treatment A: 12 − 7 = 5 days A1 Treatment B: 14 − 8 = 6 days
✓ 2
Part (b)
R1 Treatment A, since its IQR (5 days) is smaller than that of B (6 days)
✓ 1
Part (c)
M1 median of B = 12 days = Q3 of A A1 approximately 25%
✓ 2
Common error: Treatment B has the smaller range but the larger IQR. Consistency judged from the IQR is not affected by extreme values.
113IBMA2S-050-00-064.6 Combined events and diagramsPaper 1Medium7 marks
No calculator
An astronomy club has 60 members. 35 members own a telescope, 28 own a pair of binoculars and 9 own neither. Let \(x\) be the number of members who own both.
Diagram not to scale
(a)
Find the value of \(x\).
(2)
(b)
A member is chosen at random. Write down the probability that they own a telescope but not binoculars.
(1)
(c)
A member who owns binoculars is chosen at random. Find the probability that they also own a telescope.
(2)
(d)
Two different members are chosen at random. Find the probability that both own neither a telescope nor binoculars.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \((35-x)+x+(28-x)+9=60\) or equivalent
✓ 1
A1 \(x=12\)
✓ 1
Part (b)
A1 \(\frac{23}{60}\)
✓ 1
Part (c)
M1 \(\frac{12}{28}\)
✓ 1
A1 \(\frac{3}{7}\)
✓ 1
Part (d)
M1 \(\frac{9}{60}\times\frac{8}{59}\)
✓ 1
A1 \(\frac{6}{295}\)
✓ 1
Common error: In (d) the members are different people, so the second probability uses 8 out of 59, not 9 out of 60.
114IBMA2S-050-00-094.6 Combined events and diagramsPaper 1Medium7 marks
No calculator
A spacecraft carries two sensors, A and B, which operate independently. During a micrometeoroid impact, sensor A detects it with probability \(p\) and sensor B detects it with probability \(p+0.1\). The probability that at least one sensor detects an impact is 0.7.
(a)
Show that \(p^2-1.9p+0.6=0\) and hence find the value of \(p\).
(4)
(b)
Given that at least one sensor detects an impact, find the probability that both sensors detect it.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(1-(1-p)(0.9-p)=0.7\) or \(p+(p+0.1)-p(p+0.1)=0.7\)
✓ 1
A1 expanding correctly to obtain \(p^2-1.9p+0.6=0\)
✓ 1
A1 \(p=0.4\) or \(p=1.5\)
✓ 1
R1 reject \(p=1.5\) since a probability cannot exceed 1, so \(p=0.4\)
✓ 1
Part (b)
M1 \(P(\text{both})=0.4\times0.5=0.2\)
✓ 1
M1 dividing by 0.7
✓ 1
A1 \(\frac{2}{7}\)
✓ 1
Exam tip: "At least one" is fastest via the complement: \(1-P(\text{neither})\). Always state why an invalid root is rejected.
115IBMA2S-050-00-124.7 Discrete random variablesPaper 1Medium12 marks
No calculator
In a video game, a player attempts Level 1. The probability of passing Level 1 at the first attempt is \(\frac{3}{5}\). A player who fails gets one retry, with probability \(\frac{1}{2}\) of passing; a player who fails the retry is out of the game. Every player who passes Level 1 then attempts Level 2 once. The probability of passing Level 2 is \(\frac{2}{3}\) for a player who passed Level 1 at the first attempt, and \(\frac{1}{3}\) for a player who needed the retry.
(a)
Find the probability that a player passes Level 1.
(2)
(b)
Show that the probability that a player passes Level 2 is \(\frac{7}{15}\).
(3)
(c)
Given that a player passes Level 2, find the probability that they passed Level 1 at the first attempt.
(3)
(d)
Determine whether the events "passes Level 1 at the first attempt" and "passes Level 2" are independent.
(2)
(e)
Two players play the game independently. Find the probability that exactly one of them passes Level 2.
A1 \(\frac{3}{5}\times\frac{7}{15}=\frac{7}{25}\) compared with \(P(\text{both})=\frac{2}{5}\) (accept comparison of \(\frac{6}{7}\) with \(\frac{3}{5}\))
✓ 1
R1 not equal, so the events are not independent
✓ 1
Part (e)
M1 \(2\times\frac{7}{15}\times\frac{8}{15}\)
✓ 1
A1 \(\frac{112}{225}\)
✓ 1
Exam tip: Draw the full tree including the retry branch; the Level 2 branch for retry players hangs off a two-step path with probability \(\frac{2}{5}\times\frac{1}{2}\).
116IBMA2S-050-06-054.2 Presentation of dataPaper 1Medium7 marks
No calculator
An ultrasonic parking sensor takes 20 distance readings, x cm, of a stationary obstacle. For these readings, \(\sum x = 360\) and \(\sum x^2 = 6680\).
(a)
Find the mean reading.
(1)
(b)
Find the standard deviation of the readings, giving your answer in exact form.
(3)
(c)
A 21st reading of exactly 18 cm is added. Find the mean and the variance of all 21 readings, giving the variance as a fraction.
M1 new \(\sum x^2 = 6680 + 324 = 7004\), or \(\sum (x-\bar{x})^2 = 200\) unchanged
✓ 1
A1 variance \(= \frac{200}{21}\) cm2
✓ 1
Exam tip: A value equal to the mean adds nothing to \(\sum (x-\bar{x})^2\) but increases n, so the variance falls even though the mean is unchanged.
117IBMA2S-050-06-084.2 Presentation of dataPaper 1Medium5 marks
No calculator
A screen-time app records the number of hours a student spends on their phone on each of 7 days. Written in non-decreasing order, the values are 2, 4, p, q, 11, 12, 14 The median is 9 hours and the mean is 8 hours.
(a)
Write down the value of q.
(1)
(b)
Find the value of p.
(2)
(c)
Find the interquartile range.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 q = 9 (the 4th value is the median)
✓ 1
Part (b)
M1 2 + 4 + p + 9 + 11 + 12 + 14 = 7 × 8 = 56
✓ 1
A1 p = 4
✓ 1
Part (c)
M1 Q1 = 4 (median of 2, 4, 4) and Q3 = 12 (median of 11, 12, 14)
✓ 1
A1 IQR = 8 hours
✓ 1
Exam tip: When unknowns sit inside an ordered list, use the position of the median first: it often gives one unknown directly.
118IBMA2S-050-06-094.6 Combined events and diagramsPaper 1Medium7 marks
No calculator
The numbers of faulty sensors found in five batches from an electronics plant are 2, 4, 6, x, y where x and y are integers with x < y. The mean of the five values is 5 and the variance is 4.
(a)
Show that x + y = 13.
(2)
(b)
Show that x2 + y2 = 89.
(2)
(c)
Find the values of x and y.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 2 + 4 + 6 + x + y = 5 × 5 = 25
✓ 1
A1 12 + x + y = 25 so x + y = 13 (AG)
✓ 1
Part (b)
M1 uses \(\frac{\sum x^2}{5} - 5^2 = 4\) or an equivalent correct method, e.g. \(\frac{\sum (x-5)^2}{5} = 4\) with expansion
A1 (x − 5)(x − 8) = 0; since x < y, x = 5 and y = 8
✓ 1
Common error: Writing the variance as \(\frac{\sum x^2}{n}\) without subtracting \(\bar{x}^2\) leads to an impossible system. Always subtract the square of the mean.
119IBMA2S-050-03-054.4 Correlation and regression of y on xPaper 1Medium6 marks
No calculator
A coach monitors 8 distance runners over a season. For each runner she records the mean number of hours of interval training per week, x, and the percentage improvement in their 5 km time, y. The values of x ranged from 2 to 11 and \(\sum x = 48\). The regression line of y on x is \(y = 0.75x + 1.5\).
(a)
Find \(\bar{y}\).
(3)
(b)
Hence find \(\sum y\).
(1)
(c)
Estimate the percentage improvement for a runner who does 9 hours of interval training per week.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\bar{x} = \frac{48}{8} = 6\)
✓ 1
M1 substitutes \(\bar{x}\) into the regression line, since it passes through \((\bar{x}, \bar{y})\)
✓ 1
A1 \(\bar{y} = 0.75(6) + 1.5 = 6\)
✓ 1
Part (b)
A1 \(\sum y = 8 \times 6 = 48\)
✓ 1
Part (c)
M1 substitutes \(x = 9\): \(0.75(9) + 1.5\)
✓ 1
A1 8.25%
✓ 1
Exam tip: The mean point links the regression line to the summary statistics. Combined with the number of data values, it lets you recover totals such as Σx or Σy.
120IBMA2S-050-03-084.10 Regression of x on yPaper 1Medium6 marks
No calculator
An engineer tests a steel suspension cable for a footbridge. For loads, x kN, between 10 kN and 60 kN, she measures the extension of the cable, y mm. The regression line of y on x is \(y = 0.35x + 2.0\) and \(r = 0.97\).
A colleague decides to record the extension in centimetres instead, denoted by Y.
(a)
State the value of r for the relationship between x and Y, giving a reason.
(2)
(b)
Find the equation of the regression line of Y on x.
(2)
(c)
Hence estimate the extension, in centimetres, when the load is 40 kN.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(r = 0.97\)
✓ 1
R1 every y-value is divided by the same positive constant (10), which does not change the correlation coefficient
Common error: When the y-values are divided by 10, both the gradient and the intercept are divided by 10. Dividing only the gradient gives the wrong line.
121IBMA2S-050-03-104.10 Regression of x on yPaper 1Medium7 marks
No calculator
A technician takes 12 readings of Wi-Fi download speed, y Mbit/s, at different distances, x m, from a home router. The distances ranged from 2 m to 18 m. The regression line of y on x is \(y = 120 - 4.5x\) and the mean download speed was 75 Mbit/s.
(a)
Find the mean distance, \(\bar{x}\).
(2)
(b)
Hence find the sum of the 12 distances.
(1)
(c)
Estimate the download speed at a distance of 16 m.
(2)
(d)
Find the value given by the regression line for a distance of 30 m, and comment on your answer.
R1 a negative speed is impossible; 30 m is outside the data range (2 m to 18 m), so the extrapolation is invalid
✓ 1
Exam tip: An impossible prediction, such as a negative speed, is clear evidence that a linear model has been used outside the range of the data.
122IBMA2S-050-09-084.1 Sampling, data and outliersPaper 1Medium6 marks
No calculator
An e-scooter company has 12 000 registered users and wants their opinions on a new pricing plan. It shows a pop-up questionnaire to every user who opens the app on one Saturday.
(a)
State the population for this survey.
(1)
(b)
Name the sampling technique used.
(1)
(c)
Explain why the results may be biased.
(2)
(d)
The users are in three age groups: 4200 aged 18–25, 5400 aged 26–40 and 2400 aged over 40. The company instead takes a stratified sample of 80 users. Find the number of users aged over 40 in this sample.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 all 12 000 registered users of the e-scooter company
R1 a valid reason why some users are excluded or over-represented . Any one of these earns the mark: only users who open the app on that one Saturday can respond, so users who ride at other times or rarely open the app are left out; users choose whether to answer the pop-up, so those with strong opinions are more likely to respond (self-selection or non-response).
✓ 1
R1 so the sample may not be representative of all 12 000 registered users, and their views on the pricing plan may differ from those of the whole population
✓ 1
Part (d)
M1 \(\frac{2400}{12000}\times 80\)
✓ 1
A1 16
✓ 1
Exam tip: When you explain bias, say which group is over- or under-represented and why, then link it to the sample not representing the whole population.
123IBMA2S-050-09-114.3 Measures of central tendency and dispersionPaper 1Medium5 marks
No calculator
A health authority studies waiting times at two clinics. At clinic A, 30 patients had a mean waiting time of 24 minutes. At clinic B, n patients had a mean waiting time of 18 minutes. The mean waiting time for all the patients from both clinics is 21.6 minutes.
(a)
Find the value of n.
(3)
(b)
A researcher wants a simple random sample of 10 patients from all the patients at the two clinics. Describe how this could be done.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 total for clinic A = 30 × 24 = 720
✓ 1
M1 720 + 18n = 21.6(30 + n)
✓ 1
A1 n = 20
✓ 1
Part (b)
R1 list all 50 patients from both clinics and give each a unique number from 1 to 50 (follow through from their total 30 + n)
✓ 1
R1 use a random number generator to choose 10 different numbers, ignoring repeats, and select those patients. Accept any equivalent random selection, e.g. drawing 10 numbered slips from a hat without replacement
✓ 1
Common error: Averaging the two means, (24 + 18) ÷ 2 = 21, ignores the group sizes. A combined mean must be found from the combined totals.
124IBMA2S-050-08-034.10 Regression of x on yPaper 1Medium7 marks
No calculator
A technician measures the distance x metres of a laptop from a Wi-Fi router and the download speed y Mbps at a number of positions in an office. The mean distance is 12 m and the mean download speed is 64 Mbps. The regression line of x on y is \(x = -0.15y + c\). The regression line of y on x is \(y = -4x + d\).
(a)
Find the value of c.
(2)
(b)
Find the value of d.
(2)
(c)
At a new position the download speed is 40 Mbps. Estimate the distance of this position from the router.
(2)
(d)
State, with a reason, whether the correlation coefficient r is positive or negative.
R1 negative, because both regression lines have negative gradients
✓ 1
Common error: Substituting the mean of x where y belongs (or vice versa) is a frequent slip. Check which variable is the subject of each line before substituting.
125IBMA2S-050-08-074.10 Regression of x on yPaper 1Medium6 marks
No calculator
A cycling coach collects data on riders' average power output x watts and average speed y km h−1 on a flat time-trial course. The mean power is 200 W and the mean speed is 32 km h−1. The regression line of y on x is \(y = 0.05x + d\). The regression line of x on y is \(x = 16y + c\).
(a)
Find the value of d and the value of c.
(3)
(b)
A rider averages 36 km h−1. Use the appropriate line to estimate the rider's average power output.
(2)
(c)
Another rider rearranges \(y = 0.05x + d\) to make x the subject and obtains an estimate of 280 W for the same rider. Explain why your estimate from the previous part is preferred.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitutes \((200, 32)\) into either line
✓ 1
A1 \(d = 22\)
✓ 1
A1 \(c = -312\)
✓ 1
Part (b)
M1 uses the x on y line: \(x = 16(36) - 312\)
✓ 1
A1 264 W
✓ 1
Part (c)
R1 the y on x line minimises errors in y, not in x; to estimate x from a given y, the x on y line must be used (the two lines differ because \(|r| \neq 1\))
✓ 1
Common error: Making x the subject of the y on x line is algebraically fine but statistically wrong for prediction. Always choose the line whose subject is the quantity being estimated.
126IBMA2S-050-08-094.4 Correlation and regression of y on xPaper 1Medium5 marks
No calculator
A 3D-printing workshop investigates how the nozzle temperature x °C affects the tensile strength y MPa of printed parts. To choose a temperature setting for a required strength, the regression line of x on y is found to be \(x = 2.5y + 150\). A customer's printer displays temperatures in degrees Fahrenheit, F, where \(F = 1.8x + 32\).
(a)
Estimate the nozzle temperature, in °C, needed for a tensile strength of 40 MPa.
(2)
(b)
Find the equation of the regression line of F on y.
(2)
(c)
State, with a reason, the effect on the correlation coefficient of measuring temperature in °F instead of °C.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitutes \(y = 40\): \(x = 2.5(40) + 150\)
✓ 1
A1 250 °C
✓ 1
Part (b)
M1 substitutes the x on y line into \(F = 1.8x + 32\): \(F = 1.8(2.5y + 150) + 32\)
✓ 1
A1 \(F = 4.5y + 302\)
✓ 1
Part (c)
R1 no change, since \(F\) is a linear function of \(x\) with a positive gradient, so the strength and direction of the linear relationship are unchanged
✓ 1
Exam tip: Changing units by a linear rule \(F = ax + b\) with \(a > 0\) never changes r; a negative a would change only its sign.
127IBMA3S-4-16r5-054.6 Combined events and diagramsPaper 1Medium7 marks
No calculator
A music school has 60 members. 36 members play piano, 28 play violin and 10 play neither instrument. One member is chosen at random. Let P be the event that the member plays piano and V the event that the member plays violin.
(a)
Find the number of members who play both instruments.
(3)
(b)
Find P(V | P).
(2)
(c)
Determine whether P and V are independent.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 number playing at least one instrument = 60 − 10 = 50
✓ 1
M1 36 + 28 − x = 50
✓ 1
A1 14 members
✓ 1
Part (b)
M1 use of P(V | P) = P(V ∩ P)/P(P) (or 14 out of 36)
R1 7/25 ≠ 7/30 so not independent (or P(V | P) = 7/18 ≠ P(V) = 7/15)
✓ 1
Exam tip: With exact values you can compare fractions without a calculator: a common denominator or cross-multiplying is enough to show two probabilities differ.
128IBMA3S-4-16r5-064.2 Presentation of dataPaper 1Medium6 marks
No calculator
An airline pays compensation, X euros, to a delayed passenger. The probability distribution of X is: x: 0, 40, 100, 300 P(X = x): 0.60, 0.25, p, 0.05
(a)
Find the value of p.
(1)
(b)
Find E(X).
(2)
(c)
The airline changes the largest payment from €300 to €w so that the expected compensation per delayed passenger is €30. Find w.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 p = 1 − (0.60 + 0.25 + 0.05) = 0.10
✓ 1
Part (b)
M1 attempt to use ΣxP(X = x)
✓ 1
A1 E(X) = 10 + 10 + 15 = 35
✓ 1
Part (c)
M1 0.25 × 40 + 0.1 × 100 + 0.05w = 30
✓ 1
A1 20 + 0.05w = 30
✓ 1
A1 w = 200
✓ 1
Exam tip: When a value in the distribution changes, keep the other terms of the expectation and write a new equation for the unknown.
129IBMA3S-4-16r8-054.7 Discrete random variablesPaper 1Medium7 marks
No calculator
A food laboratory tests samples of a sauce. Let A be the event that a sample contains traces of an allergen and B the event that a sample fails a taste test. P(A) = 0.4, P(B) = 0.35 and P(A ∪ B) = 0.61.
(a)
Find P(A ∩ B).
(2)
(b)
Show that A and B are independent.
(2)
(c)
Find P(A ∪ B').
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
✓ 1
A1 P(A ∩ B) = 0.4 + 0.35 − 0.61 = 0.14
✓ 1
Part (b)
M1 attempt to compare P(A ∩ B) with P(A) × P(B)
✓ 1
R1 0.4 × 0.35 = 0.14 = P(A ∩ B), so A and B are independent
✓ 1
Part (c)
M1 P(A ∪ B') = P(A) + P(B') − P(A ∩ B')
✓ 1
A1 P(A ∩ B') = 0.4 × 0.65 = 0.26 (independence of A and B')
✓ 1
A1 0.4 + 0.65 − 0.26 = 0.79
✓ 1
Common error: If A and B are independent, so are A and B'. Do not assume independence before you have shown it.
130IBMA3S-4-16r9-054.6 Combined events and diagramsPaper 1Medium7 marks
No calculator
A robot arm tries to grip a component. The probability that it succeeds on the first attempt is 3/4. If the first attempt fails, it tries once more and the probability of success on this second attempt is 2/3. If both attempts fail, the component is rejected.
Tree diagram: first attempt, then second attempt
(a)
Find the probability that the component is gripped on the second attempt.
(2)
(b)
Find the probability that the component is gripped successfully.
(2)
(c)
Given that the component was gripped successfully, find the probability that this happened on the first attempt.
(2)
(d)
In a batch of 240 components, estimate how many are gripped successfully.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 1/4 × 2/3
✓ 1
A1 1/6
✓ 1
Part (b)
M1 3/4 + 1/6
✓ 1
A1 11/12
✓ 1
Part (c)
M1 (3/4) / (11/12)
✓ 1
A1 9/11
✓ 1
Part (d)
A1 240 × 11/12 = 220
✓ 1
Exam tip: For a conditional probability, put P(both events) on top and the probability of the given event underneath.
131IBMA3S-4-16r9-104.2 Presentation of dataPaper 1Medium7 marks
No calculator
The times, in minutes, taken to finish 24 garments in a workshop have minimum 14, lower quartile 23, median 27, upper quartile 35 and maximum 58. The next largest time after 58 is 49.
(a)
Find the interquartile range.
(1)
(b)
Show that 58 is an outlier, using the rule that a value above Q3 + 1.5 × IQR is an outlier.
(3)
(c)
Use the quartiles to describe the skewness of the distribution.
(2)
(d)
A box-and-whisker diagram is drawn with 58 shown as an outlier. State the value at the end of the upper whisker.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 35 − 23 = 12
✓ 1
Part (b)
M1 35 + 1.5 × 12
✓ 1
A1 53
✓ 1
R1 58 > 53, so 58 is an outlier
✓ 1
Part (c)
M1 upper quartile − median = 8 and median − lower quartile = 4
✓ 1
R1 the upper part is more spread out, so positive skew
✓ 1
Part (d)
A1 49
✓ 1
Common error: The whisker ends at the largest value that is not an outlier, not at the fence value itself.
132IBMA3S-4-16r21-144.3 Measures of central tendency and dispersionPaper 1Medium10 marks
No calculator
The temperature X °C of water at a treatment works is normally distributed with mean 15 and standard deviation 2. Temperatures in °F are given by F = 1.8X + 32.
(a)
Find the z-score of a water temperature of 19 °C. [2]
(2)
(b)
Write down the mean and the standard deviation of F. [3]
(3)
(c)
Given that P(13 < X < 17) = 0.68, find P(X > 17). [2]
(2)
(d)
A water sample has a z-score of −1.5. Find its temperature in °C and in °F. [3]
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (19 − 15)/2
✓ 1
A1 z = 2
✓ 1
Part (b)
M1 mean = 1.8 × 15 + 32
✓ 1
A1 mean = 59 °F
✓ 1
A1 standard deviation = 1.8 × 2 = 3.6 °F (the +32 does not change the spread)
✓ 1
Part (c)
M1 symmetry about the mean: P(X < 13) = P(X > 17) and (1 − 0.68)/2
✓ 1
A1 0.16
✓ 1
Part (d)
A1 X = 15 − 1.5 × 2 = 12 °C
✓ 1
M1 F = 1.8 × 12 + 32
✓ 1
A1 53.6 °F
✓ 1
Common error: Adding a constant to every value shifts the mean but leaves the standard deviation unchanged; only multiplying changes the spread.
133IBMA3S-4-16r22-084.9 The normal distributionPaper 1Medium6 marks
No calculator
The random variable X is normally distributed with mean μ and standard deviation σ. It is given that P(X < 12) = P(X > 20) = 0.2.
Normal distribution curve, not to scale
(a)
Find the value of μ.
(2)
(b)
Write down P(12 < X < 20).
(1)
(c)
Find P(X > 16 | X < 20), giving your answer as a fraction.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of symmetry: μ is halfway between 12 and 20
✓ 1
A1 μ = 16
✓ 1
Part (b)
A1 0.6
✓ 1
Part (c)
M1 P(16 < X < 20) ÷ P(X < 20)
✓ 1
A1 P(16 < X < 20) = 0.8 − 0.5 = 0.3 and P(X < 20) = 0.8
✓ 1
A1 3/8
✓ 1
Exam tip: For a normal distribution, P(X < μ) = 0.5. Use this with symmetry and a sketch instead of a calculator.
134IBMA3S-4-16r24-084.11 Conditional probability and independencePaper 1Medium7 marks
No calculator
In a hospital, let A be the event that a patient receives a scan and B the event that a patient is admitted. Events A and B are independent, P(B) = 0.4 and P(A ∪ B) = 0.76.
(a)
Find P(A).
(4)
(b)
Find P(A | B′).
(2)
(c)
Write down P(A′ ∩ B′).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(A∩B) = P(A) × P(B) = 0.4p, where p = P(A)
✓ 1
M1 p + 0.4 − 0.4p = 0.76
✓ 1
A1 0.6p = 0.36
✓ 1
A1 P(A) = 0.6
✓ 1
Part (b)
R1 A and B are independent, so A and B′ are independent and P(A | B′) = P(A)
✓ 1
A1 0.6
✓ 1
Part (c)
A1 1 − 0.76 = 0.24
✓ 1
Exam tip: For independent events, substitute P(A∩B) = P(A)P(B) into the union formula to get an equation in one unknown.
135IBMA3S-4-16r24-094.4 Correlation and regression of y on xPaper 1Medium6 marks
No calculator
A hospital records, for several wards, x, the number of nurses on duty, and y, the mean time in minutes that patients wait for medication. The regression line of y on x is y = −1.8x + c. The mean point is (x̄, ȳ) = (8, 15.6).
(a)
Find the value of c.
(3)
(b)
Interpret the gradient −1.8 in this context.
(2)
(c)
Estimate the mean waiting time when there are 10 nurses on duty.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 the regression line passes through the mean point (8, 15.6)
✓ 1
A1 15.6 = −1.8 × 8 + c = −14.4 + c
✓ 1
A1 c = 30
✓ 1
Part (b)
A1 Mean waiting time decreases by 1.8 minutes
✓ 1
A1 for each additional nurse on duty
✓ 1
Part (c)
A1 −1.8 × 10 + 30 = 12 minutes
✓ 1
Exam tip: The regression line of y on x always passes through (x̄, ȳ), which gives a quick way to find a missing constant.
136IBMA3S-4-17r33-074.1 Sampling, data and outliersPaper 1Medium7 marks
No calculator
A hospital employs 420 nurses, 120 doctors and 60 administrative staff. The management wants a sample of 50 staff for a survey about working hours.
(a)
The management uses stratified sampling, with each group represented in proportion to its size. Find the number of doctors and the number of administrative staff in the sample.
(3)
(b)
Instead, the manager asks the first 50 staff who enter the canteen at lunchtime. State the name of this sampling method and give one reason why it may produce a biased sample.
(2)
(c)
A third plan is to select every 12th name from an alphabetical list of all 600 staff, starting from a randomly chosen name among the first 12. State the name of this method and find the size of the sample it gives.
R1 any valid reason, e.g. staff on night shifts or who do not use the canteen cannot be selected, so the sample is not representative
✓ 1
Part (c)
A1 systematic sampling
✓ 1
A1 600 ÷ 12 = 50
✓ 1
Exam tip: Name the sampling method exactly (simple random, systematic, stratified, quota or convenience) and link any bias to who is left out of the sample.
137IBMA3S-4-17r33-094.11 Conditional probability and independencePaper 1Medium7 marks
No calculator
At a hospital, let A be the event that a patient needs surgery and B the event that the patient stays more than five days. It is known that P(A) = 0.4 and P(A ∪ B) = 0.7, and that A and B are independent.
Exam tip: If A and B are independent then A and B′ are also independent, so P(A | B′) = P(A) without further working.
138IBMA3S-4-18r36-034.11 Conditional probability and independencePaper 1Medium6 marks
No calculator
For a randomly chosen building designed by an architectural firm, let A be the event that it has an atrium and B the event that it has a green roof. It is known that P(A) = 3/8, P(B) = 1/2 and P(A ∪ B) = 3/4.
(a)
Find P(A ∩ B).
(2)
(b)
Show that A and B are not independent.
(2)
(c)
Find P(A | B).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
✓ 1
A1 1/8
✓ 1
Part (b)
M1 P(A) × P(B) = 3/8 × 1/2 = 3/16
✓ 1
R1 3/16 ≠ 1/8 = P(A ∩ B), so not independent
✓ 1
Part (c)
M1 P(A ∩ B) / P(B) = (1/8)/(1/2)
✓ 1
A1 1/4
✓ 1
Exam tip: To test independence, compare P(A ∩ B) with P(A) × P(B); equal values mean independent, anything else does not.
139IBMA3S-4-18r36-064.2 Presentation of dataPaper 1Medium6 marks
No calculator
The heights, in metres, of buildings in two districts are summarised below.
District P: minimum 12, lower quartile 20, median 28, upper quartile 40, maximum 62 District Q: minimum 15, lower quartile 22, median 30, upper quartile 38, maximum 52
Box-and-whisker diagrams
(a)
Find the interquartile range for each district.
(2)
(b)
State which district has the more consistent building heights. Give two pieces of evidence.
(2)
(c)
District P contains 80 buildings. Estimate the number with a height of at least 40 m.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 District P: 20
✓ 1
A1 District Q: 16
✓ 1
Part (b)
A1 District Q
✓ 1
R1 smaller IQR (16 < 20) and smaller range (37 < 50)
✓ 1
Part (c)
M1 recognising that about 25% of values are at or above the upper quartile
✓ 1
A1 20
✓ 1
Exam tip: When comparing spread, quote the actual values of the IQR or range rather than just saying one is bigger.
140IBMA3S-4-18r36-084.7 Discrete random variablesPaper 1Medium7 marks
No calculator
A research vessel deploys three sensors on each trip. Let X be the number of sensors that fail on a randomly chosen trip. The probability distribution of X is shown below.
x: 0, 1, 2, 3 P(X = x): 0.1, a, b, 0.2
It is given that E(X) = 1.7.
(a)
Find the value of a and the value of b.
(3)
(b)
Find Var(X).
(2)
(c)
Each failed sensor costs €50 to replace and there is a fixed cost of €20 per trip. Find the expected cost of a trip.
Common error: When finding a variance, subtract the square of the mean from E(X²); do not square the mean of the probabilities or forget to square the x-values.
141IBMA3S-4-18r36-124.7 Discrete random variablesPaper 1Medium11 marks
No calculator
Steel connectors for a footbridge are tested in stages. Each connector is first given a stress test, which it passes with probability 3/4. A connector that passes the stress test is then given a fatigue test, which it passes with probability 5/6. A connector that fails the stress test is repaired and given a single retest, which it passes with probability 1/3; it then goes straight to the supply bin without a fatigue test. A connector is accepted if it passes both the stress and fatigue tests, or if it passes the retest.
Tree diagram: stress test (first column), then fatigue test or retest
(a)
Find the probability that a connector passes both the stress test and the fatigue test.
(2)
(b)
Find the probability that a connector is accepted.
(3)
(c)
Given that a connector is accepted, find the probability that it failed the stress test.
(3)
(d)
The company makes a profit of €80 on a connector accepted without a retest, a profit of €50 on a connector accepted after a retest, and a loss of €30 on a rejected connector. Find the expected profit per connector.
Exam tip: Label each branch of a tree clearly and write the product for each route before adding; this makes conditional probability questions much easier to follow.
142IBMA3S-4-17r38-074.3 Measures of central tendency and dispersionPaper 1Medium6 marks
No calculator
The tensions in the 40 suspension cables of a bridge, x kN, have mean 48 and standard deviation 5. Each tension is converted to a scaled value y = ax + b, where a > 0.
(a)
First, 6 is added to every tension. Write down the new mean and the new standard deviation.
(2)
(b)
The scaled values y have mean 75.6 and standard deviation 9. Find the value of a and the value of b.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 mean 54
✓ 1
A1 standard deviation 5
✓ 1
Part (b)
M1 a × 5 = 9
✓ 1
A1 a = 1.8
✓ 1
M1 75.6 = 1.8 × 48 + b
✓ 1
A1 b = −10.8
✓ 1
Common error: Adding a constant changes the mean but not the spread; multiplying scales both the mean and the standard deviation.
143IBMA3S-4-18r39-054.6 Combined events and diagramsPaper 1Medium7 marks
No calculator
The weather at an airport on any day is either clear, with probability 0.7, or poor. When the weather is clear, a flight departs on time with probability 0.9. When the weather is poor, it departs on time with probability 0.6. The weather and punctuality on different days are independent.
Tree diagram
(a)
Find the probability that a flight on a given day departs on time.
(3)
(b)
Given that a flight departs on time, find the probability that the weather was clear.
(2)
(c)
Flights on two different days are considered. Find the probability that exactly one of them departs on time.
Exam tip: For "exactly one of two", count both orders; forgetting the factor of 2 halves the answer.
144IBMA3S-4-18r39-094.11 Conditional probability and independencePaper 1Medium7 marks
No calculator
For two events \(A\) and \(B\) concerning a flight, \(A\) is the event that it is delayed by weather and \(B\) is the event that it is delayed by a technical fault. It is given that \(P(A)=0.5\), \(P(B)=0.4\) and \(P(A\mid B)=0.25\).
(a)
Find \(P(A\cap B)\).
(2)
(b)
Find the probability that a flight is delayed by neither cause.
(2)
(c)
Find \(P(B\mid A')\).
(2)
(d)
Determine whether \(A\) and \(B\) are independent.
R1 \(P(A)P(B)=0.2\ne0.1=P(A\cap B)\), so not independent
✓ 1
Exam tip: A Venn diagram with the overlap filled in first makes every other region a simple subtraction.
145IBMA3S-4-18r39-114.6 Combined events and diagramsPaper 1Medium7 marks
No calculator
A box holds 9 spare guitar strings: 5 steel and 4 nylon. Two strings are taken at random from the box, without replacement.
(a)
Find the probability that both strings are steel.
(2)
(b)
Find the probability that the two strings are of different types.
(2)
(c)
Given that at least one string is steel, find the probability that both are steel.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac59\times\frac48\)
✓ 1
A1 \(\frac{5}{18}\)
✓ 1
Part (b)
M1 \(\frac59\times\frac48+\frac49\times\frac58\)
✓ 1
A1 \(\frac59\)
✓ 1
Part (c)
M1 \(P(\text{at least one steel})=1-\frac49\times\frac38=\frac56\)
✓ 1
M1 \(\frac{5/18}{5/6}\)
✓ 1
A1 \(\frac13\) (= 0.3333 to 4 d.p.)
✓ 1
Common error: Without replacement, the second probability changes: use 8 strings, not 9, in the second factor.
146IBMA3S-4-18r45-054.6 Combined events and diagramsPaper 1Medium7 marks
No calculator
A mountaineering team's chance of reaching a summit depends on the weather. The probability of good weather on a given day is 0.6. If the weather is good, the probability of reaching the summit is 0.8. If the weather is not good, the probability of reaching the summit is 0.3.
Tree diagram
(a)
Find the probability that the team reaches the summit on a given day.
(3)
(b)
Given that the team reached the summit, find the probability that the weather was good.
(2)
(c)
On two different days, with the outcomes independent, the team has the probability found in (a) of reaching the summit each day. Find the probability that the team reaches the summit on exactly one of the two days.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(good and summit) = 0.6 × 0.8 = 0.48
✓ 1
M1 P(not good and summit) = 0.4 × 0.3 = 0.12
✓ 1
A1 0.60
✓ 1
Part (b)
M1 P(good | summit) = 0.48 ÷ 0.60
✓ 1
A1 0.8
✓ 1
Part (c)
M1 two disjoint outcomes, 2 × 0.6 × 0.4 (using the answer to (a))
✓ 1
A1 0.48
✓ 1
Common error: In a conditional probability the denominator is the probability of the condition itself, here P(summit), not 1 and not P(good).
147IBMA3S-4-18r45-084.6 Combined events and diagramsPaper 1Medium6 marks
No calculator
Events A and B are independent. For a mountaineering team, A is the event that the team starts its climb before 05:00 and B is the event that the team reaches the col by noon. It is given that P(A) = 0.4 and P(A ∪ B) = 0.76.
(a)
Find P(B).
(3)
(b)
Find the probability that the team does not start before 05:00 and reaches the col by noon.
(2)
(c)
State, with a reason, whether A and B are mutually exclusive.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
✓ 1
M1 independence: P(A ∩ B) = 0.4 P(B), so 0.76 = 0.4 + 0.6 P(B)
✓ 1
A1 P(B) = 0.6
✓ 1
Part (b)
M1 P(A′ ∩ B) = P(A′) × P(B) = 0.6 × 0.6
✓ 1
A1 0.36
✓ 1
Part (c)
R1 P(A ∩ B) = 0.4 × 0.6 = 0.24 ≠ 0, so they are not mutually exclusive
✓ 1
Common error: Independent events are not mutually exclusive: if both have non-zero probability, their intersection is not empty.
148IBMA3S-4-18r46-054.11 Conditional probability and independencePaper 1Medium6 marks
No calculator
Each week, the probability that a volcano erupts is \(\frac15\). A monitoring system raises an alarm with probability \(\frac{9}{10}\) in a week when the volcano erupts, and with probability \(\frac18\) in a week when it does not erupt. Let \(E\) be the event that the volcano erupts and \(A\) the event that the alarm is raised.
Tree diagram
(a)
Find \(P(A)\).
(3)
(b)
Given that the alarm is raised, find the probability that the volcano erupts.
M1 \(P(E'\cap A)=\frac45\times\frac18=\frac1{10}\) and adding the two products
✓ 1
A1 \(\frac{7}{25}\)
✓ 1
Part (b)
M1 \(P(E\mid A)=\frac{P(E\cap A)}{P(A)}\)
✓ 1
A1 \(\frac{9/50}{7/25}\) (ECF from (a))
✓ 1
A1 \(\frac{9}{14}\)
✓ 1
Exam tip: In conditional probability the denominator is the probability of the event you are given. Here it is \(P(A)\), not \(P(E)\).
149IBMA3S-4-18r46-094.11 Conditional probability and independencePaper 1Medium7 marks
No calculator
A fleet has \(80\) vessels. Of these, \(48\) carry refrigerated containers (event \(R\)), \(35\) sail the Atlantic route (event \(S\)) and \(12\) do neither. A vessel is chosen at random from the fleet.
(a)
Find the number of vessels that both carry refrigerated containers and sail the Atlantic route.
(2)
(b)
Show that \(R\) and \(S\) are not independent events.
(3)
(c)
Find \(P(R\mid S)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(n(R\cup S)=80-12=68\) and \(48+35-n(R\cap S)=68\)
R1 \(P(R\cap S)=\frac{15}{80}\neq\frac{21}{80}\), so the events are not independent (ECF from (a))
✓ 1
Part (c)
M1 \(\frac{n(R\cap S)}{n(S)}=\frac{15}{35}\) or \(\frac{P(R\cap S)}{P(S)}\)
✓ 1
A1 \(\frac37\) (ECF from (a))
✓ 1
Common error: Events with \(P(A\cap B)\neq P(A)P(B)\) are dependent. Do not conclude independence just because the two events can happen together.
150IBMA3S-4-18r48-014.7 Discrete random variablesPaper 1Medium7 marks
No calculator
A beekeeper inspects hives in spring. The number of queen cells, \(X\), found in a randomly chosen hive has probability distribution \(P(X = x) = k(5 - x)\) for \(x = 0, 1, 2, 3\), where \(k\) is a constant.
Common error: Variance is \(\mathrm{E}(X^2) - [\mathrm{E}(X)]^2\); subtracting \(\mathrm{E}(X)\) without squaring it is a frequent slip.
151AASL4-q00534.6 Combined events and diagramsPaper 1Hard9 marks
In a large college, a proportion \(p\) of the students take a course online and the rest take it in a classroom. The probability that a student passes the final test is 0.6 for a student who takes the course online and 0.8 for a student who takes it in a classroom.
Diagram not to scale
(a)
Show that the probability that a student chosen at random passes the final test is \(0.8-0.2p\).
(2)
(b)
Given that a student passes the final test, the probability that this student took the course online is \(\dfrac{1}{3}\). Find the value of \(p\).
(3)
(c)
A student chosen at random fails the final test. Find the probability that this student took the course online.
(2)
(d)
Two students are chosen at random, independently. Find the probability that exactly one of them passes the final test.
(2)
Show mark scheme
Working
Mark
Notes
(a)
tree diagram or \(0.6p+0.8(1-p)\)
M1
\(0.6p+0.8-0.8p\)
A1
\(=0.8-0.2p\)
AG
(b)
\(\dfrac{0.6p}{0.8-0.2p}=\dfrac{1}{3}\)
M1
\(1.8p=0.8-0.2p\), or \(2p=0.8\)
A1
\(p=0.4\)
A1
(c)
\(\mathrm{P}(\text{fail})=1-0.72=0.28\) and \(\mathrm{P}(\text{online}\cap\text{fail})=0.4\times0.4=0.16\)
(M1)
FT from (b).
\(\dfrac{0.16}{0.28}=\dfrac{4}{7}\)
A1
(d)
\(2\times0.72\times0.28\)
(M1)
Award M0 for \(0.72\times0.28\) without the factor 2. FT from (b).
153AASL4-q00564.6 Combined events and diagramsPaper 1Hard15 marks
A bag contains 5 red counters and 3 blue counters. Two counters are taken at random from the bag, one after the other, without replacement. The tree diagram shows the possible outcomes.
R = red, B = blue
(a)
Write down the value of \(a\) and the value of \(b\).
(2)
(b)
Find the probability that both counters are the same colour.
(2)
(c)
Given that both counters are the same colour, find the probability that they are both red.
(3)
(d)(i)
Let \(X\) be the number of blue counters taken.
Find the probability distribution of \(X\).
(3)
(d)(ii)
Show that \(\mathrm{E}(X)=\dfrac{3}{4}\).
(2)
(e)
In a game, a player pays $\(k\) to take two counters in this way and then receives $3 for each blue counter taken. Find the value of \(k\) for which the game is fair.
154AASL4-q00654.7 Discrete random variablesPaper 1Hard8 marks
In a game, a player rolls a fair six-sided die. The player may either keep this score or roll the die once more, in which case the score of the second roll is the final score. A player decides to use the strategy “roll again if the first score is \(m\) or less”, where \(m\in\{0,1,2,3,4,5\}\).
(a)
Write down the expected final score when \(m=0\) (the player never rolls again).
(1)
(b)
Show that the expected final score when \(m=2\) is \(\dfrac{25}{6}\).
(3)
(c)
Determine the value of \(m\) that gives the greatest expected final score, and state this expected score.
(3)
(d)
Explain, without further calculation, why a player should never roll again after scoring 4.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(3.5\)
A1
(b)
first score 3, 4, 5 or 6 is kept: \(\dfrac{3+4+5+6}{6}\)
M1
first score 1 or 2 (probability \(\tfrac{2}{6}\)) followed by an expected 3.5: \(\dfrac{2}{6}\times3.5\)
M1
\(\dfrac{18}{6}+\dfrac{7}{6}\)
A1
\(=\dfrac{25}{6}\)
AG
(c)
general expression or at least two further values: \(m=1:\dfrac{23.5}{6}\), \(m=3:\dfrac{15+10.5}{6}\), \(m=4:\dfrac{11+14}{6}\)
M1
\(m=3\)
A1
\(\dfrac{25.5}{6}=\dfrac{17}{4}\)
A1
(d)
rolling again gives an expected score of 3.5, which is less than the 4 already obtained
Since \(A\subset A\cup B\), \(A\cap(A\cup B)=A\). FT from their \(\mathrm{P}(A)\).
Answers: (b)(i) \(\mathrm{P}(A)=0.3\) · (ii) \(\tfrac{5}{12}\)
157AASL4-q01104.11 Conditional probability and independencePaper 1Hard13 marks
A supermarket buys bags of rice from two suppliers, A and B. Supplier A provides 60% of the bags and supplier B provides the rest. Of the bags from supplier A, 20% are underweight; of the bags from supplier B, 5% are underweight. The tree diagram shows this information.
U = underweight, N = not underweight
(a)
Write down the value of \(p\), of \(q\) and of \(r\).
(2)
(b)
A bag is chosen at random. Find the probability that it is underweight.
(2)
(c)
Given that a bag is underweight, find the probability that it came from supplier A.
(2)
(d)
Determine whether the events “from supplier A” and “underweight” are independent.
(2)
(e)
Two bags are chosen at random. Find the probability that exactly one of them is underweight.
(2)
(f)(i)
The masses, \(X\) grams, of bags from supplier B are normally distributed with mean \(\mu\). A bag is underweight if its mass is less than 490 g, and it is known that \(\mathrm{P}(X\gt 510)=0.05\).
Write down the value of \(\mu\).
(1)
(f)(ii)
A bag from supplier B is not underweight. Find the probability that its mass is more than 510 g.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(p=0.4\)
A1
\(q=0.8\) and \(r=0.95\)
A1
Both required.
(b)
\(0.6(0.2)+0.4(0.05)=0.12+0.02\)
(M1)
\(0.14\)
A1
(c)
\(\dfrac{0.12}{0.14}\)
(M1)
\(\dfrac{6}{7}\)
A1
(d)
\(\mathrm{P}(\text{underweight}\mid A)=0.2\) and \(\mathrm{P}(\text{underweight})=0.14\), or \(\mathrm{P}(A)\,\mathrm{P}(U)=0.084\ne0.12=\mathrm{P}(A\cap U)\)
M1
not independent
A1
Must follow from a correct comparison.
(e)
\(2\times0.14\times0.86\)
(M1)
Award for 0.14 × 0.86 seen.
\(0.2408\)
A1
Accept \(\dfrac{301}{1250}\).
(f)(i)
\(\mu=500\)
A1
\(\mathrm{P}(X\lt 490)=0.05=\mathrm{P}(X\gt 510)\), so by symmetry \(\mu\) is midway between 490 and 510.
158IBMA2S-050-07-114.6 Combined events and diagramsPaper 1Hard14 marks
No calculator
A technician has a box containing n working fuses and 2 faulty fuses, where n ≥ 1. She tests fuses one at a time, chosen at random without replacement, until she finds a working fuse. Let X be the number of fuses she tests, so X ∈ {1, 2, 3}.
(a)
For the case n = 3, find the probability distribution of X and hence find E(X).
(4)
(b)
For general n, show that \(P(X=2)=\dfrac{2n}{(n+1)(n+2)}\) and \(P(X=3)=\dfrac{2}{(n+1)(n+2)}\).
(3)
(c)
Show that \(E(X)=\dfrac{n+3}{n+1}\).
(3)
(d)
The expected number of fuses tested is 1.1. Find n.
(2)
(e)
For this value of n, given that the first fuse tested is faulty, find the probability that she tests exactly three fuses.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 multiplies conditional probabilities without replacement
Exam tip: Check a general formula against a case you have already worked out: with n = 3, (n + 3)/(n + 1) = 6/4 = 3/2, matching part (a).
159IBMA2S-050-01-104.6 Combined events and diagramsPaper 1Hard11 marks
No calculator
A space probe carries 10 identical sensors. During a pass through a dust cloud, each sensor is damaged with probability \(p\), where \(0 < p < 1\), independently of the others. Let \(X\) be the number of damaged sensors. It is known that \(P(X = 1) = P(X = 2)\).
(a)
Show that \(p = \frac{2}{11}\).
(4)
(b)
By considering \(\dfrac{P(X=k+1)}{P(X=k)}\) for \(k = 0, 1, \ldots, 9\), show that 1 and 2 are the most likely numbers of damaged sensors.
M1 divides by \(p(1-p)^8\) (valid as \(0<p<1\)) to get \(10(1-p) = 45p\)
✓ 1
A1 \(55p = 10\) so \(p = \frac{2}{11}\)
✓ 1
Part (b)
M1 ratio \(= \frac{\binom{10}{k+1}}{\binom{10}{k}}\cdot\frac{p}{1-p}\)
✓ 1
A1 \(= \frac{10-k}{k+1}\cdot\frac{2}{9}\)
✓ 1
R1 \(k=0\): ratio \(= \frac{20}{9} > 1\) so \(P(X=1) > P(X=0)\); \(k=1\): ratio \(= 1\) so \(P(X=2) = P(X=1)\)
✓ 1
R1 \(\frac{10-k}{k+1}\) decreases as \(k\) increases and at \(k=2\) the ratio is \(\frac{16}{27} < 1\), so probabilities decrease for \(k \ge 2\); hence 1 and 2 are most likely
Exam tip: When cancelling \(p\) or \((1-p)\) from both sides, state that they are non-zero; this justifies the division and is part of a complete argument.
160IBMA2S-050-04-124.6 Combined events and diagramsPaper 1Hard14 marks
No calculator
A spacecraft carries two independent star-tracking sensors, A and B. Let \(A\) and \(B\) also denote the events that the respective sensors work during a manoeuvre, with \(P(A)>P(B)\). It is known that \(P(A\cap B)=\frac{1}{6}\) and \(P(A\cup B)=\frac{2}{3}\).
(a)
Show that \(P(A)+P(B)=\frac{5}{6}\).
(2)
(b)
Hence find \(P(A)\) and \(P(B)\).
(4)
(c)
Find the probability that exactly one sensor works.
(2)
(d)
Given that at least one sensor works, find the probability that sensor A works.
(2)
(e)
A control unit C works with probability \(q\), independently of the sensors. The navigation system works only if C works and at least one sensor works. The probability that the navigation system works is \(\frac{1}{2}\). Find \(q\).
(2)
(f)
Given that the navigation system fails, find the probability that the control unit failed.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), i.e. \(\frac{2}{3}=P(A)+P(B)-\frac{1}{6}\)
M1 uses independence \(P(A)P(B)=\frac{1}{6}\) with (a) to form a quadratic
✓ 1
A1 \(6t^2-5t+1=0\) (or equivalent)
✓ 1
A1 \(t=\frac{1}{2}\) or \(t=\frac{1}{3}\)
✓ 1
R1 since \(P(A)>P(B)\): \(P(A)=\frac{1}{2}\), \(P(B)=\frac{1}{3}\)
✓ 1
Part (c)
M1 \(P(A\cup B)-P(A\cap B)\) or \(\frac{1}{2}\times\frac{2}{3}+\frac{1}{2}\times\frac{1}{3}\)
✓ 1
A1 \(\frac{1}{2}\)
✓ 1
Part (d)
M1 \(\frac{P(A)}{P(A\cup B)}=\frac{1/2}{2/3}\)
✓ 1
A1 \(\frac{3}{4}\)
✓ 1
Part (e)
M1 \(\frac{2}{3}q=\frac{1}{2}\)
✓ 1
A1 \(q=\frac{3}{4}\)
✓ 1
Part (f)
M1 recognises C failing implies system fails, so \(\frac{P(C')}{P(\text{fail})}=\frac{1/4}{1/2}\)
✓ 1
A1 \(\frac{1}{2}\)
✓ 1
Exam tip: When you know the sum and product of two probabilities, treat them as roots of a quadratic \(t^2-(\text{sum})t+(\text{product})=0\); then use the given condition to decide which root is which.
161IBMA2S-050-05-114.2 Presentation of dataPaper 1Hard7 marks
No calculator
The delays, in minutes, of seven trains on one morning were 10, 13, 15, 16, 18, 21 and x, where x is a positive integer. The quartiles are found as the medians of the lower and upper halves of the ordered data, excluding the median. An outlier is a value more than 1.5 × IQR beyond the nearer quartile.
(a)
Given that x < 10, show that x cannot be an outlier.
(3)
(b)
Given that x > 21, find the smallest value of x for which x is an outlier.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 ordered data x, 10, 13, 15, 16, 18, 21 gives Q1 = 10, Q3 = 18 A1 IQR = 8, lower boundary = 10 − 12 = −2 R1 x is positive so x > −2 and cannot be an outlier
✓ 3
Part (b)
M1 ordered data 10, 13, 15, 16, 18, 21, x A1 Q1 = 13, Q3 = 21, IQR = 8 M1 upper boundary = 21 + 1.5 × 8 = 33, requiring x > 33 A1 x = 34 minutes
✓ 4
Common error: The position of x in the ordered list changes the quartiles. Recalculate them for each case rather than reusing them.
162IBMA2S-050-09-094.2 Presentation of dataPaper 1Hard12 marks
No calculator
A cyclist's fitness tracker records the distances, in km, of seven rides. In ascending order they are 3, 4, a, 7, 8, b, 20 The mean distance is 8 km and the interquartile range is 6 km.
(a)
Show that a + b = 14.
(2)
(b)
By considering the quartiles, find the value of a and the value of b.
(4)
(c)
Show that the ride of 20 km is an outlier.
(3)
(d)
The 20 km ride was logged when the tracker was accidentally left on in a car, so it is removed. Find the median and the mean of the remaining distances.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 total = 7 × 8 = 56
✓ 1
A1 a + b = 56 − (3 + 4 + 7 + 8 + 20) = 14
✓ 1
Part (b)
R1 Q1 is the median of 3, 4, a, and since a ≥ 4, Q1 = 4
✓ 1
R1 Q3 is the median of 8, b, 20, so Q3 = b
✓ 1
M1 b − 4 = 6
✓ 1
A1 b = 10, a = 4
✓ 1
Part (c)
M1 upper fence = Q3 + 1.5 × IQR = 10 + 1.5 × 6
✓ 1
A1 upper fence = 19
✓ 1
R1 20 > 19, so 20 km is an outlier
✓ 1
Part (d)
A1 median = (4 + 7) ÷ 2 = 5.5 km
✓ 1
M1 (3 + 4 + 4 + 7 + 8 + 10) ÷ 6 = 36 ÷ 6
✓ 1
A1 mean = 6 km
✓ 1
Common error: With 7 values, the quartiles are the medians of the three values on each side of the median. The median itself is not included in either half, so Q₁ = 4, not (4 + a)/2.
163IBMA2S-050-08-114.10 Regression of x on yPaper 1Hard11 marks
No calculator
Engineers testing a planetary rover record the slope angle x degrees of the ground and the wheel slip y% at a number of test sites. The regression line of y on x is \(y = 1.5x + p\). The regression line of x on y is \(x = 0.6y + q\). The mean slope angle is 12°. When the wheel slip is 30%, the regression line of x on y gives an estimated slope angle of 15°.
(a)
Find the value of q.
(2)
(b)
Find the mean wheel slip \(\bar{y}\).
(2)
(c)
Find the value of p.
(2)
(d)
Estimate the wheel slip on a slope of 10°.
(2)
(e)
(i) State what would be true of the two regression lines if the correlation were perfect, \(r = 1\). (ii) Assuming the regression line of y on x is unchanged, find the gradient that the regression line of x on y would then have.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitutes: \(15 = 0.6(30) + q\)
✓ 1
A1 \(q = -3\)
✓ 1
Part (b)
M1 uses the fact that the line passes through \((\bar{x},\bar{y})\): \(12 = 0.6\bar{y} - 3\)
✓ 1
A1 \(\bar{y} = 25\)%
✓ 1
Part (c)
M1 substitutes \((12, 25)\) into the y on x line: \(25 = 1.5(12) + p\)
✓ 1
A1 \(p = 7\)
✓ 1
Part (d)
M1 uses the y on x line: \(y = 1.5(10) + 7\)
✓ 1
A1 22%
✓ 1
Part (e)
R1 (i) the two lines would be identical (coincide)
✓ 1
M1 (ii) rearranges \(y = 1.5x + 7\) to \(x = \frac{y - 7}{1.5}\)
✓ 1
A1 gradient \(\frac{2}{3}\)
✓ 1
Exam tip: Any information about a mean point or a prediction gives a linear equation in the unknown constants; work through them in the order the line equations make possible.
164IBMA3S-4-16r5-124.7 Discrete random variablesPaper 1Hard13 marks
No calculator
At a music festival, the number of encore songs X played by a band has probability distribution P(X = x) = k(6 − x) for x = 1, 2, 3, 4, 5.
(a)
Show that k = 1/15.
(2)
(b)
Find E(X).
(2)
(c)
Find Var(X).
(3)
(d)
Find P(X ≥ 3 | X ≥ 2).
(3)
(e)
The band plays at two festivals, with independent numbers of encores X1 and X2. Find P(X1 + X2 = 3).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 5k + 4k + 3k + 2k + k = 1
✓ 1
A1 15k = 1 so k = 1/15 (AG)
✓ 1
Part (b)
M1 attempt to use ΣxP(X = x)
✓ 1
A1 E(X) = (5 + 8 + 9 + 8 + 5)/15 = 7/3
✓ 1
Part (c)
M1 E(X2) = (5 + 16 + 27 + 32 + 25)/15
✓ 1
A1 E(X2) = 7
✓ 1
A1 Var(X) = 7 − (7/3)2 = 14/9
✓ 1
Part (d)
M1 use of P(A | B) = P(A ∩ B)/P(B), where X ≥ 3 is within X ≥ 2
✓ 1
A1 P(X ≥ 3) = 6/15 and P(X ≥ 2) = 10/15
✓ 1
A1 3/5
✓ 1
Part (e)
M1 two cases: (1, 2) and (2, 1)
✓ 1
M1 2 × (5/15) × (4/15)
✓ 1
A1 8/45
✓ 1
Common error: Var(X) = E(X²) − [E(X)]². Squaring the values in E(X²) before multiplying by the probabilities is the step that is most often skipped.
165IBMA3S-4-16r6-124.10 Regression of x on yPaper 1Hard7 marks
No calculator
A study of two variables x and y gives the regression lines L1: 2x − 5y = −25 and L2: 5x − 8y = −22. One line is the regression line of y on x and the other is the regression line of x on y. You may use the fact that, if the y on x line is y = mx + c and the x on y line is x = ny + d, then mn = r2, where r is the correlation coefficient.
(a)
Determine which line is the regression line of y on x.
(2)
(b)
Find the mean of x and the mean of y.
(3)
(c)
Estimate the value of x when y = 12.5.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 L1 as y on x gives m = 0.4; L2 as x on y gives n = 1.6, so r2 = 0.64 (valid)
✓ 1
R1 the other way round gives m = 0.625 and n = 2.5, so r2 = 1.5625 > 1, impossible; so L1 is y on x
✓ 1
Part (b)
M1 the lines intersect at (x̄, ȳ); attempt to solve simultaneously
✓ 1
A1 ȳ = 9
✓ 1
A1 x̄ = 10
✓ 1
Part (c)
M1 use the x on y line: x = 1.6y − 4.4
✓ 1
A1 x = 15.6
✓ 1
Exam tip: The regression line passes through the point of means, so two regression lines meet exactly there.
166IBMA3S-4-16r8-124.8 The binomial distributionPaper 1Hard11 marks
No calculator
Two independent telecoms masts, A and B, each try to connect a handset. The probability that mast A connects is p and the probability that mast B connects is p + 0.1. The probability that at least one mast connects is 0.88.
(a)
Show that p2 − 1.9p + 0.78 = 0.
(3)
(b)
Solve the equation to find the value of p.
(3)
(c)
Let E be the event that at least one mast connects. Determine whether A and E are independent.
(2)
(d)
Given that at least one mast connects, find the probability that only mast A connects.
Common error: 'At least one' events are not independent of the individual events that make them up. Compare P(A ∩ E) with P(A)P(E) rather than assuming.
167IBMA3S-4-16r9-074.2 Presentation of dataPaper 1Hard7 marks
No calculator
Six response delays, in milliseconds, of a robot's sensor are 3, 5, 6, x, 10 and y, where x < y. The mean delay is 7 and the variance is 20/3.
(a)
Show that x + y = 18.
(2)
(b)
Show that x2 + y2 = 164.
(2)
(c)
Find the value of x and the value of y.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (3 + 5 + 6 + x + 10 + y) / 6 = 7, so the sum is 42
✓ 1
A1 x + y = 18
✓ 1
Part (b)
M1 variance = Σx²/6 − 7² = 20/3, so Σx² = 334
✓ 1
A1 334 − (9 + 25 + 36 + 100) = 164
✓ 1
Part (c)
M1 substitute y = 18 − x (or use (x + y)² to find xy = 80)
✓ 1
A1 x² − 18x + 80 = 0
✓ 1
A1 x = 8, y = 10
✓ 1
Exam tip: When asked to show a result, write every line of the working, including the formula for variance with the mean subtracted.
168IBMA3S-4-17r12-084.11 Conditional probability and independencePaper 1Hard8 marks
No calculator
In a food-science taste test, event \(A\) is that a taster rates a sauce as 'sweet' and event \(B\) is that the taster rates it as 'salty'. The events \(A\) and \(B\) are independent, with \(\mathrm{P}(A) = x\) and \(\mathrm{P}(B) = x + 0.3\). It is given that \(\mathrm{P}(A \cup B) = 0.82\).
(a)
Show that \(x^2 - 1.7x + 0.52 = 0\).
(3)
(b)
Hence find the value of \(x\), justifying your choice.
(3)
(c)
Find the probability that a taster rates the sauce as exactly one of 'sweet' or 'salty'.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 independence: \(\mathrm{P}(A\cap B) = x(x+0.3)\)
✓ 1
M1 \(\mathrm{P}(A\cup B) = x + (x+0.3) - x(x+0.3) = 0.82\)
✓ 1
A1 expand and rearrange to reach \(x^2 - 1.7x + 0.52 = 0\) (AG)
✓ 1
Part (b)
M1 attempt to solve, e.g. \((x-0.4)(x-1.3)=0\)
✓ 1
A1 \(x = 0.4\) or \(x = 1.3\)
✓ 1
R1 reject 1.3 because a probability cannot exceed 1, so \(x = 0.4\)
✓ 1
Part (c)
M1 \(\mathrm{P}(A\cup B) - \mathrm{P}(A\cap B) = 0.82 - 0.4 \times 0.7\) (ECF from (b))
✓ 1
A1 0.54
✓ 1
Common error: For independent events use \(\mathrm{P}(A\cap B)=\mathrm{P}(A)\mathrm{P}(B)\); do not assume they are mutually exclusive.
169IBMA3S-4-18r36-094.6 Combined events and diagramsPaper 1Hard7 marks
No calculator
Late deliveries are caused by road works (event L) or by a sorting error (event S). The events L and S are independent, P(L) = 0.4 and P(L ∪ S) = 0.7.
(a)
Find P(S).
(3)
(b)
Find P(L′ ∩ S).
(2)
(c)
Find the probability that exactly one of the two causes occurs.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(L ∩ S) = P(L) × P(S)
✓ 1
M1 0.7 = 0.4 + P(S) − 0.4 P(S)
✓ 1
A1 P(S) = 0.5
✓ 1
Part (b)
M1 0.6 × 0.5
✓ 1
A1 0.3
✓ 1
Part (c)
M1 P(L ∩ S′) + P(L′ ∩ S) = 0.4 × 0.5 + 0.3
✓ 1
A1 0.5
✓ 1
Exam tip: Independence lets you replace P(A ∩ B) by P(A) × P(B), which turns the union formula into an equation in one unknown.
170IBMA3S-4-17r38-124.11 Conditional probability and independencePaper 1Hard9 marks
No calculator
A box holds 10 sensor chips for a space probe, of which 3 are defective. Two chips are chosen at random without replacement.
(a)
Find the probability that exactly one chip is defective.
(3)
(b)
Given that at least one chip is defective, find the probability that both are defective.
(3)
(c)
Given that at least one chip is defective, find the probability that the first chip chosen is defective.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (3/10)(7/9) + (7/10)(3/9)
✓ 1
A1 42/90
✓ 1
A1 7/15
✓ 1
Part (b)
A1 P(both defective) = (3/10)(2/9) = 1/15 and P(at least one) = 8/15
✓ 1
M1 P(both | at least one) = (1/15)/(8/15)
✓ 1
A1 1/8
✓ 1
Part (c)
M1 the event "first defective" lies inside "at least one defective", so the intersection is P(first defective) = 3/10
✓ 1
M1 (3/10)/(8/15)
✓ 1
A1 9/16
✓ 1
Exam tip: Without replacement, the second probability changes: use 9 chips remaining on the second draw.
171AASL4-q00134.2 Presentation of dataPaper 2Easy7 marks
The frequency histogram shows the time, \(h\) hours per week, that a group of students spend on homework. The class intervals are \(0\le h\lt 5\), \(5\le h\lt 10\), …, \(30\le h\lt 35\).
Graph drawn to scale
(a)
Write down the number of students in the group.
(1)
(b)
Write down the modal class.
(1)
(c)
Find an estimate for the mean time spent on homework.
(2)
(d)
Find an estimate for the standard deviation of the time spent on homework.
(1)
(e)
Find the percentage of the students who spend at least 20 hours per week on homework.
(1)
(f)
Explain why the answer to part (c) is an estimate.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(120\)
A1
(b)
\(10\le h\lt 15\)
A1
(c)
use of mid-interval values \(2.5,\,7.5,\,\ldots,\,32.5\) with frequencies \(10,\,20,\,34,\,26,\,18,\,8,\,4\)
(M1)
e.g. \(\dfrac{2.5(10)+7.5(20)+\cdots+32.5(4)}{120}\)
\(15.1\) hours
A1
\(15.083\ldots\)
(d)
\(7.39\) hours
A1
\(7.3875\ldots\) (GDC \(\sigma_{x}\)).
(e)
\(\dfrac{18+8+4}{120}=25\%\)
A1
(f)
the exact times are not known; every value in a class is assumed to equal its mid-interval value
173AASL4-q00234.3 Measures of central tendency and dispersionPaper 2Easy8 marks
The maximum daily temperatures, in °C, in a town on ten days in May were
Day
1
2
3
4
5
6
7
8
9
10
Temperature (°C)
18.4
21.0
19.6
22.3
20.8
17.9
23.5
21.7
19.2
20.6
(a)
Find the mean temperature.
(1)
(b)(i)
Find
the standard deviation;
(1)
(b)(ii)
the variance.
(1)
(c)
Use your calculator to find the interquartile range of the temperatures.
(2)
(d)
The temperatures are converted to degrees Fahrenheit, °F, using \(F=1.8C+32\). Find the mean and the standard deviation of the temperatures in °F.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(20.5\) °C
A1
(b)(i)
\(1.66\) °C
A1
\(1.66433\ldots\) (\(\sigma_{x}\)).
(b)(ii)
\(2.77\) °C2
A1
Accept 2.76 from squaring 1.66. FT from (b)(i).
(c)
\(Q_{1}=19.2\) and \(Q_{3}=21.7\)
(A1)
GDC values.
IQR \(=2.5\) °C
A1
Other methods of finding quartiles can give different values (e.g. 19.3 and 21.525, giving 2.225); accept a correct IQR from a clearly stated alternative method.
(d)
mean \(=1.8(20.5)+32\)
(M1)
\(68.9\) °F
A1
standard deviation \(=1.8\times1.66433\ldots=3.00\) °F
A1
Award A0 for adding 32 to the standard deviation. Accept 2.99 or 3.
Answers: (a) 20.5 °C · (b)(i) 1.66 °C, (ii) 2.77 · (c) 2.5 °C · (d) 68.9 °F, 3.00 °F
174AASL4-q00244.3 Measures of central tendency and dispersionPaper 2Easy9 marks
The table shows the number of goals, \(g\), scored in each of the 46 matches played in a football league in one month.
Goals, \(g\)
0
1
2
3
4
5
6
Number of matches
8
14
11
7
3
2
1
(a)
Write down the mode.
(1)
(b)
Find the median.
(1)
(c)
Find the mean number of goals per match.
(2)
(d)
Find the interquartile range.
(2)
(e)
Find the standard deviation.
(1)
(f)
Determine whether the match with 6 goals is an outlier.
Find the equation of the regression line of \(y\) on \(x\), in the form \(y=ax+b\).
(2)
(c)(i)
In the context of the question, interpret
the value of \(a\);
(1)
(c)(ii)
the value of \(b\).
(1)
(d)
Use the regression line to estimate the volume of water in the reservoir halfway through week 6 \((x=5.5)\).
(2)
(e)
Use the regression line to find the volume it predicts at the end of week 15, and comment on your answer.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(r=-0.997\)
A1
\(-0.997281\ldots\)
(b)
\(a=-2.05\)
A1
\(-2.05238\ldots\)
\(b=26.0\)
A1
\(26.0357\ldots\) Award A1A0 if not written as an equation.
(c)(i)
the volume of water decreases by about 2.05 million m3 each week
R1
Must refer to a decrease in volume per week.
(c)(ii)
the estimated volume at the start of the summer (week 0) is about 26.0 million m3
R1
(d)
\(-2.05238(5.5)+26.0357\)
(M1)
\(14.7\) million m3
A1
\(14.7476\ldots\)
(e)
\(-4.75\) million m3
A1
Accept \(-4.74\) to \(-4.76\).
a negative volume is impossible; week 15 is far outside the data range (weeks 1 to 8), so the prediction is an unreliable extrapolation (e.g. rain may fall or the reservoir empties)
177AASL4-q00444.5 Probability and expected occurrencesPaper 2Easy7 marks
A drawing pin is dropped onto a table and lands either point up or point down. Selin drops the pin repeatedly and records the total number of times it has landed point up.
Total number of drops
50
100
200
500
1000
Total number landing point up
21
38
74
182
371
(a)
Find the relative frequency of the pin landing point up after 100 drops.
(1)
(b)
Write down the best estimate of the probability that the pin lands point up. Give a reason for your choice.
(2)
(c)
The pin is dropped another 1200 times. Use your answer to part (b) to find the expected number of times the pin lands point up.
(2)
(d)
Two of these pins are dropped together, and they land independently. Estimate the probability that both land point down.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac{38}{100}=0.38\)
A1
(b)
\(\dfrac{371}{1000}=0.371\)
A1
it is based on the greatest number of trials (relative frequencies settle down as the number of trials increases)
R1
Award R1 only with 0.371.
(c)
\(1200\times0.371\)
(M1)
FT from (b).
\(445\)
A1
\(445.2\) Accept 445.2.
(d)
\(\mathrm{P}(\text{point down})=1-0.371=0.629\), and \(0.629^{2}\)
179AASL4-q00704.8 The binomial distributionPaper 2Easy6 marks
The probability that a seed of a certain type fails to germinate is 0.12, independently of other seeds. A packet contains 20 of these seeds. Let \(X\) be the number of seeds in the packet that fail to germinate.
(a)
Find \(\mathrm{P}(X=3)\).
(2)
(b)
Find the probability that at most two seeds fail to germinate.
(2)
(c)
Find the mean and the variance of \(X\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(X\sim\mathrm{B}(20,\,0.12)\) and attempt to find \(\mathrm{P}(X=3)\)
181AASL4-q00834.9 The normal distributionPaper 2Easy7 marks
The lengths, \(L\) cm, of fish in a lake are normally distributed with mean 18.0 cm and standard deviation 1.5 cm. The shaded region on the diagram represents \(\mathrm{P}(17\lt L\lt 19)\).
Graph drawn to scale
(a)
Find \(\mathrm{P}(L\gt 20)\).
(2)
(b)
Find \(\mathrm{P}(17\lt L\lt 19)\).
(2)
(c)
A sample of 400 fish is caught. Find the expected number of these fish that are longer than 20 cm.
(2)
(d)
Write down the approximate percentage of fish with lengths between 15.0 cm and 21.0 cm.
(1)
Show mark scheme
Working
Mark
Notes
(a)
normalcdf with lower 20, upper large, \(\mu=18\), \(\sigma=1.5\)
(M1)
\(0.0912\)
A1
\(0.0912112\ldots\)
(b)
normalcdf(17, 19, 18, 1.5)
(M1)
\(0.495\)
A1
\(0.495014\ldots\)
(c)
\(400\times0.0912112\ldots\)
(M1)
\(36.5\)
A1
\(36.484\ldots\) An expected number need not be a whole number: award A0 for 36 or 37.
182AASL4-q00854.9 The normal distributionPaper 2Easy6 marks
The time, \(T\) minutes, that adults take to complete a particular logic puzzle is normally distributed with mean 42 minutes and standard deviation 3.2 minutes.
Diagram not to scale
(a)
Find the probability that a randomly chosen adult takes more than 45 minutes.
(2)
(b)
Hence, or otherwise, find \(\mathrm{P}(39\lt T\lt 45)\).
(2)
(c)
Find the time \(t\) such that 95% of adults complete the puzzle in less than \(t\) minutes.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{P}(T\gt 45)\) with \(\mu=42,\ \sigma=3.2\), e.g. normalcdf(45, large, 42, 3.2)
(M1)
\(0.174\)
A1
\(0.174250\ldots\)
(b)
\(1-2\times0.174250\ldots\) (symmetry about 42), or normalcdf(39, 45, 42, 3.2)
(M1)
\(0.651\)
A1
\(0.651498\ldots\) FT from (a) for the symmetry method.
(c)
\(\mathrm{P}(T\lt t)=0.95\), inverse normal
(M1)
\(t=47.3\) minutes
A1
\(47.2635\ldots\)
Answers: (a) 0.174 · (b) 0.651 · (c) 47.3 min
183AASL4-q00874.9 The normal distributionPaper 2Easy8 marks
The lengths, \(L\) cm, of leaves on a type of shrub are modelled by a normal distribution with mean 16.5 cm and standard deviation 2.2 cm. A biologist collects 200 leaves and uses the model to find the expected number of leaves in each length class.
Length, \(L\) cm
\(L\lt 13\)
\(13\le L\lt 15\)
\(15\le L\lt 17\)
\(17\le L\lt 19\)
\(19\le L\lt 21\)
\(L\ge 21\)
Expected frequency
\(a\)
38.4
\(b\)
56.4
21.5
\(c\)
(a)
Find \(\mathrm{P}(L\lt 13)\).
(2)
(b)
Hence find the value of \(a\).
(1)
(c)
Find the value of \(b\) and the value of \(c\).
(3)
(d)
Find the length exceeded by 5% of the leaves, according to the model.
(2)
Show mark scheme
Working
Mark
Notes
(a)
normalcdf(−large, 13, 16.5, 2.2)
(M1)
\(0.0558\)
A1
\(0.0558150\ldots\)
(b)
\(a=200\times0.0558150\ldots=11.2\)
A1
\(11.1630\ldots\) FT from (a).
(c)
\(200\times\mathrm{P}(15\le L\lt 17)\) or \(200\times\mathrm{P}(L\ge 21)\)
(M1)
\(b=68.4\)
A1
\(68.4434\ldots\)
\(c=4.08\)
A1
\(4.08100\ldots\) Accept 4.1. Award A1 also for \(c=200-(11.2+38.4+68.4+56.4+21.5)=4.1\).
184AASL4-q00984.10 Regression of x on yPaper 2Easy6 marks
A forensic scientist records the height, \(x\) cm, and the shoe size, \(y\) (European sizes), of ten adults.
\(x\) (cm)
158
163
167
170
172
175
178
181
184
190
\(y\)
42.1
43.9
44.6
45.8
45.9
47.2
47.6
48.9
49.1
51.0
(a)
Write down the value of Pearson’s product-moment correlation coefficient, \(r\).
(1)
(b)
Find the equation of the regression line of \(x\) on \(y\).
(2)
(c)
A shoe print of size 46.5 is found. Estimate the height of the person who made it.
(2)
(d)
Explain why the regression line of \(x\) on \(y\), rather than \(y\) on \(x\), is used in part (c).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(r=0.996\)
A1
\(0.996104\ldots\)
(b)
gradient \(3.64\)
A1
\(3.64120\ldots\)
\(x=3.64y+4.08\)
A1
\(4.08329\ldots\)
(c)
\(x=3.64120(46.5)+4.08329\)
(M1)
\(173\) cm
A1
\(173.399\ldots\) Accept 173.4.
(d)
the shoe size \(y\) is known and the height \(x\) is to be estimated, so the line of \(x\) on \(y\) is required
R1
Answers: (a) 0.996 · (b) \(x=3.64y+4.08\) · (c) 173 cm
185IBMA2S-050-02-014.9 The normal distributionPaper 2Easy5 marks
Calculator
The charge capacity, X mAh, of a lithium-ion battery made for a smartphone is normally distributed with mean 4500 mAh and standard deviation 120 mAh.
Diagram not to scale
(a)
Find P(X < 4300).
(2)
(b)
Find P(4400 < X < 4700).
(2)
(c)
Find the expected number of batteries in a batch of 2000 with capacity less than 4300 mAh.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 correct set-up, e.g. z = (4300 − 4500)/120 = −1.667 or GDC normalcdf with correct bounds ; A1 0.0478
✓ 2
Part (b)
M1 correct bounds 4400 and 4700 (or z = −0.833 and 1.667) used ; A1 0.750
✓ 2
Part (c)
A1 2000 × 0.04779… = 95.6
✓ 1
Exam tip: An expected number does not need to be a whole number. Leave it as 95.6 rather than rounding to 96 unless the question asks for an integer.
186IBMA2S-050-02-034.9 The normal distributionPaper 2Easy5 marks
Calculator
In a maternity unit, the birth mass, X kg, of full-term babies is modelled by a normal distribution with mean 3.40 kg and standard deviation 0.45 kg. A baby is flagged for extra monitoring if its birth mass is in the lowest 10% of the distribution.
Diagram not to scale
(a)
Find the mass m such that P(X < m) = 0.10.
(2)
(b)
Find the probability that exactly one of five babies is flagged.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 inverse normal with area 0.10, e.g. (m − 3.40)/0.45 = −1.28155 ; A1 m = 2.82 kg
✓ 2
Part (b)
M1 recognises binomial distribution Y ~ B(5, 0.1) ; M1 5 × 0.1 × 0.94 or GDC binompdf(5, 0.1, 1) ; A1 0.328
✓ 3
Common error: In part (a), using +1.28 instead of −1.28 gives the 90th percentile, 3.98 kg. A lower-tail area of 0.10 must give a value below the mean.
187IBMA2S-050-02-044.9 The normal distributionPaper 2Easy6 marks
Calculator
The distance, D metres, that a javelin thrower achieves is normally distributed with mean 62.0 m and standard deviation 4.5 m.
Diagram not to scale
(a)
Find P(D > 70).
(2)
(b)
Find d such that P(D > d) = 0.25.
(2)
(c)
Find P(D > 70 | D > 62).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 correct set-up, e.g. z = (70 − 62)/4.5 = 1.778 ; A1 0.0377
✓ 2
Part (b)
M1 inverse normal with upper-tail area 0.25 (lower area 0.75), z = 0.67449 ; A1 d = 65.0 m
✓ 2
Part (c)
M1 P(D > 70)/P(D > 62) = 0.03772/0.5 ; A1 0.0754
✓ 2
Exam tip: For conditional probability, the event 'exceeds 70 m' lies entirely inside 'exceeds 62 m'. The intersection is therefore just P(D > 70).
188IBMA2S-050-07-014.7 Discrete random variablesPaper 2Easy5 marks
Calculator
An astronomy club records the number of bright satellite passes, N, visible from its observatory on a randomly chosen clear night. The probability distribution of N is shown below.
n: 0, 1, 2, 3 P(N = n): 0.1, a, b, 0.2
It is known that E(N) = 1.6.
(a)
Find the value of a and the value of b.
(4)
(b)
Find the probability that at least two bright satellite passes are visible on a randomly chosen clear night.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses the fact that the probabilities sum to 1, \(0.1+a+b+0.2=1\)
✓ 1
A1 \(a+b=0.7\)
✓ 1
M1 forms the expectation equation \(0(0.1)+1(a)+2b+3(0.2)=1.6\), i.e. \(a+2b=1\)
✓ 1
A1 \(a=0.4\), \(b=0.3\)
✓ 1
Part (b)
A1 \(P(N\ge 2)=b+0.2=0.3+0.2=0.5\) (follow through from their value of \(b\), provided \(0 \le b \le 0.8\))
✓ 1
Exam tip: When a table has two unknowns, you need two equations: one from the probabilities summing to 1 and one from the given expected value.
189IBMA2S-050-01-024.8 The binomial distributionPaper 2Easy6 marks
Calculator
A small satellite carries 6 solar panels. During its first year in orbit, each panel fails with probability 0.07, independently of the others. Let \(X\) be the number of panels that fail during the first year.
(a)
State the distribution of \(X\), including its parameters.
(1)
(b)
Find the probability that exactly one panel fails.
(2)
(c)
The satellite loses its main power supply if two or more panels fail. Find the probability that this happens during the first year.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(X \sim \text{B}(6,\,0.07)\)
✓ 1
Part (b)
M1 \(\binom{6}{1}(0.07)(0.93)^5\) or correct use of binomial pdf
✓ 1
A1 0.292
✓ 1
Part (c)
M1 recognises \(P(X \ge 2) = 1 - P(X \le 1)\)
✓ 1
M1 \(1 - (0.93^6 + 0.292189\ldots)\) or \(1 - 0.939179\ldots\)
✓ 1
A1 0.0608
✓ 1
Common error: Using \(1 - P(X \le 2)\) for \(P(X \ge 2)\) removes the outcome you want. Write the complement explicitly before using the cumulative function.
190IBMA2S-050-01-044.8 The binomial distributionPaper 2Easy6 marks
Calculator
A vaccine causes mild side effects in 12% of patients who receive it. On one day a clinic vaccinates 80 patients. Assume that patients experience side effects independently. Let \(X\) be the number of these patients who experience mild side effects.
(a)
Find the expected number of patients who experience mild side effects.
(2)
(b)
Find the standard deviation of \(X\).
(2)
(c)
Find the probability that exactly 10 patients experience mild side effects.
M1 \(\binom{80}{10}(0.12)^{10}(0.88)^{70}\) or binomial pdf with \(n=80\), \(p=0.12\), \(x=10\)
✓ 1
A1 0.132
✓ 1
Common error: \(np(1-p)\) is the variance, not the standard deviation. Take the square root before writing your final answer.
191IBMA2S-050-05-024.2 Presentation of dataPaper 2Easy6 marks
Calculator
The reaction times, t milliseconds, of 50 sprinters leaving the blocks were recorded. 100 ≤ t < 120: 4 120 ≤ t < 140: 11 140 ≤ t < 160: 18 160 ≤ t < 180: 12 180 ≤ t < 200: 5
(a)
Write down the cumulative frequencies at t = 120, 140, 160, 180 and 200.
(2)
(b)
Write down the modal class.
(1)
(c)
State, with a reason, the class that contains the median reaction time.
(1)
(d)
Find the percentage of sprinters whose reaction time was at least 160 ms.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 at least three correct values from 4, 15, 33, 45, 50 A1 all five correct: 4, 15, 33, 45, 50
✓ 2
Part (b)
A1 140 ≤ t < 160
✓ 1
Part (c)
A1 140 ≤ t < 160, with a reason: the median lies between the 25th and 26th values (accept 25th or 25.5th), and the cumulative frequency rises from 15 to 33 in this class
Common error: 'At least 160 ms' includes every class starting at 160, so count both 160 ≤ t < 180 and 180 ≤ t < 200, not just the top class.
192IBMA2S-050-05-124.2 Presentation of dataPaper 2Easy5 marks
Calculator
A cumulative frequency curve for the ages of 160 runners in a city marathon gives the following information: youngest 19, 10th percentile 26, lower quartile 31, median 38, upper quartile 46, 90th percentile 55, oldest 72 (all in years). An outlier is a value more than 1.5 × IQR beyond the nearer quartile.
(a)
Find the interquartile range.
(1)
(b)
Write down the number of runners older than 46 years.
(1)
(c)
Find the number of runners aged between 26 and 55 years.
(1)
(d)
Determine whether the oldest runner is an outlier.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 46 − 31 = 15 years
✓ 1
Part (b)
A1 25% of 160 = 40 runners
✓ 1
Part (c)
A1 80% of 160 = 128 runners
✓ 1
Part (d)
M1 upper boundary = 46 + 1.5 × 15 = 68.5 years R1 72 > 68.5, so the oldest runner is an outlier
✓ 2
Exam tip: The 10th and 90th percentiles enclose the middle 80% of the data, just as the quartiles enclose the middle 50%.
193IBMA2S-050-00-024.9 The normal distributionPaper 2Easy4 marks
Calculator
The flight time, \(T\) minutes, of a fully charged delivery drone is normally distributed with mean 24.5 minutes and standard deviation 1.8 minutes.
(a)
Find the standardized value (\(z\)-score) corresponding to a flight time of 27.2 minutes.
(2)
(b)
Hence find the probability that a randomly chosen flight lasts longer than 27.2 minutes.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(z=\frac{27.2-24.5}{1.8}\)
✓ 1
A1 \(z=1.5\)
✓ 1
Part (b)
M1 \(P(Z>1.5)\) or \(P(T>27.2)\) seen
✓ 1
A1 0.0668
✓ 1
Exam tip: Writing the probability statement, e.g. \(P(Z>1.5)\), before reaching for the GDC secures the method mark even if a keying error follows.
194IBMA2S-050-00-104.9 The normal distributionPaper 2Easy5 marks
Calculator
In a maternity unit, the birth masses of newborn babies are normally distributed with mean 3.4 kg and standard deviation 0.45 kg.
(a)
Find the mass below which 15% of babies' birth masses lie.
(2)
(b)
Find the interquartile range of the birth masses.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(P(M<m)=0.15\) or \(z=-1.03643\ldots\)
✓ 1
A1 \(2.93\) kg (2.93361…)
✓ 1
Part (b)
M1 attempt to find a quartile, e.g. \(Q_3=3.4+0.67449\times0.45=3.7035\ldots\), or use of \(z=\pm0.67449\)
✓ 1
M1 \(Q_3-Q_1\) with \(Q_1=3.0965\ldots\), OR by symmetry \(2\times0.67449\times0.45\)
✓ 1
A1 \(0.607\) kg (0.60704…)
✓ 1
Exam tip: For a normal distribution the quartiles are symmetric about the mean, so the IQR is \(2\times0.67449\sigma\).
195IBMA2S-050-06-014.9 The normal distributionPaper 2Easy6 marks
Calculator
A consumer technology website tests the battery life of eight smartphones of the same model under identical video-streaming conditions. The battery lives, in hours, are 11.2, 12.5, 10.8, 13.1, 12.0, 11.7, 12.9, 10.4
(a)
Find the median battery life.
(2)
(b)
Find the mean battery life.
(2)
(c)
Find the standard deviation of the battery lives.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 orders the data and averages the 4th and 5th values, (11.7 + 12.0) ÷ 2
✓ 1
A1 median = 11.85 h
✓ 1
Part (b)
M1 sum = 94.6, divided by 8
✓ 1
A1 mean = 11.825 ≈ 11.8 h
✓ 1
Part (c)
M1 uses the GDC or \(\sqrt{\frac{\sum (x-\bar{x})^2}{n}}\) with \(\sum (x-\bar{x})^2 = 6.755\)
✓ 1
A1 σ = 0.919 h
✓ 1
Exam tip: In IB AA the standard deviation of a data set is the population value σ (divide by n). On the GDC use σx, not sx.
196IBMA2S-050-06-044.2 Presentation of dataPaper 2Easy5 marks
Calculator
A hospital emergency department records the waiting time, t minutes, of 100 patients on one evening. 0 ≤ t < 20: 14 patients 20 ≤ t < 40: 26 patients 40 ≤ t < 60: 31 patients 60 ≤ t < 80: 19 patients 80 ≤ t < 120: 10 patients
(a)
Write down the modal class.
(1)
(b)
Find an estimate of the mean waiting time.
(2)
(c)
Find an estimate of the standard deviation of the waiting times.
M1 uses midpoints with frequencies on GDC, or Σft2 = 295400 and 2954 − 482
✓ 1
A1 σ ≈ 25.5 minutes
✓ 1
Common error: The last class is twice as wide as the others: its midpoint is 100, not 90. Using 90 changes both estimates.
197IBMA2S-050-03-024.4 Correlation and regression of y on xPaper 2Easy7 marks
Calculator
On a test bench, an engineer measures the speed of a small engine, x (thousands of revolutions per minute), and its fuel flow rate, y (litres per hour).
x: 1, 2, 3, 4, 5 y: 2.1, 3.9, 6.2, 7.8, 10.0
(a)
Find the value of Pearson's product-moment correlation coefficient, r.
(2)
(b)
Describe the correlation between engine speed and fuel flow rate.
(1)
(c)
Write down the equation of the regression line of y on x.
(2)
(d)
Use your regression line to estimate the fuel flow rate when the engine speed is 3500 revolutions per minute.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 evidence of entering the data into a GDC for linear regression
✓ 1
A1 \(r = 0.999\) (0.998829…)
✓ 1
Part (b)
A1 very strong positive linear correlation
✓ 1
Part (c)
A1 gradient 1.97
✓ 1
A1 intercept 0.09, giving \(y = 1.97x + 0.09\)
✓ 1
Part (d)
M1 substitutes \(x = 3.5\) (not 3500) into the regression line
✓ 1
A1 \(y = 6.99\) litres per hour (6.985)
✓ 1
Common error: Check the units of x before substituting. Here x is measured in thousands of rpm, so 3500 rpm corresponds to x = 3.5, not x = 3500.
198IBMA2S-050-08-024.10 Regression of x on yPaper 2Easy5 marks
Calculator
In a clinical trial, six volunteers are each given a dose of a new drug and the concentration in their blood is measured one hour later. Dose x (mg): 10, 20, 30, 40, 50, 60 Concentration y (mg L−1): 2.1, 3.9, 6.2, 7.8, 10.1, 12.3 A pharmacologist wants to estimate the dose needed to achieve a chosen blood concentration.
(a)
Write down the value of Pearson's product-moment correlation coefficient, r.
(1)
(b)
Write down the equation of the regression line of x on y, in the form \(x = ay + b\).
(2)
(c)
Estimate the dose required to give a blood concentration of 9.0 mg L−1.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(r = 0.999\) (0.99895…)
✓ 1
Part (b)
A1 \(a = 4.91\) (4.90538…)
✓ 1
A1 \(b = 0.335\) (0.3353…)
✓ 1
Part (c)
M1 substitutes \(y = 9\) into the x on y line: \(4.905…\times 9 + 0.335…\)
✓ 1
A1 44.5 mg
✓ 1
Exam tip: Enter the variable you want to predict as the 'response' on your GDC. Here that is the dose, so the dose list must be treated as the dependent variable.
199IBMA2S-050-08-124.10 Regression of x on yPaper 2Easy4 marks
Calculator
In a hospital, the gestational age of a fetus is estimated from the femur length measured on an ultrasound scan. Using scans of pregnancies of known age, with femur lengths between 40 mm and 75 mm, the regression line of gestational age x weeks on femur length y mm is \(x = 0.36y + 9.5\), with \(r = 0.97\).
(a)
Estimate the gestational age of a fetus with femur length 60 mm.
(2)
(b)
A student suggests using this equation to estimate the femur length of a fetus whose gestational age is known. Explain why this is not appropriate.
(1)
(c)
Comment on the reliability of using the equation for a femur length of 20 mm.
(1)
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Notes
Part (a)
M1 substitutes \(y = 60\): \(x = 0.36(60) + 9.5\)
✓ 1
A1 31.1 weeks
✓ 1
Part (b)
R1 to estimate femur length (y) from age (x) the regression line of y on x is required, not the x on y line
✓ 1
Part (c)
R1 unreliable: 20 mm is outside the data range of 40–75 mm, so it is extrapolation
✓ 1
Exam tip: A reliability comment earns its mark when it names the data range explicitly and uses the word extrapolation or interpolation.
200IBMA3S-4-16r5-034.2 Presentation of dataPaper 2Easy5 marks
Calculator
A music streaming service lists the lengths, in minutes, of 12 tracks on an album: 3.2, 3.5, 3.8, 4.0, 4.1, 4.3, 4.5, 4.8, 5.0, 5.2, 5.9, 9.5
(a)
Find the median, the lower quartile and the upper quartile.
(3)
(b)
A value is an outlier if it is more than 1.5 × IQR above the upper quartile. Show that the track of length 9.5 minutes is an outlier.
A1 upper limit = 34.0 g, and 38 > 34.0 so 38 g is an outlier
✓ 1
Exam tip: Quartiles for a small data set are found from the medians of the lower and upper halves; state the values you use before combining them.
202IBMA3S-4-16r6-024.1 Sampling, data and outliersPaper 2Easy4 marks
Calculator
Each of 15 cyclists on a club ride independently suffers a puncture with probability 0.12. Let X be the number of cyclists who suffer a puncture.
(a)
Find P(X = 2).
(2)
(b)
Find the probability that at least 2 cyclists suffer a puncture.
(2)
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Notes
Part (a)
M1 attempt to use X ~ B(15, 0.12), e.g. \(\binom{15}{2}(0.12)^2(0.88)^{13}\)
✓ 1
A1 0.287
✓ 1
Part (b)
M1 attempt to use 1 − P(X ≤ 1)
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A1 0.552
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Common error: 'At least 2' means 1 − P(X ≤ 1), not 1 − P(X ≤ 2); check which value is the last one you exclude.
203IBMA3S-4-16r6-044.9 The normal distributionPaper 2Easy5 marks
Calculator
The working life, X months, of a hard drive in a data centre is modelled by a normal distribution with mean 54 and standard deviation 6.5.
Normal distribution curve, not to scale
(a)
Find the probability that a hard drive fails before 48 months.
(2)
(b)
The data centre replaces a hard drive at time t months, chosen so that 90% of hard drives are still working at that time. Find t.
(3)
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Notes
Part (a)
M1 attempt to find P(X < 48) using X ~ N(54, 6.52)
✓ 1
A1 0.178
✓ 1
Part (b)
M1 P(X > t) = 0.9, so P(X < t) = 0.1
✓ 1
M1 attempt to use inverse normal, e.g. z = −1.2816, t = 54 − 1.2816 × 6.5
✓ 1
A1 t = 45.7 (months)
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Common error: For the value exceeded by 90% of items, the area to the left is 0.1; using 0.9 gives a value above the mean.
204IBMA3S-4-16r8-024.7 Discrete random variablesPaper 2Easy4 marks
Calculator
A telecoms network records that each call is dropped independently with probability 0.04. A technician monitors 25 calls. Let X be the number of calls dropped.
(a)
Find P(X = 2). Give your answer to 3 significant figures.
(2)
(b)
Find the probability that at least one call is dropped. Give your answer to 3 significant figures.
(2)
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Notes
Part (a)
M1 recognise X ~ B(25, 0.04) and attempt P(X = 2)
✓ 1
A1 0.188
✓ 1
Part (b)
M1 P(X ≥ 1) = 1 − P(X = 0) = 1 − 0.9625
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A1 0.640
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Exam tip: 'At least one' is easiest as 1 − P(none). Write the complement before using your calculator.
205IBMA3S-4-16r8-034.9 The normal distributionPaper 2Easy5 marks
Calculator
The mass of jam in a jar is normally distributed with mean 454 g and standard deviation 3.5 g.
Normal distribution curve, not to scale
(a)
Find the probability that a randomly chosen jar contains less than 450 g of jam. Give your answer to 3 significant figures.
(2)
(b)
The heaviest 2.5% of jars are rejected for re-filling. Find the smallest mass of a rejected jar, correct to 1 decimal place.
(3)
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Notes
Part (a)
M1 attempt to use the normal distribution with μ = 454, σ = 3.5
✓ 1
A1 0.127
✓ 1
Part (b)
M1 attempt to find z with P(Z < z) = 0.975
✓ 1
A1 z = 1.9600
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A1 454 + 1.9600 × 3.5 = 460.9 g
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Exam tip: For an upper percentage, convert to a probability below the value (here 0.975) before using the inverse normal.
206IBMA3S-4-16r8-044.3 Measures of central tendency and dispersionPaper 2Easy5 marks
Calculator
The download speeds, in Mbps, recorded at 11 locations in a mountain village are: 25, 17, 56, 22, 19, 30, 15, 23, 28, 12, 31 For this question, the lower quartile is the median of the values below the median and the upper quartile is the median of the values above the median.
(a)
Find the median, the lower quartile and the upper quartile.
(3)
(b)
A value is an outlier if it is more than 1.5 × IQR above the upper quartile. Show that 56 is an outlier.
(2)
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Notes
Part (a)
A1 median = 23
✓ 1
A1 lower quartile = 17
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A1 upper quartile = 30
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Part (b)
M1 IQR = 30 − 17 = 13 and attempt 30 + 1.5 × 13
✓ 1
A1 upper boundary = 49.5, and 56 > 49.5 so 56 is an outlier
✓ 1
Exam tip: Sort the data first. State the boundary value as well as the comparison so the reasoning is complete.
207IBMA3S-4-16r9-034.4 Correlation and regression of y on xPaper 2Easy7 marks
Calculator
A robot carries loads of different mass x kg across a test course. The time y seconds taken is recorded. x: 1, 2, 3, 4, 5, 6 y: 12.3, 14.1, 17.2, 18.4, 21.5, 22.9
Scatter diagram of the data
(a)
Find the Pearson product-moment correlation coefficient r, to 3 significant figures.
(2)
(b)
Find the equation of the regression line of y on x, giving the coefficients to 3 significant figures.
(2)
(c)
Interpret the gradient of the line in this context.
(1)
(d)
Estimate the time taken when the load is 3.5 kg. Give your answer to 3 significant figures.
(2)
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Notes
Part (a)
M1 use of technology to find r
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A1 r = 0.994
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Part (b)
A1 gradient 2.18
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A1 y = 2.18x + 10.1
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Part (c)
R1 each extra kilogram of load adds about 2.18 s to the time
✓ 1
Part (d)
M1 substitute x = 3.5 into the regression equation
✓ 1
A1 17.7 s
✓ 1
Exam tip: Use the regression line of y on x only to estimate y from x, and only within the range of the data.
208IBMA3S-4-17r12-014.4 Correlation and regression of y on xPaper 2Easy4 marks
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The air temperature \(T\) °C was recorded at several points on a mountain at the same moment. The altitude of each point is \(h\) hundred metres above sea level, where \(5 \le h \le 30\). The regression line of \(T\) on \(h\) is \(T = 15.4 - 0.64h\).
(a)
Explain what the value \(-0.64\) tells you about the temperature on the mountain.
(1)
(b)
Use the regression line to estimate the temperature at an altitude of 4200 m.
(2)
(c)
Explain why the estimate in part (b) may be unreliable.
(1)
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Notes
Part (a)
A1 the temperature falls by 0.64 °C for every additional 100 m of altitude
R1 4200 m (\(h=42\)) lies outside the range of the data (\(5 \le h \le 30\)), so this is extrapolation and the linear trend may not continue
✓ 1
Common error: The gradient is a rate: state the change in the dependent variable per one unit of the independent variable, using the units given in the question.
209IBMA3S-4-17r12-024.11 Conditional probability and independencePaper 2Easy8 marks
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A telecoms company surveys its customers. The probability that a randomly chosen customer has a fibre broadband contract, \(F\), is 0.62. The probability that the customer has a mobile plan, \(M\), is 0.45. The probability that the customer has neither is 0.18.
(a)
Find \(\mathrm{P}(F \cap M)\).
(2)
(b)
Find the probability that a randomly chosen customer has fibre broadband but no mobile plan.
(2)
(c)
Determine whether \(F\) and \(M\) are independent. Justify your answer.
(2)
(d)
The company has 2400 customers. Estimate how many have both a fibre contract and a mobile plan.
(2)
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Notes
Part (a)
M1 \(\mathrm{P}(F \cup M) = 1 - 0.18 = 0.82\) and use \(\mathrm{P}(F\cup M)=\mathrm{P}(F)+\mathrm{P}(M)-\mathrm{P}(F\cap M)\)
R1 \(0.279 \ne 0.25 = \mathrm{P}(F\cap M)\), so the events are not independent (ECF from (a))
✓ 1
Part (d)
M1 \(2400 \times 0.25\) (ECF from (a))
✓ 1
A1 600 customers
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Exam tip: Drawing a Venn diagram and filling in the 'neither' region first makes the intersection easy to find.
210IBMA3S-4-17r12-034.9 The normal distributionPaper 2Easy6 marks
Calculator
The mass, \(M\) grams, of jam in a jar is normally distributed with mean 350 and standard deviation 4.
Normal distribution curve, not to scale
(a)
Find the probability that a jar contains less than 343 g of jam.
(2)
(b)
Find \(\mathrm{P}(M > 356)\).
(2)
(c)
A batch contains 1000 jars. Estimate the number of jars in the batch with a mass between 343 g and 356 g.
(2)
Show mark scheme
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Notes
Part (a)
M1 \(z = \frac{343-350}{4} = -1.75\) or normal cdf with correct parameters
✓ 1
A1 0.0401 (3 s.f. 0.0401)
✓ 1
Part (b)
M1 \(z = 1.5\) or normal cdf with correct parameters
✓ 1
A1 0.0668 (3 s.f. 0.0668)
✓ 1
Part (c)
M1 \(1 - 0.0401 - 0.0668 = 0.893\) (or the equivalent probability, ECF from (a) and (b))
✓ 1
A1 \(1000 \times 0.893 = 893\) jars
✓ 1
Exam tip: Sketch the normal curve and shade the required region; a 'between' probability is the total minus the two tails.
211IBMA3S-4-17r12-044.4 Correlation and regression of y on xPaper 2Easy6 marks
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A telecoms engineer measures the download speed \(s\) (Mbps) of a mobile connection at seven locations, each a distance \(d\) km from the nearest mast. \(d\) (km): 0.5, 1.0, 1.5, 2.5, 3.0, 4.0, 5.0 \(s\) (Mbps): 92, 85, 81, 68, 60, 49, 38
Scatter diagram of the data
(a)
Write down the Pearson correlation coefficient, \(r\), for these data.
(1)
(b)
Describe the correlation between \(d\) and \(s\).
(1)
(c)
Write down the equation of the regression line of \(s\) on \(d\), with coefficients given to 3 significant figures.
(2)
(d)
Use your equation from part (c) to estimate the download speed at a distance of 2.0 km. Give your answer to the nearest whole number.
(2)
Show mark scheme
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Notes
Part (a)
A1 \(r = -0.999\)
✓ 1
Part (b)
A1 strong negative linear correlation (speed decreases as distance increases)
✓ 1
Part (c)
A1 gradient -12.1
✓ 1
A1 \(s = 97.9 - 12.1d\) (intercept 97.9)
✓ 1
Part (d)
M1 substitute \(d = 2.0\) into the equation from (c) (ECF from (c))
✓ 1
A1 74 Mbps
✓ 1
Common error: Use the regression line of y on x only to estimate y from x, and only within the range of the data.
212IBMA3S-4-16r21-044.12 Standardization and inverse normalPaper 2Easy5 marks
Calculator
The mass M mg of active ingredient in a tablet is normally distributed with mean 250 mg and standard deviation σ mg. It is known that 5% of tablets contain less than 244 mg.
Normal distribution curve, not to scale
(a)
Find σ, giving your answer to 3 significant figures. [3]
(3)
(b)
Use the unrounded value of σ to find the probability that a tablet contains more than 255 mg. Give your answer to 3 significant figures. [2]
(2)
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Notes
Part (a)
M1 \(P(M<244)=0.05\) so \(\frac{244-250}{\sigma}=z\) where \(P(Z<z)=0.05\)
✓ 1
A1 z = −1.645 (−1.6449...)
✓ 1
A1 σ = 3.65 (3.6477...)
✓ 1
Part (b)
M1 attempt to find P(M > 255) using the normal distribution with mean 250 and standard deviation 3.6477...
✓ 1
A1 0.0852 (0.085232...)
✓ 1
Common error: An inverse normal value for a lower-tail probability is negative; forgetting the sign gives a negative standard deviation.
213IBMA3S-4-16r22-014.1 Sampling, data and outliersPaper 2Easy6 marks
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A retail chain has 350 employees in four stores: store A has 154 employees, store B has 98, store C has 63 and store D has 35. The company plans to survey 50 employees, using a sample stratified by store.
(a)
State one reason why a stratified sample is more suitable than a simple random sample here.
(1)
(b)
Calculate the number of employees that should be sampled from store C.
(3)
(c)
Describe how a simple random sample of 50 employees could be selected from all 350 employees.
(2)
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Notes
Part (a)
R1 it guarantees that every store is represented in proportion to its size (a simple random sample might miss a small store)
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Part (b)
M1 total number of employees 350
✓ 1
M1 63 ÷ 350 × 50
✓ 1
A1 9
✓ 1
Part (c)
M1 number the employees 1 to 350
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A1 use a random number generator to select 50 different numbers, and choose those employees
✓ 1
Exam tip: In a stratified sample, the number from each group is the group's share of the population multiplied by the sample size.
214IBMA3S-4-16r22-044.4 Correlation and regression of y on xPaper 2Easy5 marks
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In a study of 12 regions, the vaccination rate x (% of the population) and the number of infections per 1000 people, y, are recorded. The values of x range from 45 to 90. The regression line of y on x is y = −0.42x + 48.5, and the mean vaccination rate is x̄ = 65.0.
(a)
Interpret the gradient of the regression line in this context.
(1)
(b)
Explain why it would be unreliable to use the line to estimate y when x = 20.
(1)
(c)
A health officer uses the line to estimate the vaccination rate needed for y = 15. Explain why this is not appropriate.
(1)
(d)
Find the mean value of y.
(2)
Show mark scheme
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Notes
Part (a)
R1 each 1 percentage point increase in vaccination rate is associated with 0.42 fewer infections per 1000 people
✓ 1
Part (b)
R1 20 is outside the range of the data (45 to 90), so this is extrapolation
✓ 1
Part (c)
R1 the line of y on x is for estimating y from x; a regression line of x on y should be used to estimate x
✓ 1
Part (d)
M1 ȳ = −0.42 × 65.0 + 48.5, because the line passes through (x̄, ȳ)
✓ 1
A1 21.2
✓ 1
Exam tip: The regression line always passes through the mean point (x̄, ȳ), which gives a quick way to find a missing mean.
215IBMA3S-4-16r22-094.9 The normal distributionPaper 2Easy5 marks
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At a clothing store, the amount S spent by a customer on one visit, in euros (€), is normally distributed with mean 34.20 and standard deviation 6.50.
Normal distribution curve, not to scale
(a)
Find the probability that a customer spends more than €45.00.
(2)
(b)
Find the standardized value (z-score) for a spend of €45.00.
(2)
(c)
Interpret your answer to (b).
(1)
Show mark scheme
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Notes
Part (a)
M1 attempt to use the normal distribution with correct parameters
✓ 1
A1 0.0483
✓ 1
Part (b)
M1 (45 − 34.2) ÷ 6.5
✓ 1
A1 1.66
✓ 1
Part (c)
R1 a spend of €45.00 is 1.66 standard deviations above the mean
✓ 1
Exam tip: A z-score tells you how many standard deviations a value is from the mean; positive means above.
216IBMA3S-4-16r24-024.4 Correlation and regression of y on xPaper 2Easy5 marks
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A Formula 1 team models the lap time t seconds of a car after n laps on the same set of tyres by the regression line of t on n: t = 0.045n + 91.2. The line was found using data for 3 ≤ n ≤ 25.
(a)
Interpret the value 0.045 in this context.
(2)
(b)
Use the model to estimate the lap time when n = 22. Give your answer to 3 significant figures.
(2)
(c)
Explain why the model should not be used to estimate the lap time when n = 60.
(1)
Show mark scheme
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Notes
Part (a)
A1 Lap time increases by 0.045 seconds
✓ 1
A1 for each additional lap on the same tyres (on average)
✓ 1
Part (b)
M1 0.045 × 22 + 91.2
✓ 1
A1 92.19, so 92.2 seconds
✓ 1
Part (c)
R1 60 is outside the range of the data (3 to 25), so this is extrapolation and may be unreliable
✓ 1
Common error: Interpret a gradient with units and the phrase 'for each additional' or 'per', and say the change is an average.
217IBMA3S-4-16r24-034.3 Measures of central tendency and dispersionPaper 2Easy6 marks
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The waiting times, in minutes, of 11 patients in a hospital emergency department are: 12, 15, 18, 20, 22, 25, 27, 29, 33, 38, 95.
(a)
Find the median, the lower quartile and the upper quartile.
(3)
(b)
A value is an outlier if it is more than 1.5 × IQR above the upper quartile. Show that 95 is an outlier.
(2)
(c)
Find the mean waiting time of the other ten patients.
(1)
Show mark scheme
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Notes
Part (a)
A1 median = 25
✓ 1
A1 lower quartile = 18
✓ 1
A1 upper quartile = 33
✓ 1
Part (b)
M1 IQR = 33 − 18 = 15 and 33 + 1.5 × 15 = 55.5
✓ 1
R1 95 > 55.5, so 95 is an outlier
✓ 1
Part (c)
A1 (sum 239) ÷ 10 = 23.9 minutes
✓ 1
Exam tip: For an odd number of values, exclude the median when finding the quartiles by hand; a calculator gives the same result.
218IBMA3S-4-17r33-014.4 Correlation and regression of y on xPaper 2Easy6 marks
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A beekeeper records the mean daytime temperature, x °C, and the mass of honey collected from one hive, y kg, in each of seven weeks. x (°C): 14, 16, 18, 19, 21, 23, 25 y (kg): 3.1, 3.9, 4.4, 5.2, 5.6, 6.5, 6.9
Scatter diagram of the data
(a)
Write down the Pearson correlation coefficient r and the equation of the regression line of y on x, giving all values to three significant figures.
(3)
(b)
Interpret the gradient of the regression line in this context.
(1)
(c)
Use your equation, with coefficients as given to three significant figures, to estimate the mass of honey in a week with mean temperature 20 °C. Give your answer to one decimal place.
(2)
Show mark scheme
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Notes
Part (a)
A1 r = 0.993
✓ 1
A1 gradient = 0.354
✓ 1
A1 intercept = −1.79, so y = 0.354x − 1.79
✓ 1
Part (b)
A1 each extra 1 °C of mean temperature is associated with about 0.354 kg more honey collected
✓ 1
Part (c)
M1 substitute x = 20 into the regression equation
✓ 1
A1 5.3 kg
✓ 1
Exam tip: State the interpretation of a gradient with units and the words 'associated with', not 'causes'.
219IBMA3S-4-17r33-024.3 Measures of central tendency and dispersionPaper 2Easy6 marks
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The waiting times, in minutes, of twelve patients in a hospital emergency department were recorded as follows. 8, 12, 15, 17, 19, 22, 24, 27, 31, 35, 38, 96
(a)
Find the median, the lower quartile and the upper quartile.
(3)
(b)
Find the interquartile range.
(1)
(c)
A value is classed as an outlier if it is greater than Q3 + 1.5 × IQR. Show that the waiting time of 96 minutes is an outlier.
(2)
Show mark scheme
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Notes
Part (a)
A1 median = 23 minutes
✓ 1
A1 lower quartile = 16 minutes
✓ 1
A1 upper quartile = 33 minutes
✓ 1
Part (b)
A1 IQR = 33 − 16 = 17 minutes
✓ 1
Part (c)
M1 Q3 + 1.5 × IQR = 33 + 1.5 × 17 = 58.5 minutes
✓ 1
R1 96 > 58.5, so 96 is an outlier
✓ 1
Common error: Compare the value with the upper fence, not with the upper quartile itself; the fence is Q₃ + 1.5 × IQR.
220IBMA3S-4-17r33-034.7 Discrete random variablesPaper 2Easy6 marks
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A hospital records whether each of 200 patients had surgery (S) and whether the patient was discharged within three days (D). Surgery: 18 discharged within 3 days, 42 discharged after 3 days. No surgery: 84 discharged within 3 days, 56 discharged after 3 days. A patient is chosen at random.
(a)
Find P(D).
(2)
(b)
Find P(D | S).
(2)
(c)
Determine whether the events S and D are independent. Justify your answer.
(2)
Show mark scheme
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Notes
Part (a)
M1 (18 + 84) / 200
✓ 1
A1 0.51
✓ 1
Part (b)
M1 18 / 60
✓ 1
A1 0.3
✓ 1
Part (c)
M1 compare P(D | S) = 0.3 with P(D) = 0.51
✓ 1
R1 the probabilities are not equal, so S and D are not independent
✓ 1
Exam tip: To test independence, compare P(A | B) with P(A), or P(A ∩ B) with P(A)P(B); state the conclusion in words.
221IBMA3S-4-17r33-044.2 Presentation of dataPaper 2Easy6 marks
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The time, T seconds, taken by a Formula 1 team for a pit stop is modelled by a normal distribution with mean 2.60 and standard deviation 0.30.
Normal distribution curve, not to scale
(a)
Find the probability that a pit stop takes less than 2.2 seconds.
(2)
(b)
Find the probability that a pit stop takes between 2.3 and 3.0 seconds.
(2)
(c)
In a season the team makes 60 pit stops. Use the unrounded probability from part (a) to estimate how many take less than 2.2 seconds. Give your answer to three significant figures.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 attempt to use the normal distribution with mean 2.60 and standard deviation 0.30 (or z = −1.33)
✓ 1
A1 P(T < 2.2) = 0.0912
✓ 1
Part (b)
M1 P(2.3 < T < 3.0) with correct limits (or P(T < 3.0) − P(T < 2.3))
✓ 1
A1 0.750
✓ 1
Part (c)
M1 60 × P(T < 2.2)
✓ 1
A1 5.47
✓ 1
Common error: Do not subtract the two limits before standardising; find each cumulative probability (or use the calculator with both limits) and then subtract.
222IBMA3S-4-18r36-024.2 Presentation of dataPaper 2Easy6 marks
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A courier company weighs 50 parcels. The masses, w kg, are summarised below.
0 ≤ w < 2: 6 parcels 2 ≤ w < 4: 14 parcels 4 ≤ w < 6: 18 parcels 6 ≤ w < 8: 8 parcels 8 ≤ w < 10: 4 parcels
(a)
Write down the modal class.
(1)
(b)
Estimate the mean mass of a parcel.
(2)
(c)
Estimate the median mass of a parcel, giving your answer to 3 significant figures.
(3)
Show mark scheme
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Notes
Part (a)
A1 4 ≤ w < 6
✓ 1
Part (b)
M1 use of midpoints 1, 3, 5, 7, 9 in Σfx/50
✓ 1
A1 4.60 kg
✓ 1
Part (c)
M1 median is the 25th value, which lies in the class 4 ≤ w < 6 (cumulative frequency 20 before it)
✓ 1
M1 interpolation 4 + (25 − 20)/18 × 2
✓ 1
A1 4.56 kg
✓ 1
Common error: Use the class midpoints for the mean but cumulative frequencies for the median; mixing the two methods gives the wrong class or value.
223IBMA3S-4-18r36-044.7 Discrete random variablesPaper 2Easy6 marks
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A logistics company records the number of attempts, X, needed to deliver a parcel. The probability distribution is P(X = x) = k(6 − x) for x = 1, 2, 3, 4, where k is a constant.
(a)
Find the value of k.
(2)
(b)
Find P(X ≥ 3).
(2)
(c)
Find E(X), giving your answer to 3 significant figures.
Exam tip: Probabilities in a distribution must sum to 1; use this to find an unknown constant before anything else.
224IBMA3S-4-18r36-074.1 Sampling, data and outliersPaper 2Easy6 marks
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A warehouse stores 2400 crates in three sections: section A has 900 crates, section B has 1100 crates and section C has 400 crates. A manager wants a sample of 120 crates in which each section is represented in proportion to its size.
(a)
Name the sampling technique that achieves this.
(1)
(b)
Calculate the number of crates that should be sampled from section B.
(2)
(c)
In section C, 20 crates are to be chosen using simple random sampling. Describe how this can be done.
(2)
(d)
State one advantage of this technique over taking a simple random sample of all 2400 crates.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 stratified sampling
✓ 1
Part (b)
M1 120 × 1100/2400
✓ 1
A1 55
✓ 1
Part (c)
Number the crates 1 to 400
✓ 1
Use a random number generator to select 20 different numbers and choose the crates with those numbers
✓ 1
Part (d)
R1 it guarantees every section is represented (in proportion), so the sample is more representative
✓ 1
Common error: Stratified sample sizes must be proportional to the size of each group and must add up to the total sample size; check that yours do.
225IBMA3S-4-17r38-014.4 Correlation and regression of y on xPaper 2Easy7 marks
Calculator
A coffee roaster records the roasting time, x minutes, and the percentage of mass lost, y, for seven batches of green beans. x: 8, 9, 10, 11, 12, 13, 14 y: 11.4, 12.1, 13.9, 13.8, 15.6, 16.2, 17.1
Scatter diagram of the data
(a)
Find Pearson's product-moment correlation coefficient, r, to 3 significant figures, and describe the correlation.
(3)
(b)
Find the equation of the regression line of y on x, with coefficients to 3 significant figures.
(2)
(c)
Use your equation from (b), with the coefficients to 3 significant figures, to estimate the percentage mass lost for a roasting time of 11 minutes. Give your answer to 1 decimal place.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to find r using technology
✓ 1
A1 r = 0.986
✓ 1
A1 strong positive correlation
✓ 1
Part (b)
A1 gradient 0.964
✓ 1
A1 intercept 3.69, so y = 0.964x + 3.69
✓ 1
Part (c)
M1 substitute x = 11 into the equation
✓ 1
A1 0.964×11 + 3.69 = 14.3 (%) (ECF from (b))
✓ 1
Common error: Give the regression line for the variable you are predicting: here y on x, because mass lost is predicted from time.
226IBMA3S-4-17r38-034.2 Presentation of dataPaper 2Easy6 marks
Calculator
An engineer measures the mid-span deflection, in mm, of 12 steel girders under the same test load: 22, 15, 41, 19, 24, 17, 15, 25, 21, 14, 22, 18 A value is an outlier if it is more than 1.5 × IQR above the upper quartile or more than 1.5 × IQR below the lower quartile.
(a)
Find the median, the lower quartile and the upper quartile.
(3)
(b)
Show that the largest value is an outlier.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 median = 20
✓ 1
A1 lower quartile = 16
✓ 1
A1 upper quartile = 23
✓ 1
Part (b)
A1 IQR = 23 − 16 = 7
✓ 1
M1 upper fence = 23 + 1.5 × 7
✓ 1
A1 33.5, and 41 > 33.5 so it is an outlier
✓ 1
Exam tip: State the fence value and compare it with the data value; a bare "yes" earns nothing.
227IBMA3S-4-18r39-024.11 Conditional probability and independencePaper 2Easy7 marks
Calculator
A survey of 60 farms found that 35 grow wheat, 28 grow barley and 12 grow both crops. One farm is chosen at random. Let \(W\) be the event that it grows wheat and \(B\) the event that it grows barley.
(a)
Find the number of farms that grow neither crop.
(2)
(b)
Find \(P(W\mid B)\).
(2)
(c)
Determine whether \(W\) and \(B\) are independent, justifying your answer.
R1 the values differ, so \(W\) and \(B\) are not independent
✓ 1
Common error: Overlapping events are not the same as dependent events: independence is tested by comparing \(P(W\cap B)\) with \(P(W)P(B)\).
228IBMA3S-4-19r41-024.3 Measures of central tendency and dispersionPaper 2Easy7 marks
Calculator
In a mobile game, 50 players each attempt a daily challenge. The number of levels, x, cleared by each player is summarised below.
x
0
1
2
3
4
5
Frequency
3
7
12
15
9
4
Bar chart of the data
(a)
Find the mean number of levels cleared.
(2)
(b)
Find the population standard deviation of the number of levels cleared, giving your answer to 3 significant figures.
(2)
(c)
The game awards points P = 40x + 25 to each player. Using the unrounded standard deviation from part (b), find the mean and the standard deviation of P. Give the mean exactly and the standard deviation to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use Σfx / Σf (or GDC statistics mode)
✓ 1
A1 mean = 132/50 = 2.64
✓ 1
Part (b)
M1 use of σ from GDC, or √(Σfx2/n − mean2) with Σfx2 = 434
✓ 1
A1 σ = 1.31
✓ 1
Part (c)
M1 mean of P = 40 × 2.64 + 25
✓ 1
A1 mean = 130.6
✓ 1
A1 standard deviation = 40 × 1.3078 = 52.3
✓ 1
Common error: When data are transformed by P = 40x + 25, the standard deviation is multiplied by 40 but the +25 has no effect on it.
229IBMA3S-4-19r41-034.7 Discrete random variablesPaper 2Easy7 marks
Calculator
The daily water demand, X megalitres (ML), of a town is modelled by a normal distribution with mean 480 ML and standard deviation 35 ML.
Normal distribution curve, not to scale
(a)
Find the probability that the demand on a given day exceeds 540 ML.
(2)
(b)
Find the probability that the demand on a given day is between 450 ML and 520 ML.
(2)
(c)
The supply company wants to be able to meet the demand on 95% of days. Find the smallest daily supply, in ML, that achieves this, giving your answer to the nearest ML.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use normal distribution with X > 540 (or z = 1.714...)
✓ 1
A1 probability = 0.0432
✓ 1
Part (b)
M1 attempt to find P(450 < X < 520)
✓ 1
A1 probability = 0.678
✓ 1
Part (c)
M1 z = 1.645 (invNorm(0.95)) used with x = μ + zσ
✓ 1
A1 x = 537.57...
✓ 1
A1 supply = 538 ML
✓ 1
Exam tip: Write the probability statement, such as P(X > 540), before using the calculator so that the method mark is secure even if the number is wrong.
230IBMA3S-4-19r41-044.10 Regression of x on yPaper 2Easy6 marks
Calculator
A pharmacist records how many days, x, a syrup stays effective after opening when stored at different temperatures, y °C.
y (°C)
4
8
12
16
20
24
x (days)
41
37
30
26
18
13
Scatter diagram of the data
(a)
Find the Pearson correlation coefficient, r, and describe the correlation.
(2)
(b)
Find the equation of the regression line of x on y, giving the coefficients to 3 significant figures.
(2)
(c)
Use your equation from (b), with the unrounded coefficients, to estimate the number of days the syrup stays effective at 10 °C. Give your answer to 1 decimal place.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 r = −0.996
✓ 1
A1 strong negative correlation
✓ 1
Part (b)
M1 use of GDC regression of x on y (x as the dependent variable)
✓ 1
A1 x = −1.44y + 47.6
✓ 1
Part (c)
M1 substitute y = 10 into the regression line of x on y
✓ 1
A1 x = 33.2 days
✓ 1
Common error: To estimate x from a given y, use the regression line of x on y; rearranging the line of y on x gives a different, less suitable estimate.
231IBMA3S-4-18r45-024.2 Presentation of dataPaper 2Easy6 marks
Calculator
A food scientist measures the sugar content, in grams per 100 g, of ten cereal bars from one batch: 14.2, 15.1, 13.8, 16.0, 14.7, 15.5, 14.9, 13.5, 15.8, 14.4.
(a)
Find the mean sugar content.
(2)
(b)
Find the median sugar content.
(2)
(c)
Find the population standard deviation of the sugar content, correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sum of values ÷ 10 = 147.9 ÷ 10
✓ 1
A1 14.79 g per 100 g
✓ 1
Part (b)
M1 ordered data: 13.5, 13.8, 14.2, 14.4, 14.7, 14.9, 15.1, 15.5, 15.8, 16.0 and the median is the mean of the 5th and 6th values
✓ 1
A1 (14.7 + 14.9) ÷ 2 = 14.8 g per 100 g
✓ 1
Part (c)
M1 use of GDC (or the formula) for the population standard deviation of the ten values
✓ 1
A1 0.791 g per 100 g
✓ 1
Common error: Make sure your calculator is giving the population standard deviation (σ), not the sample standard deviation, when the question asks for population.
232IBMA3S-4-18r45-034.8 The binomial distributionPaper 2Easy5 marks
Calculator
On a telecoms network, each data packet sent across a link is lost independently with probability 0.04. A batch of 25 packets is sent. Let X be the number of packets in the batch that are lost, so X ~ B(25, 0.04).
(a)
Find P(X = 2).
(2)
(b)
Find the probability that at least one packet in the batch is lost.
(2)
(c)
Find E(X).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use the binomial probability with n = 25, p = 0.04, k = 2 (e.g. 300 × 0.042 × 0.9623)
✓ 1
A1 0.188
✓ 1
Part (b)
M1 1 − P(X = 0) = 1 − 0.9625
✓ 1
A1 0.640
✓ 1
Part (c)
A1 25 × 0.04 = 1.00
✓ 1
Exam tip: 'At least one' is almost always quickest as 1 − P(none).
233IBMA3S-4-18r45-044.2 Presentation of dataPaper 2Easy5 marks
Calculator
The times taken by 80 climbers to ascend a steep section of a mountain path are summarised in the table.
Time t (minutes) : frequency 20 ≤ t < 30 : 6 30 ≤ t < 40 : 14 40 ≤ t < 50 : 26 50 ≤ t < 60 : 22 60 ≤ t < 70 : 12
(a)
Write down the modal class.
(1)
(b)
Find the class that contains the median. Show the working that supports your answer.
(2)
(c)
Estimate the mean time taken.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 40 ≤ t < 50
✓ 1
Part (b)
M1 cumulative frequencies 6, 20, 46, … and the median is between the 40th and 41st values
✓ 1
A1 40 ≤ t < 50
✓ 1
Part (c)
M1 use of midpoints 25, 35, 45, 55, 65 in (25×6 + 35×14 + 45×26 + 55×22 + 65×12) ÷ 80
✓ 1
A1 47.5 minutes
✓ 1
Exam tip: For grouped data use class midpoints, and remember that the result is only an estimate.
234IBMA3S-4-19r45-014.2 Presentation of dataPaper 2Easy5 marks
Calculator
The departure delays, in minutes, of eight flights from a regional airport one morning were 12, 7, 15, 9, 22, 11, 7, 17.
(a)
Find the mean delay.
(1)
(b)
Find the median delay.
(2)
(c)
Calculate the population standard deviation of the delays, giving your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 mean = 12.5 minutes
✓ 1
Part (b)
M1 values placed in order (7, 7, 9, 11, 12, 15, 17, 22) or middle two values identified
✓ 1
A1 median = 11.5 minutes
✓ 1
Part (c)
M1 attempt to find the population standard deviation, e.g. sqrt(192/8) or use of a calculator statistics mode
✓ 1
A1 standard deviation = 4.90 minutes
✓ 1
Common error: Calculators give both a sample and a population standard deviation; read the question to see which one is wanted.
235IBMA3S-4-19r45-024.2 Presentation of dataPaper 2Easy4 marks
Calculator
A farmer records the mass of tomatoes harvested from each of 40 plants.
Mass, m (kg)
0 ≤ m < 2
2 ≤ m < 4
4 ≤ m < 6
6 ≤ m < 8
Number of plants
4
13
15
8
(a)
Write down the modal class.
(1)
(b)
Estimate the mean mass of tomatoes per plant.
(2)
(c)
Explain why your answer to part (b) is only an estimate.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 4 ≤ m < 6
✓ 1
Part (b)
M1 use of mid-interval values 1, 3, 5, 7 with frequencies, e.g. (1×4 + 3×13 + 5×15 + 7×8)/40
✓ 1
A1 mean = 4.35 kg
✓ 1
Part (c)
R1 the exact mass of each plant is not known; midpoints are used as representative values for each class
✓ 1
Exam tip: For grouped data use the midpoint of each class, not the class boundary, and divide by the total frequency.
236IBMA3S-4-19r45-044.8 The binomial distributionPaper 2Easy5 marks
Calculator
A grower plants seeds in trays of 10. Each seed independently germinates with probability 0.8. Let X be the number of seeds in a tray that germinate, so X ~ B(10, 0.8).
(a)
Write down E(X).
(1)
(b)
Find the probability that exactly 8 seeds in a tray germinate.
(2)
(c)
Find the probability that at least 9 seeds in a tray germinate.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 E(X) = 10 × 0.8 = 8
✓ 1
Part (b)
M1 use of the binomial probability, e.g. 45 × 0.8^8 × 0.2^2
✓ 1
A1 0.302
✓ 1
Part (c)
M1 P(X = 9) + P(X = 10) or 1 − P(X ≤ 8)
✓ 1
A1 0.376
✓ 1
Exam tip: "At least 9" means 9 or 10 here; counting the values that are included is quicker than using the complement.
237IBMA3S-4-18r46-014.7 Discrete random variablesPaper 2Easy5 marks
Calculator
On any day, the probability that a seismometer on a volcano records a tremor is \(0.3\), independently of other days. Let \(X\) be the number of days with a tremor in a period of 12 days.
(a)
Find \(P(X=4)\).
(2)
(b)
Find \(P(X\geq 3)\).
(2)
(c)
Find the expected number of days with a tremor in the period.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 recognises \(X\sim B(12,0.3)\) and attempts \(P(X=4)\), e.g. \(\binom{12}{4}(0.3)^4(0.7)^8\)
✓ 1
A1 \(0.231\)
✓ 1
Part (b)
M1 \(1-P(X\leq 2)\)
✓ 1
A1 \(0.747\)
✓ 1
Part (c)
A1 \(E(X)=12\times0.3=3.6\) days
✓ 1
Exam tip: For 'at least' probabilities, use the complement: \(P(X\geq 3)=1-P(X\leq 2)\) is much quicker than adding many terms.
238IBMA3S-4-18r46-064.8 The binomial distributionPaper 2Easy6 marks
Calculator
A theatre sells tickets for three areas. A randomly chosen ticket is for the stalls with probability \(0.55\), for the circle with probability \(0.30\) and for the balcony with probability \(k\).
(a)
Find the value of \(k\).
(2)
(b)
A total of \(1200\) tickets are sold for a performance. Find the expected number of balcony tickets.
(2)
(c)
Two tickets are chosen at random, independently of each other. Find the probability that both are for the stalls.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(0.55+0.30+k=1\)
✓ 1
A1 \(k=0.15\)
✓ 1
Part (b)
M1 \(1200\times0.15\)
✓ 1
A1 \(180\) (ECF from (a))
✓ 1
Part (c)
M1 \(0.55\times0.55\)
✓ 1
A1 \(0.3025\)
✓ 1
Exam tip: The probabilities of all the possible outcomes must add up to \(1\). Use this to find a missing probability before anything else.
239IBMA3S-4-18r48-104.4 Correlation and regression of y on xPaper 2Easy6 marks
Calculator
A hospital studies \(x\), the mean hours of nursing care per patient per day, and \(y\), the mean length of stay in days, for several wards. The regression line of \(y\) on \(x\) is \(y = 14.2 - 1.8x\). The mean of the values of \(x\) is 3.5.
(a)
Interpret the gradient of the regression line in context.
(1)
(b)
Find the mean of the values of \(y\).
(2)
(c)
Estimate the mean length of stay for a ward with 4.2 hours of nursing care per patient per day.
(2)
(d)
A manager wants to estimate the nursing hours needed for a mean stay of 10 days. Explain why the equation above should not be used for this.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 each extra hour of nursing care per patient per day is associated with a mean stay about 1.8 days shorter
✓ 1
Part (b)
M1 the regression line passes through the mean point, so \(\bar{y} = 14.2 - 1.8(3.5)\)
✓ 1
A1 \(\bar{y} = 7.9\) days
✓ 1
Part (c)
M1 substitute \(x = 4.2\): \(14.2 - 1.8(4.2)\)
✓ 1
A1 6.64 days
✓ 1
Part (d)
R1 the line is a regression of \(y\) on \(x\), so it is only valid for estimating \(y\) from \(x\), not \(x\) from \(y\)
✓ 1
Exam tip: The regression line always passes through the point \((\bar{x}, \bar{y})\), which gives the missing mean directly.
240AASL4-q00044.1 Sampling, data and outliersPaper 2Medium8 marks
The masses, in grams, of 15 apples picked from one tree were recorded.
Apple
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Mass (g)
238
251
212
263
244
397
231
256
270
238
247
284
225
263
259
(a)
Write down the median and the lower and upper quartiles of the masses.
(2)
(b)
Show that the mass of 397 g is an outlier.
(2)
(c)
It is discovered that the value 397 was a recording error and the mass of that apple was 297 g. Using the corrected data, find the mean and the standard deviation of the masses.
(2)
(d)
Explain why correcting the error does not change the median or the quartiles.
(1)
(e)
Determine whether 297 g is an outlier.
(1)
Show mark scheme
Working
Mark
Notes
(a)
median \(=251\) g
A1
\(Q_{1}=238\) g and \(Q_{3}=263\) g
A1
Both required.
(b)
\(Q_{3}+1.5\times\text{IQR}=263+1.5\times25\)
M1
Award for 1.5 × their IQR added to their upper quartile.
\(=300.5\) and \(397\gt 300.5\)
A1
Must see 300.5 and a comparison.
(c)
mean \(=251.87\) g, i.e. \(252\) g
A1
\(251.866\ldots\)
standard deviation \(=21.5\) g
A1
\(21.4968\ldots\) (GDC \(\sigma_{x}\)). Accept \(22.3\) (\(s_{x}\)) only if the data are explicitly treated as a sample.
(d)
397 and 297 are both the largest value, so the order (positions) of all the data values is unchanged
R1
(e)
\(297\lt 300.5\), so 297 g is not an outlier
A1
FT from their upper boundary in (b).
Answers: (a) median 251 g, \(Q_{1}=238\) g, \(Q_{3}=263\) g · (c) mean 252 g, sd 21.5 g · (e) not an outlier
241AASL4-q00054.1 Sampling, data and outliersPaper 2Medium7 marks
The table shows the members of a tennis club, classified by age group and sex.
Age group
Female
Male
Under 18
46
39
18 to 59
118
97
60 and over
63
52
The club secretary wants a stratified sample of 50 members, using all six age-group and sex categories as strata.
(a)
Find the number of female members aged 18 to 59 who should be in the sample.
(2)
(b)
Show that, if the number for each of the six strata is rounded to the nearest integer, the total sample size is not 50.
(2)
(c)
Suggest how the secretary could adjust the numbers so that the sample size is exactly 50.
(1)
(d)
Once the number from each stratum is known, state how the members from that stratum should be chosen.
(1)
(e)
State one advantage of a stratified sample compared with a simple random sample of 50 members.
(1)
Show mark scheme
Working
Mark
Notes
(a)
total \(=415\); \(\dfrac{118}{415}\times50\)
(M1)
\(14.2\ldots\), so \(14\) members
A1
\(14.2168\ldots\) Award A1 for 14 only.
(b)
\(5.54,\ 4.70,\ 14.22,\ 11.69,\ 7.59,\ 6.27\) (at least four correct)
M1
Female under 18, male under 18, female 18–59, male 18–59, female 60+, male 60+.
rounded values \(6+5+14+12+8+6=51\neq50\)
A1
Must see the sum 51.
(c)
reduce by one the stratum whose rounding up was greatest relative to its value, e.g. female under 18: \(5.54\to5\) (so \(5,5,14,12,8,6\))
R1
Accept any sensible adjustment that removes one member and is justified, e.g. reduce the value with the smallest decimal part among those rounded up.
(d)
by simple random sampling within the stratum (e.g. a random number generator applied to a numbered list of that stratum)
A1
(e)
each age group and sex is represented in proportion to its size in the club, so the sample is more representative
R1
Answers: (a) 14 · (b) total 51 · (c) e.g. take 5 females under 18
242AASL4-q00064.1 Sampling, data and outliersPaper 2Medium8 marks
Eleven students recorded the number of hours they slept on one night. The data were entered into a spreadsheet as follows.
Student
1
2
3
4
5
6
7
8
9
10
11
Hours
7.5
8
6.5
9
75
7
8.5
6
7
8
5.5
(a)
Find the mean of the values in the spreadsheet.
(1)
(b)
Explain why the value 75 must be an error.
(1)
(c)(i)
The student who recorded 75 confirms that the correct value is 7.5.
Find the mean of the corrected data.
(1)
(c)(ii)
Find the median of the corrected data.
(1)
(d)
Instead of correcting the value, a second analyst removes it. Find the mean and the median of the remaining ten values.
(2)
(e)
State which of the mean and the median of the data in the spreadsheet was less affected by the error, giving a reason.
(1)
(f)
A twelfth student did not answer the question. Explain why this student’s value should not be entered as 0.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(13.5\) hours
A1
\(13.4545\ldots\)
(b)
75 hours is more than the 24 hours in a day (it is likely 7.5 with the decimal point missing)
R1
(c)(i)
\(7.32\) hours
A1
\(7.31818\ldots\)
(c)(ii)
\(7.5\) hours
A1
(d)
mean \(=7.3\) hours
A1
median \(=7.25\) hours
A1
(e)
the median: with the error the median is still 7.5, the same as for the corrected data, whereas the mean almost doubles (13.5 compared with 7.32)
R1
Must give the median with a reason referring to the values or to the median not depending on the size of extreme values.
(f)
0 would be treated as a real value (a student who did not sleep) and would lower the mean; the missing value should be left out of the calculations
R1
Answers: (a) 13.5 · (c)(i) 7.32 (ii) 7.5 · (d) mean 7.3, median 7.25 · (e) median
243AASL4-q00084.1 Sampling, data and outliersPaper 2Medium9 marks
The resting heart rates, in beats per minute (bpm), of 18 members of a cycling team are shown.
bpm
64
71
58
67
73
62
69
75
41
bpm
66
70
63
68
77
60
72
65
79
(a)
Find the median and the interquartile range of the heart rates.
(2)
(b)
Show that the heart rate of 41 bpm is an outlier.
(2)
(c)
The heart rate of 41 bpm belongs to an Olympic cyclist. Very fit athletes often have resting heart rates below 45 bpm. State, with a reason, whether this value should be removed from the data.
(1)
(d)
Find the mean and the standard deviation of all 18 heart rates.
(2)
(e)
Find the standard deviation of the heart rates when the value 41 is removed.
(1)
(f)
Comment on the effect of the value 41 on the standard deviation and on the interquartile range.
(1)
Show mark scheme
Working
Mark
Notes
(a)
median \(=67.5\) bpm
A1
\(\text{IQR}=72-63=9\) bpm
A1
\(Q_{1}=63,\ Q_{3}=72\)
(b)
\(Q_{1}-1.5\times\text{IQR}=63-1.5\times9\)
M1
\(=49.5\) and \(41\lt 49.5\)
A1
Must see 49.5 and a comparison.
(c)
it should not be removed: it is a genuine (valid) value for a member of the team, not an error
R1
Award R0 for “remove it because it is an outlier”.
(d)
mean \(=66.7\) bpm
A1
\(66.666\ldots\)
\(\sigma=8.36\) bpm
A1
\(8.35995\ldots\)
(e)
\(5.74\) bpm
A1
\(5.74185\ldots\)
(f)
the standard deviation is much larger with 41 included (8.36 compared with 5.74), whereas the IQR is unchanged (9 in both cases), because quartiles do not depend on the extreme values
R1
Accept IQR \(72.5-63.5=9\) without 41. Must compare both measures.
Answers: (a) 67.5, 9 · (d) 66.7, 8.36 · (e) 5.74
244AASL4-q00154.2 Presentation of dataPaper 2Medium9 marks
The lengths, \(l\) cm, of 200 leaves from a tree were measured. The cumulative frequency graph shows the results. The shortest leaf is 5 cm long and the longest is 16 cm long.
Graph drawn to scale
(a)
Use the graph to write down the median length.
(1)
(b)
Use the graph to find the interquartile range.
(2)
(c)
Find the 15th percentile of the lengths.
(2)
(d)
Estimate the number of leaves that are longer than 13 cm.
(2)
(e)
Determine whether any of the leaves have lengths that are outliers.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(10.5\) cm
A1
Accept \(10.4\)–\(10.6\).
(b)
\(Q_{1}=9\) and \(Q_{3}=12\) (reading at 50 and 150)
(M1)
Accept \(8.9\)–\(9.1\) and \(11.9\)–\(12.1\).
\(\text{IQR}=3\) cm
A1
(c)
reading at cumulative frequency \(0.15\times200=30\)
(M1)
\(8\) cm
A1
Accept \(7.9\)–\(8.1\).
(d)
cumulative frequency at 13 cm is about 171
(M1)
Accept \(169\)–\(174\).
\(200-171=29\) leaves
A1
Accept \(26\)–\(31\).
(e)
boundaries \(9-1.5\times3=4.5\) and \(12+1.5\times3=16.5\)
M1
FT from their quartiles.
all lengths are between 5 cm and 16 cm, which lie inside \(4.5\) to \(16.5\), so there are no outliers
A1
Answers: (a) 10.5 cm · (b) 3 cm · (c) 8 cm · (d) about 29 leaves · (e) no outliers
245AASL4-q00164.2 Presentation of dataPaper 2Medium9 marks
The amounts, \(x\) euros (€), spent by 200 customers in a supermarket on one day are summarised below.
Amount, \(x\) (€)
\(0\le x\lt 10\)
\(10\le x\lt 20\)
\(20\le x\lt 30\)
\(30\le x\lt 40\)
\(40\le x\lt 50\)
\(50\le x\lt 60\)
\(60\le x\lt 70\)
Frequency
8
30
52
46
34
20
10
The cumulative frequency table is
Amount, \(x\) (€)
\(x\lt 10\)
\(x\lt 20\)
\(x\lt 30\)
\(x\lt 40\)
\(x\lt 50\)
\(x\lt 60\)
\(x\lt 70\)
Cumulative frequency
8
38
90
\(a\)
170
\(b\)
200
(a)
Write down the value of \(a\) and the value of \(b\).
(2)
(b)
Write down the class that contains the median amount.
(1)
(c)
Find the percentage of the customers who spent at least €50.
(1)
(d)
Find an estimate of the mean and an estimate of the standard deviation of the amounts spent.
(3)
(e)
The 25% of customers who spent the most are sent a discount voucher. Determine the class that contains the least amount spent by a customer who is sent a voucher.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(a=136\)
A1
\(b=190\)
A1
(b)
\(30\le x\lt 40\)
A1
100th/100.5th value; cumulative frequency passes 100 in this class.
(c)
\(\dfrac{20+10}{200}=15\%\)
A1
(d)
use of mid-interval values \(5,15,\ldots,65\) with the frequencies
(M1)
mean \(=\)€33.4
A1
\(\sigma=\)€14.9
A1
\(14.8808\ldots\)
(e)
\(25\%\) of \(200=50\) customers, so the boundary is at the 150th/151st amount (cumulative frequency \(150\)), i.e. the upper quartile
(M1)
\(40\le x\lt 50\) (cumulative frequency \(136\) for \(x\lt 40\) and \(170\) for \(x\lt 50\))
246AASL4-q00274.3 Measures of central tendency and dispersionPaper 2Medium7 marks
The number of people, \(n\), living in each of 85 households in a village is shown in the table.
Number of people, \(n\)
1
2
3
4
5
Number of households
23
31
18
9
4
(a)
Find the mean of \(n\).
(1)
(b)
Find the standard deviation and the variance of \(n\).
(2)
(c)
Find the number of households for which \(n\) is within one standard deviation of the mean.
(2)
(d)
Find the median of \(n\).
(1)
(e)
With reference to the shape of the distribution, explain why the mean is greater than the median.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(2.29\)
A1
\(\tfrac{195}{85}=2.29411\ldots\)
(b)
\(\sigma=1.11\)
A1
\(1.11485\ldots\)
variance \(=1.24\)
A1
\(1.24290\ldots\)
(c)
\(2.294-1.115=1.18\) to \(2.294+1.115=3.41\), so \(n=2\) or \(n=3\)
(M1)
\(31+18=49\)
A1
(d)
\(2\)
A1
43rd value; cumulative frequencies 23, 54.
(e)
the distribution is positively skewed (most households have 1 or 2 people, with a long tail up to 5), so the few large values pull the mean above the median
247AASL4-q00324.4 Correlation and regression of y on xPaper 2Medium8 marks
The table shows the altitude, \(x\) metres, of ten weather stations in a mountain region and the mean July temperature, \(y\) °C, recorded at each station.
248AASL4-q00354.4 Correlation and regression of y on xPaper 2Medium10 marks
Nine teenagers recorded their average daily screen time, \(x\) hours, and their average night’s sleep, \(y\) hours, over a month.
\(x\) (hours)
1.5
2
2.5
3
3.5
4.5
5
6
6.5
\(y\) (hours)
8.2
7.6
8
7.1
7.4
6.8
6.2
6.5
5.9
(a)
Find the value of \(r\) and describe the correlation between \(x\) and \(y\).
(2)
(b)
Find the equation of the regression line of \(y\) on \(x\).
(2)
(c)
Use your equation to estimate the average sleep of a teenager whose average daily screen time is 4 hours.
(2)
(d)
A teenager sleeps an average of 7.5 hours per night. Explain why your equation should not be used to estimate this teenager’s screen time.
(1)
(e)
The sleep times are converted to minutes, \(Y=60y\). Write down the equation of the regression line of \(Y\) on \(x\).
(2)
(f)
A newspaper states that “screen time causes teenagers to lose sleep”. Comment on whether the data support this claim.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(r=-0.937\)
A1
\(-0.936939\ldots\)
strong negative (linear) correlation
A1
(b)
\(a=-0.421\)
A1
\(-0.421333\ldots\)
\(y=-0.421x+8.69\)
A1
\(8.69289\ldots\)
(c)
\(-0.421333(4)+8.69289\)
(M1)
\(7.01\) hours
A1
\(7.00756\ldots\) Accept 7.006 or 7.0 from 3 s.f. coefficients.
(d)
the line of \(y\) on \(x\) is for predicting \(y\) from \(x\); it cannot reliably be used to predict \(x\) from a value of \(y\)
R1
(e)
gradient \(-25.3\)
A1
\(60\times(-0.421333)=-25.28\)
\(Y=-25.3x+522\)
A1
\(521.573\ldots\) FT from (b).
(f)
no: the strong correlation shows an association, but correlation does not imply causation (other factors, such as stress or schoolwork, could affect both)
250AASL4-q00384.4 Correlation and regression of y on xPaper 2Medium8 marks
For nine towns, the table shows the number of public libraries, \(L\), the number of car thefts reported in one year, \(C\), and the population, \(P\) thousand.
Town
A
B
C
D
E
F
G
H
I
\(L\)
3
5
4
8
6
11
9
14
12
\(C\)
410
620
540
1080
700
1390
1210
1720
1450
\(P\)
48
71
60
118
85
152
131
190
163
(a)
Find Pearson’s product-moment correlation coefficient between \(L\) and \(C\).
(1)
(b)
A politician claims that closing libraries would reduce car thefts. Comment on this claim.
(2)
(c)
Find Pearson’s product-moment correlation coefficient between \(P\) and \(C\).
(1)
(d)
Find the equation of the regression line of \(C\) on \(P\).
(2)
(e)
Hence estimate the number of car thefts in a town with a population of 100 thousand.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(r=0.994\)
A1
\(0.994420\ldots\)
(b)
correlation does not imply causation
R1
both \(L\) and \(C\) increase with the size of the population: larger towns have more libraries and more cars to steal
R1
Must identify population (town size) as a third variable.
251AASL4-q00454.5 Probability and expected occurrencesPaper 2Medium7 marks
A town has 8000 residents. The table shows the number of residents in each age group and, from previous years, the probability that a resident in that group catches influenza during a winter.
Age group
Number of residents
Probability of catching influenza
0–17
1 200
0.084
18–64
5 300
0.031
65 and over
1 500
0.047
(a)
Find the expected number of residents aged 18–64 who catch influenza during the winter.
(2)
(b)
Find the expected total number of residents who catch influenza during the winter.
(2)
(c)
Hence find the probability that a resident chosen at random from the town catches influenza during the winter.
(1)
(d)
During one winter, 362 residents caught influenza. Find the relative frequency of catching influenza that winter and comment on your answer, with reference to part (c).
this is higher than the expected probability of 0.0420, so more residents than expected caught influenza that winter (the past probabilities may underestimate that winter’s risk)
R1
Accept any valid comment that compares the two values.
252AASL4-q00554.6 Combined events and diagramsPaper 2Medium8 marks
A system contains three components, \(A\), \(B\) and \(C\), which work independently of each other. The system works if \(A\) works and at least one of \(B\) and \(C\) works. The diagram shows the probability that each component works.
Probability that each component works is shown above it
(a)
Find the probability that both \(B\) and \(C\) fail.
(1)
(b)
Find the probability that the system works.
(2)
(c)
Component \(C\) is replaced by a new component, which works with probability \(q\). The system now works with probability 0.92. Find the value of \(q\).
(3)
(d)
Three of these systems operate independently. Find the probability that at least one of them does not work.
254AASL4-q00754.8 The binomial distributionPaper 2Medium8 marks
The probability that a randomly chosen visitor to a website clicks on an advertisement is 0.08, independently of other visitors.
(a)
Find the probability that at least one of 20 visitors clicks on the advertisement.
(2)
(b)
Find the least number of visitors needed for the probability that at least one of them clicks on the advertisement to be at least 0.99.
(3)
(c)
Find the least number of visitors needed for the probability that at least two of them click on the advertisement to be greater than 0.9.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(1-0.92^{20}\)
(M1)
\(0.811\)
A1
\(0.811306\ldots\)
(b)
\(1-0.92^{n}\ge 0.99\), or \(0.92^{n}\le 0.01\)
(M1)
\(n\ge 55.2\), or \(n=55:\ 0.98981\), \(n=56:\ 0.99062\)
(A1)
\(56\)
A1
(c)
\(1-\mathrm{P}(X\le 1)\gt 0.9\) for \(X\sim\mathrm{B}(n,\,0.08)\), using a table of values
(M1)
\(n=47:\ 0.8990\), \(n=48:\ 0.9055\)
(A1)
\(48\)
A1
Answers: (a) 0.811 · (b) 56 · (c) 48
255AASL4-q00764.8 The binomial distributionPaper 2Medium8 marks
At a supermarket, 36% of customers use a loyalty card, independently of each other. A random sample of 25 customers is taken. Let \(X\) be the number of customers in the sample who use a loyalty card.
(a)
Find the mean and the standard deviation of \(X\).
(3)
(b)
Find the probability that \(X\) is equal to its mean.
(2)
(c)
Find the probability that \(X\) lies within one standard deviation of its mean.
Answers: (a) mean 9, standard deviation 2.4 · (b) 0.164 · (c) 0.703
256AASL4-q00844.9 The normal distributionPaper 2Medium7 marks
The battery life, \(H\) hours, of a model of mobile phone is normally distributed with mean 14.2 hours and standard deviation 2.6 hours.
(a)
Find the probability that a randomly chosen phone has a battery life of less than 10 hours.
(2)
(b)
Find the lower quartile and the upper quartile of \(H\).
(2)
(c)
A battery life is classed as an outlier if it is more than \(1.5\times\) IQR from the nearest quartile. Find the probability that a randomly chosen phone has a battery life that is an outlier.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{P}(H\lt 10)\) with \(\mu=14.2,\ \sigma=2.6\)
(M1)
\(0.0531\)
A1
\(0.0531137\ldots\)
(b)
\(Q_{1}=12.4\) hours
A1
\(12.4463\ldots\), from invNorm(0.25, 14.2, 2.6)
\(Q_{3}=16.0\) hours
A1
\(15.9537\ldots\)
(c)
boundaries \(Q_{1}-1.5\,\text{IQR}\) and \(Q_{3}+1.5\,\text{IQR}\), with IQR \(=3.5073\ldots\)
257AASL4-q00864.9 The normal distributionPaper 2Medium7 marks
The times, \(R\) minutes, taken by runners to complete a 10 km charity run are normally distributed with mean 52 minutes and standard deviation 6.5 minutes.
Diagram not to scale
(a)
Find the probability that a randomly chosen runner finishes in less than 45 minutes.
(2)
(b)
Find the value of \(k\) such that \(\mathrm{P}(52-k\lt R\lt 52+k)=0.9\).
(3)
(c)
The slowest 10% of runners are given a free training plan. Find the shortest time for which a runner receives the plan.
(2)
Show mark scheme
Working
Mark
Notes
(a)
normalcdf(−large, 45, 52, 6.5)
(M1)
\(0.141\)
A1
\(0.140757\ldots\)
(b)
by symmetry \(\mathrm{P}(R\lt 52+k)=0.95\) (or \(\mathrm{P}(R\lt 52-k)=0.05\))
\(\mathrm{P}(R\gt t)=0.1\), i.e. invNorm(0.9, 52, 6.5)
(M1)
\(60.3\) minutes
A1
\(60.3300\ldots\) Award (M1)A0 for 43.7 (from invNorm(0.1, …)).
Answers: (a) 0.141 · (b) \(k=10.7\) · (c) 60.3 min
258AASL4-q00884.9 The normal distributionPaper 2Medium9 marks
The masses, \(M\) grams, of apricots picked on a farm are normally distributed with mean 210 g and standard deviation 25 g. Apricots are graded and sold as follows.
Grade
Small
Medium
Large
Mass (g)
\(M\lt 195\)
\(195\le M\le 230\)
\(M\gt 230\)
Price per apricot (€)
0.25
0.35
0.50
Diagram not to scale
(a)
Find the probability that a randomly chosen apricot is graded medium.
(2)
(b)
Find the probability that a randomly chosen apricot is graded small and the probability that it is graded large.
(2)
(c)
Find the expected income from 1000 apricots. Give your answer to the nearest euro.
(3)
(d)
The heaviest 15% of apricots are also labelled “premium”. Find the minimum mass of a premium apricot.
259AASL4-q00894.9 The normal distributionPaper 2Medium8 marks
The delivery times, \(D\) minutes, of a pizza company are normally distributed with mean 26 minutes and standard deviation 4 minutes. A pizza is free if its delivery takes more than 32 minutes. Assume that delivery times are independent.
Diagram not to scale
(a)
Find the probability that a delivery is free.
(2)
(b)
On one evening there are 20 deliveries. Find the probability that at least two of them are free.
(2)
(c)
Find the probability that the first free delivery of the evening is the fifth delivery.
(2)
(d)
The company wants only 2% of deliveries to be free, without changing the distribution of delivery times. Find the new time limit, in minutes.
260AASL4-q00904.9 The normal distributionPaper 2Medium7 marks
The lifetimes of two brands of LED lamp are normally distributed. Brand A lamps have mean 1200 hours and standard deviation 150 hours. Brand B lamps have mean 1300 hours and standard deviation 250 hours.
(a)
For each brand, find the probability that a lamp lasts more than 1500 hours.
(2)
(b)
For each brand, find the probability that a lamp lasts more than 1000 hours.
(2)
(c)
Show that the probability that a lamp lasts more than \(t\) hours is the same for both brands when \(t=1050\).
(2)
(d)
A hospital needs lamps that are as likely as possible to last at least 1000 hours. State, with a reason, which brand it should buy.
(1)
Show mark scheme
Working
Mark
Notes
(a)
A: \(0.0228\)
A1
\(0.0227501\ldots\)
B: \(0.212\)
A1
\(0.211855\ldots\)
(b)
A: \(0.909\)
A1
\(0.908789\ldots\)
B: \(0.885\)
A1
\(0.884930\ldots\)
(c)
equal probabilities when the standardized values are equal: \(\dfrac{t-1200}{150}=\dfrac{t-1300}{250}\)
Find the equation of the regression line of \(y\) on \(x\).
(2)
(c)
Use your equation to estimate the height of the son of a father who is 172 cm tall.
(1)
(d)
Find the equation of the regression line of \(x\) on \(y\).
(2)
(e)
A son is 165 cm tall. Estimate the height of his father.
(2)
(f)
Explain why you did not use the equation from part (b) in part (e).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(r=0.982\)
A1
\(0.98182\ldots\)
(b)
gradient \(0.937\)
A1
\(0.93690\ldots\)
\(y=0.937x+11.0\)
A1
\(11.0011\ldots\)
(c)
\(172\) cm
A1
\(172.148\ldots\) Accept 172.1 or 172.2.
(d)
gradient \(1.03\)
A1
\(1.02890\ldots\)
\(x=1.03y-5.21\)
A1
\(-5.21099\ldots\) Award A0A0 for rearranging the equation of \(y\) on \(x\).
(e)
\(x=1.02890(165)-5.21099\)
(M1)
FT from their line of x on y only.
\(165\) cm
A1
\(164.558\ldots\) Accept answers from 164.5 to 164.8 (from rounded coefficients). Award M0A0 for \(164.4\), from rearranging the line of \(y\) on \(x\).
(f)
the line of \(y\) on \(x\) should only be used to predict \(y\) from \(x\); to predict \(x\) from \(y\) the regression line of \(x\) on \(y\) must be used
R1
Answers: (a) 0.982 · (b) \(y=0.937x+11.0\) · (c) 172 cm · (d) \(x=1.03y-5.21\) · (e) 165 cm (164.6)
262AASL4-q00954.10 Regression of x on yPaper 2Medium9 marks
On a long journey, a driver records the distance, \(x\) km, travelled since the fuel tank was filled, and the volume of fuel, \(y\) litres, remaining in the tank.
Distance, \(x\) (km)
50
120
180
240
310
370
450
520
Fuel, \(y\) (litres)
50.4
47.5
40.6
38.3
30.9
28.8
20.5
17.6
(a)
Find Pearson’s product-moment correlation coefficient, \(r\), and interpret its value in context.
(2)
(b)
Find the equation of the regression line of \(x\) on \(y\).
(2)
(c)
On a later journey, the trip counter fails but the fuel gauge shows 34 litres remaining. Use your answer to part (b) to estimate the distance travelled since the tank was filled.
(2)
(d)
Find the equation of the regression line of \(y\) on \(x\), and use it to estimate the fuel remaining after 400 km.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(r=-0.995\)
A1
\(-0.994907\ldots\)
very strong negative linear correlation: the further the car has travelled, the less fuel remains
R1
(b)
gradient \(-13.6\)
A1
\(-13.5522\ldots\)
\(x=-13.6y+745\)
A1
\(745.181\ldots\)
(c)
\(x=-13.5522(34)+745.181\)
(M1)
Award M0 for rearranging the line of \(y\) on \(x\).
\(284\) km
A1
\(284.404\ldots\) Accept 282 to 285 (from rounded coefficients).
263AASL4-q00974.10 Regression of x on yPaper 2Medium8 marks
The scatter diagram and table show the floor area, \(x\) m2, and the monthly energy bill, \(y\) euros, for ten apartments in one building.
Graph drawn to scale
\(x\) (m2)
62
75
80
88
95
104
112
120
135
148
\(y\) (€)
48
55
61
63
70
74
81
84
140
101
(a)
Find the value of Pearson’s product-moment correlation coefficient, \(r\), for all ten apartments.
(1)
(b)
One bill was recorded incorrectly. Identify this apartment, giving a reason.
(1)
(c)
The correct bill for this apartment was €94. Using the corrected data for all ten apartments, find the new value of \(r\) and the equation of the regression line of \(x\) on \(y\).
(3)
(d)
Another apartment in the building has a monthly bill of €90. Use your equation from part (c) to estimate its floor area.
(2)
(e)
Comment on whether your equation would give a reliable estimate of the floor area of an apartment with a bill of €30.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(r=0.868\)
A1
\(0.867525\ldots\)
(b)
the apartment with \(x=135,\ y=140\); its bill is far above the pattern of the other nine points
R1
Must identify the point and give a reason.
(c)
\(r=0.998\)
A1
\(0.997817\ldots\)
gradient \(1.60\)
A1
\(1.59673\ldots\)
\(x=1.60y-14.8\)
A1
\(-14.8210\ldots\)
(d)
\(x=1.59673(90)-14.8210\)
(M1)
Must use their line of x on y.
\(129\) m2
A1
\(128.884\ldots\) Accept 129.2 from rounded coefficients.
(e)
not reliable: €30 is outside the range of the bills in the data (€48 to €101), so this would be extrapolation
R1
Answers: (a) 0.868 · (b) area 135 m2, bill €140 · (c) \(r=0.998\), \(x=1.60y-14.8\) · (d) 129 m2
264AASL4-q00994.10 Regression of x on yPaper 2Medium7 marks
On ten days, an engineer records the mean wind speed, \(x\) m s−1, at a small wind turbine and its mean power output, \(y\) kW.
\(x\) (m s−1)
3.2
4.0
4.5
5.1
5.8
6.3
7.0
7.6
8.4
9.1
\(y\) (kW)
21
30
33
42
47
55
61
66
77
82
(a)
Find the equation of the regression line of \(x\) on \(y\).
(2)
(b)
Interpret the gradient of this line in context.
(1)
(c)
On one day the anemometer failed, but the mean power output was 50 kW. Estimate the mean wind speed on that day.
(2)
(d)
The wind speed \(K\) in km h−1 is given by \(K=3.6x\). Find the equation of the regression line of \(K\) on \(y\), giving your answer in the form \(K=ay+b\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
gradient \(0.0950\)
A1
\(0.0950139\ldots\)
\(x=0.0950y+1.22\)
A1
\(1.21628\ldots\)
(b)
each additional kW of power output is associated with an increase of about \(0.095\) m s−1 in the mean wind speed
R1
Accept equivalent, e.g. 10 kW more output corresponds to about 0.95 m s⁻¹ more wind speed.
(c)
\(x=0.0950139(50)+1.21628\)
(M1)
Award M0 for rearranging a line of y on x.
\(5.97\) m s−1
A1
\(5.96698\ldots\) Accept 5.96 to 5.97.
(d)
\(K=3.6(0.0950139y+1.21628)\)
(M1)
Or by finding the line of K on y directly with technology.
Answers: (a) 0.998 · (b) \(y=2.61x+7.91\) · (c) 60.1 cm · (d) \(x=0.381y-2.89\) · (e)(i) 0.0912 (ii) 89.2 cm (iii) 31.1 years · (f) 0.273
266AASL4-q01084.11 Conditional probability and independencePaper 2Medium7 marks
A survey asked 200 students in Grades 11 and 12 how they usually travel to school. The results are shown in the table.
Walk (W)
Bus (B)
Car (C)
Total
Grade 11 (G)
30
50
20
100
Grade 12
30
40
30
100
Total
60
90
50
200
A student is chosen at random from the 200 students.
(a)
Find the probability that the student travels by bus.
(1)
(b)
Given that the student travels by car, find the probability that the student is in Grade 12.
(2)
(c)
Show that the events \(W\) and \(G\) are independent.
(2)
(d)
Determine whether the events \(B\) and \(G\) are independent.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac{90}{200}=0.45\)
A1
(b)
\(\dfrac{30}{50}\)
(M1)
\(0.6\)
A1
(c)
\(\mathrm{P}(W\mid G)=\dfrac{30}{100}=0.3\) and \(\mathrm{P}(W)=\dfrac{60}{200}=0.3\)
M1
Or \(\mathrm{P}(W\cap G)=0.15=\mathrm{P}(W)\,\mathrm{P}(G)=0.3\times0.5\).
\(\mathrm{P}(W\mid G)=\mathrm{P}(W)\), so \(W\) and \(G\) are independent
R1
(d)
\(\mathrm{P}(B\mid G)=\dfrac{50}{100}=0.5\) and \(\mathrm{P}(B)=0.45\)
M1
Or \(\mathrm{P}(B\cap G)=0.25\) and \(\mathrm{P}(B)\,\mathrm{P}(G)=0.225\).
\(0.5\neq0.45\), so \(B\) and \(G\) are not independent
A1
Answers: (a) 0.45 · (b) 0.6 · (d) not independent
267AASL4-q01134.12 Standardization and inverse normalPaper 2Medium8 marks
In a year group, the marks in a mathematics test are normally distributed with mean 62 and standard deviation 8, and the marks in a physics test are normally distributed with mean 55 and standard deviation 6. Ana scored 74 in mathematics and 65 in physics.
(a)
Find Ana’s standardized score, \(z\), for each test.
(2)
(b)
Hence state, with a reason, in which test Ana performed better relative to the year group.
(1)
(c)
Find the proportion of the year group who scored more than Ana in physics.
(2)
(d)
The marks in a chemistry test are normally distributed with mean 60. Exactly 10% of the year group scored more than 75. Find the standard deviation of the chemistry marks.
(3)
Show mark scheme
Working
Mark
Notes
(a)
mathematics: \(z=\dfrac{74-62}{8}=1.5\)
A1
physics: \(z=\dfrac{65-55}{6}=1.67\)
A1
\(1.666\ldots\)
(b)
physics, because her \(z\)-score is higher (she is more standard deviations above the mean)
R1
FT from (a).
(c)
\(\mathrm{P}(Z\gt 1.666\ldots)\) or normalcdf(65, large, 55, 6)
268AASL4-q01154.12 Standardization and inverse normalPaper 2Medium8 marks
A machine fills boxes with cereal. The mass of cereal in a box is normally distributed with mean \(\mu\) grams and standard deviation 4 grams. The label states 500 g, and only 1% of boxes may contain less than 500 g.
(a)
Find the least value of \(\mu\) that meets this requirement.
(3)
(b)
With \(\mu\) equal to the value found in part (a), find the probability that a box contains more than 515 g.
(2)
(c)
A new machine has mean 506 g. Find the greatest standard deviation for which it meets the 1% requirement.
Answers: (a) 40.6 mm · (b) 0.890 mm · (c) 0.0123 · (d) 2.25
270AASL4-q01194.12 Standardization and inverse normalPaper 2Medium7 marks
At a regional athletics meeting, the 50 m swimming times of competitors are normally distributed with mean 31.5 seconds and standard deviation 2.2 seconds, and the shot-put distances are normally distributed with mean 9.4 m and standard deviation 1.1 m. Kofi swims 50 m in 28.9 seconds and puts the shot 10.8 m.
(a)
Find Kofi’s standardized value, \(z\), for each event.
(2)
(b)
Justify which of Kofi’s two performances is better relative to the other competitors.
(1)
(c)
Find the shot-put distance that represents the same relative standing as Kofi’s swimming time.
(2)
(d)
Find the proportion of swimmers who are faster than Kofi.
(2)
Show mark scheme
Working
Mark
Notes
(a)
swimming: \(z=\dfrac{28.9-31.5}{2.2}=-1.18\)
A1
\(-1.18181\ldots\)
shot put: \(z=\dfrac{10.8-9.4}{1.1}=1.27\)
A1
\(1.27272\ldots\)
(b)
the shot put: a lower time is better in swimming, and his shot put is \(1.27\) standard deviations better than the mean, more than the \(1.18\) for swimming
R1
Must refer to the direction of a good performance in each event. FT from (a).
(c)
\(9.4+1.18181\ldots(1.1)\)
(M1)
Sign must reflect that a fast time corresponds to a long distance.
\(10.7\) m
A1
\(10.7\) exactly.
(d)
\(\mathrm{P}(S\lt 28.9)\) with \(\mu=31.5,\ \sigma=2.2\)
(M1)
\(0.119\)
A1
\(0.118641\ldots\)
Answers: (a) −1.18 and 1.27 · (b) shot put · (c) 10.7 m · (d) 0.119
271IBMA2S-050-02-054.9 The normal distributionPaper 2Medium7 marks
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A pharmaceutical machine fills vials with a liquid vaccine. The volume, V mL, in a vial is normally distributed with mean 10.0 mL and standard deviation σ mL. It is found that 2% of vials contain less than 9.8 mL.
(a)
Find σ.
(3)
(b)
Find P(V > 10.15).
(2)
(c)
Find the probability that a box of 12 contains at least one vial with less than 9.8 mL.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 standardising, (9.8 − 10.0)/σ = z ; A1 z = −2.05375 ; A1 σ = 0.0974 mL (0.097383…)
✓ 3
Part (b)
M1 correct use of the normal distribution with their σ, e.g. z = 0.15/0.09738 = 1.540 ; A1 0.0617 (accept 0.0618 from σ = 0.0974)
✓ 2
Part (c)
M1 1 − 0.9812 or 1 − P(Y = 0) with Y ~ B(12, 0.02) ; A1 0.215
✓ 2
Common error: Rounding σ too early, for example to 0.097, gives about 0.0610 in part (b) instead of 0.0617. Store the unrounded σ in your calculator and use it in later parts.
272IBMA2S-050-02-064.4 Correlation and regression of y on xPaper 2Medium7 marks
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A broadband provider measures the download speed, S Mbps, at a home router many times. The speeds are modelled by a normal distribution with mean μ and standard deviation σ. 15% of the measurements are below 42 Mbps and 10% are above 71 Mbps.
(a)
Find the values of μ and σ.
(5)
(b)
Find the speed s such that P(S ≥ s) = 0.99.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 forms an equation using an inverse normal z-value ; A1 μ − 1.03643σ = 42 ; A1 μ + 1.28155σ = 71 ; A1 σ = 12.5 ; A1 μ = 55.0
✓ 5
Part (b)
M1 s = μ − 2.32635σ or inverse normal with lower-tail area 0.01 using their μ, σ ; A1 25.9 Mbps
✓ 2
Exam tip: When two tail percentages are given, write both equations in the form μ + zσ = x before solving. The second z-value comes from a lower-tail area of 0.90, not 0.10.
273IBMA2S-050-02-074.12 Standardization and inverse normalPaper 2Medium11 marks
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A machine cuts steel rods for a bridge. The length, L mm, of a rod is normally distributed with mean 250 mm and standard deviation 0.8 mm. A rod is accepted if its length is between 248.5 mm and 251.5 mm. Otherwise it is rejected.
Diagram not to scale
(a)
Find P(248.5 < L < 251.5).
(2)
(b)
Find the probability that at least 14 of 15 rods are accepted.
M1 numerator P(251 < L < 251.5) = 0.07526 ; M1 divides by P(accepted) = 0.93921 ; A1 0.0801
✓ 3
Common error: In part (d), the numerator is not P(L > 251). An accepted rod cannot be longer than 251.5 mm, so the intersection is P(251 < L < 251.5).
274IBMA2S-050-02-094.12 Standardization and inverse normalPaper 2Medium5 marks
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A company buys glass phone screens from two factories. 60% of the screens come from factory A and 40% from factory B. The thickness of a screen from factory A is normally distributed with mean 0.70 mm and standard deviation 0.02 mm. The thickness of a screen from factory B is normally distributed with mean 0.71 mm and standard deviation 0.03 mm. A screen is rejected if its thickness exceeds 0.74 mm.
(a)
Find P(rejected | A) and P(rejected | B).
(2)
(b)
Find the probability that a randomly chosen screen is rejected.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 P(rejected | A) = 0.0228 ; A1 P(rejected | B) = 0.159
✓ 2
Part (b)
M1 multiplies each conditional probability by its factory proportion, e.g. 0.6 × 0.02275 ; M1 adds the two branches ; A1 0.0771
✓ 3
Common error: Averaging the two rejection probabilities, (0.0228 + 0.159)/2, ignores the 60 : 40 split. Each branch must be weighted by its factory proportion.
275IBMA2S-050-02-104.2 Presentation of dataPaper 2Medium12 marks
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In sprint races, a starting block sensor measures each athlete's reaction time, R seconds. For a group of elite sprinters, R is normally distributed with mean 0.165 s and standard deviation 0.020 s. A reaction time of less than 0.100 s counts as a false start.
(a)
Find P(R < 0.100).
(2)
(b)
Find the interquartile range of R.
(3)
(c)
Find the probability that at least 3 of 8 sprinters have R < 0.150.
M1 uses inverse normal for the 25th or 75th percentile (z = ±0.67449) ; A1 Q1 = 0.1515, Q3 = 0.1785 ; A1 IQR = 0.0270 s
✓ 3
Part (c)
M1 finds p = P(R < 0.150) ; A1 p = 0.2266 ; M1 Y ~ B(8, 0.2266), uses 1 − P(Y ≤ 2) ; A1 0.264
✓ 4
Part (d)
M1 (0.100 − μ)/0.020 = z with lower-tail area 0.01 ; A1 z = −2.32635 ; A1 μ = 0.147 s
✓ 3
Exam tip: 'At least 3' means 1 − P(Y ≤ 2). Calculating 1 − P(Y ≤ 3) wrongly leaves out the case Y = 3.
276IBMA2S-050-07-054.7 Discrete random variablesPaper 2Medium7 marks
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A small rocket company has a launch window that allows at most three launch attempts. Each attempt succeeds with probability p, independently of other attempts. Attempts stop as soon as a launch succeeds, and no more than three attempts are made in total.
Let X be the number of attempts made in the window, so X ∈ {1, 2, 3}.
(a)
Write down P(X = 1), P(X = 2) and P(X = 3) in terms of p.
(2)
(b)
Show that \(E(X)=p^2-3p+3\).
(2)
(c)
The company wants the expected number of attempts to be 1.2. Find the required value of p.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(P(X=1)=p\), \(P(X=2)=p(1-p)\)
✓ 1
A1 \(P(X=3)=(1-p)^2\)
✓ 1
Part (b)
M1 \(E(X)=p+2p(1-p)+3(1-p)^2\)
✓ 1
A1 expands \(p+2p-2p^2+3-6p+3p^2\) to obtain \(p^2-3p+3\)
✓ 1
Part (c)
M1 solves \(p^2-3p+3=1.2\), i.e. \(p^2-3p+1.8=0\)
✓ 1
A1 \(p=0.829\)
✓ 1
R1 rejects \(p=2.17\) because a probability cannot exceed 1
✓ 1
Common error: P(X = 3) is (1 − p)², not (1 − p)²p: the third attempt is made whether or not it succeeds, so X = 3 only requires the first two to fail.
277IBMA2S-050-07-074.7 Discrete random variablesPaper 2Medium12 marks
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A company offers phone-screen insurance. For a randomly chosen customer, the number of claims, C, made in one year has the following distribution.
c: 0, 1, 2 P(C = c): 0.8, 0.15, 0.05
Each claim costs the company USD 150. Customers behave independently.
(a)
Find E(C).
(2)
(b)
Find the expected cost of claims per customer per year.
(2)
(c)
The company charges an annual premium of USD P per customer. Find P so that the company's expected profit per customer is USD 10.
(2)
(d)
Two customers are chosen at random. Find the probability that they make a total of exactly 2 claims in a year.
(3)
(e)
A shop has 12 customers with this insurance. Find the probability that more than 2 of them make at least one claim in a year.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(0(0.8)+1(0.15)+2(0.05)\)
✓ 1
A1 E(C) = 0.25
✓ 1
Part (b)
M1 \(150\times0.25\)
✓ 1
A1 USD 37.50
✓ 1
Part (c)
M1 \(P-37.5=10\)
✓ 1
A1 P = 47.50
✓ 1
Part (d)
M1 identifies the cases (0, 2), (2, 0) and (1, 1)
✓ 1
M1 multiplies probabilities using independence: \(2(0.8)(0.05)+0.15^2\)
✓ 1
A1 0.1025
✓ 1
Part (e)
M1 P(at least one claim) = 0.2 and \(Y\sim B(12,\,0.2)\)
✓ 1
M1 \(P(Y>2)=1-P(Y\le 2)\)
✓ 1
A1 0.442
✓ 1
Exam tip: In the last part, the binomial success probability is P(C ≥ 1) = 0.2, not P(C = 1) = 0.15. Read carefully what counts as a 'success'.
278IBMA2S-050-07-094.7 Discrete random variablesPaper 2Medium7 marks
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A factory tests each microchip it makes. A chip passes test A with probability 0.9. Only chips that pass test A take test B, which they pass with probability 0.8, independently. • A chip failing test A is scrapped, giving a profit of −$2. • A chip passing A but failing B is sold as standard grade, giving a profit of $3. • A chip passing both tests is sold as top grade, giving a profit of $10.
Let Y be the profit, in dollars, on a randomly chosen chip.
Diagram not to scale
(a)
Find the probability distribution of Y.
(3)
(b)
Find E(Y).
(2)
(c)
The profit on a top-grade chip is changed to $s, with all other profits and probabilities unchanged. Find s so that the expected profit per chip is $9.00.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 multiplies along branches of a tree, e.g. \(0.9\times0.2\)
✓ 1
A1 \(P(Y=3)=0.18\)
✓ 1
A1 \(P(Y=-2)=0.1\) and \(P(Y=10)=0.72\)
✓ 1
Part (b)
M1 \(-2(0.1)+3(0.18)+10(0.72)\)
✓ 1
A1 \(E(Y)=7.54\)
✓ 1
Part (c)
M1 \(-0.2+0.54+0.72s=9\)
✓ 1
A1 \(s=12.0\) (12.03)
✓ 1
Common error: The probability of top grade is 0.9 × 0.8 = 0.72, not 0.8: test B is only reached after passing test A.
279IBMA2S-050-07-104.7 Discrete random variablesPaper 2Medium13 marks
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An archer's score, S, on a single arrow has the following distribution.
s: 10, 8, 5, 0 P(S = s): 0.3, 0.45, 0.2, 0.05
Scores on different arrows are independent.
(a)
Find E(S).
(2)
(b)
The archer shoots two arrows. Find the probability that her total score is at least 18.
(3)
(c)
Given that her total score from two arrows is at least 18, find the probability that both arrows scored 10.
(2)
(d)
The archer shoots 6 arrows. Find the probability that at least 4 of them score 10.
(3)
(e)
After coaching, the probability of scoring 10 becomes p and the probability of scoring 8 becomes \(0.75-p\); the probabilities of scoring 5 and 0 are unchanged. Her new expected score per arrow is 7.9. Find p.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(10(0.3)+8(0.45)+5(0.2)+0(0.05)\)
✓ 1
A1 \(E(S)=7.6\)
✓ 1
Part (b)
M1 identifies (10, 10), (10, 8) and (8, 10) as the only possibilities
✓ 1
M1 \(0.3^2+2(0.3)(0.45)\)
✓ 1
A1 \(0.36\)
✓ 1
Part (c)
M1 \(\frac{0.09}{0.36}\)
✓ 1
A1 \(0.25\)
✓ 1
Part (d)
M1 recognises \(T\sim B(6,\,0.3)\) for the number of 10s
✓ 1
M1 \(P(T\ge4)=P(T=4)+P(T=5)+P(T=6)\) or \(1-P(T\le3)\)
✓ 1
A1 \(0.0705\)
✓ 1
Part (e)
M1 \(10p+8(0.75-p)+5(0.2)=7.9\)
✓ 1
A1 \(2p+7=7.9\)
✓ 1
A1 \(p=0.45\)
✓ 1
Exam tip: In part (e), check the new probabilities still sum to 1: p + (0.75 − p) + 0.2 + 0.05 = 1 for any p, so only the expectation fixes p.
280IBMA2S-050-01-054.8 The binomial distributionPaper 2Medium6 marks
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In a microchip factory, each chip produced is defective with probability 0.015, independently of other chips.
(a)
A sample of \(n\) chips is tested. Find the smallest value of \(n\) for which the probability that the sample contains at least one defective chip is greater than 0.5.
(4)
(b)
A batch of 200 chips is tested. Find the probability that it contains at most 2 defective chips.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(at least one defective) \(= 1 - 0.985^n\)
✓ 1
M1 sets up \(1 - 0.985^n > 0.5\) or \(0.985^n < 0.5\)
✓ 1
A1 \(n > 45.86\), or values \(n=45\): 0.493 and \(n=46\): 0.501
✓ 1
A1 \(n = 46\)
✓ 1
Part (b)
M1 recognises \(X \sim \text{B}(200,\,0.015)\) and requires \(P(X \le 2)\)
✓ 1
A1 0.421
✓ 1
Exam tip: When solving \(0.985^n < 0.5\) with logarithms, dividing by \(\ln 0.985\) (a negative number) reverses the inequality. Checking the two values of \(n\) on either side of the boundary avoids this trap.
281IBMA2S-050-01-074.8 The binomial distributionPaper 2Medium7 marks
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A football player takes penalty kicks. For each kick, she aims to the left with probability 0.6 and to the right otherwise. When she aims left she scores with probability 0.85; when she aims right she scores with probability 0.7. Different kicks are independent.
Diagram not to scale
(a)
Show that the probability that she scores from a randomly chosen penalty is 0.79.
(2)
(b)
In a season she takes 45 penalties. Find the expected number of penalties she misses.
(2)
(c)
Find the probability that she misses exactly 2 of her next 5 penalties.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(0.6 \times 0.85 + 0.4 \times 0.7\)
✓ 1
A1 \(0.51 + 0.28 = 0.79\)
✓ 1
Part (b)
M1 \(45 \times 0.21\)
✓ 1
A1 9.45
✓ 1
Part (c)
M1 recognises binomial with \(n = 5\), \(p = 0.21\)
✓ 1
M1 \(\binom{5}{2}(0.21)^2(0.79)^3\)
✓ 1
A1 0.217
✓ 1
Common error: In part (b), using 0.79 instead of \(1 - 0.79\) gives the expected number of goals, not misses. Check which outcome the question counts.
282IBMA2S-050-01-094.7 Discrete random variablesPaper 2Medium14 marks
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A suspension bridge is supported by 12 steel cables. At each annual inspection, each cable is found to need replacement with probability 0.08, independently of the other cables. Let \(X\) be the number of cables on a bridge that need replacement at an inspection.
(a)
Find the probability that no cables need replacement.
(2)
(b)
Find the probability that at least two cables need replacement.
(3)
(c)
Replacing one cable costs 4500 dollars. Find the expected cost of cable replacement at one inspection.
(2)
(d)
An engineering firm inspects \(m\) such bridges, which behave independently. Find the least value of \(m\) for which the probability that at least one bridge needs two or more cables replaced exceeds 0.99.
(4)
(e)
The firm inspects 5 such bridges in one year. Find the probability that exactly 3 of them need no cables replaced.
Exam tip: Parts (d) and (e) need a second binomial model whose success probability comes from an earlier part. Keep the unrounded value in your calculator memory to avoid accuracy errors.
283IBMA2S-050-01-124.7 Discrete random variablesPaper 2Medium12 marks
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At a charity fair, a player pays 2 dollars to throw 3 darts at a target. Each dart hits the target with probability 0.35, independently. Let \(X\) be the number of darts that hit the target.
(a)
Construct the probability distribution table for \(X\), giving each probability as an exact decimal.
(4)
(b)
During the fair, 400 games are played. Find the expected number of games in which at least 2 darts hit the target.
(2)
(c)
A player wins 5 dollars for exactly 2 hits and 20 dollars for 3 hits; otherwise nothing. Find the player's expected net gain per game, and state whether the game favours the player or the organiser.
(3)
(d)
The organiser keeps the 5-dollar prize for 2 hits but changes the prize for 3 hits to \(k\) dollars so that the game is fair. Find \(k\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 recognises \(X \sim \text{B}(3,\,0.35)\)
✓ 1
A1 \(P(X=0) = 0.274625\)
✓ 1
A1 \(P(X=1) = 0.443625\) and \(P(X=2) = 0.238875\)
R1 positive expected gain, so the game favours the player
✓ 1
Part (d)
M1 sets expected net gain to zero: \(5(0.238875) + 0.042875k - 2 = 0\)
✓ 1
A1 \(0.042875k = 0.805625\)
✓ 1
A1 \(k = 18.8\) (accept 18.79)
✓ 1
Common error: A fair game means expected net gain is zero. Forgetting to subtract the 2-dollar entry fee gives the expected winnings, not the expected gain.
284IBMA2S-050-04-054.7 Discrete random variablesPaper 2Medium7 marks
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An ultrasonic scanner inspects welds on a pipeline. 3% of welds are defective. The scanner flags 92% of defective welds and also flags 5% of welds that are not defective. A weld is chosen at random and scanned.
Diagram not to scale
(a)
Find the probability that the weld is flagged.
(2)
(b)
Given that the weld is flagged, find the probability that it is defective.
(2)
(c)
Every flagged weld is scanned a second time. For a weld whose condition (defective or not defective) is known, the result of the second scan is independent of the first and has the same flagging probabilities. Find the probability that a weld is defective, given that it is flagged on both scans.
Exam tip: Draw a tree with defective/not defective as the first branch. The conditional probability is the relevant branch product divided by the sum of all branch products that give the observed result.
285IBMA2S-050-04-074.7 Discrete random variablesPaper 2Medium7 marks
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A tennis player's first serve lands in with probability \(0.62\). If the first serve lands in, she wins the point with probability \(0.75\). If the first serve is out, she takes a second serve, which lands in with probability \(0.9\); if the second serve lands in she wins the point with probability \(0.52\). If both serves are out (a double fault) she loses the point.
(a)
Find the probability that she serves a double fault.
(2)
(b)
Find the probability that she wins the point.
(3)
(c)
Given that she wins the point, find the probability that her first serve landed in.
Exam tip: Keep unrounded values (0.64284) from earlier parts when dividing; rounding to 0.643 first can still give 0.723 here, but early rounding often costs the accuracy mark.
286IBMA2S-050-05-044.3 Measures of central tendency and dispersionPaper 2Medium6 marks
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A karting driver recorded 15 lap times, in seconds: 42.1, 43.0, 41.8, 42.6, 47.8, 42.4, 43.5, 41.2, 42.9, 42.0, 44.0, 42.7, 41.5, 43.2, 42.3 An outlier is a value more than 1.5 × IQR beyond the nearer quartile.
(a)
Find the median, lower quartile and upper quartile of the lap times.
(3)
(b)
Show that the lap time of 47.8 s is an outlier.
(2)
(c)
The driver spun off on that lap, so the 47.8 s time is removed. Find the median of the remaining 14 lap times.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 median = 42.6 s A1 Q1 = 42.0 s A1 Q3 = 43.2 s
✓ 3
Part (b)
M1 upper boundary = 43.2 + 1.5 × 1.2 = 45.0 s R1 47.8 > 45.0, so 47.8 s is an outlier
✓ 2
Part (c)
A1 (42.4 + 42.6)/2 = 42.5 s
✓ 1
Common error: Put the data in order before finding the median and quartiles; reading the middle value of the unordered list gives a wrong answer.
287IBMA2S-050-05-064.3 Measures of central tendency and dispersionPaper 2Medium7 marks
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The number of goals, x, scored by a hockey team in each match of a season is summarised below. x: 0, 1, 2, 3, 4, 5 Frequency: 6, 11, k, 9, 5, 2 The mean number of goals per match is 2.05.
(a)
Find the value of k.
(3)
(b)
Find the median number of goals.
(2)
(c)
Find the interquartile range.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 Σfx = 68 + 2k and Σf = 33 + k M1 (68 + 2k)/(33 + k) = 2.05 A1 k = 7
✓ 3
Part (b)
M1 40 matches, so median is mean of 20th and 21st values; cumulative frequencies 6, 17, 24 A1 median = 2
✓ 2
Part (c)
A1 Q1 = 1 and Q3 = 3 A1 IQR = 2 goals
✓ 2
Common error: The median is a value of x, not a frequency; stating the frequency of the middle row is a frequent mistake.
288IBMA2S-050-05-074.2 Presentation of dataPaper 2Medium7 marks
Calculator
A mobile network surveyed 200 users about their daily data usage, d GB. The cumulative frequencies are: d ≤ 1: 18 d ≤ 2: 52 d ≤ 3: 110 d ≤ 4: 158 d ≤ 5: 186 d ≤ 6: 200
(a)
Find the frequency for each class 0 < d ≤ 1, 1 < d ≤ 2, …, 5 < d ≤ 6.
(2)
(b)
Write down the class containing the median and the classes containing the lower and upper quartiles.
(2)
(c)
Write down the class containing the 90th percentile.
(1)
(d)
Find the percentage of users who use more than 4 GB per day.
A1 median (100th value): 2 < d ≤ 3 A1 Q1: 1 < d ≤ 2 and Q3: 3 < d ≤ 4
✓ 2
Part (c)
A1 180th value lies in 4 < d ≤ 5
✓ 1
Part (d)
M1 200 − 158 = 42 users A1 21%
✓ 2
Exam tip: The pth percentile of n values is located at position p% of n in the cumulative frequencies; find that position first, then the class.
289IBMA2S-050-05-104.2 Presentation of dataPaper 2Medium12 marks
Calculator
Two shot-put athletes each made 11 throws in training. The distances, in metres, are: Athlete P: 17.2, 18.1, 17.8, 18.4, 16.9, 18.0, 17.5, 18.6, 17.9, 14.3, 18.2 Athlete Q: 17.0, 17.6, 18.9, 16.8, 19.3, 17.4, 18.5, 16.5, 19.0, 17.7, 18.2 An outlier is a value more than 1.5 × IQR beyond the nearer quartile.
(a)
For Athlete P, find the median, lower quartile and upper quartile.
(3)
(b)
Show that the throw of 14.3 m is an outlier.
(2)
(c)
For Athlete Q, find the median and the interquartile range.
(3)
(d)
Compare the distributions of the two athletes' throws in two ways.
(2)
(e)
The 14.3 m throw was a foul and is removed. Find the mean distance of Athlete P's remaining throws.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 median = 17.9 m A1 Q1 = 17.2 m A1 Q3 = 18.2 m
✓ 3
Part (b)
M1 lower boundary = 17.2 − 1.5 × 1.0 = 15.7 m R1 14.3 < 15.7, so 14.3 m is an outlier
✓ 2
Part (c)
A1 median = 17.7 m A1 Q1 = 17.0 m and Q3 = 18.9 m A1 IQR = 1.9 m
✓ 3
Part (d)
R1 P has the higher median (17.9 m vs 17.7 m), so P typically throws further R1 P has the smaller IQR (1.0 m vs 1.9 m), so P is more consistent
✓ 2
Part (e)
M1 178.6 ÷ 10 A1 17.9 m (17.86)
✓ 2
Exam tip: A comparison should use a measure of centre and a measure of spread, each with a statement in context about the athletes.
290IBMA2S-050-00-034.12 Standardization and inverse normalPaper 2Medium5 marks
Calculator
An engineering firm produces steel bolts whose diameters, \(D\) mm, are normally distributed with mean 8.00 mm and standard deviation \(\sigma\) mm. It is found that 3% of bolts have a diameter greater than 8.15 mm.
(a)
Find the value of \(\sigma\).
(3)
(b)
Find the probability that a randomly chosen bolt has a diameter less than 7.90 mm.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(z\) for upper 3%: \(z=1.88079\ldots\) (accept invNorm(0.97))
✓ 1
M1 \(\frac{8.15-8.00}{\sigma}=1.88079\ldots\)
✓ 1
A1 \(\sigma=0.0798\) mm (0.079754…)
✓ 1
Part (b)
M1 \(P(D<7.90)\) with \(\sigma=0.07975\ldots\), or \(P(Z<-1.2539\ldots)\)
✓ 1
A1 0.105
✓ 1
Common error: Using invNorm(0.03) for an upper-tail percentage gives a negative \(z\) and hence a negative standard deviation. The area to the left of 8.15 is 0.97.
291IBMA2S-050-00-054.12 Standardization and inverse normalPaper 2Medium7 marks
Calculator
The burn time, \(T\) seconds, of a test rocket engine is normally distributed with mean \(\mu\) and standard deviation \(\sigma\). It is known that 10% of burns last less than 148.0 s and 5% of burns last more than 156.0 s.
(a)
Write down two equations in \(\mu\) and \(\sigma\).
(2)
(b)
Find the value of \(\mu\) and the value of \(\sigma\).
(3)
(c)
Find the probability that a randomly chosen burn lasts between 150 s and 153 s.
(2)
Show mark scheme
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Mark
Notes
Part (a)
A1 \(\frac{148-\mu}{\sigma}=-1.28155\ldots\)
✓ 1
A1 \(\frac{156-\mu}{\sigma}=1.64485\ldots\)
✓ 1
Part (b)
M1 attempt to solve the equations simultaneously, e.g. \(2.92640\sigma=8\)
✓ 1
A1 \(\sigma=2.73\) s (2.73373…)
✓ 1
A1 \(\mu=152\) s (151.503…)
✓ 1
Part (c)
M1 \(P(150<T<153)\) using unrounded \(\mu\) and \(\sigma\)
✓ 1
A1 0.417
✓ 1
Common error: Rounding \(\mu\) and \(\sigma\) before part (c) shifts the answer; keep full GDC values stored for later parts.
292IBMA2S-050-00-074.12 Standardization and inverse normalPaper 2Medium5 marks
Calculator
A café's coffee machine dispenses volumes, \(V\) ml, that are normally distributed with mean \(\mu\) ml and standard deviation 6 ml. The owner wants at least 99% of cups to contain at least 250 ml.
(a)
Find the minimum value of \(\mu\) that achieves this, giving your answer to 3 significant figures.
(3)
(b)
The machine is set with \(\mu=264\). Find the volume that is exceeded by 25% of cups.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(z=-2.32635\ldots\) (invNorm(0.01))
✓ 1
M1 \(\frac{250-\mu}{6}=-2.32635\ldots\)
✓ 1
A1 \(\mu=264\) ml (263.958…)
✓ 1
Part (b)
M1 \(P(V>v)=0.25\), e.g. \(v=264+0.67449\times6\)
✓ 1
A1 \(268\) ml (268.047…)
✓ 1
Exam tip: "At least 99% of cups contain at least 250 ml" means 250 ml must be the lower 1% point, so the area to its left is 0.01.
293IBMA2S-050-06-064.1 Sampling, data and outliersPaper 2Medium10 marks
Calculator
A mobile network records the daily data usage, in GB, of 15 randomly chosen customers. The values, in increasing order, are 1.2, 1.8, 2.1, 2.3, 2.5, 2.6, 2.8, 3.0, 3.1, 3.3, 3.6, 3.9, 4.2, 4.8, 9.5 An outlier is defined as a value more than 1.5 × IQR below the lower quartile or above the upper quartile.
(a)
Write down the median, the lower quartile and the upper quartile.
(3)
(b)
Show that 9.5 is the only outlier.
(3)
(c)
The value 9.5 is found to be a recording fault and is removed. Find the median and the mean of the remaining 14 values.
(3)
(d)
State, with a reason, which of the mean and the median was more affected by removing this value.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 median = 3.0 GB
✓ 1
A1 Q1 = 2.3 GB
✓ 1
A1 Q3 = 3.9 GB
✓ 1
Part (b)
M1 IQR = 1.6 and uses Q3 + 1.5 × 1.6
✓ 1
A1 upper boundary 6.3 and lower boundary −0.1
✓ 1
R1 9.5 > 6.3 and no other value lies outside −0.1 to 6.3, so 9.5 is the only outlier
✓ 1
Part (c)
A1 median = (2.8 + 3.0) ÷ 2 = 2.9 GB
✓ 1
M1 (50.7 − 9.5) ÷ 14
✓ 1
A1 mean = 2.94 GB
✓ 1
Part (d)
R1 the mean (falls from 3.38 to 2.94 while the median only falls from 3.0 to 2.9), since the mean uses the size of every value
✓ 1
Exam tip: For an odd number of values, the quartiles are the medians of the lower and upper halves excluding the overall median, matching the GDC method.
294IBMA2S-050-06-104.8 The binomial distributionPaper 2Medium7 marks
Calculator
A tennis analyst records the number of aces served by a player in each of a set of matches. 0 aces: 4 matches 1 ace: 7 matches 2 aces: k matches 3 aces: 9 matches 4 aces: 5 matches 5 aces: 3 matches The mean number of aces per match is 2.25.
(a)
Find the value of k.
(3)
(b)
Find the median number of aces.
(2)
(c)
Find the standard deviation of the number of aces per match.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 Σfx = 69 + 2k and Σf = 28 + k
✓ 1
M1 \(\frac{69 + 2k}{28 + k} = 2.25\)
✓ 1
A1 k = 24
✓ 1
Part (b)
M1 52 matches, median is mean of 26th and 27th values; cumulative frequencies 4, 11, 35
Exam tip: Enter the frequencies as a second list on the GDC; leaving them out treats each value as occurring once and gives a completely wrong standard deviation.
295IBMA2S-050-06-124.3 Measures of central tendency and dispersionPaper 2Medium10 marks
Calculator
In a clinical trial, the systolic blood pressure (mmHg) of 10 patients is measured before and after a course of medication. Before: 142, 150, 138, 161, 147, 155, 144, 158, 149, 152 After: 135, 146, 136, 150, 140, 151, 141, 149, 145, 147 For each patient the reduction d = before − after is calculated.
(a)
Find the mean reduction.
(2)
(b)
Treating the 10 reductions as the whole data set, find the (population) standard deviation σ of the reductions.
(2)
(c)
Write down the median and the mode of the reductions.
(2)
(d)
It is later found that the monitor used for every 'after' reading gave values 2 mmHg too high. Write down the corrected mean and standard deviation of the reductions, giving a reason for your answer for the standard deviation.
(2)
(e)
Using the original (uncorrected) reductions, an 11th patient joins the trial and the mean reduction of all 11 patients becomes 6 mmHg. Find this patient's reduction.
R1 standard deviation stays 2.69 mmHg, since every reduction increases by the same constant 2
✓ 1
Part (e)
M1 \(\frac{56 + d}{11} = 6\)
✓ 1
A1 d = 10 mmHg
✓ 1
Common error: If the 'after' readings were too high, the true reductions are larger, not smaller: lowering each 'after' value increases before − after.
296IBMA2S-050-03-034.4 Correlation and regression of y on xPaper 2Medium10 marks
Calculator
In a clinical trial, six groups of patients are given different daily doses, x mg, of a new blood-pressure drug. The mean reduction in systolic blood pressure, y mmHg, is recorded for each group.
x: 10, 20, 30, 40, 50, 60 y: 4, 9, 11, 17, 20, 23
(a)
Find the value of Pearson's product-moment correlation coefficient, r, and interpret it in context.
(3)
(b)
Find the equation of the regression line of y on x.
(2)
(c)
Interpret the gradient of the regression line in context.
(1)
(d)
Estimate the mean reduction in systolic blood pressure for a daily dose of 45 mg.
(2)
(e)
A doctor suggests using the regression line to predict the effect of a daily dose of 150 mg. Comment on this suggestion.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 evidence of using a GDC for linear regression
✓ 1
A1 \(r = 0.993\) (0.993273…)
✓ 1
A1 very strong positive linear correlation: larger doses are associated with larger reductions in blood pressure
✓ 1
Part (b)
A1 gradient 0.383 (0.382857…)
✓ 1
A1 intercept 0.6, giving \(y = 0.383x + 0.6\)
✓ 1
Part (c)
A1 each extra 1 mg of daily dose is associated with an extra reduction of about 0.383 mmHg
✓ 1
Part (d)
M1 substitutes \(x = 45\) into the regression line
✓ 1
A1 17.8 mmHg (17.8285…)
✓ 1
Part (e)
R1 150 mg lies well outside the range of doses in the data (10 mg to 60 mg), so this is extrapolation
✓ 1
R1 the linear relationship may not continue at this dose (and the dose may not be safe), so the prediction is unreliable
✓ 1
Exam tip: When interpreting a gradient, give both the size and the units in context: 'for each extra 1 mg, y increases by about 0.383 mmHg'. A purely algebraic statement does not earn the mark.
297IBMA2S-050-03-074.4 Correlation and regression of y on xPaper 2Medium7 marks
Calculator
A weather balloon rises through the lower atmosphere. Its altitude, x km, and the air temperature, y °C, are recorded.
x: 0, 2, 4, 6, 8, 10 y: 15, 4, −12, −22, −37, −48
(a)
Find the value of Pearson's product-moment correlation coefficient, r.
(2)
(b)
Find the equation of the regression line of y on x.
(2)
(c)
Interpret the gradient of the regression line in context.
(1)
(d)
Estimate the air temperature at an altitude of 7.5 km.
A1 the air temperature falls by about 6.4 °C for each 1 km increase in altitude
✓ 1
Part (d)
M1 substitutes \(x = 7.5\) into the regression line
✓ 1
A1 −32.7 °C (−32.666…)
✓ 1
Common error: Take care with negative values when entering data into a GDC. A sign slip in one temperature changes both the gradient and r.
298IBMA2S-050-03-114.4 Correlation and regression of y on xPaper 2Medium7 marks
Calculator
A football is kicked from ground level. A tracking system records the height of the ball, h metres, at time t seconds after the kick.
t: 0, 1, 2, 3, 4 h: 0, 15, 20, 15, 0
(a)
Find the value of Pearson's product-moment correlation coefficient, r, between t and h.
(2)
(b)
Write down the equation of the regression line of h on t, in the form \(h = at + b\).
(2)
(c)
A student concludes that there is no relationship between time and height. Explain why this conclusion is incorrect.
(2)
(d)
Explain why the regression line should not be used to estimate the height of the ball at \(t = 1.5\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 valid method, e.g. GDC linear regression, or showing \(\sum (t-\bar{t})(h-\bar{h}) = 20-5+0+5-20 = 0\)
✓ 1
A1 \(r = 0\)
✓ 1
Part (b)
A1 \(a = 0\)
✓ 1
A1 \(b = 10\), so \(h = 10\)
✓ 1
Part (c)
R1 r measures only the strength of linear correlation
✓ 1
R1 the data show a clear non-linear (quadratic) relationship: the height rises and then falls symmetrically, e.g. \(h = 5t(4-t)\) fits exactly
✓ 1
Part (d)
R1 the line gives 10 m at every time and does not fit the curved pattern; the data suggest a height well above 10 m at \(t = 1.5\) (between 15 m and 20 m)
✓ 1
Common error: A value of r close to 0 means there is no linear correlation, not that there is no relationship. Always think about the shape of the data before relying on r.
299IBMA2S-050-09-054.1 Sampling, data and outliersPaper 2Medium10 marks
Calculator
The resting heart rates, in beats per minute, of 10 patients in a clinical trial are listed below. 58, 62, 64, 65, 67, 68, 70, 71, 73, 104
(a)
Find the mean heart rate.
(2)
(b)
Write down the lower quartile and the upper quartile.
(2)
(c)
Show that 104 is an outlier.
(3)
(d)
The reading of 104 is found to be caused by a faulty sensor and is removed. For the remaining 9 patients, find (i) the mean heart rate; (ii) the standard deviation of the heart rates.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sum = 702, divided by 10
✓ 1
A1 70.2 bpm
✓ 1
Part (b)
A1 Q1 = 64
✓ 1
A1 Q3 = 71
✓ 1
Part (c)
M1 IQR = 71 − 64 = 7
✓ 1
M1 upper fence = 71 + 1.5 × 7 = 81.5
✓ 1
R1 104 > 81.5, so 104 is an outlier
✓ 1
Part (d)
A1 (i) 598 ÷ 9 = 66.4 bpm
✓ 1
M1 (ii) attempt to find the standard deviation of the 9 remaining values, e.g. using a GDC
✓ 1
A1 σ = 4.45 bpm
✓ 1
Exam tip: An outlier should only be removed if there is a reason, such as a measurement error. If it is a genuine reading, it stays in the data and you comment on its effect.
300IBMA2S-050-09-064.3 Measures of central tendency and dispersionPaper 2Medium7 marks
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A javelin coach records the best throw, in metres, of 8 athletes in a training session: 62.4, 65.1, 58.7, x, 66.3, 63.8, 61.2, 64.5 The mean of the 8 throws is 62.0 m.
(a)
Find the value of x.
(2)
(b)
Find the lower quartile and the upper quartile of the 8 throws.
(2)
(c)
The coach thinks the throw of x metres is an outlier. Determine whether the coach is correct.
R1 54.0 > 52.7, so it is not an outlier and the coach is not correct
✓ 1
Common error: A value that looks far from the rest is not automatically an outlier. Test it against the 1.5 × IQR fence before deciding.
301IBMA2S-050-09-104.2 Presentation of dataPaper 2Medium11 marks
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An amateur astronomer counts the number of visible satellite passes, x, on each of 40 clear nights. The results are: x: 1, 2, 3, 4, 5, 12 number of nights: 4, 9, 12, 8, 6, 1
(a)
State whether x is discrete or continuous.
(1)
(b)
Find the median, the lower quartile and the upper quartile of x.
(3)
(c)
Show that the night with 12 passes is an outlier.
(2)
(d)
Find the mean number of passes per night.
(2)
(e)
The night with 12 passes coincided with a cluster of new satellite launches and is excluded. Find the new mean.
(2)
(f)
State which of the mean and the median is unaffected by excluding this night.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 discrete
✓ 1
Part (b)
A1 median = 3
✓ 1
A1 Q1 = 2
✓ 1
A1 Q3 = 4
✓ 1
Part (c)
M1 upper fence = 4 + 1.5 × 2 = 7
✓ 1
R1 12 > 7, so it is an outlier
✓ 1
Part (d)
M1 Σfx = 4 + 18 + 36 + 32 + 30 + 12 = 132
✓ 1
A1 mean = 132 ÷ 40 = 3.3
✓ 1
Part (e)
M1 (132 − 12) ÷ 39
✓ 1
A1 3.08
✓ 1
Part (f)
R1 the median, which remains 3
✓ 1
Exam tip: In a frequency table, use the running total to find which value holds each position. For example, the 10th and 11th values both fall in the block of 2s, which covers positions 5 to 13.
302IBMA2S-050-09-124.1 Sampling, data and outliersPaper 2Medium10 marks
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Two machines produce bolts with a target length of 50.0 mm. For machine A, the lengths of a sample of bolts have minimum 48.2 mm, lower quartile 49.6 mm, median 50.0 mm, upper quartile 50.4 mm and maximum 52.1 mm. In this question, an outlier is a value that is more than 1.5 × IQR below the lower quartile or more than 1.5 × IQR above the upper quartile.
Diagram not to scale
(a)
Find the interquartile range for machine A.
(1)
(b)
Find the boundaries beyond which a length from machine A is an outlier.
(2)
(c)
Determine whether the minimum and the maximum lengths for machine A are outliers.
(2)
(d)
For machine B, the lower quartile is 49.1 mm and the maximum is 52.0 mm, which is not an outlier. Find the least possible value of the upper quartile for machine B.
(3)
(e)
The bolts were sampled by taking the bolt produced on the hour, every hour, during a shift. Name this sampling technique and give one reason why it is suitable here.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 50.4 − 49.6 = 0.8 mm
✓ 1
Part (b)
M1 49.6 − 1.5 × 0.8 or 50.4 + 1.5 × 0.8
✓ 1
A1 48.4 mm and 51.6 mm
✓ 1
Part (c)
R1 48.2 < 48.4, so the minimum is an outlier
✓ 1
R1 52.1 > 51.6, so the maximum is an outlier
✓ 1
Part (d)
M1 Q3 + 1.5(Q3 − 49.1) ≥ 52.0
✓ 1
M1 2.5Q3 ≥ 125.65
✓ 1
A1 least Q3 = 50.26 mm
✓ 1
Part (e)
A1 systematic sampling
✓ 1
R1 any one valid reason, for example: it spreads the sample evenly over the whole shift, so drift in the machine over time would be detected; or it is quick and simple to carry out on a production line
✓ 1
Exam tip: When an unknown quartile appears in the outlier condition, remember that the IQR also contains it. Write the inequality in terms of Q₃ before you solve it.
303IBMA2S-050-08-044.10 Regression of x on yPaper 2Medium6 marks
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An engineer plans to use a steel wire as a simple force sensor. Known loads x newtons are hung from the wire and the extension y mm is measured. Load x (N): 50, 100, 150, 200, 250 Extension y (mm): 1.2, 2.5, 3.5, 5.0, 6.1 In use, the extension will be measured and the load will be estimated from it.
(a)
Explain why the regression line of x on y is the appropriate line for this purpose.
(1)
(b)
Find the equation of the regression line of x on y, in the form \(x = ay + b\).
(2)
(c)
The extension under an unknown load is 4.2 mm. Estimate the load.
(2)
(d)
Comment on the reliability of using this line to estimate the load when the extension is 12 mm.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 the load (x) is the quantity to be estimated from a known extension (y), so x must be regressed on y
✓ 1
Part (b)
A1 \(a = 40.5\) (40.535…)
✓ 1
A1 \(b = 1.64\) (1.6412…)
✓ 1
Part (c)
M1 substitutes \(y = 4.2\) into the x on y line
✓ 1
A1 172 N (171.88…)
✓ 1
Part (d)
R1 unreliable: 12 mm is far outside the data range (1.2–6.1 mm), so this is extrapolation and the wire may no longer behave linearly
✓ 1
Exam tip: Which variable was controlled in the experiment does not decide the line; the line is chosen by which quantity you are trying to estimate.
304IBMA2S-050-08-084.10 Regression of x on yPaper 2Medium12 marks
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A long-jump coach records the approach speed x m s−1 and the jump distance y m for eight athletes. Speed x (m s−1): 8.2, 8.6, 9.0, 9.3, 9.5, 9.8, 10.1, 10.4 Distance y (m): 6.10, 6.45, 6.70, 6.95, 7.20, 7.35, 7.60, 7.90
(a)
Find the value of the correlation coefficient r and interpret its value.
(2)
(b)
Find the equation of the regression line of x on y.
(2)
(c)
An athlete is aiming for a jump of 7.50 m. Estimate the approach speed needed.
(2)
(d)
Find the equation of the regression line of y on x and use it to estimate the jump distance of an athlete with approach speed 9.7 m s−1.
(3)
(e)
Use your regression line of x on y to estimate the approach speed needed for a jump of 8.50 m, and comment on the reliability of your estimate.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(r = 0.998\) (0.99805…)
✓ 1
R1 very strong positive linear correlation between speed and distance
✓ 1
Part (b)
A1 gradient 1.24 (1.2390…)
✓ 1
A1 intercept 0.651 (0.6505…), i.e. \(x = 1.24y + 0.651\)
✓ 1
Part (c)
M1 substitutes \(y = 7.50\) into the x on y line
✓ 1
A1 9.94 m s−1 (9.9433…; accept 9.95 from 3 s.f. coefficients)
✓ 1
Part (d)
A1 \(y = 0.804x - 0.496\)
✓ 1
M1 substitutes \(x = 9.7\)
✓ 1
A1 7.30 m (7.3025…)
✓ 1
Part (e)
M1 substitutes \(y = 8.50\) into the x on y line
✓ 1
A1 11.2 m s−1 (11.18…)
✓ 1
R1 unreliable: 8.50 m lies outside the data range (6.10–7.90 m), so this is extrapolation
✓ 1
Exam tip: Store the regression coefficients in your GDC rather than retyping rounded values; this avoids small rounding differences in the final answer.
305IBMA2S-050-08-104.10 Regression of x on yPaper 2Medium10 marks
Calculator
A rower uses an ergometer to find the stroke rate needed to produce a target power. The stroke rate x strokes per minute and the power output y watts are recorded for six trials. Stroke rate x: 20, 22, 24, 26, 28, 30 Power y (W): 180, 205, 230, 250, 400, 300 One of the power readings is later found to be caused by a sensor fault.
(a)
Using all six trials, find the equation of the regression line of x on y and the value of the correlation coefficient r.
(3)
(b)
Identify the faulty reading, giving a reason.
(1)
(c)
Remove the faulty reading. Find the new equation of the regression line of x on y and the new value of r.
(3)
(d)
Estimate the stroke rate needed to produce 260 W, justifying your choice of line.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 gradient 0.0381 (0.038107…)
✓ 1
A1 intercept 15.1 (15.060…), i.e. \(x = 0.0381y + 15.1\)
✓ 1
A1 \(r = 0.810\)
✓ 1
Part (b)
R1 (28, 400): the power of 400 W is far above the trend of the other readings, which suggests roughly 275 W at 28 strokes per minute
✓ 1
Part (c)
A1 gradient 0.0840 (0.084009…)
✓ 1
A1 intercept 4.83 (4.8258…), i.e. \(x = 0.0840y + 4.83\)
✓ 1
A1 \(r = 0.9995\) (accept 1.00)
✓ 1
Part (d)
M1 substitutes \(y = 260\) into the regression line of x on y found with the faulty reading removed
✓ 1
A1 26.7 strokes per minute (26.668…)
✓ 1
R1 the line without the faulty reading is used, because that reading distorts the six-point line and r is much closer to 1 without it
✓ 1
Common error: Remember to delete both the x and y values of a removed point from your GDC lists; deleting only one shifts every later pairing.
306IBMA3S-4-16r5-044.8 The binomial distributionPaper 2Medium6 marks
Calculator
A farm uses seeds from two batches. 60% of the seeds come from batch A and 40% come from batch B. A seed from batch A germinates with probability 0.92 and a seed from batch B germinates with probability 0.85. A seed is chosen at random.
Tree diagram
(a)
Find the probability that the seed germinates.
(3)
(b)
Given that the seed does not germinate, find the probability that it came from batch B.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use the two routes (tree diagram or table)
✓ 1
A1 0.6 × 0.92 + 0.4 × 0.85
✓ 1
A1 0.892
✓ 1
Part (b)
M1 use of P(B | not germinate) = P(B ∩ not germinate)/P(not germinate)
✓ 1
A1 0.4 × 0.15/(1 − 0.892) (ECF from (a))
✓ 1
A1 0.556
✓ 1
Common error: In a conditional probability, the denominator is the probability of the condition (here, not germinating), not the probability of the batch.
307IBMA3S-4-16r5-074.2 Presentation of dataPaper 2Medium7 marks
Calculator
A farmer records the masses of 100 apples: Mass (g): 100 ≤ m < 120: 6; 120 ≤ m < 140: 14; 140 ≤ m < 160: 30; 160 ≤ m < 180: 28; 180 ≤ m < 200: 15; 200 ≤ m < 220: 7.
Histogram of the data
(a)
Write down the cumulative frequency for each class.
(2)
(b)
Estimate the mean mass of an apple.
(3)
(c)
Estimate the number of apples with mass less than 150 g, assuming the masses in each class are evenly spread.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 6, 20, 50
✓ 1
A1 78, 93, 100
✓ 1
Part (b)
M1 use of mid-interval values 110, 130, 150, 170, 190, 210
M1 20 apples below 140 g plus half of the 140 ≤ m < 160 class
✓ 1
A1 20 + 15 = 35 apples
✓ 1
Exam tip: Use the class midpoints for the mean, but use the class boundaries and the 'evenly spread' assumption when estimating how many values lie below a point inside a class.
308IBMA3S-4-16r5-084.6 Combined events and diagramsPaper 2Medium6 marks
Calculator
An aircraft has two independent hydraulic systems. During a flight, system A fails with probability 0.02 and system B fails with probability 0.05.
(a)
Find the probability that at least one system fails.
(3)
(b)
Given that at least one system fails, find the probability that both fail.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(neither fails) = 0.98 × 0.95
✓ 1
A1 0.931
✓ 1
A1 P(at least one fails) = 0.069
✓ 1
Part (b)
M1 use of P(both | at least one) = P(both)/P(at least one)
✓ 1
A1 P(both) = 0.02 × 0.05 = 0.001
✓ 1
A1 0.0145
✓ 1
Common error: 'At least one' is easiest through the complement, 1 − P(none). Do not add 0.02 and 0.05, which counts the case where both fail twice.
309IBMA3S-4-16r5-104.2 Presentation of dataPaper 2Medium7 marks
Calculator
The wheat yields, in tonnes per hectare, of plots on two farms are summarised below. Farm A: minimum 12, Q1 18, median 22, Q3 27, maximum 40. Farm B: minimum 15, Q1 19, median 21, Q3 24, maximum 29.
Box-and-whisker diagrams
(a)
Find the interquartile range for each farm.
(2)
(b)
A value is an outlier if it is more than 1.5 × IQR above Q3. Show that the maximum yield on Farm A is not an outlier.
(3)
(c)
A farmer says that the yields on Farm A are more consistent than on Farm B. Comment on this statement.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 Farm A: IQR = 9
✓ 1
A1 Farm B: IQR = 5
✓ 1
Part (b)
M1 27 + 1.5 × 9 (ECF from (a))
✓ 1
A1 upper boundary = 40.5
✓ 1
R1 40 < 40.5 so it is not an outlier
✓ 1
Part (c)
R1 the IQR of Farm A (9) is larger than that of Farm B (5), so Farm A is more spread out
✓ 1
R1 the statement is not supported: Farm B is more consistent
✓ 1
Exam tip: Check an extreme value against the boundary by writing the calculation down. A value just below the boundary is not an outlier.
310IBMA3S-4-16r5-114.11 Conditional probability and independencePaper 2Medium7 marks
Calculator
200 listeners were asked which genre they prefer. The results are in the table. Under 25: rock 38, pop 52, jazz 10 (100 listeners). 25 and over: rock 45, pop 30, jazz 25 (100 listeners). One listener is chosen at random.
(a)
Find the probability that the listener prefers rock.
(1)
(b)
Given that the listener does not prefer rock, find the probability that the listener is under 25.
(3)
(c)
Let J be the event that the listener prefers jazz and O the event that the listener is 25 or over. Determine whether J and O are independent.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 83/200 = 0.415
✓ 1
Part (b)
M1 use of conditional probability with 200 − 83 = 117 listeners not preferring rock
✓ 1
A1 52 + 10 = 62 listeners under 25 who do not prefer rock
✓ 1
A1 62/117 = 0.530
✓ 1
Part (c)
M1 P(J) = 35/200 = 0.175 and P(O) = 0.5, so the product = 0.0875
✓ 1
A1 P(J ∩ O) = 25/200 = 0.125
✓ 1
R1 0.125 ≠ 0.0875 so not independent
✓ 1
Common error: In a two-way table, a condition such as 'does not prefer rock' changes the total you divide by. Do not divide by 200.
311IBMA3S-4-16r6-054.10 Regression of x on yPaper 2Medium7 marks
Calculator
A coach records the weekly training time, x hours, and the time y minutes taken to ride a 20 km time trial for eight club cyclists.
x
3
5
6
8
9
11
12
14
y
58.4
55.1
54.0
50.2
49.5
46.3
45.8
42.1
Scatter diagram of the data
(a)
Write down the Pearson correlation coefficient, r.
(1)
(b)
Find the equation of the regression line of x on y. Give the coefficients correct to 4 significant figures.
(3)
(c)
A cyclist wants a time of 48 minutes for the time trial. Use your equation from part (b) to estimate the weekly training time needed.
(2)
(d)
Comment on the reliability of the estimate in part (c).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 r = −0.997
✓ 1
Part (b)
M1 attempt to find regression of x on y (not y on x)
✓ 1
A1 gradient −0.6872
✓ 1
A1 x = 42.98 − 0.6872y
✓ 1
Part (c)
M1 substitute y = 48 into the equation
✓ 1
A1 9.99 hours
✓ 1
Part (d)
R1 reliable: 48 lies within the range of the data (interpolation) and |r| is close to 1
✓ 1
Exam tip: When the unknown is x and y is given, use the regression line of x on y; rearranging the y on x line gives a different, less suitable estimate.
312IBMA3S-4-16r6-064.9 The normal distributionPaper 2Medium5 marks
Calculator
The lengths, L cm, of thigh bones recovered from an ancient cemetery are normally distributed. 10% of the bones are shorter than 41.2 cm and 20% are longer than 47.5 cm. Find the mean and the standard deviation of L.
Normal distribution curve, not to scale
(a)
Find the mean and the standard deviation of L. Give your answers correct to 3 significant figures.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use inverse normal for either probability: z = −1.2816 for 0.10 or z = 0.8416 for 0.80
✓ 1
M1 attempt to form two equations: 41.2 = μ − 1.2816σ and 47.5 = μ + 0.8416σ
✓ 1
M1 attempt to solve simultaneously, e.g. 2.1232σ = 6.3
✓ 1
A1 σ = 2.97 cm
✓ 1
A1 μ = 45.0 cm
✓ 1
Exam tip: Standardise each given probability separately, giving two linear equations in μ and σ, then solve them together.
313IBMA3S-4-16r6-074.3 Measures of central tendency and dispersionPaper 2Medium7 marks
Calculator
Each server in a data centre fails during a given month independently with probability 0.04. The data centre has 50 servers. Let X be the number of servers that fail in the month.
(a)
Find P(X = 3).
(2)
(b)
The data centre could instead install n servers. Find the smallest value of n for which the probability that at least one server fails is greater than 0.95.
(3)
(c)
For the original 50 servers, find the probability that at most one server fails.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use X ~ B(50, 0.04)
✓ 1
A1 0.184
✓ 1
Part (b)
M1 1 − 0.96n > 0.95, i.e. 0.96n < 0.05
✓ 1
M1 attempt to solve, e.g. n > ln 0.05 ÷ ln 0.96 = 73.4
✓ 1
A1 n = 74
✓ 1
Part (c)
M1 attempt to find P(X = 0) + P(X = 1)
✓ 1
A1 0.400
✓ 1
Common error: After solving an inequality involving n, round up to the next whole number that keeps the inequality true.
314IBMA3S-4-16r6-084.2 Presentation of dataPaper 2Medium5 marks
Calculator
An archaeologist measures the lengths, in centimetres, of 54 fragments of iron pin from a settlement. The results are grouped as shown.
Length, l (cm)
20 ≤ l < 25
25 ≤ l < 30
30 ≤ l < 35
35 ≤ l < 40
40 ≤ l < 45
45 ≤ l < 50
Frequency
4
11
18
12
6
3
(a)
Write down the modal class.
(1)
(b)
Estimate the mean length of a fragment.
(2)
(c)
Estimate the standard deviation of the lengths.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 30 ≤ l < 35
✓ 1
Part (b)
M1 attempt to use mid-interval values 22.5, 27.5, …
✓ 1
A1 33.8 cm
✓ 1
Part (c)
M1 attempt to use mid-interval values and frequencies in the standard deviation calculation
✓ 1
A1 6.32 cm
✓ 1
Exam tip: For grouped data, use the mid-interval value of each class; your answers are estimates, so say so.
315IBMA3S-4-16r6-094.4 Correlation and regression of y on xPaper 2Medium7 marks
Calculator
A data centre engineer compares the average processor load, x (%), of a rack with its temperature, y (°C). Over a sample of readings: x̄ = 62, ȳ = 31.5, standard deviation of x = 9.0, standard deviation of y = 3.2, and the covariance of x and y is 24.2.
(a)
Calculate the correlation coefficient, r.
(2)
(b)
Find the equation of the regression line of x on y. Give the coefficients correct to 4 significant figures.
(3)
(c)
A rack has temperature 34 °C. Use your equation from part (b) to estimate its processor load.
M1 gradient = \(\frac{s_{xy}}{s_y^{2}}=\frac{24.2}{3.2^2}\) and line passes through (ȳ, x̄)
✓ 1
A1 gradient = 2.363
✓ 1
A1 x = 2.363y − 12.44
✓ 1
Part (c)
M1 substitute y = 34
✓ 1
A1 67.9 %
✓ 1
Common error: The gradient of the x on y line uses the variance of y in the denominator, not the variance of x.
316IBMA3S-4-16r6-104.9 The normal distributionPaper 2Medium7 marks
Calculator
In a cycling time trial the finishing times are normally distributed with mean 3.20 hours and standard deviation 0.25 hours.
Normal distribution curve, not to scale
(a)
Find the probability that a randomly chosen cyclist finishes in under 3.0 hours.
(2)
(b)
The fastest 15% of cyclists qualify for the final. Find the slowest qualifying time, in hours, correct to 3 significant figures.
(2)
(c)
Eight cyclists are chosen at random. Using the unrounded probability from part (a), find the probability that exactly two of them finish in under 3.0 hours.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to find P(T < 3.0) with T ~ N(3.20, 0.252)
✓ 1
A1 0.212
✓ 1
Part (b)
M1 attempt to use inverse normal with probability 0.15
✓ 1
A1 2.94 hours
✓ 1
Part (c)
M1 X ~ B(8, p) where p = 0.2119 (unrounded value from (a))
✓ 1
M1 attempt to find P(X = 2)
✓ 1
A1 0.301
✓ 1
Exam tip: Keep unrounded answers in your calculator memory when a later part builds on an earlier probability.
317IBMA3S-4-16r6-134.3 Measures of central tendency and dispersionPaper 2Medium10 marks
Calculator
A data centre records the response time, in milliseconds (ms), of ten requests: 212, 198, 245, 230, 205, 219, 261, 208, 226, 215.
(a)
Find the mean and the standard deviation of the response times.
(2)
(b)
Find the median and the lower and upper quartiles.
(3)
(c)
A value is an outlier if it is more than 1.5 × IQR below Q1 or above Q3. Determine whether any of the response times is an outlier.
(3)
(d)
After an upgrade, every response time is multiplied by 0.80. Find the new mean and the new standard deviation, correct to 3 significant figures.
R1 all values lie between 175 and 263 (largest is 261), so there is no outlier
✓ 1
Part (d)
A1 new mean = 178 ms (177.52)
✓ 1
A1 new standard deviation = 14.6 ms (14.64)
✓ 1
Exam tip: Multiplying every value by a constant multiplies both the mean and the standard deviation by that constant; adding a constant changes only the mean.
318IBMA3S-4-16r8-064.8 The binomial distributionPaper 2Medium7 marks
Calculator
A mountaineering club inspects 12 carabiners. Each carabiner independently fails the inspection with probability 0.08. Let X be the number of carabiners that fail.
(a)
Find E(X) and Var(X).
(2)
(b)
Find P(X = 2). Give your answer to 3 significant figures.
(2)
(c)
Given that at least one carabiner fails, find the probability that at least two fail. Give your answer to 3 significant figures.
Common error: In a conditional probability with 'given at least one', the denominator is P(X ≥ 1), not 1.
319IBMA3S-4-16r8-074.12 Standardization and inverse normalPaper 2Medium6 marks
Calculator
The sugar content, S grams, of a cereal bar is normally distributed with mean μ and standard deviation σ. It is known that P(S < 12) = 0.1 and P(S > 18) = 0.2.
Normal distribution curve, not to scale
(a)
Find the z-values and write down two equations connecting μ and σ.
(3)
(b)
Find σ, correct to 3 significant figures.
(2)
(c)
Hence find μ, correct to 3 significant figures.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use the inverse normal for P(Z < z) = 0.1 and P(Z < z) = 0.8
✓ 1
A1 z = −1.2816 and z = 0.8416
✓ 1
A1 (12 − μ)/σ = −1.2816 and (18 − μ)/σ = 0.8416
✓ 1
Part (b)
M1 subtract the equations to get 6 = (0.8416 + 1.2816)σ
✓ 1
A1 σ = 2.83
✓ 1
Part (c)
A1 μ = 12 + 1.2816 × 6 / (1.2816 + 0.8416) = 15.6
✓ 1
Exam tip: Keep the unrounded value of σ on your calculator when finding μ, so the final answer is accurate.
320IBMA3S-4-16r8-084.2 Presentation of dataPaper 2Medium5 marks
Calculator
The durations of 100 calls to a telecoms helpline are summarised in the table. Duration t (minutes): 0 < t ≤ 2: 18 calls; 2 < t ≤ 4: 35 calls; 4 < t ≤ 6: 27 calls; 6 < t ≤ 10: 14 calls; 10 < t ≤ 20: 6 calls.
(a)
Write down the modal class.
(1)
(b)
Use the midpoints of the classes to estimate the mean duration.
(2)
(c)
Estimate the population standard deviation of the durations. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 2 < t ≤ 4
✓ 1
Part (b)
M1 midpoints 1, 3, 5, 8, 15 used with the frequencies
✓ 1
A1 mean = 460 / 100 = 4.6 minutes
✓ 1
Part (c)
M1 attempt to use the midpoints and frequencies in the standard deviation formula (calculator)
✓ 1
A1 3.37 minutes
✓ 1
Common error: With grouped data the mean and standard deviation are only estimates, because every value is replaced by its class midpoint.
321IBMA3S-4-16r8-094.2 Presentation of dataPaper 2Medium7 marks
Calculator
A telecoms company has 200 customers on a survey list. The table shows each customer's plan and whether they used roaming last month. Contract plan: 45 used roaming, 75 did not. Prepaid plan: 15 used roaming, 65 did not. Let C be the event that a customer is on a contract plan and R the event that the customer used roaming.
(a)
A customer who used roaming is chosen at random. Find the probability that the customer is on a contract plan.
(2)
(b)
Show that C and R are not independent.
(3)
(c)
Two customers are chosen at random without replacement from the 200. Find the probability that exactly one of them used roaming. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(C | R) = 45 / 60
✓ 1
A1 0.75
✓ 1
Part (b)
M1 P(R) = 60/200 = 0.3
✓ 1
M1 P(R | C) = 45/120 = 0.375
✓ 1
R1 P(R | C) ≠ P(R), so C and R are not independent
✓ 1
Part (c)
M1 2 × (60/200) × (140/199)
✓ 1
A1 0.422
✓ 1
Common error: When choosing without replacement, the second probability uses the reduced total (199), not 200.
322IBMA3S-4-16r8-104.8 The binomial distributionPaper 2Medium6 marks
Calculator
In a food factory, each packaged tray of ready meals is independently contaminated with probability 0.03.
(a)
Find the smallest number of trays, n, that must be inspected so that the probability of finding at least one contaminated tray is greater than 0.9.
(4)
(b)
A batch of 20 trays is inspected. Find the probability that at least one is contaminated. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 1 − 0.97n > 0.9
✓ 1
M1 0.97n < 0.1, so n > ln 0.1 / ln 0.97
✓ 1
A1 n > 75.6
✓ 1
A1 n = 76
✓ 1
Part (b)
M1 1 − 0.9720
✓ 1
A1 0.456
✓ 1
Exam tip: When solving for the exponent, take logarithms and remember that dividing by ln 0.97 (a negative number) reverses the inequality.
323IBMA3S-4-16r8-114.3 Measures of central tendency and dispersionPaper 2Medium10 marks
Calculator
A mountaineer records the distance, in km, covered on each of eight days: 12, 14, 15, 15, 16, 18, 19, k, where k > 19. For this question, the lower quartile is the median of the lower four values and the upper quartile is the median of the upper four values. A value is an outlier if it is more than 1.5 × IQR above the upper quartile.
(a)
Find the lower quartile, the upper quartile and the interquartile range.
(3)
(b)
Given that k is an outlier and is an integer, find the smallest possible value of k.
(3)
(c)
Find the mean distance when k = 25.
(2)
(d)
The outlier is removed from the data. Find the decrease in the mean, correct to 3 significant figures.
Exam tip: Check that the quartiles do not depend on k: because k is the largest value, it only changes the upper half's last entry, not its median.
324IBMA3S-4-16r9-044.2 Presentation of dataPaper 2Medium7 marks
Calculator
A fish farm measures the lengths l cm of 60 fish. 20 ≤ l < 30: 6 fish 30 ≤ l < 40: 14 fish 40 ≤ l < 50: 22 fish 50 ≤ l < 60: 12 fish 60 ≤ l < 70: 6 fish
Histogram of the data
(a)
Estimate the mean length, giving your answer to 3 significant figures.
(2)
(b)
Estimate the standard deviation, giving your answer to 3 significant figures.
(2)
(c)
Use the cumulative frequencies to estimate the median length, giving your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of mid-interval values 25, 35, 45, 55, 65 with the frequencies
✓ 1
A1 44.7 cm
✓ 1
Part (b)
M1 use of technology with mid-interval values and frequencies
✓ 1
A1 11.1 cm
✓ 1
Part (c)
A1 cumulative frequencies 6, 20, 42, 54, 60
✓ 1
M1 30th value lies in the class 40 ≤ l < 50, with 20 fish below it
✓ 1
A1 40 + (10/22) × 10 = 44.5 cm
✓ 1
Common error: For grouped data every value is replaced by its mid-interval value, so the results are only estimates.
325IBMA3S-4-16r9-064.3 Measures of central tendency and dispersionPaper 2Medium7 marks
Calculator
A fashion label sells 50 different shirts. The prices p in euros have mean €18.40 and standard deviation €2.20. Treat the 50 shirts as the whole population. The label changes every price to a new price 1.25p + 3.
(a)
Find the new mean price.
(2)
(b)
Find the new standard deviation.
(2)
(c)
A 51st shirt is added to the range with a new price of €26.00. Find the mean and the standard deviation of the 51 new prices, giving the standard deviation to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 1.25 × 18.40 + 3
✓ 1
A1 €26.00
✓ 1
Part (b)
M1 1.25 × 2.20 (the +3 has no effect)
✓ 1
A1 €2.75
✓ 1
Part (c)
A1 the mean remains €26.00
✓ 1
M1 sum of squared deviations = 50 × 2.752 = 378.125 and the new shirt adds 0 to this
✓ 1
A1 sd = sqrt(378.125 / 51) = €2.72
✓ 1
Common error: Adding a constant changes the mean but not the standard deviation; multiplying changes both.
326IBMA3S-4-16r9-084.4 Correlation and regression of y on xPaper 2Medium7 marks
Calculator
The water temperature x °C and the mass y kg of a day's catch are recorded on a number of days. The mean temperature is 14.5 °C and the mean catch is 50.2 kg. The regression line of y on x is y = a + bx, where b = −2.6. The correlation coefficient is r = −0.91. All temperatures recorded were between 10 °C and 19 °C.
(a)
Find the value of a.
(2)
(b)
Estimate the mass of the catch when the water temperature is 12 °C.
(2)
(c)
Interpret the value of b in this context.
(1)
(d)
Describe the correlation between temperature and catch.
(1)
(e)
A student uses this line to estimate the temperature on a day when the catch was 40 kg. Explain why this is not appropriate.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 the line passes through (14.5, 50.2): 50.2 = a − 2.6 × 14.5
✓ 1
A1 a = 87.9
✓ 1
Part (b)
M1 87.9 − 2.6 × 12
✓ 1
A1 56.7 kg
✓ 1
Part (c)
R1 for each 1 °C rise in temperature the catch decreases by 2.6 kg on average
✓ 1
Part (d)
R1 strong negative correlation
✓ 1
Part (e)
R1 the line of y on x should only be used to predict y from x; a regression of x on y is needed
✓ 1
Common error: The regression line of y on x cannot be rearranged to predict x, because it minimises the vertical distances only.
327IBMA3S-4-16r9-094.2 Presentation of dataPaper 2Medium7 marks
Calculator
A stockroom contains 12 scarves: 5 are silk and 7 are cotton. Two scarves are chosen at random without replacement.
(a)
Find the probability that both scarves are silk. Give your answer as a fraction and to 3 significant figures.
(2)
(b)
Find the probability that the two scarves are of different materials, to 3 significant figures.
(2)
(c)
Find the probability that at least one scarf is cotton, to 3 significant figures.
(2)
(d)
The two scarves are returned and the experiment is repeated 99 times, each time choosing two scarves at random. Find the expected number of times both scarves are silk.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 5/12 × 4/11
✓ 1
A1 5/33 = 0.152
✓ 1
Part (b)
M1 5/12 × 7/11 + 7/12 × 5/11
✓ 1
A1 35/66 = 0.530
✓ 1
Part (c)
M1 1 − P(both silk)
✓ 1
A1 28/33 = 0.848
✓ 1
Part (d)
A1 99 × 5/33 = 15
✓ 1
Exam tip: With 'at least one' wording, the complement is usually quicker than adding several cases.
328IBMA3S-4-16r9-114.4 Correlation and regression of y on xPaper 2Medium12 marks
Calculator
A fish farm records the oxygen concentration x mg/L in a tank and the percentage y of fish that survive a season. x: 4.0, 4.8, 5.5, 6.1, 6.9, 7.4, 8.2, 9.0 y: 62, 68, 71, 79, 83, 85, 91, 94
Scatter diagram of the data
(a)
Find the correlation coefficient r, to 3 significant figures.
(2)
(b)
Describe the correlation.
(1)
(c)
Find the equation of the regression line of y on x, with coefficients to 3 significant figures.
(2)
(d)
Interpret the gradient in this context.
(1)
(e)
Estimate the survival percentage when the oxygen concentration is 7.0 mg/L, to 3 significant figures.
(2)
(f)
The line is used to estimate the survival when x = 12. Explain why this is unreliable.
(2)
(g)
The farm stocks 12 000 fry at 7.0 mg/L. Using your answer to part (e), estimate the number that survive, to 3 significant figures.
(2)
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Notes
Part (a)
M1 use of technology
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A1 r = 0.992
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Part (b)
R1 very strong positive linear correlation
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Part (c)
A1 gradient 6.57
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A1 y = 6.57x + 36.5
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Part (d)
R1 each extra 1 mg/L of oxygen raises the survival percentage by about 6.57 on average
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Part (e)
M1 6.57 × 7.0 + 36.5
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A1 82.5 %
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Part (f)
R1 12 is outside the range of the data (extrapolation)
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R1 the line gives about 115 %, which is impossible
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Part (g)
M1 12 000 × 82.5 / 100
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A1 9900
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Exam tip: Check that a prediction makes sense in context; a percentage above 100 shows the model has broken down.
329IBMA3S-4-17r12-054.9 The normal distributionPaper 2Medium8 marks
Calculator
The burning time, \(T\) hours, of the oxygen cylinders used by a mountaineering expedition is normally distributed with mean 8.0 hours. It is known that 8% of the cylinders burn for less than 6.0 hours.
Normal distribution curve, not to scale
(a)
Find the standard deviation of \(T\). Give your answer to 3 significant figures.
(3)
(b)
Find the probability that a cylinder burns for more than 9.0 hours. Use \(\sigma = 1.42\).
(2)
(c)
The supplier guarantees that 90% of cylinders burn for at least \(t\) hours. Find \(t\) to 3 significant figures, using \(\sigma = 1.42\).
(3)
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Notes
Part (a)
M1 $z = $ inverse normal of 0.08 \(= -1.405\)
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M1 \(\frac{6.0 - 8.0}{\sigma} = -1.405\)
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A1 \(\sigma = 1.42\)
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Part (b)
M1 \(z = \frac{9-8}{1.42} = 0.704\) or normal cdf with correct parameters
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A1 0.241
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Part (c)
M1 \(\mathrm{P}(T<t) = 0.1\)
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M1 \(t = 8 + (-1.282)\times 1.42\) or inverse normal with mean 8 and standard deviation 1.42
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A1 \(t = 6.18\) hours
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Exam tip: When a probability is given and a parameter is unknown, convert to a standard normal value with inverse normal first, then solve for the parameter.
330IBMA3S-4-17r12-064.4 Correlation and regression of y on xPaper 2Medium7 marks
Calculator
During an expedition the mass \(y\) kg of food remaining in a camp store was recorded each day for the first 12 days, where \(x\) is the number of days since the start. The regression line of \(y\) on \(x\) is \(y = 84.6 - 5.8x\) and the mean value of \(x\) is 6.5.
(a)
Find the mean value of \(y\).
(2)
(b)
Estimate the mass of food remaining after 10 days.
(2)
(c)
A porter uses the line to predict the day on which the store will be empty. Find this value of \(x\) to 3 significant figures and comment on whether the prediction is reliable.
(3)
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Notes
Part (a)
M1 the regression line passes through the mean point \((\bar{x}, \bar{y})\): \(84.6 - 5.8 \times 6.5\)
R1 unreliable: 14.6 days is outside the range of the data (extrapolation) and daily use may not stay linear
✓ 1
Common error: A regression line always passes through \((\bar{x}, \bar{y})\); this is a quick way to recover a missing mean.
331IBMA3S-4-17r12-074.11 Conditional probability and independencePaper 2Medium8 marks
Calculator
A telecoms provider monitors its phone lines with an automatic test. On any test, the probability that a line has a fault is 0.04. If a line has a fault, the alarm sounds with probability 0.95. If a line has no fault, the alarm sounds (a false alarm) with probability 0.03.
Tree diagram
(a)
Find the probability that the alarm sounds on a randomly chosen test.
(3)
(b)
Given that the alarm sounds, find the probability that the line has a fault. Give your answer to 3 significant figures.
(3)
(c)
In 5000 tests, estimate the number of false alarms.
Exam tip: A conditional probability has the 'given' event in the denominator; write the formula before substituting.
332IBMA3S-4-17r12-094.6 Combined events and diagramsPaper 2Medium8 marks
Calculator
On a mountaineering trip the weather on each of two consecutive days is either good or poor. The probability that day 1 is good is 0.6. If day 1 is good, the probability that day 2 is good is 0.8. If day 1 is poor, the probability that day 2 is good is 0.35.
Tree diagram: day 1, then day 2
(a)
Find the probability that both days are good.
(2)
(b)
Find the probability that exactly one of the two days is good.
(3)
(c)
Given that exactly one day is good, find the probability that day 1 was the good day. Give your answer to 3 significant figures.
(3)
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Notes
Part (a)
M1 \(0.6\times 0.8\)
✓ 1
A1 0.48
✓ 1
Part (b)
M1 good then poor: \(0.6\times 0.2 = 0.12\)
✓ 1
M1 poor then good: \(0.4 \times 0.35 = 0.14\)
✓ 1
A1 0.26
✓ 1
Part (c)
M1 conditional probability with denominator 0.26 (ECF from (b))
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A1 \(\frac{0.12}{0.26}\) (or \(\frac{6}{13}\))
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A1 0.462
✓ 1
Exam tip: List every route through the tree that satisfies the condition before adding; check that the branches from each node sum to 1.
333IBMA3S-4-17r12-104.9 The normal distributionPaper 2Medium7 marks
Calculator
The mass, \(X\) grams, of a sachet of baking yeast is normally distributed with mean 7.00 and standard deviation 0.15. A sachet passes quality control if \(6.8 \le X \le 7.3\).
Normal distribution curve, not to scale
(a)
Find the probability that a sachet passes quality control. Give your answer to 3 significant figures.
(3)
(b)
In a production run of 10 000 sachets, estimate the number that fail quality control.
(2)
(c)
The producer wants the lightest 1% of sachets to be labelled 'under-weight'. Find the mass below which a sachet is labelled under-weight. Give your answer to 3 significant figures.
(2)
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Notes
Part (a)
M1 standardise: \(z = \frac{6.8-7.0}{0.15}\) and \(z=\frac{7.3-7.0}{0.15}\) (or normal cdf with correct parameters)
Exam tip: 'Fail' is the complement of 'pass': subtract from 1 before multiplying by the number of items.
334IBMA3S-4-17r12-124.2 Presentation of dataPaper 2Medium7 marks
Calculator
A food scientist dries apple slices and records the percentage moisture loss \(y\) after drying for \(x\) minutes. Using data for \(20 \le x \le 120\), she finds the regression line of \(y\) on \(x\) to be \(y = 0.31x + 0.1\).
(a)
Estimate the moisture loss after 38 minutes.
(2)
(b)
Interpret the value 0.31 in this context.
(1)
(c)
The scientist prefers to measure time in hours, \(t\), where \(x = 60t\). Write down the gradient of the line of \(y\) against \(t\).
(2)
(d)
The line is used to estimate the moisture loss after 6 hours. Find this estimate and explain why it is not valid.
(2)
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Notes
Part (a)
M1 \(0.31\times38 + 0.1\)
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A1 11.9 %
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Part (b)
A1 each extra minute of drying increases the moisture loss by 0.31 percentage points (on average)
R1 a moisture loss above 100% is impossible; 6 hours is far outside the data range (extrapolation)
✓ 1
Exam tip: Check units before substituting: the line is in minutes, so 6 hours must be converted to 360 minutes first.
335IBMA3S-4-17r12-144.2 Presentation of dataPaper 2Medium9 marks
Calculator
200 people taste two new snack recipes, A and B, and say which they prefer, or that they have no preference. The results are shown below. Prefers A / prefers B / no preference: Under 30: 46 / 38 / 16 30 or over: 28 / 52 / 20
(a)
Find the probability that a randomly chosen person prefers recipe B.
(1)
(b)
Given that a person prefers recipe B, find the probability that the person is under 30. Give your answer to 3 significant figures.
(2)
(c)
Find the probability that a randomly chosen person prefers recipe A or is aged 30 or over.
(2)
(d)
Determine whether 'prefers recipe A' and 'is under 30' are independent events. Justify your answer.
(2)
(e)
The recipes are later tested on 1500 people. Assuming the same proportions, estimate how many prefer recipe B and are aged 30 or over.
(2)
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Notes
Part (a)
A1 \(\frac{90}{200} = 0.45\)
✓ 1
Part (b)
M1 \(\frac{38}{90}\) (conditional on the 90 who prefer B)
Exam tip: An expected number of days is probability × number of days; round only at the end.
337IBMA3S-4-16r21-074.10 Regression of x on yPaper 2Medium7 marks
Calculator
In a trial, a drug is given in a dose of x mg to each of 7 patients and a response score y is recorded. x: 10, 20, 30, 40, 50, 60, 70 y: 14, 25, 33, 47, 58, 66, 79
Scatter diagram of the data
(a)
Find the equation of the regression line of y on x. Give the coefficients to 3 significant figures. [2]
(2)
(b)
Find the equation of the regression line of x on y. Give the coefficients to 3 significant figures. [2]
(2)
(c)
A clinician wants to estimate the dose that gives a response score of 45. State which of the two lines should be used and find the estimated dose, to 3 significant figures. [3]
(3)
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Notes
Part (a)
A1 gradient 1.08 (1.0786...)
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A1 y = 1.08x + 2.86 (intercept 2.857...)
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Part (b)
A1 gradient 0.924 (0.92411...)
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A1 x = 0.924y − 2.51 (intercept −2.5091...)
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Part (c)
R1 the regression line of x on y, because the dose x is being estimated from a given y
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M1 substitute y = 45 into the equation from (b)
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A1 39.1 mg (39.076...)
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Exam tip: Choose the regression line by asking which variable you are estimating: it must be the one on the left-hand side of the equation.
338IBMA3S-4-16r21-094.1 Sampling, data and outliersPaper 2Medium7 marks
Calculator
The ping times, in milliseconds (ms), recorded by a gamer in 10 online matches are: 38, 41, 36, 44, 40, 39, 42, 37, 44, 98
(a)
Find the mean and the population standard deviation of the ping times, each to 3 significant figures. [2]
(2)
(b)
A value is an outlier if it is more than 2 standard deviations from the mean. Show that 98 ms is an outlier. [2]
(2)
(c)
The 98 ms value is caused by a server fault and is removed. Find the mean and population standard deviation of the remaining 9 values, to 3 significant figures. [2]
(2)
(d)
Describe the effect of the outlier on the standard deviation. [1]
(1)
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Notes
Part (a)
A1 mean = 45.9
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A1 standard deviation = 17.6 (17.558...)
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Part (b)
M1 mean + 2 × standard deviation = 45.9 + 2 × 17.558...
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R1 = 81.0 and 98 > 81.0, so 98 is an outlier
✓ 1
Part (c)
A1 mean = 40.1 (361/9 = 40.111...)
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A1 standard deviation = 2.73 (2.7262...)
✓ 1
Part (d)
R1 the outlier increases the standard deviation greatly (17.6 compared with 2.73)
✓ 1
Common error: Population and sample standard deviation differ on a calculator; use the one stated in the question.
339IBMA3S-4-16r21-104.9 The normal distributionPaper 2Medium7 marks
Calculator
A machine fills bottles with water. The volume V mL in a bottle is normally distributed with mean 500.0 mL and standard deviation 2.1 mL.
Normal distribution curve, not to scale
(a)
Find the probability that a bottle contains less than 495 mL. Give your answer to 3 significant figures. [2]
(2)
(b)
Bottles with less than 495 mL are rejected. A crate contains 24 bottles chosen independently. Find the probability that at least one bottle in the crate is rejected. Use the unrounded answer to (a); give your answer to 3 significant figures. [3]
(3)
(c)
The mean is reset to μ mL with the same standard deviation so that only 0.1% of bottles contain less than 495 mL. Find μ, giving your answer to the nearest 0.1 mL. [2]
(2)
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Notes
Part (a)
M1 attempt to find P(V < 495)
✓ 1
A1 0.00863 (0.0086339...)
✓ 1
Part (b)
M1 recognises binomial, X ~ B(24, 0.0086339...)
✓ 1
M1 attempt to find 1 − P(X = 0), i.e. 1 − (1 − 0.0086339...)24
✓ 1
A1 0.188 (0.18789...)
✓ 1
Part (c)
M1 z = −3.0902 for P(Z < z) = 0.001 and (495 − μ)/2.1 = −3.0902
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A1 μ = 501.5 (501.4895...)
✓ 1
Exam tip: 'At least one' is almost always easiest as 1 minus the probability of none.
340IBMA3S-4-16r22-034.10 Regression of x on yPaper 2Medium7 marks
Calculator
A wind farm records the average wind speed x (m s−1) and the average power output y (MW) of one turbine on six days. Wind speed x: 4, 6, 7, 9, 10, 12 Power output y: 0.8, 1.9, 2.4, 3.9, 4.4, 6.3
Scatter diagram of the data
(a)
Find the Pearson product-moment correlation coefficient, r.
(2)
(b)
Find the equation of the regression line of y on x.
(2)
(c)
Interpret the gradient of the regression line in this context.
(1)
(d)
Use your regression line to estimate the power output when the wind speed is 8 m s−1. Give your answer to 3 significant figures.
(2)
Show mark scheme
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Notes
Part (a)
M1 use of GDC statistics mode to find r
✓ 1
A1 r = 0.994
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Part (b)
A1 gradient 0.679
✓ 1
A1 y = 0.679x − 2.15
✓ 1
Part (c)
R1 each extra 1 m s−1 of wind speed is associated with about 0.679 MW more power output
✓ 1
Part (d)
M1 substitute x = 8 into the equation
✓ 1
A1 3.28 MW
✓ 1
Common error: Quote the gradient with its units and context: it is a change in y per one unit of x, not a total.
341IBMA3S-4-16r22-054.9 The normal distributionPaper 2Medium6 marks
Calculator
A supermarket sells rice in bags labelled 2.00 kg. The mass M kg of a bag is normally distributed with mean 2.02 and standard deviation 0.03.
Normal distribution curve, not to scale
(a)
Find the probability that a randomly chosen bag has a mass less than 2.00 kg.
(2)
(b)
Find P(2.00 < M < 2.05).
(2)
(c)
The supermarket receives 800 bags. Find the expected number of bags with a mass less than 2.00 kg, to the nearest whole number.
(2)
Show mark scheme
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Notes
Part (a)
M1 attempt to use the normal distribution with mean 2.02 and standard deviation 0.03
✓ 1
A1 0.252
✓ 1
Part (b)
M1 correct bounds with the correct parameters
✓ 1
A1 0.589
✓ 1
Part (c)
M1 800 × 0.2525 (using the answer to (a))
✓ 1
A1 202
✓ 1
Exam tip: When the question gives the mean and standard deviation, use them directly on the calculator rather than converting to z first; this avoids rounding errors.
342IBMA3S-4-16r22-064.12 Standardization and inverse normalPaper 2Medium7 marks
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The daily energy output X kWh of a rooftop solar panel is normally distributed with mean 18.5 and standard deviation 4.2.
Normal distribution curve, not to scale
(a)
Find the probability that the output on a given day is greater than 25 kWh.
(2)
(b)
On 10% of days the output is greater than k kWh. Find the value of k.
(2)
(c)
Over 5 days, find the probability that the output exceeds 25 kWh on exactly 2 days. Assume the days are independent and use the unrounded value from (a).
(3)
Show mark scheme
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Mark
Notes
Part (a)
M1 attempt to use the normal distribution with correct parameters
✓ 1
A1 0.0609
✓ 1
Part (b)
M1 use of inverse normal with probability 0.9 (or 0.10 in the upper tail) and correct parameters
✓ 1
A1 23.9
✓ 1
Part (c)
M1 identify a binomial model B(5, 0.0609)
✓ 1
M1 10 × 0.06092 × (1 − 0.0609)3
✓ 1
A1 0.0307
✓ 1
Common error: For 'greater than k' with an upper-tail probability of 0.10, the inverse normal needs the area to the left, 0.90.
343IBMA3S-4-16r22-104.9 The normal distributionPaper 2Medium10 marks
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A hospital network compares how unusual a result is across two laboratories. In laboratory A, antibody levels are normally distributed with mean 140 and standard deviation 22. In laboratory B, antibody levels are normally distributed with mean 95 and standard deviation 14. A patient tested in laboratory A has a level of 172. A different patient tested in laboratory B has a level of 121.
(a)
Find the z-score for each patient and state which patient's result is further from the mean of their own laboratory, in standard deviations.
(3)
(b)
Find the probability that a patient in laboratory A has a level greater than 172.
(2)
(c)
Laboratory B flags the highest 5% of levels. Find the lowest level that is flagged.
(2)
(d)
A laboratory A patient has the same z-score as the laboratory B patient with a level of 121. Find the laboratory A level, to 3 significant figures.
(3)
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Notes
Part (a)
A1 laboratory A: z = (172 − 140) ÷ 22 = 1.45
✓ 1
A1 laboratory B: z = (121 − 95) ÷ 14 = 1.86
✓ 1
R1 the laboratory B patient is further from the mean
✓ 1
Part (b)
M1 attempt to use the normal distribution with mean 140 and standard deviation 22
✓ 1
A1 0.0729
✓ 1
Part (c)
M1 use of inverse normal with probability 0.95 and mean 95, standard deviation 14
Common error: Compare results by z-score, not by raw difference from the mean, when the spreads of the two distributions differ.
344IBMA3S-4-16r22-114.2 Presentation of dataPaper 2Medium10 marks
Calculator
A health authority records a town's number of new infections each day for 15 days: 12, 15, 9, 14, 18, 11, 13, 16, 10, 14, 42, 12, 17, 13, 15
(a)
Find the median, the lower quartile and the upper quartile.
(3)
(b)
A value is classed as an outlier if it is more than 1.5 × IQR above the upper quartile or below the lower quartile. Show that 42 is an outlier and that it is the only outlier.
(3)
(c)
Find the mean number of new infections (i) for all 15 days, (ii) after the outlier is removed.
(2)
(d)
After removing 42, the median of the remaining 14 values is 13.5. Compare how the median and the mean changed, and explain which is a better measure of the typical daily count.
(2)
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Notes
Part (a)
A1 median = 14
✓ 1
A1 lower quartile = 12
✓ 1
A1 upper quartile = 16
✓ 1
Part (b)
M1 IQR = 16 − 12 = 4, so 1.5 × IQR = 6
✓ 1
M1 limits are 12 − 6 = 6 and 16 + 6 = 22
✓ 1
R1 42 > 22, while every other value (9 to 18) lies between 6 and 22
✓ 1
Part (c)
A1 (i) 231 ÷ 15 = 15.4
✓ 1
A1 (ii) 189 ÷ 14 = 13.5
✓ 1
Part (d)
A1 median changed by 0.5 (14 to 13.5); mean changed by 1.9 (15.4 to 13.5)
✓ 1
R1 the mean is affected by extreme values, so the median is the better measure of the typical count
✓ 1
Exam tip: In a data set with an outlier, the median and IQR resist the extreme value, while the mean and standard deviation do not.
345IBMA3S-4-16r22-124.4 Correlation and regression of y on xPaper 2Medium10 marks
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A retail chain records the weekly advertising spend x (hundreds of euros) and the weekly sales y (thousands of euros) for seven stores. Store A: (2, 14); Store B: (3, 17); Store C: (4, 19); Store D: (5, 24); Store E: (6, 26); Store F: (8, 31); Store G: (9, 18)
Scatter diagram of the data
(a)
Find the Pearson product-moment correlation coefficient, r, for all seven stores.
(2)
(b)
Describe the correlation between advertising spend and sales for these seven stores.
(1)
(c)
Store G had a supply problem during the week, so its sales are not representative. Remove store G and find the equation of the regression line of y on x for the remaining six stores.
(3)
(d)
For the six stores, find r and comment on whether a linear model is appropriate.
(2)
(e)
Use your equation from (c) to estimate the sales for a store with an advertising spend of 7 hundred euros.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 use of GDC to find r
✓ 1
A1 r = 0.585
✓ 1
Part (b)
A1 moderate (weak to moderate) positive linear correlation
✓ 1
Part (c)
M1 use of GDC with the six remaining points
✓ 1
A1 gradient 2.90
✓ 1
A1 y = 2.90x + 8.30
✓ 1
Part (d)
A1 r = 0.993
✓ 1
R1 very strong positive linear correlation, so a linear model is appropriate
✓ 1
Part (e)
M1 substitute x = 7
✓ 1
A1 28.6 thousand euros
✓ 1
Common error: Do not remove a data point just because it is far from the line; there must be a reason, such as a known fault, for treating it as unrepresentative.
346IBMA3S-4-16r24-054.4 Correlation and regression of y on xPaper 2Medium7 marks
Calculator
A beekeeper records, for seven weeks, the mean daily temperature x (°C) and the mass of honey collected per hive y (kg). x: 14, 16, 18, 19, 21, 23, 25 y: 3.1, 3.9, 4.4, 5.2, 5.6, 6.5, 6.8
Scatter diagram of the data
(a)
Find the Pearson correlation coefficient r. Give your answer to 3 significant figures.
(2)
(b)
Find the equation of the regression line of y on x. Give the coefficients to 3 significant figures.
(2)
(c)
Use the unrounded regression line to estimate the mass of honey per hive when x = 20. Give your answer to 3 significant figures.
(2)
(d)
Comment on the strength and direction of the correlation, in context.
(1)
Show mark scheme
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Mark
Notes
Part (a)
M1 use of calculator for r
✓ 1
A1 r = 0.991
✓ 1
Part (b)
A1 gradient 0.348 (0.347611…)
✓ 1
A1 intercept −1.68 (−1.68217…), so y = 0.348x − 1.68
✓ 1
Part (c)
M1 0.347611 × 20 − 1.68217
✓ 1
A1 5.27 kg
✓ 1
Part (d)
R1 strong positive linear correlation: warmer weeks are associated with more honey per hive
✓ 1
Exam tip: For predictions, use the unrounded coefficients stored in the calculator; rounded coefficients can change the third significant figure.
347IBMA3S-4-16r24-064.2 Presentation of dataPaper 2Medium7 marks
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A hospital surveys 80 patients. Let A be the event that a patient has high blood pressure and B the event that a patient has diabetes. 34 patients have high blood pressure, 22 have diabetes and 30 have neither condition.
(a)
Find the number of patients who have both conditions.
(3)
(b)
Find P(B | A). Give your answer to 3 significant figures.
(2)
(c)
Determine whether A and B are independent. Justify your answer.
(2)
Show mark scheme
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Notes
Part (a)
M1 n(A∪B) = 80 − 30 = 50
✓ 1
M1 34 + 22 − n(A∩B) = 50
✓ 1
A1 n(A∩B) = 6
✓ 1
Part (b)
M1 6 ÷ 34 (ECF from (a))
✓ 1
A1 0.176
✓ 1
Part (c)
M1 P(A) × P(B) = (34/80)(22/80) = 0.117
✓ 1
R1 P(A∩B) = 6/80 = 0.075, which is not equal to 0.117, so A and B are not independent
✓ 1
Common error: Independence requires P(A∩B) = P(A)P(B) exactly; being 'close' is not enough, and mutually exclusive is a different idea.
348IBMA3S-4-16r24-074.1 Sampling, data and outliersPaper 2Medium6 marks
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A hospital employs 240 nurses, 90 doctors and 70 administrators. A stratified sample of 40 staff is taken, in proportion to the number in each group.
(a)
Find the number of nurses, doctors and administrators in the sample.
(3)
(b)
State one reason why a stratified sample is more suitable than a simple random sample here.
(1)
(c)
The hospital instead asks the first 40 staff who enter the canteen. Explain why this sample may be biased.
(2)
Show mark scheme
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Notes
Part (a)
M1 sampling fraction 40 ÷ 400 = 1/10
✓ 1
A1 nurses 24 and doctors 9
✓ 1
A1 administrators 7
✓ 1
Part (b)
R1 It ensures each group is represented in proportion to its size (so the sample is more representative)
✓ 1
Part (c)
R1 This is a convenience sample, not random, so not every member of staff has an equal chance of selection
✓ 1
R1 Staff who do not use the canteen at that time (for example night staff) are missed, so the sample may not be representative
✓ 1
Common error: In a stratified sample the numbers must be proportional to group sizes and must add to the sample size.
349IBMA3S-4-16r24-104.6 Combined events and diagramsPaper 2Medium7 marks
Calculator
A box contains 12 jars of honey: 7 clover and 5 heather. Two jars are taken at random from the box, without replacement.
(a)
Find the probability that both jars are clover.
(2)
(b)
Find the probability that exactly one jar is clover.
(2)
(c)
Given that at least one of the jars is clover, find the probability that both are clover.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 7/12 × 6/11
✓ 1
A1 7/22
✓ 1
Part (b)
M1 2 × (7/12 × 5/11)
✓ 1
A1 35/66
✓ 1
Part (c)
M1 P(at least one clover) = 1 − 5/12 × 4/11 = 28/33
✓ 1
M1 P(both clover) ÷ P(at least one clover) = (7/22) ÷ (28/33)
✓ 1
A1 3/8
✓ 1
Common error: Without replacement, the second probability changes: the denominator falls from 12 to 11 and the numerator falls only if the first jar was of that type.
350IBMA3S-4-16r24-114.3 Measures of central tendency and dispersionPaper 2Medium7 marks
Calculator
During a Formula 1 race, nine consecutive lap times (seconds) of a driver are: 90.8, 91.1, 90.6, 91.4, 90.9, 91.0, 91.3, 90.7, 98.6. A lap time is an outlier if it is more than 2 standard deviations above the mean. Use the population standard deviation.
(a)
Find the mean and the standard deviation of the nine lap times. Give each to 3 significant figures.
(2)
(b)
Show that 98.6 is an outlier.
(2)
(c)
The outlier is removed. Find the mean of the remaining eight lap times. Give your answer to 3 significant figures.
(1)
(d)
State whether the mean or the median is less affected by removing the outlier, and give a reason.
R1 The median depends on the middle values, not the size of the extreme value, whereas the mean uses every value
✓ 1
Exam tip: Use unrounded mean and standard deviation in an outlier rule, then round only the final comparison.
351IBMA3S-4-17r33-054.12 Standardization and inverse normalPaper 2Medium6 marks
Calculator
The mass of honey, X grams, in jars filled by a beekeeping co-operative is normally distributed with mean μ and standard deviation σ. It is found that 15% of jars contain less than 442 g and 8% of jars contain more than 458 g.
Normal distribution curve, not to scale
(a)
Find the values of z such that P(Z < z1) = 0.15 and P(Z > z2) = 0.08, where Z ~ N(0, 1). Give each to three significant figures.
(2)
(b)
Hence find μ and σ, giving each to three significant figures. Use unrounded z-values in your working.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 z1 = −1.04
✓ 1
A1 z2 = 1.41
✓ 1
Part (b)
M1 (442 − μ)/σ = z1 and (458 − μ)/σ = z2 (standardising both conditions)
✓ 1
M1 attempt to solve simultaneously, e.g. 16 = (z2 − z1)σ
✓ 1
A1 σ = 6.55
✓ 1
A1 μ = 449
✓ 1
Exam tip: Use the unrounded z-values from your calculator in the simultaneous equations; rounding them early can change the third significant figure.
352IBMA3S-4-17r33-064.10 Regression of x on yPaper 2Medium7 marks
Calculator
During a Formula 1 practice run, a team records the age of the tyres, x laps, and the lap time, y seconds, at eight points in the run. x (laps): 2, 5, 8, 11, 14, 17, 20, 23 y (s): 91.2, 91.5, 91.9, 92.0, 92.6, 92.9, 93.4, 93.6
Scatter diagram of the data
(a)
Write down the value of the correlation coefficient r, and find the equation of the regression line of y on x. Give all values to three significant figures.
(3)
(b)
Interpret the value of the gradient in this context.
(1)
(c)
Estimate the lap time when the tyres are 12 laps old, using coefficients to three significant figures. Give your answer to one decimal place.
(2)
(d)
The engineer proposes using the same line to estimate the tyre age at which the lap time reaches 95.0 s. Give one reason why this estimate would be unreliable.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 r = 0.994
✓ 1
A1 gradient = 0.119
✓ 1
A1 intercept = 90.9, so y = 0.119x + 90.9
✓ 1
Part (b)
A1 the lap time increases by about 0.119 s for each additional lap of tyre age
✓ 1
Part (c)
M1 substitute x = 12 into the equation
✓ 1
A1 92.3 s
✓ 1
Part (d)
R1 any one valid reason, e.g. 95.0 s is outside the range of lap times recorded (extrapolation), or the regression line of y on x should not be used to estimate x from y
✓ 1
Common error: The line of y on x predicts y from x only; solving it backwards for x gives an unreliable estimate.
353IBMA3S-4-17r33-084.7 Discrete random variablesPaper 2Medium6 marks
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In a region, 20% of beehives are infested with a parasitic mite. A test for the mite gives a positive result for 90% of infested hives and for 8% of hives that are not infested. A hive is chosen at random and tested.
Tree diagram
(a)
Find the probability that the test result is positive.
(3)
(b)
Given that the test result is positive, find the probability that the hive is infested. Give your answer to three significant figures.
Common error: P(positive | infested) = 0.9 is not the same as P(infested | positive); the second needs the total probability of a positive result in the denominator.
354IBMA3S-4-17r33-104.12 Standardization and inverse normalPaper 2Medium7 marks
Calculator
A driver's qualifying lap time, T seconds, on a particular circuit is normally distributed with mean 78.4 and standard deviation 0.40.
Normal distribution curve, not to scale
(a)
Find the probability that the driver sets a lap time of less than 78.0 seconds.
(2)
(b)
The fastest 5% of the driver's laps are faster than k seconds. Find k to three significant figures.
(2)
(c)
The central 80% of the driver's lap times lie between 78.4 − a and 78.4 + a. Find a to three significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use the normal distribution with mean 78.4 and standard deviation 0.40 (or z = −1)
✓ 1
A1 0.159
✓ 1
Part (b)
M1 z = −1.64… for a lower-tail probability of 0.05 and k = 78.4 + z × 0.40
Exam tip: For a symmetric central probability of 0.80, 0.10 is left in each tail, so use the inverse normal at 0.90 (or 0.10), not 0.80.
355IBMA3S-4-17r33-114.4 Correlation and regression of y on xPaper 2Medium6 marks
Calculator
Over 30 days a hospital records the number of nurses on sick leave, x, and the mean waiting time, y minutes, in its emergency department. The values of x ranged from 0 to 14. The mean of x is 6.4, the correlation coefficient is r = 0.83, and the regression line of y on x is y = 12.5 + 3.8x.
(a)
Describe the correlation between x and y.
(1)
(b)
Find the mean waiting time over the 30 days.
(2)
(c)
Estimate the mean waiting time on a day when 9 nurses are on sick leave.
(2)
(d)
Explain why the line should not be used to estimate the waiting time when 25 nurses are on sick leave.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 strong positive correlation
✓ 1
Part (b)
M1 the regression line passes through (x̄, ȳ), so substitute x = 6.4
✓ 1
A1 ȳ = 36.8 minutes
✓ 1
Part (c)
M1 12.5 + 3.8 × 9
✓ 1
A1 46.7 minutes
✓ 1
Part (d)
R1 25 lies outside the range of the data (0 to 14), so this is extrapolation and may be unreliable
✓ 1
Common error: The regression line always passes through the mean point (x̄, ȳ); use this instead of searching for a data point.
356IBMA3S-4-18r36-054.6 Combined events and diagramsPaper 2Medium7 marks
Calculator
A marine biologist plans dolphin surveys. On any survey day the sea is calm with probability 0.7 and rough otherwise. When the sea is calm the probability of sighting dolphins is 0.8, and when it is rough the probability is 0.35. Give non-exact answers to 3 significant figures.
Tree diagram
(a)
Find the probability that dolphins are sighted on a given day.
(3)
(b)
Given that dolphins are sighted, find the probability that the sea was calm.
(2)
(c)
Surveys on two different days are independent. Find the probability that dolphins are sighted on both days.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to add products of branch probabilities
✓ 1
A1 0.7 × 0.8 = 0.56 and 0.3 × 0.35 = 0.105
✓ 1
A1 0.665
✓ 1
Part (b)
M1 P(calm and sighted)/P(sighted) = 0.56/0.665
✓ 1
A1 0.842
✓ 1
Part (c)
M1 0.665² (using the unrounded value of the answer to (a))
✓ 1
A1 0.442
✓ 1
Common error: For 'given that' questions the denominator is the probability of the condition, not the probability of the event you are finding.
357IBMA3S-4-18r36-104.2 Presentation of dataPaper 2Medium15 marks
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A survey of 80 reef sites records whether coral bleaching (B) and invasive algae (A) are present. Algae are present at 40 sites, bleaching at 36 sites, and neither is present at 12 sites. Give non-exact answers to 3 significant figures.
(a)
Find the number of sites with both algae and bleaching.
(3)
(b)
Write down the number of sites with algae only and the number with bleaching only.
(2)
(c)
A site is chosen at random. Find the probability that it has algae, given that it has bleaching.
(2)
(d)
Determine whether having algae and having bleaching are independent.
(2)
(e)
Two different sites are chosen at random. Find the probability that both have bleaching.
(3)
(f)
Two different sites are chosen at random. Find the probability that exactly one of them has algae.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 n(A ∪ B) = 80 − 12 = 68
✓ 1
M1 68 = 40 + 36 − n(A ∩ B)
✓ 1
A1 8
✓ 1
Part (b)
A1 algae only: 32
✓ 1
A1 bleaching only: 28
✓ 1
Part (c)
M1 8/36
✓ 1
A1 2/9 (= 0.222)
✓ 1
Part (d)
M1 P(A) × P(B) = 0.5 × 0.45 = 0.225 compared with P(A ∩ B) = 0.1
✓ 1
R1 not equal, so not independent
✓ 1
Part (e)
M1 36/80
✓ 1
M1 × 35/79
✓ 1
A1 0.199
✓ 1
Part (f)
M1 40/80 × 40/79
✓ 1
M1 multiplied by 2 (two orders)
✓ 1
A1 0.506
✓ 1
Exam tip: For selection without replacement, reduce both the numerator and the denominator on the second pick, and count every order that gives the required outcome.
358IBMA3S-4-18r36-144.3 Measures of central tendency and dispersionPaper 2Medium14 marks
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On a whale-watching trip, the number of whales spotted is the random variable X, with the probability distribution below.
M1 P(at least one trip with no whales) = 1 − P(whales on every trip)
✓ 1
M1 1 − 0.85³
✓ 1
A1 0.386
✓ 1
Exam tip: Adding a constant to X does not change the variance, and multiplying X by a constant multiplies the variance by the square of that constant.
359IBMA3S-4-18r37-034.11 Conditional probability and independencePaper 2Medium6 marks
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A packing shed sorts 300 crates of pears by variety and quality grade. Variety X: grade A 90, grade B 60, grade C 30 Variety Y: grade A 50, grade B 45, grade C 25 One crate is chosen at random.
(a)
Find the probability that the crate is grade A.
(1)
(b)
Determine whether variety X and grade B are independent.
(3)
(c)
Given that the crate is grade C, find the probability that it is variety Y.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(\dfrac{140}{300}=\dfrac{7}{15}\)
✓ 1
Part (b)
M1 \(P(X)=\dfrac{180}{300}=0.6\) and \(P(B)=\dfrac{105}{300}=0.35\), so \(P(X)P(B)=0.21\)
✓ 1
A1 \(P(X\cap B)=\dfrac{60}{300}=0.2\)
✓ 1
R1 \(0.2\neq0.21\), so they are not independent
✓ 1
Part (c)
M1 \(\dfrac{25}{55}\)
✓ 1
A1 \(\dfrac{5}{11}\) (0.455)
✓ 1
Exam tip: For a conditional probability from a table, the denominator is the total of the given condition, not the grand total.
360IBMA3S-4-18r37-064.6 Combined events and diagramsPaper 2Medium6 marks
Calculator
A playlist contains 20 tracks: 8 rock, 7 pop and 5 jazz. Two different tracks are played at random, one after the other, with no track repeated.
(a)
Find the probability that both tracks are of the same genre.
(4)
(b)
A radio show plays a new random pair on each of 50 evenings. Find the expected number of evenings on which both tracks are of the same genre. Use the exact value from (a) and give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to find the probability for one genre, for example \(P(\text{both rock})=\dfrac{8}{20}\times\dfrac{7}{19}\)
✓ 1
A1 correct products for all three genres: \(\dfrac{56}{380}\), \(\dfrac{42}{380}\), \(\dfrac{20}{380}\)
✓ 1
M1 adding the three probabilities
✓ 1
A1 \(\dfrac{118}{380}=\dfrac{59}{190}\) (0.311)
✓ 1
Part (b)
M1 \(50\times\dfrac{59}{190}\)
✓ 1
A1 15.5 (15.526...)
✓ 1
Common error: With no replacement, the second probability has denominator 19, not 20.
361IBMA3S-4-18r37-074.11 Conditional probability and independencePaper 2Medium10 marks
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A survey of 150 passengers on a long-haul flight records whether each passenger ordered a drink (\(D\)) and whether each ordered a meal (\(M\)). 90 passengers ordered a drink, 60 ordered a meal and 25 ordered neither.
(a)
Find the number of passengers who ordered both a drink and a meal.
(3)
(b)
A passenger is chosen at random. Given that the passenger ordered a drink, find the probability that the passenger also ordered a meal.
(2)
(c)
On similar flights there are 600 passengers. Estimate the number who order a meal but no drink.
R1 \(0.24\neq0.167\), so \(D\) and \(M\) are not independent
✓ 1
Exam tip: Draw a Venn diagram and fill in the intersection first; the region outside both sets is part of the total.
362IBMA3S-4-18r37-094.1 Sampling, data and outliersPaper 2Medium7 marks
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A grower weighs 25 apples. The mean mass is 180 g and the (population) standard deviation is 12 g.
(a)
Find \(\sum x^2\).
(2)
(b)
Each apple is peeled, leaving 80% of its original mass. Write down the new mean and the new standard deviation.
(2)
(c)
One further apple of mass 180 g is added to the original 25 apples. Find the standard deviation of the 26 masses, to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\sigma^2=\dfrac{\sum x^2}{n}-\mu^2\), so \(144=\dfrac{\sum x^2}{25}-180^2\)
✓ 1
A1 813 600
✓ 1
Part (b)
A1 mean 144 g
✓ 1
A1 standard deviation 9.6 g
✓ 1
Part (c)
M1 mean stays 180 g, and \(\sum x^2=813\,600+180^2=846\,000\)
✓ 1
M1 \(\sigma^2=\dfrac{846\,000}{26}-180^2\)
✓ 1
A1 11.8 g (11.77...)
✓ 1
Exam tip: Multiplying every value by a constant multiplies the mean and the standard deviation by that constant; adding a constant changes only the mean.
363IBMA3S-4-18r37-104.2 Presentation of dataPaper 2Medium6 marks
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The delays of 80 flights at an airport are summarised in the table. Delay \(t\) (minutes): \(0\le t<10\), \(10\le t<20\), \(20\le t<30\), \(30\le t<40\), \(40\le t<60\) Frequency: 14, 32, 20, 9, 5
(a)
Write down the modal class.
(1)
(b)
Find the class that contains the median.
(2)
(c)
Estimate the mean delay.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(10\le t<20\)
✓ 1
Part (b)
M1 the median lies between the 40th and 41st values; cumulative frequencies 14, 46
Common error: Use the midpoint of each class, including the last class (50 minutes here), not the lower boundary.
364IBMA3S-4-18r37-124.10 Regression of x on yPaper 2Medium7 marks
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A farmer records the fertiliser applied, \(x\) kg/ha, and the wheat yield, \(y\) t/ha, on eight fields. \(x\): 310, 420, 380, 505, 450, 350, 480, 395 \(y\): 4.4, 5.6, 4.7, 6.7, 5.6, 4.6, 6.0, 5.8
Scatter diagram of the data
(a)
Find the Pearson product-moment correlation coefficient, \(r\).
(2)
(b)
Find the equation of the regression line of \(x\) on \(y\), in the form \(x=ay+b\).
(2)
(c)
The farmer wants to estimate the fertiliser needed for a target yield. Explain why the regression line of \(x\) on \(y\) is the appropriate one to use.
(1)
(d)
Estimate the fertiliser needed to give a yield of 6.0 t/ha.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to find \(r\) using the GDC
✓ 1
A1 0.924
✓ 1
Part (b)
A1 \(a=76.6\)
✓ 1
A1 \(b=-4.38\)
✓ 1
Part (c)
R1 the value of \(x\) is being estimated from a given value of \(y\), so \(x\) is the variable being predicted
✓ 1
Part (d)
M1 substitute \(y=6.0\) into \(x=76.6y-4.38\)
✓ 1
A1 455 kg/ha (using unrounded coefficients also gives 455)
✓ 1
Common error: To estimate \(x\) from a value of \(y\), use the line of \(x\) on \(y\); rearranging the line of \(y\) on \(x\) gives a less reliable estimate.
365IBMA3S-4-18r37-134.4 Correlation and regression of y on xPaper 2Medium6 marks
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A music teacher models weekly practice time, \(x\) hours, against audition score, \(y\) marks, for students whose scores range from 45 to 95. The regression line of \(x\) on \(y\) is \(x=0.15y-5.1\). The mean audition score is 64.
(a)
Interpret the gradient 0.15 in context.
(1)
(b)
Find the mean weekly practice time, given that the regression line passes through the mean point.
(2)
(c)
A student scores 20 marks. Explain why the regression line should not be used to estimate this student's practice time.
(1)
(d)
Estimate the weekly practice time of a student who scores 82 marks.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 each extra mark in the audition score is associated with about 0.15 extra hours of weekly practice
✓ 1
Part (b)
M1 substitute \(y=64\): \(0.15\times64-5.1\)
✓ 1
A1 4.5 hours
✓ 1
Part (c)
R1 a score of 20 is outside the data range 45 to 95, so the estimate is an unreliable extrapolation (and it would give a negative time)
✓ 1
Part (d)
M1 \(0.15\times82-5.1\)
✓ 1
A1 7.2 hours
✓ 1
Exam tip: Only estimate within the range of the data; say that an estimate outside the range is unreliable.
366IBMA3S-4-17r38-044.4 Correlation and regression of y on xPaper 2Medium7 marks
Calculator
A space probe records the temperature, T °C, of an instrument housing at seven distances, d million km, from a star. d: 60, 80, 100, 120, 140, 160, 180 T: 452, 401, 368, 322, 281, 243, 196
Scatter diagram of the data
(a)
Find the equation of the regression line of T on d, with coefficients to 3 significant figures.
(2)
(b)
Interpret the gradient of the line in this context.
(2)
(c)
The probe will reach a distance of 220 million km. Use your equation from (a), with the coefficients to 3 significant figures, to estimate the temperature there, to the nearest degree, and comment on the reliability of the estimate.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 gradient -2.09
✓ 1
A1 intercept 574, so T = -2.09d + 574
✓ 1
Part (b)
A1 the housing temperature falls
✓ 1
A1 by about 2.09 °C for each additional million km from the star
✓ 1
Part (c)
M1 substitute d = 220
✓ 1
A1 574.0 + (-2.09)×220 = 114 °C (ECF from (a))
✓ 1
R1 220 lies outside the range of the data (extrapolation) so the estimate may be unreliable
✓ 1
Common error: A gradient needs units and a direction: say what changes, by how much, and per what.
367IBMA3S-4-17r38-064.2 Presentation of dataPaper 2Medium6 marks
Calculator
The lengths, l mm, of 60 roasted coffee beans are summarised in the table. 8 ≤ l < 10: 6 10 ≤ l < 12: 14 12 ≤ l < 14: 22 14 ≤ l < 16: 12 16 ≤ l < 18: 6
(a)
Write down the modal class, and find the class that contains the median.
(2)
(b)
Use the midpoints of the classes to estimate the mean length.
(2)
(c)
Estimate the population standard deviation of the lengths, to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 modal class 12 ≤ l < 14
✓ 1
A1 cumulative frequencies 6, 20, 42, … so the 30th and 31st values lie in 12 ≤ l < 14
✓ 1
Part (b)
M1 midpoints 9, 11, 13, 15, 17 used with frequencies
✓ 1
A1 776/60 = 12.9 mm
✓ 1
Part (c)
M1 attempt to find the standard deviation using midpoints (technology)
✓ 1
A1 2.22 mm
✓ 1
Exam tip: Grouped-data mean and standard deviation are estimates because every value is replaced by its class midpoint.
368IBMA3S-4-17r38-084.6 Combined events and diagramsPaper 2Medium6 marks
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A space agency records the outcome of 240 probe launches, grouped by launch window. Early window: 64 successes, 26 failures Middle window: 70 successes, 20 failures Late window: 50 successes, 10 failures A launch is chosen at random.
(a)
Find the probability that the launch was a success. Give your answer to 3 significant figures.
(2)
(b)
Determine whether "late window" and "success" are independent events.
(2)
(c)
Given that the launch was a success, find the probability that it was in the middle window. Give your answer to 3 significant figures.
R1 P(late ∩ success) = 50/240 = 0.208 ≠ 0.192 (or 0.833 ≠ 0.767), so not independent
✓ 1
Part (c)
M1 70/184
✓ 1
A1 0.380
✓ 1
Exam tip: To test independence, compare P(A ∩ B) with P(A)P(B), or P(A | B) with P(A).
369IBMA3S-4-17r38-094.9 The normal distributionPaper 2Medium7 marks
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A roaster produces batches that are light (30%), medium (50%) or dark (20%). The probability that a batch is rated premium is 0.2 for a light roast, 0.5 for a medium roast and 0.7 for a dark roast.
Tree diagram
(a)
Find the probability that a randomly chosen batch is rated premium.
(3)
(b)
Given that a batch is rated premium, find the probability that it was a dark roast. Give your answer to 3 significant figures.
(2)
(c)
Two batches are chosen independently. Find the probability that exactly one of them is rated premium.
Common error: In (b) the denominator is P(premium), not 1 and not P(dark).
370IBMA3S-4-17r38-104.11 Conditional probability and independencePaper 2Medium7 marks
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During an inspection of a bridge, event A is "a crack is found" and event B is "corrosion is found". P(A) = 0.35, P(B) = 0.28 and the probability that neither is found is 0.50.
(a)
Find P(A ∩ B).
(3)
(b)
Determine whether A and B are independent.
(2)
(c)
Find P(A | B). Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 P(A ∪ B) = 1 − 0.5 = 0.5
✓ 1
M1 P(A ∩ B) = P(A) + P(B) − P(A ∪ B)
✓ 1
A1 0.13
✓ 1
Part (b)
M1 P(A)P(B) = 0.35 × 0.28 = 0.098
✓ 1
R1 0.098 ≠ 0.13 so not independent (ECF from (a))
✓ 1
Part (c)
M1 0.13/0.28
✓ 1
A1 0.464
✓ 1
Exam tip: "Neither" is the complement of A ∪ B, not of A ∩ B.
371IBMA3S-4-18r39-064.7 Discrete random variablesPaper 2Medium11 marks
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A farmer receives deliveries of three crates of fruit. The random variable \(X\) is the number of damaged crates in a delivery, with the following distribution.
Estimate the population standard deviation of the masses, giving your answer to 3 significant figures.
(2)
(c)
Write down the modal class.
(1)
(d)
A second batch of 30 potatoes has mean mass 155 g. Use the value 144.4 g from part (a) to find the mean mass of all 80 potatoes, giving your answer to 1 decimal place.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 midpoints 110, 130, 150, 170, 190 with \(\sum fx=7220\)
✓ 1
A1 \(144.4\) g
✓ 1
Part (b)
M1 use of midpoints and frequencies in a calculator or the formula (\(\sigma^2=400.64\))
✓ 1
A1 \(20.0\) g
✓ 1
Part (c)
A1 \(140\le m<160\)
✓ 1
Part (d)
M1 \(\frac{50(144.4)+30(155)}{80}\)
✓ 1
A1 \(148.4\) g
✓ 1
Exam tip: Combine means through totals, not by averaging the two means: the batches are different sizes.
373IBMA3S-4-18r39-104.7 Discrete random variablesPaper 2Medium7 marks
Calculator
At a music festival raffle, a ticket wins a prize of €100 with probability 0.02, a prize of €20 with probability 0.08, a prize of €5 with probability 0.2, and nothing otherwise. Let \(X\) be the value in euros of the prize on one ticket.
(a)
Write down the probability that a ticket wins nothing.
(1)
(b)
Find \(E(X)\).
(2)
(c)
Find the standard deviation of \(X\), giving your answer to 3 significant figures.
(3)
(d)
Each ticket is sold for €6, and 1500 tickets are sold. Find the organisers' expected profit from prizes and ticket sales.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(0.7\)
✓ 1
Part (b)
M1 \(100(0.02)+20(0.08)+5(0.2)\)
✓ 1
A1 €4.60
✓ 1
Part (c)
M1 \(E(X^2)=10000(0.02)+400(0.08)+25(0.2)=237\)
✓ 1
M1 \(\operatorname{Var}(X)=237-4.6^2=215.84\)
✓ 1
A1 \(\sigma=14.7\)
✓ 1
Part (d)
A1 \(1500\times(6-4.6)=\) €2100
✓ 1
Exam tip: Work with the expected prize per ticket first; expected profit is then the price minus this value, multiplied by the number of tickets.
374IBMA3S-4-19r41-054.2 Presentation of dataPaper 2Medium7 marks
Calculator
A water company records the daily water use of 120 households, as shown in the table.
Daily use, L litres
100 ≤ L < 150
150 ≤ L < 200
200 ≤ L < 250
250 ≤ L < 300
300 ≤ L < 350
350 ≤ L < 400
Households
8
25
38
30
14
5
Histogram of the data
(a)
Use the midpoints of the classes to estimate the mean daily water use.
(2)
(b)
Estimate the standard deviation of the daily water use, giving your answer to 3 significant figures.
(2)
(c)
Assuming that households are spread evenly within each class, estimate the number of households that use fewer than 190 litres per day.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of midpoints 125, 175, 225, 275, 325, 375 with frequencies
✓ 1
A1 mean = 238.3 litres
✓ 1
Part (b)
M1 use of GDC or formula for σ with midpoints and frequencies
✓ 1
A1 σ = 61.1 litres
✓ 1
Part (c)
M1 cumulative frequency up to 150 is 8
✓ 1
M1 (190 − 150)/50 × 25 = 20 households in the class 150 – 190 (or equivalent proportion)
✓ 1
A1 8 + 20 = 28 households
✓ 1
Exam tip: 'Assuming that households are spread evenly' signals linear interpolation within a class; state the fraction of the class you are using.
375IBMA3S-4-19r41-064.9 The normal distributionPaper 2Medium8 marks
Calculator
The time, T minutes, that a player takes to finish a game level is modelled by a normal distribution with mean μ and standard deviation σ. It is known that P(T < 12) = 0.2 and P(T > 20) = 0.1.
Normal distribution curve, not to scale
(a)
Find the standardised value, z, that corresponds to a cumulative probability of (i) 0.2, (ii) 0.9. Give each to 3 significant figures.
(2)
(b)
Hence find the values of μ and σ, giving each to 3 significant figures.
(4)
(c)
Using the unrounded values of μ and σ, find the probability that a player finishes the level in less than 10 minutes. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 z = −0.842 for cumulative probability 0.2
✓ 1
A1 z = 1.28 for cumulative probability 0.9
✓ 1
Part (b)
M1 forming (12 − μ)/σ = −0.8416 and (20 − μ)/σ = 1.282 (accept 3 s.f. values or unrounded)
✓ 1
M1 attempt to solve the two equations simultaneously (e.g. subtract to eliminate μ)
✓ 1
A1 σ = 3.77
✓ 1
A1 μ = 15.2
✓ 1
Part (c)
M1 attempt to find P(T < 10) using the normal distribution
✓ 1
A1 probability = 0.0850
✓ 1
Common error: Both z-values must keep their signs: a cumulative probability below 0.5 gives a negative z.
376IBMA3S-4-19r41-074.12 Standardization and inverse normalPaper 2Medium9 marks
Calculator
The mass, M mg, of a tablet produced by a machine is normally distributed with mean 250 mg and standard deviation σ mg. A tablet is accepted if its mass is between 245 mg and 255 mg. The probability that a tablet is accepted is 0.92.
Normal distribution curve, not to scale
(a)
Find P(M > 255).
(2)
(b)
Find the value of σ, giving your answer to 3 significant figures.
(3)
(c)
A batch contains 2500 tablets. Find the expected number of tablets that are rejected.
(1)
(d)
The machine is adjusted so that the mean is still 250 mg but 98% of tablets are accepted. Find the new standard deviation, giving your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of symmetry about the mean: P(M < 245) = P(M > 255)
✓ 1
A1 (1 − 0.92) ÷ 2 = 0.04
✓ 1
Part (b)
M1 P(M < 255) = 0.96 and z = invNorm(0.96) = 1.7507...
✓ 1
M1 (255 − 250)/σ = 1.7507...
✓ 1
A1 σ = 2.86
✓ 1
Part (c)
A1 0.08 × 2500 = 200
✓ 1
Part (d)
M1 P(M < 255) = 0.99 and z = invNorm(0.99) = 2.3263...
✓ 1
M1 (255 − 250)/σ = 2.3263...
✓ 1
A1 σ = 2.15 mg
✓ 1
Exam tip: For a symmetric interval about the mean, split the tail probability equally and use a single inverse-normal calculation.
377IBMA3S-4-19r41-084.3 Measures of central tendency and dispersionPaper 2Medium9 marks
Calculator
Water pressure is measured at 30 hydrants. The readings, x kPa, satisfy Σx = 1260 and Σx2 = 53 190.
(a)
Find the mean and the population standard deviation of the readings.
(3)
(b)
Two readings, of 36 kPa and 48 kPa, are found to be faulty and are removed. Show that the mean of the remaining readings is still 42 kPa.
(2)
(c)
Find the population standard deviation of the remaining 28 readings, giving your answer to 3 significant figures.
(3)
(d)
Explain why the standard deviation decreases when these two readings are removed.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 mean = 1260 ÷ 30 = 42 kPa
✓ 1
M1 variance = 53 190/30 − 422
✓ 1
A1 σ = √9 = 3 kPa
✓ 1
Part (b)
M1 new sum = 1260 − 36 − 48 = 1176
✓ 1
A1 1176 ÷ 28 = 42
✓ 1
Part (c)
M1 new Σx2 = 53 190 − 362 − 482 = 49 590
✓ 1
M1 variance = 49 590/28 − 422
✓ 1
A1 σ = 2.66 kPa
✓ 1
Part (d)
R1 each removed reading was 6 kPa from the mean, which is twice the original standard deviation, so they were further from the mean than typical; removing them reduces the spread (or equivalent valid reason)
✓ 1
Common error: When removing data, update Σx² as well as Σx; the variance cannot be found from the new mean alone.
378IBMA3S-4-19r41-104.2 Presentation of dataPaper 2Medium7 marks
Calculator
The daily volumes of water drawn, in ML, from two reservoirs over a year are summarised below.
Minimum
Q1
Median
Q3
Maximum
Reservoir A
18
29
35
44
68
Reservoir B
20
31
37
42
55
An observation is an outlier if it is greater than Q3 + 1.5 × IQR or less than Q1 − 1.5 × IQR.Box-and-whisker diagrams
(a)
Write down the range and the interquartile range of Reservoir A.
(2)
(b)
Show that the maximum for Reservoir B is not an outlier, and find the upper boundary for outliers for Reservoir A.
(3)
(c)
Compare the daily volumes drawn from the two reservoirs, referring to the median and the interquartile range.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 range = 68 − 18 = 50
✓ 1
A1 IQR = 44 − 29 = 15
✓ 1
Part (b)
A1 IQR of B = 42 − 31 = 11, so upper boundary for B = 42 + 16.5 = 58.5 and 55 < 58.5, so not an outlier
✓ 1
M1 upper boundary for A = 44 + 1.5 × 15
✓ 1
A1 66.5 ML (so the maximum of 68 is an outlier for A)
✓ 1
Part (c)
A1 B has the higher median (37 > 35), so typically more water is drawn from B
✓ 1
A1 B has the smaller IQR (11 < 15), so the volumes drawn from B are more consistent
✓ 1
Exam tip: A comparison needs a statement about a measure of centre and a measure of spread, each written in the context of water drawn.
379IBMA3S-4-19r41-134.9 The normal distributionPaper 2Medium8 marks
Calculator
The latency, X ms, of a game server is modelled by a normal distribution with mean 42 ms and standard deviation 6.5 ms. A player experiences lag when the latency exceeds 55 ms.
Normal distribution curve, not to scale
(a)
Find the probability that a single ping causes lag. Give your answer to 3 significant figures.
(2)
(b)
Twelve pings are sent independently. Using the unrounded probability from (a), find the probability that at least one causes lag. Give your answer to 3 significant figures.
(3)
(c)
After an upgrade the mean is unchanged but the standard deviation becomes σ ms, so that P(X > 55) = 0.005. Find σ, giving your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to find P(X > 55) (z = 2)
✓ 1
A1 probability = 0.0228
✓ 1
Part (b)
M1 recognising P(at least one) = 1 − P(none)
✓ 1
M1 1 − (1 − 0.02275...)12
✓ 1
A1 probability = 0.241
✓ 1
Part (c)
M1 z = invNorm(0.995) = 2.5758...
✓ 1
M1 (55 − 42)/σ = 2.5758...
✓ 1
A1 σ = 5.05 ms
✓ 1
Common error: 'At least one' is 1 minus the probability of none; adding 12 copies of the single-ping probability overestimates it.
380IBMA3S-4-18r45-064.8 The binomial distributionPaper 2Medium7 marks
Calculator
A food manufacturer finds that each jar from a production line is independently contaminated with probability 0.02.
(a)
Find the probability that at least one jar in a random sample of 30 jars is contaminated.
(2)
(b)
A sample of n jars is taken. Find the smallest value of n for which the probability that at least one jar is contaminated is greater than 0.9.
(3)
(c)
Using your answer to (b), find the expected number of contaminated jars in a sample of that size.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 1 − 0.9830
✓ 1
A1 0.455
✓ 1
Part (b)
M1 1 − 0.98n > 0.9, i.e. 0.98n < 0.1
✓ 1
M1 attempt to solve (logarithms, a table of values or a graph on the GDC), giving n > 113.97
✓ 1
A1 n = 114
✓ 1
Part (c)
M1 E(X) = np = 114 × 0.02
✓ 1
A1 2.28
✓ 1
Common error: When solving an inequality for a whole number of trials, check that the next smaller integer fails the condition.
381IBMA3S-4-18r45-074.2 Presentation of dataPaper 2Medium7 marks
Calculator
A telecoms company records the number of dropped calls on one tower on each of 40 days.
Dropped calls : number of days 0 : 6 1 : 13 2 : 12 3 : 6 4 : 2 5 : 1
Bar chart of the data
(a)
Find the mean number of dropped calls per day.
(2)
(b)
Find the population standard deviation, correct to 3 significant figures.
(2)
(c)
A value is an outlier if it is more than 1.5 × IQR above the upper quartile. Determine which numbers of dropped calls are outliers.
Exam tip: Find the quartiles from the cumulative frequencies before applying the 1.5 × IQR rule.
382IBMA3S-4-18r45-094.2 Presentation of dataPaper 2Medium10 marks
Calculator
The moisture content, m %, of 100 samples of dried fruit is summarised in the cumulative frequency table.
Moisture content : cumulative frequency m < 12 : 4 m < 14 : 15 m < 16 : 40 m < 18 : 76 m < 20 : 94 m < 22 : 100
(a)
Write down the number of samples with 14 ≤ m < 16.
(1)
(b)
Estimate the median moisture content, assuming that values are evenly spread within each class. Give your answer to 3 significant figures.
(3)
(c)
Estimate the lower quartile and the upper quartile, using the same assumption. Give your answers to 3 significant figures.
(4)
(d)
Find an estimate of the interquartile range. Use the unrounded quartiles from (c). Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 25
✓ 1
Part (b)
M1 the median is the 50th value, which lies in the class 16 ≤ m < 18
✓ 1
M1 16 + 2 × (50 − 40) ÷ 36
✓ 1
A1 16.6
✓ 1
Part (c)
M1 lower quartile is the 25th value, in the class 14 ≤ m < 16: 14 + 2 × (25 − 15) ÷ 25
✓ 1
A1 14.8
✓ 1
M1 upper quartile is the 75th value, in the class 16 ≤ m < 18: 16 + 2 × (75 − 40) ÷ 36
✓ 1
A1 17.9
✓ 1
Part (d)
M1 IQR = 17.944… − 14.8
✓ 1
A1 3.14
✓ 1
Exam tip: Locate the quartile by its position in the cumulative frequency (25th, 50th, 75th of 100) and then interpolate within that class.
383IBMA3S-4-18r45-124.3 Measures of central tendency and dispersionPaper 2Medium7 marks
Calculator
A food scientist is explaining how the mean and standard deviation of a data set change when it is modified. The 20 temperatures, in °C, recorded in a cold store have mean −12.5 and population standard deviation 4.0.
(a)
The temperatures are converted to °F using F = 1.8C + 32. Find the mean of the converted temperatures.
(2)
(b)
Find the standard deviation of the converted temperatures.
(2)
(c)
A 21st reading, equal to −12.5 °C, is added to the original 20 readings. Find the new population standard deviation, in °C, correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 1.8 × (−12.5) + 32
✓ 1
A1 9.5 °F
✓ 1
Part (b)
M1 multiply by 1.8 (the addition of 32 does not change the spread)
✓ 1
A1 7.2 °F
✓ 1
Part (c)
M1 sum of squared deviations = 20 × 4.02 = 320 (the new reading adds 0)
✓ 1
M1 new variance = 320 ÷ 21 = 15.238…
✓ 1
A1 3.90 °C
✓ 1
Exam tip: Adding a constant moves the mean but leaves the standard deviation unchanged; multiplying scales both.
384IBMA3S-4-18r45-144.6 Combined events and diagramsPaper 2Medium7 marks
Calculator
In a taste test of 60 people, 35 like a sweet flavour, 28 like a savoury flavour and 12 like neither flavour.
(a)
Find the number of people who like both flavours.
(3)
(b)
One of the people who like the sweet flavour is chosen at random. Find the probability that this person also likes the savoury flavour. Give your answer to 3 significant figures.
(2)
(c)
Two of the 60 people are chosen at random without replacement. Find the probability that both of them like both flavours. Give your answer to 3 significant figures.
Common error: Without replacement, the second probability uses one fewer person in both the numerator and the denominator.
385IBMA3S-4-19r45-054.4 Correlation and regression of y on xPaper 2Medium8 marks
Calculator
An airline records the take-off ground roll y (km) of a loaded jet at six different air temperatures x (°C).
x
5
12
18
24
30
36
y
1.62
1.78
1.95
2.09
2.31
2.42
Use the unrounded values from your calculator in all later calculations.Scatter diagram of the data
(a)
Write down the Pearson correlation coefficient r for the data, to 3 significant figures.
(1)
(b)
Find the equation of the regression line of y on x, giving the coefficients to 3 significant figures.
(2)
(c)
Interpret the gradient of the regression line in this context.
(1)
(d)
Use the regression line to estimate the ground roll at a temperature of 28 °C, giving your answer to 3 significant figures.
(2)
(e)
The airline also wants an estimate for 45 °C. Comment on the reliability of using the regression line at 45 °C.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 r = 0.997
✓ 1
Part (b)
M1 attempt to find the regression line using a calculator
✓ 1
A1 y = 0.0266x + 1.47
✓ 1
Part (c)
R1 each 1 °C increase in temperature is associated with an increase of about 0.0266 km (26.6 m) in the ground roll
✓ 1
Part (d)
M1 substitute x = 28 into the unrounded regression equation
✓ 1
A1 2.22 km
✓ 1
Part (e)
R1 45 °C lies outside the range of the data (5 to 36 °C), so this is extrapolation
✓ 1
R1 the linear relationship may not continue, so the estimate is less reliable, even though r is close to 1
✓ 1
Common error: Do not copy rounded coefficients into the next calculation if the question asks for unrounded values; keep the full calculator values.
386IBMA3S-4-19r45-064.6 Combined events and diagramsPaper 2Medium7 marks
Calculator
An orchestra has 120 members, of whom 60 play string instruments and 60 play brass. Members were asked whether they practise every day. 45 string players and 18 brass players said yes. Let S be the event that a randomly chosen member plays a string instrument and D the event that the member practises every day.
(a)
Find P(S | D), giving your answer to 3 significant figures.
(2)
(b)
Determine whether S and D are independent. Justify your answer.
(3)
(c)
Two different members are chosen at random. Find the probability that both practise every day, giving your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(S | D) = 45/63 or P(S ∩ D)/P(D)
✓ 1
A1 0.714
✓ 1
Part (b)
M1 P(S) × P(D) = (60/120) × (63/120)
✓ 1
A1 = 0.2625 and P(S ∩ D) = 45/120 = 0.375
✓ 1
R1 these are not equal, so S and D are not independent
✓ 1
Part (c)
M1 (63/120) × (62/119)
✓ 1
A1 0.274
✓ 1
Exam tip: In sampling without replacement, reduce both the numerator and the denominator for the second selection.
387IBMA3S-4-19r45-074.6 Combined events and diagramsPaper 2Medium7 marks
Calculator
On a busy route, each flight independently departs late with probability 0.15. Over 25 flights, let X be the number that depart late.
(a)
Find P(X = 3), giving your answer to 3 significant figures.
(2)
(b)
On a different route with the same probability of 0.15 for each flight, the airline monitors m flights. Find the smallest value of m for which the probability that at least one flight departs late is greater than 0.95.
(3)
(c)
Find P(X ≥ 2), giving your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of the binomial distribution B(25, 0.15), e.g. 2300 × 0.15^3 × 0.85^22
✓ 1
A1 0.217
✓ 1
Part (b)
M1 1 − 0.85^m > 0.95 (or 0.85^m < 0.05)
✓ 1
M1 solving, e.g. m > ln 0.05 / ln 0.85 = 18.4 or table of values
✓ 1
A1 m = 19
✓ 1
Part (c)
M1 1 − P(X ≤ 1) or 1 − P(X = 0) − P(X = 1)
✓ 1
A1 0.907
✓ 1
Exam tip: For "at least one" problems with n unknown, use the complement 1 − (1 − p)^n and check the values on either side of the boundary.
388IBMA3S-4-19r45-084.3 Measures of central tendency and dispersionPaper 2Medium7 marks
Calculator
The numbers of tracks on five albums are 3, x, 8, 2x and 14, where x is a positive integer. The mean number of tracks is 11.
(a)
Find the value of x.
(2)
(b)
Find the median number of tracks.
(2)
(c)
Each album is re-released as a double edition with one bonus track, so that the new number of tracks is 2t + 1, where t is the original number. Find the new mean number of tracks.
(1)
(d)
Find the new population standard deviation, giving your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (3 + x + 8 + 2x + 14)/5 = 11
✓ 1
A1 x = 10
✓ 1
Part (b)
M1 data placed in order 3, 8, 10, 14, 20
✓ 1
A1 median = 10
✓ 1
Part (c)
A1 new mean = 2 × 11 + 1 = 23
✓ 1
Part (d)
M1 original population standard deviation sqrt(32.8) = 5.727… and a factor of 2 applied (adding 1 does not change the spread)
✓ 1
A1 11.5
✓ 1
Exam tip: Adding a constant to every value shifts the mean but leaves the standard deviation unchanged; multiplying by k multiplies both by k.
389IBMA3S-4-19r45-094.2 Presentation of dataPaper 2Medium7 marks
Calculator
The daily milk yields, in litres, of eleven cows on a dairy farm are 18.2, 21.5, 22.0, 23.4, 24.1, 25.0, 25.8, 26.3, 27.9, 28.6, 41.0. For this question, the lower quartile is the median of the five values below the median and the upper quartile is the median of the five values above the median.
(a)
Find the lower quartile and the upper quartile.
(2)
(b)
An outlier is a value more than 1.5 × IQR above Q3 or below Q1. Show that 41.0 is an outlier and that there is no outlier at the lower end.
(3)
(c)
The outlier is removed. Find the new mean and new median, and state, with a reason, which of the two measures is less affected by the removal.
R1 41.0 > 36.75 so it is an outlier, and 18.2 > 13.15 so there is no lower outlier
✓ 1
Part (c)
A1 new mean = 24.28 and new median = 24.55 litres
✓ 1
R1 the median changed from 25.0 to 24.55 (by 0.45) while the mean changed from 25.8 to 24.28 (by 1.52), so the median is less affected
✓ 1
Common error: The median of an even number of values is the mean of the middle two, so removing one value can shift it by half a gap.
390IBMA3S-4-19r45-104.11 Conditional probability and independencePaper 2Medium6 marks
Calculator
A flight is delayed with probability 0.2. If a flight is delayed, the probability that a passenger's bag is misrouted is 0.15. If the flight is not delayed, this probability is 0.04. Let D be the event that the flight is delayed and M the event that the bag is misrouted.
Tree diagram
(a)
Find P(M).
(2)
(b)
Given that a bag is misrouted, find the probability that the flight was delayed. Give your answer to 3 significant figures.
(2)
(c)
A passenger makes two journeys that are independent of each other. Find the probability that the bag is misrouted on exactly one of them. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(M) = 0.2 × 0.15 + 0.8 × 0.04
✓ 1
A1 0.062
✓ 1
Part (b)
M1 P(D | M) = 0.03/0.062
✓ 1
A1 0.484
✓ 1
Part (c)
M1 2 × 0.062 × 0.938
✓ 1
A1 0.116
✓ 1
Exam tip: For a reverse conditional probability, put the branch you are given in the denominator: the total probability P(M).
391IBMA3S-4-19r45-114.10 Regression of x on yPaper 2Medium7 marks
Calculator
A farmer models the barley yield y (quintals per hectare) as a function of the rainfall x (mm) in the growing month. The regression line of y on x for data collected over several years is y = 0.45x + 12.6. The rainfall values in the data ranged from 30 mm to 90 mm, and the mean rainfall was 60 mm.
(a)
Find the mean yield.
(2)
(b)
Interpret the value 0.45 in this context.
(1)
(c)
Explain why the line is not reliable for estimating the yield when the rainfall is 150 mm.
(2)
(d)
Use the line to estimate the yield when the rainfall is 48 mm.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitute x = 60 (the mean point lies on the line)
✓ 1
A1 mean yield = 39.6 quintals per hectare
✓ 1
Part (b)
R1 each extra 1 mm of rainfall is associated with an increase of 0.45 quintals per hectare in yield
✓ 1
Part (c)
R1 150 mm lies well outside the range of the data (30 mm to 90 mm), so this is extrapolation
✓ 1
R1 the linear relationship may not continue beyond the data (for example, too much rain can damage crops)
✓ 1
Part (d)
M1 substitute x = 48
✓ 1
A1 34.2 quintals per hectare
✓ 1
Common error: The mean point (x̄, ȳ) always lies on the regression line, which lets you find a missing mean without the raw data.
392IBMA3S-4-18r46-034.9 The normal distributionPaper 2Medium6 marks
Calculator
The mass \(M\) tonnes of a loaded shipping container is normally distributed with mean \(24.0\) and standard deviation \(1.8\).
Normal distribution curve, not to scale
(a)
Find the probability that a container has mass greater than \(26\) tonnes.
(2)
(b)
A container is classed as extra heavy if its mass is in the heaviest \(5\%\) of all containers. Find the minimum mass of an extra heavy container, correct to 3 significant figures.
(2)
(c)
A vessel carries \(200\) containers chosen at random. Find the expected number of these containers with mass less than \(22\) tonnes, correct to 3 significant figures. Use the unrounded probability.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use \(M\sim N(24,1.8^2)\) with \(P(M>26)\)
✓ 1
A1 \(0.133\)
✓ 1
Part (b)
M1 attempt to find \(m\) with \(P(M<m)=0.95\)
✓ 1
A1 \(27.0\) tonnes
✓ 1
Part (c)
M1 \(200\times P(M<22)\) with \(P(M<22)=0.13326\ldots\)
✓ 1
A1 \(26.7\)
✓ 1
Exam tip: Sketch a quick bell curve and shade the required area before using the calculator; it shows whether you want a tail or a middle region.
393IBMA3S-4-18r46-044.9 The normal distributionPaper 2Medium6 marks
Calculator
The running time \(T\) minutes of a play is normally distributed with mean \(148\) and standard deviation \(\sigma\). The probability that a performance lasts longer than \(155\) minutes is \(0.12\).
Normal distribution curve, not to scale
(a)
Find the value of \(\sigma\), correct to 3 significant figures.
(4)
(b)
Using \(\sigma=5.96\), find the running time that is exceeded by only \(1\%\) of performances. Give your answer correct to 1 decimal place.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(P(T<155)=0.88\)
✓ 1
M1 standardising, \(\frac{155-148}{\sigma}=z\)
✓ 1
A1 \(z=1.1749\ldots\)
✓ 1
A1 \(\sigma=5.96\)
✓ 1
Part (b)
M1 \(z=2.3263\ldots\) for \(P(T>t)=0.01\), e.g. \(t=148+2.326\times5.96\)
✓ 1
A1 \(161.9\) minutes
✓ 1
Common error: Use the correct tail. 'Longer than \(155\)' with probability \(0.12\) means the area to the left is \(0.88\), so the \(z\)-value is positive.
394IBMA3S-4-18r46-074.8 The binomial distributionPaper 2Medium6 marks
Calculator
A port inspects containers one at a time. Each container independently has a defect with probability \(0.08\).
(a)
Find the probability that, in a sample of \(15\) containers, at least one has a defect.
(2)
(b)
Find the smallest sample size \(n\) for which the probability that at least one container has a defect is greater than \(0.95\).
(3)
(c)
For a sample of the size found in (b), find the expected number of containers with a defect.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(1-0.92^{15}\) or \(1-P(X=0)\) with \(X\sim B(15,0.08)\)
✓ 1
A1 \(0.714\)
✓ 1
Part (b)
M1 \(1-0.92^{n}>0.95\) or \(0.92^{n}<0.05\)
✓ 1
M1 attempt to solve, e.g. \(n>\frac{\ln 0.05}{\ln 0.92}=35.9\ldots\) or a table showing \(n=35\) gives \(0.946\) and \(n=36\) gives \(0.950\)
✓ 1
A1 \(n=36\)
✓ 1
Part (c)
A1 \(36\times0.08=2.88\) (ECF from (b))
✓ 1
Common error: When solving \(0.92^n<0.05\) by logarithms, the inequality reverses on dividing by \(\ln 0.92\), which is negative. Always check the answer with the nearest whole numbers.
395IBMA3S-4-18r46-084.12 Standardization and inverse normalPaper 2Medium7 marks
Calculator
The time \(T\) seconds taken for a scene change in a theatre is normally distributed with mean \(75\) and standard deviation \(9\).
Normal distribution curve, not to scale
(a)
Find the probability that a scene change takes less than \(60\) seconds.
(2)
(b)
In one performance there are \(10\) scene changes, with times independent of each other. Find the probability that exactly \(2\) of them take less than \(60\) seconds. Use the unrounded value from (a).
(3)
(c)
Find the probability that at least one of the \(10\) scene changes takes less than \(60\) seconds.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use \(T\sim N(75,9^2)\) with \(P(T<60)\)
✓ 1
A1 \(0.0478\)
✓ 1
Part (b)
M1 \(Y\sim B(10,p)\) with \(p=0.04779\ldots\)
✓ 1
M1 \(\binom{10}{2}p^2(1-p)^8\)
✓ 1
A1 \(0.0695\)
✓ 1
Part (c)
M1 \(1-(1-p)^{10}\) or \(1-P(Y=0)\)
✓ 1
A1 \(0.387\)
✓ 1
Exam tip: When one part feeds the next, keep the unrounded value in your calculator memory; it avoids small rounding differences in the final answer.
396IBMA3S-4-18r46-134.6 Combined events and diagramsPaper 2Medium12 marks
Calculator
A theatre sells \(70\%\) of its tickets online and the rest at the door. A ticket sold online is not used (a no-show) with probability \(0.04\). A ticket sold at the door is a no-show with probability \(0.10\). Let \(O\) be the event that a ticket was sold online and \(N\) the event that a ticket is a no-show.
Tree diagram
(a)
Find \(P(N)\).
(3)
(b)
Given that a ticket is a no-show, find the probability that it was sold online.
(2)
(c)
Determine whether \(O\) and \(N\) are independent events. Justify your answer.
(2)
(d)
For a performance, \(450\) tickets are sold. Find the expected number of no-shows.
(2)
(e)
A group of \(20\) tickets is chosen at random from all tickets sold. Assuming that each is a no-show independently with probability \(0.058\), find the probability that exactly \(2\) of the tickets are no-shows.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(P(O\cap N)=0.7\times0.04=0.028\)
✓ 1
M1 \(P(O'\cap N)=0.3\times0.10=0.03\) and adding the two products
M1 compares \(P(O)P(N)=0.7\times0.058=0.0406\) with \(P(O\cap N)=0.028\)
✓ 1
R1 \(0.0406\neq0.028\), so not independent (ECF from (a))
✓ 1
Part (d)
M1 \(450\times0.058\)
✓ 1
A1 \(26.1\) (ECF from (a))
✓ 1
Part (e)
M1 \(X\sim B(20,0.058)\)
✓ 1
M1 \(\binom{20}{2}(0.058)^2(0.942)^{18}\)
✓ 1
A1 \(0.218\)
✓ 1
Common error: Independence needs \(P(A\cap B)=P(A)P(B)\). Do not argue from the fact that two events seem unrelated; always test with the numbers.
397IBMA3S-4-18r48-024.7 Discrete random variablesPaper 2Medium7 marks
Calculator
In a Formula 1 race, the number of pit stops \(X\) made by a driver has the probability distribution below, where \(p\) and \(q\) are constants. \(x\): 1, 2, 3, 4 \(P(X = x)\): 0.2, \(p\), 0.35, \(q\) It is known that \(\mathrm{E}(X) = 2.5\).
(a)
Write down two equations in \(p\) and \(q\).
(2)
(b)
Find the value of \(p\) and the value of \(q\).
(2)
(c)
Find \(P(X \ge 3)\).
(1)
(d)
Find \(\mathrm{Var}(X)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(0.2 + p + 0.35 + q = 1\), so \(p + q = 0.45\)
Exam tip: 'At least 10' means \(X \ge 10\), which is \(1 - P(X \le 9)\) on a calculator that gives cumulative probabilities.
399IBMA3S-4-18r48-054.3 Measures of central tendency and dispersionPaper 2Medium8 marks
Calculator
A hospital clinic books 25 appointments each day. Each patient independently fails to attend with probability 0.12. Let \(X\) be the number of patients who fail to attend on a given day, so \(X \sim B(25, 0.12)\).
(a)
Find the mean and the standard deviation of \(X\), giving the standard deviation to 3 significant figures.
(3)
(b)
Find the probability that at least 5 patients fail to attend, to 3 significant figures.
(2)
(c)
The clinic manager wants to be more than 95% sure that the number of no-shows does not exceed \(k\). Find the smallest integer \(k\) for which \(P(X \le k) > 0.95\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 mean \(= 25 \times 0.12 = 3\)
✓ 1
M1 attempt to use \(\sqrt{np(1-p)} = \sqrt{25(0.12)(0.88)}\)
✓ 1
A1 standard deviation = 1.62
✓ 1
Part (b)
M1 attempt to use \(1 - P(X \le 4)\)
✓ 1
A1 0.173
✓ 1
Part (c)
M1 attempt to compare cumulative probabilities for several values of \(k\)
✓ 1
A1 \(P(X \le 5) = 0.9291\) which is below 0.95, and \(P(X \le 6) = 0.9757\) which is above 0.95
✓ 1
A1 \(k = 6\)
✓ 1
Exam tip: For a 'smallest k' question, tabulate \(P(X \le k)\) for successive \(k\) and compare each with the target probability.
400IBMA3S-4-18r48-084.2 Presentation of dataPaper 2Medium9 marks
Calculator
A beekeeper records the honey yield, in kg, from each of 12 hives in one season: 12, 15, 17, 18, 20, 21, 23, 24, 26, 29, 31, 52
(a)
Find the median, the lower quartile \(Q_1\) and the upper quartile \(Q_3\).
(3)
(b)
Show that 52 is an outlier, using the rule that an outlier lies more than \(1.5 \times \mathrm{IQR}\) above \(Q_3\) or below \(Q_1\).
(3)
(c)
Find the mean yield for all 12 hives, and the mean yield when the outlier is removed. Give your answers to 3 significant figures.
(2)
(d)
Explain why the median is a more suitable measure of the typical yield than the mean for the full data set.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 median = 22
✓ 1
A1 \(Q_1 = 17.5\)
✓ 1
A1 \(Q_3 = 27.5\)
✓ 1
Part (b)
M1 \(\mathrm{IQR} = 27.5 - 17.5 = 10\)
✓ 1
M1 upper boundary \(= 27.5 + 1.5(10) = 42.5\)
✓ 1
R1 \(52 > 42.5\), so 52 is an outlier (and no value is below \(17.5 - 15 = 2.5\))
✓ 1
Part (c)
A1 all 12 hives: 24.0 kg
✓ 1
A1 without 52: 21.5 kg
✓ 1
Part (d)
R1 the median is not affected by the extreme value 52, whereas the mean is pulled upwards by it
✓ 1
Common error: With an even number of data values, split the data into two equal halves and take the median of each half to find the quartiles.
401IBMA3S-4-18r48-094.4 Correlation and regression of y on xPaper 2Medium8 marks
Calculator
During a Formula 1 stint, the lap time \(y\) seconds is recorded against the age of the tyres \(x\) laps, for six laps: \(x\): 2, 5, 8, 11, 14, 17 \(y\): 81.6, 82.1, 82.9, 83.2, 84.1, 84.5
Scatter diagram of the data
(a)
Find the Pearson product-moment correlation coefficient \(r\), to 3 significant figures, and describe the correlation.
(2)
(b)
Find the equation of the regression line of \(y\) on \(x\), with coefficients to 3 significant figures.
(2)
(c)
Interpret the gradient of the regression line in context.
(1)
(d)
Estimate the lap time when the tyres are 10 laps old, to 3 significant figures.
(2)
(e)
A strategist uses the line to predict the lap time for tyres 40 laps old. Give one reason why this prediction is unreliable.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(r = 0.994\)
✓ 1
A1 strong positive linear correlation
✓ 1
Part (b)
M1 attempt to use a calculator to find gradient and intercept
✓ 1
A1 \(y = 0.198x + 81.2\)
✓ 1
Part (c)
R1 lap time increases by about 0.198 seconds for each extra lap of tyre age
✓ 1
Part (d)
M1 substitute \(x = 10\) into the regression equation
✓ 1
A1 83.2 s
✓ 1
Part (e)
R1 \(x = 40\) lies far outside the data range (2 to 17), so this is extrapolation and the linear trend may not continue
✓ 1
Common error: Regression of \(y\) on \(x\) is for estimating \(y\) from \(x\) within the range of the data; do not extrapolate far beyond it.
402IBMA3S-4-18r48-114.7 Discrete random variablesPaper 2Medium6 marks
Calculator
A beekeeper is paid to collect swarms. The number of swarms \(X\) collected in one call-out has probability distribution \(P(X = x) = k(x^2 + 1)\) for \(x = 0, 1, 2, 3\), where \(k\) is a constant. The beekeeper's income in € for a call-out is \(I = 150X + 80\).
(a)
Show that \(k = \frac{1}{18}\).
(2)
(b)
Find \(\mathrm{E}(X)\).
(2)
(c)
Find the expected income per call-out, in €.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(k(1 + 2 + 5 + 10) = 1\)
✓ 1
A1 \(18k = 1\), so \(k = \frac{1}{18}\) (AG)
✓ 1
Part (b)
M1 attempt to use \(\sum xP(X=x) = \frac{0 + 2 + 10 + 30}{18}\)
✓ 1
A1 \(\mathrm{E}(X) = \frac{7}{3}\)
✓ 1
Part (c)
M1 attempt to use \(\mathrm{E}(150X + 80) = 150\mathrm{E}(X) + 80\) (ECF from (b))
✓ 1
A1 € 430
✓ 1
Exam tip: \(\mathrm{E}(aX + b) = a\mathrm{E}(X) + b\), so you do not need to build a new distribution for \(I\).
403AASL4-q00074.1 Sampling, data and outliersPaper 2Hard7 marks
The times, in minutes, that ten employees take to travel to work are
Employee
1
2
3
4
5
6
7
8
9
10
Time (min)
22
14
26
18
29
21
17
25
20
23
An eleventh employee joins the company. Their travel time is \(x\) minutes, where \(x\in\mathbb{Z}^{+}\). Quartiles are found using the method of your GDC.
(a)
Given that \(x\ge 29\), find the lower quartile and the upper quartile of the eleven times.
(2)
(b)
Hence find the least value of \(x\) for which \(x\) is an outlier.
(2)
(c)
Given instead that \(x\le 17\), find all the values of \(x\) for which \(x\) is an outlier.
(2)
(d)
Explain why \(x\) cannot be an outlier if \(17\lt x\lt 29\).
for these values \(17\le Q_{1}\le 18\) and \(25\le Q_{3}\le 26\), so \(\text{IQR}\ge 7\) and the outlier boundaries are at most \(18-10.5=7.5\) and at least \(25+10.5=35.5\); every \(x\) with \(17\lt x\lt 29\) lies between them
R1
Accept a clear argument that \(x\) lies within the interval \(\left[Q_{1}-1.5\,\text{IQR},\ Q_{3}+1.5\,\text{IQR}\right]\).
404AASL4-q00174.2 Presentation of dataPaper 2Hard16 marks
The cumulative frequency graph shows the time, \(t\) minutes, that 160 people take to travel to work.
Graph drawn to scale
(a)
Use the graph to estimate the median time.
(1)
(b)
Use the graph to estimate the lower quartile and the upper quartile, and hence find the interquartile range.
(3)
(c)
Ninety per cent of the people take less than \(k\) minutes to travel to work. Use the graph to estimate \(k\).
(2)
(d)(i)
The information is also shown in the following frequency table.
Time, \(t\) (minutes)
\(0\lt t\le 10\)
\(10\lt t\le 20\)
\(20\lt t\le 30\)
\(30\lt t\le 40\)
\(40\lt t\le 50\)
\(50\lt t\le 60\)
Frequency
12
28
52
40
\(p\)
\(q\)
Write down the value of \(p\) and the value of \(q\).
(2)
(d)(ii)
Write down the modal class.
(1)
(d)(iii)
Find an estimate of the mean and an estimate of the standard deviation of the times.
(3)
(e)(i)
A researcher suggests that the times can be modelled by a normal distribution with the mean and standard deviation found in part (d)(iii).
Use this model to find the probability that a person takes more than 50 minutes, and compare your answer with the proportion of the 160 people who took more than 50 minutes.
(2)
(e)(ii)
Use the model to find \(\mathrm{P}(T\lt 0)\) and comment on what this shows about the model.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(28\) minutes
A1
Accept 27 to 28.5. (Read at cumulative frequency 80.)
(b)
\(Q_{1}=20\) minutes
A1
Accept 19 to 21 (read at 40).
\(Q_{3}=36\) minutes
A1
Accept 35.5 to 37.5 (read at 120).
IQR \(=16\) minutes
A1
Accept 15 to 17. FT from their quartiles.
(c)
reading at cumulative frequency \(0.9\times160=144\)
(M1)
\(k=45\)
A1
Accept 44 to 46.
(d)(i)
\(p=152-132=20\)
A1
\(q=160-152=8\)
A1
(d)(ii)
\(20\lt t\le 30\)
A1
(d)(iii)
use of mid-interval values 5, 15, 25, 35, 45, 55 with their frequencies
(M1)
mean \(=28.25\) minutes
A1
Accept 28.3. FT from their p and q.
standard deviation \(=12.5\) minutes
A1
\(12.5274\ldots\) (\(\sigma_{x}\)).
(e)(i)
\(\mathrm{P}(T\gt 50)=0.0413\)
A1
\(0.041265\ldots\) Accept 0.0409 to 0.0413 (from 12.5).
from the data, \(\dfrac{8}{160}=0.05\); the model gives a slightly lower value but the two are close
R1
FT from their q. Must see 0.05 and a comparison.
(e)(ii)
\(\mathrm{P}(T\lt 0)=0.0121\)
A1
\(0.012065\ldots\) Accept 0.0119 to 0.0121.
times cannot be negative, so the model is not perfect (it gives a non-zero probability to impossible values); it is only an approximation
R1
Answers: (a) 28 min · (b) 20, 36, IQR 16 · (c) 45 · (d)(i) \(p=20,\ q=8\), (ii) \(20\lt t\le 30\), (iii) 28.25, 12.5 · (e)(i) 0.0413 vs 0.05, (ii) 0.0121
405AASL4-q00184.2 Presentation of dataPaper 2Hard13 marks
Two groups of 100 people, X and Y, completed the same task. Group X had a short training session first; group Y did not. The cumulative frequency curves show the times, \(t\) minutes, taken to complete the task. In each group the longest time was 12 minutes.
Graph drawn to scale
(a)
Write down the median time for each group.
(2)
(b)
Find the interquartile range for each group.
(2)
(c)
Compare the times taken by the two groups.
(2)
(d)
Estimate the percentage of group Y who took no longer than the median time of group X.
(2)
(e)
The slowest 10% of group Y took longer than \(k\) minutes. Find the value of \(k\).
(2)
(f)
Determine, for each group, whether the time of 12 minutes is an outlier.
(3)
Show mark scheme
Working
Mark
Notes
(a)
X: \(5.5\) minutes
A1
Accept \(5.4\)–\(5.6\).
Y: \(7.5\) minutes
A1
Accept \(7.4\)–\(7.6\).
(b)
X: \(7-4=3\) minutes
A1
Y: \(9-6=3\) minutes
A1
(c)
group X has the lower median (5.5 compared with 7.5), so the people with training generally completed the task faster
R1
Must compare a measure of centre.
the IQRs are equal (3 minutes), so the spread of the middle half of the times is the same
R1
Must compare a measure of spread.
(d)
reading the curve for Y at \(t=5.5\)
(M1)
FT from their median of X.
about \(20\%\)
A1
Accept \(18\%\)–\(23\%\).
(e)
reading the curve for Y at cumulative frequency \(90\)
X: \(12\gt 11.5\), so 12 minutes is an outlier in group X
A1
Y: \(12\lt 13.5\), so 12 minutes is not an outlier in group Y
A1
Answers: (a) X 5.5, Y 7.5 · (b) 3 and 3 · (d) about 20% · (e) \(k=10.5\) · (f) outlier in X only
406AASL4-q00264.3 Measures of central tendency and dispersionPaper 2Hard10 marks
The waiting times, in minutes, of nine patients at a clinic were
Time (min)
12
15
17
20
22
25
29
31
49
(a)
Find the standard deviation and the variance of the waiting times.
(2)
(b)
Use your GDC to write down the lower quartile and the upper quartile.
(2)
(c)
Using your answers to part (b), show that the time of 49 minutes is not an outlier.
(2)
(d)
A receptionist finds the quartiles by hand. He includes the median in both halves of the data and finds the median of each half. Find the quartiles obtained by this method, and determine whether 49 minutes is then an outlier.
(3)
(e)
Comment on your answers to parts (c) and (d).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(\sigma=10.5\) minutes
A1
\(10.49985\ldots\)
variance \(=110\) minutes2
A1
\(110.247\ldots\) Accept \(110.2\); accept 110.25 from 10.5².
407AASL4-q00284.3 Measures of central tendency and dispersionPaper 2Hard15 marks
The annual salaries, in thousands of euros, of the 11 employees of a small company are
Salary (€000)
28
31
29
34
30
27
33
32
36
30
118
The salary of 118 thousand euros is that of the company director.
(a)
Find the mean and the median salary.
(2)
(b)
Find the standard deviation of the salaries.
(1)
(c)
Show that the director’s salary is an outlier.
(3)
(d)
In an advertisement for a job at the company, the director states that the “average salary” is €38 900. State, with a reason, whether this gives a fair picture of a typical salary at the company.
(2)
(e)
Find the mean and the standard deviation of the salaries of the 10 employees other than the director.
(2)
(f)(i)
Two pay rises are proposed for these 10 employees.
Under proposal 1, each salary increases by €1500. Write down the new mean and the new standard deviation of their salaries.
(2)
(f)(ii)
Under proposal 2, each salary increases by 5%. Find the new mean and the new standard deviation of their salaries.
(2)
(g)
Using your answers to part (f), state which proposal increases the spread of the salaries, and explain why.
(1)
Show mark scheme
Working
Mark
Notes
(a)
mean \(=38.9\) thousand euros
A1
\(38.9090\ldots\)
median \(=31\) thousand euros
A1
(b)
\(25.1\) thousand euros
A1
\(25.1376\ldots\)
(c)
\(Q_{1}=29,\ Q_{3}=34\)
A1
\(34+1.5\times5\)
M1
\(=41.5\) and \(118\gt 41.5\)
A1
(d)
no, it does not give a fair picture
A1
Do not award A1 without R1.
the mean is pulled up by the director’s (outlying) salary; 10 of the 11 employees earn less than €38 900 (the median is €31 000)
R1
(e)
mean \(=31\) thousand euros
A1
standard deviation \(=2.65\) thousand euros
A1
\(2.64575\ldots\)
(f)(i)
mean \(=32.5\) thousand euros
A1
FT from (e).
standard deviation \(=2.65\) thousand euros (unchanged)
A1
FT from (e).
(f)(ii)
mean \(=1.05\times31=32.55\) thousand euros
A1
Accept 32.6. FT from (e).
standard deviation \(=1.05\times2.6458=2.78\) thousand euros
A1
\(2.77804\ldots\) FT from (e).
(g)
proposal 2: multiplying every salary by 1.05 multiplies the standard deviation by 1.05, whereas adding the same amount to every salary does not change the spread
408AASL4-q00344.4 Correlation and regression of y on xPaper 2Hard8 marks
In school A, ten students recorded the number of hours, \(x\), they spent practising for a music examination and their examination score, \(y\).
\(x\)
2
3
4
5
7
9
10
12
14
15
\(y\)
43
52
63
67
69
53
47
75
75
85
In school B, six students recorded the same information.
\(x\)
3
5
6
8
12
13
\(y\)
38
57
68
51
81
68
The table gives critical values of \(r\) for a sample of size \(n\). If \(\lvert r\rvert\) is greater than the critical value, there is significant evidence of linear correlation.
\(n\)
6
8
10
12
16
Critical value
0.811
0.707
0.632
0.576
0.497
(a)
Find the value of \(r\) for school A and for school B.
(2)
(b)
Determine, for each school, whether there is significant evidence of linear correlation between \(x\) and \(y\).
(2)
(c)
Explain why the data for school B give weaker evidence of correlation than the data for school A, even though the value of \(r\) is greater.
(1)
(d)
The data for the two schools are combined. Find \(r\) for the combined data and determine whether there is significant evidence of linear correlation.
(2)
(e)
A teacher concludes that practising more causes students to score higher. Comment on this conclusion.
(1)
Show mark scheme
Working
Mark
Notes
(a)
school A: \(r=0.679\)
A1
\(0.679290\ldots\)
school B: \(r=0.758\)
A1
\(0.758382\ldots\)
(b)
school A: \(0.679\gt 0.632\) (\(n=10\)), so there is significant evidence of correlation
R1
FT from their r.
school B: \(0.758\lt 0.811\) (\(n=6\)), so there is not significant evidence of correlation
R1
FT from their r.
(c)
school B has a smaller sample, so a strong-looking value of \(r\) is more likely to occur by chance; the critical value is larger for smaller \(n\)
R1
(d)
\(r=0.701\)
A1
\(0.701325\ldots\)
\(0.701\gt 0.497\) (\(n=16\)), so there is significant evidence of linear correlation
R1
FT from their r.
(e)
correlation does not imply causation; other factors (e.g. ability, or the teacher) may affect both practice time and score
R1
Answers: (a) A 0.679, B 0.758 · (b) A significant, B not · (d) 0.701, significant
409AASL4-q00374.4 Correlation and regression of y on xPaper 2Hard10 marks
The fuel consumption, \(y\) litres per 100 km, of a car was measured at ten different constant speeds, \(x\) km h−1.
\(x\)
30
40
50
60
70
80
90
100
110
120
\(y\)
9.1
7.6
6.5
5.8
5.5
5.7
6.2
7
8.1
9.4
Graph drawn to scale
(a)
Find the value of \(r\) for all ten data pairs.
(1)
(b)
A student says that, since \(r\) is close to zero, there is no relationship between speed and fuel consumption. Comment on this statement.
(2)
(c)
Using only the data for speeds from 70 to 120 km h−1, find the value of \(r\) and the equation of the regression line of \(y\) on \(x\).
(3)
(d)
Use the regression line from part (c) to estimate the fuel consumption at a speed of 85 km h−1.
(2)
(e)
Use the regression line from part (c) to estimate the fuel consumption at 40 km h−1. Comment on your answer.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(r=0.131\)
A1
\(0.131307\ldots\)
(b)
the scatter diagram shows a strong relationship: consumption decreases and then increases (a U-shaped curve)
R1
\(r\) only measures linear correlation, so a value near zero only shows that there is little linear relationship
R1
(c)
\(r=0.967\)
A1
\(0.967217\ldots\)
\(a=0.0786\)
A1
\(0.0785714\ldots\)
\(y=0.0786x-0.481\)
A1
\(-0.480952\ldots\)
(d)
\(0.0785714(85)-0.480952\)
(M1)
\(6.20\) litres per 100 km
A1
\(6.19761\ldots\)
(e)
\(2.66\) litres per 100 km
A1
\(2.66190\ldots\)
unreliable: 40 is outside the range 70 to 120 used for the line (extrapolation), and the data show that consumption at 40 km h−1 is actually about 7.6, so the linear model does not apply there
410AASL4-q00394.4 Correlation and regression of y on xPaper 2Hard15 marks
An online shop recorded the amount it spent on advertising, \(x\) hundred euros, and the number of orders it received, \(y\), in each of ten weeks.
\(x\) (€00)
18
20
21
23
24
26
27
29
30
32
\(y\)
64
75
73
88
91
104
103
119
121
134
(a)
Find the mean point \((\bar{x},\,\bar{y})\).
(2)
(b)
Find the value of \(r\) and describe the correlation.
(2)
(c)
Find the equation of the regression line of \(y\) on \(x\).
(2)
(d)
Interpret the gradient of the regression line in context.
(1)
(e)
Use the regression line to estimate the number of orders in a week when €2800 is spent on advertising.
(2)
(f)
Use the regression line to find the number of orders it predicts for a week when €500 is spent on advertising. Comment on your answer.
(2)
(g)(i)
In week 11 the shop spent €2500 on advertising and received 110 orders. The data for week 11 are added to the data for the first ten weeks.
Find the value of \(r\) for the eleven weeks.
(1)
(g)(ii)
Find the equation of the regression line of \(y\) on \(x\) for the eleven weeks.
(2)
(g)(iii)
Compare the regression lines in parts (c) and (g)(ii), and explain why the gradient has not changed.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(\bar{x}=25\)
A1
\(\bar{y}=97.2\)
A1
(b)
\(r=0.994\)
A1
\(0.993750\ldots\)
strong positive (linear) correlation
A1
(c)
\(a=4.98\)
A1
\(4.98421\ldots\)
\(y=4.98x-27.4\)
A1
\(-27.4052\ldots\)
(d)
for each extra €100 spent on advertising, the shop receives about 5 more orders per week
R1
(e)
\(4.98421(28)-27.4052\)
(M1)
\(112\) orders
A1
\(112.152\ldots\) Accept 112.1 or 112.2.
(f)
\(-2.48\)
A1
\(-2.48421\ldots\) Accept \(-2.5\).
a negative number of orders is impossible: €500 \((x=5)\) is far outside the range of the data (€1800 to €3200), so this is extrapolation and is not reliable
R1
(g)(i)
\(r=0.979\)
A1
\(0.978619\ldots\)
(g)(ii)
\(a=4.98\)
A1
\(4.98421\ldots\)
\(y=4.98x-26.2\)
A1
\(-26.2416\ldots\)
(g)(iii)
the new line is the line in (c) moved up by about 1.16 (same gradient, larger intercept): the new point has \(x=25=\bar{x}\), so \(\bar{x}\) is unchanged and the point does not tilt the line; it only raises the mean of \(y\) (to 98.4) through which the line must pass
R1
Must refer to the new \(x\)-value being equal to the mean of \(x\).
411AASL4-q00574.6 Combined events and diagramsPaper 2Hard16 marks
On any working day, the probability that it rains is 0.3. If it rains, the probability that Mia is late for work is 0.25. If it does not rain, the probability that Mia is late for work is 0.1. The tree diagram shows this information.
R = rains, L = Mia is late
(a)
Write down the values of \(p\), \(q\) and \(r\).
(2)
(b)
Find the probability that Mia is late for work on a given day.
(2)
(c)
Given that Mia is late for work, find the probability that it rained.
(3)
(d)(i)
Assume that whether Mia is late is independent from day to day. Let \(L\) be the number of days on which Mia is late in a period of 20 working days.
Find \(\mathrm{E}(L)\).
(1)
(d)(ii)
Find the probability that Mia is late on at most 2 of the 20 days.
(2)
(d)(iii)
Find the probability that Mia is late on more than 4 of the 20 days.
(2)
(e)
A working week has 5 days. A week is called “bad” if Mia is late on at least 3 days. Find the probability that at least one of the next 4 weeks is bad.
(3)
(f)
Suggest a reason why the assumption of independence in part (d) may not be valid.
412AASL4-q00674.7 Discrete random variablesPaper 2Hard15 marks
A bakery makes a speciality cake each morning. The daily demand, \(D\), for the cake has the probability distribution shown. Any cake not sold that day is thrown away.
\(d\)
0
1
2
3
4
\(\mathrm{P}(D=d)\)
0.1
0.2
0.35
0.25
0.1
Each cake costs $6 to make and is sold for $15. Demand is never more than 4 cakes and is independent from day to day.
(a)
Find \(\mathrm{E}(D)\).
(2)
(b)
Find the expected number of days in a 30-day month on which there is no demand for the cake.
(1)
(c)(i)
The bakery makes 3 cakes each day. Let \(Y\) be the daily profit, in dollars.
Find the probability distribution of \(Y\).
(3)
(c)(ii)
Find \(\mathrm{E}(Y)\).
(2)
(d)
Show that the expected daily profit is $12 if the bakery makes 2 cakes each day.
(2)
(e)
Find the expected daily profit if the bakery makes 4 cakes each day, and hence determine the number of cakes the bakery should make each day to maximize its expected profit.
(2)
(f)
The bakery makes 2 cakes each day for 30 days. Find the probability that it sells all of its cakes on at least 20 of these days.
413AASL4-q00714.8 The binomial distributionPaper 2Hard9 marks
A basketball player takes 15 free throws in each training session. The probability that she scores with any free throw is 0.78. Let \(X\) be the number of free throws she scores in a session.
(a)
State one assumption needed for \(X\) to be modelled by a binomial distribution.
(1)
(b)
Find the probability that she scores at least 12 free throws in a session.
(2)
(c)
Given that she scores at least 12 free throws in a session, find the probability that she scores exactly 12.
(3)
(d)
Find the probability that, in 8 sessions, she scores at least 12 free throws in exactly 5 of the sessions.
(3)
Show mark scheme
Working
Mark
Notes
(a)
the outcomes of different free throws are independent (or the probability of scoring is the same, 0.78, for every throw)
recognizing a second binomial: \(Y\sim\mathrm{B}(8,\,0.572625\ldots)\)
(M1)
\(\mathrm{P}(Y=5)\)
(M1)
\(0.269\)
A1
\(0.269133\ldots\) Using the 3 s.f. value 0.573 also gives 0.269.
Answers: (b) 0.573 · (c) 0.429 · (d) 0.269
414AASL4-q00724.8 The binomial distributionPaper 2Hard8 marks
A tour company uses a minibus with 20 passenger seats. From past records, the probability that a passenger who has bought a ticket turns up for a tour is 0.85, independently of other passengers. The company sells \(n\) tickets for each tour, where \(n\ge 20\).
(a)
For \(n=22\), find the probability that more passengers turn up than there are seats.
(2)
(b)
Find the greatest value of \(n\) for which the probability that more passengers turn up than there are seats is less than 0.25.
(3)
(c)
For \(n=22\), the company pays $40 to each passenger who turns up but cannot get a seat. Find the expected amount the company pays per tour.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(X\sim\mathrm{B}(22,\,0.85)\) and \(\mathrm{P}(X\ge 21)\)
(M1)
\(0.137\)
A1
\(0.136724\ldots\)
(b)
\(\mathrm{P}(X\ge 21)\lt 0.25\) for \(X\sim\mathrm{B}(n,\,0.85)\); table of values
416AASL4-q00784.8 The binomial distributionPaper 2Hard9 marks
In each of 80 trays, 5 seeds are planted. The number of seeds that germinate in each tray is recorded.
Number germinating
0
1
2
3
4
5
Number of trays
1
3
9
21
28
18
(a)
Find the mean number of seeds germinating per tray.
(2)
(b)
Hence estimate the probability, \(p\), that a seed germinates.
(1)
(c)
The number of seeds germinating in a tray is modelled by \(\mathrm{B}(5,\,p)\), using your value of \(p\). Find the expected number of trays, out of 80, in which all 5 seeds germinate.
(2)
(d)
Find the variance of the data and the variance predicted by the model.
(2)
(e)
Use your answers to part (d) to comment on the suitability of the binomial model, suggesting a reason in context.
417AASL4-q00794.8 The binomial distributionPaper 2Hard14 marks
Eva plays tennis matches against Lina. In each set, the probability that Eva wins the set is 0.7, independently of all other sets. In a best-of-three match, the first player to win two sets wins the match.
(a)
Find the probability that Eva wins a best-of-three match in exactly two sets.
(1)
(b)
Show that the probability that Eva wins a best-of-three match is 0.784.
(3)
(c)
Let \(Y\sim\mathrm{B}(3,\,0.7)\). Explain why the probability that Eva wins a best-of-three match is equal to \(\mathrm{P}(Y\ge 2)\).
(2)
(d)
In a best-of-five match, the first player to win three sets wins the match. Find the probability that Eva wins a best-of-five match.
(2)
(e)
Comment on your answers to parts (b) and (d).
(1)
(f)(i)
Eva and Lina play 20 best-of-three matches. Let \(W\) be the number of matches that Eva wins.
Find \(\mathrm{E}(W)\).
(1)
(f)(ii)
Find the probability that Eva wins at least 16 matches.
(2)
(g)
Find the least probability of winning a set that Eva would need for the probability that she wins a best-of-five match to be at least 0.9.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(0.7^{2}=0.49\)
A1
(b)
Eva wins in three sets: LWW or WLW
M1
\(2\times0.7^{2}\times0.3=0.294\)
A1
\(0.49+0.294\)
A1
\(=0.784\)
AG
(c)
imagine all three sets are always played; the extra set does not change who wins
R1
Eva wins the match exactly when she wins at least 2 of the 3 sets, and the number of sets she wins is \(\mathrm{B}(3,\,0.7)\)
R1
(d)
\(\mathrm{P}(Z\ge 3)\) for \(Z\sim\mathrm{B}(5,\,0.7)\)
(M1)
\(0.837\)
A1
\(0.83692\)
(e)
the longer match gives the stronger player (Eva) a greater probability of winning
418AASL4-q00914.9 The normal distributionPaper 2Hard14 marks
The masses, \(M\) grams, of eggs from a farm are normally distributed with mean 62 g and standard deviation 4.5 g. An egg is classed as large if its mass is more than 65 g.
Diagram not to scale
(a)
Find \(\mathrm{P}(55\lt M\lt 65)\).
(2)
(b)
Find the mass \(m\) such that 95% of eggs have a mass less than \(m\) grams.
(2)
(c)
An egg is chosen at random and is found to be large. Find the probability that its mass is more than 70 g.
(3)
(d)(i)
Eggs are packed in boxes of 12. Assume the masses of the eggs in a box are independent.
Find the probability that a box contains exactly 4 large eggs.
(2)
(d)(ii)
Find the probability that a box contains at least 3 large eggs.
(2)
(e)
The farmer changes the feed. The mean mass is still 62 g, but now 30% of eggs are large. Find the new standard deviation.
(3)
Show mark scheme
Working
Mark
Notes
(a)
normalcdf(55, 65, 62, 4.5)
(M1)
\(0.688\)
A1
\(0.687600\ldots\)
(b)
\(\mathrm{P}(M\lt m)=0.95\), inverse normal
(M1)
\(m=69.4\) g
A1
\(69.4018\ldots\)
(c)
\(\mathrm{P}(M\gt 65)=0.252492\ldots\) and \(\mathrm{P}(M\gt 70)=0.0377201\ldots\)
Answers: (a) 0.688 · (b) 69.4 g · (c) 0.149 · (d)(i) 0.196, (ii) 0.617 · (e) 5.72 g
419AASL4-q00964.10 Regression of x on yPaper 2Hard12 marks
A manager records, for eight sales staff, the number of years of experience, \(x\), and the number of sales made in one week, \(y\).
\(x\)
2
4
5
7
8
10
12
13
\(y\)
9
6
14
8
15
11
18
13
(a)
Find the value of Pearson’s product-moment correlation coefficient, \(r\), and describe the correlation.
(2)
(b)
Find the equation of the regression line of \(y\) on \(x\).
(2)
(c)
Find the equation of the regression line of \(x\) on \(y\).
(2)
(d)
A new member of staff made 15 sales in one week. Estimate his number of years of experience using the appropriate regression line.
(2)
(e)
A colleague estimates the experience by rearranging the equation from part (b) and obtains a different answer. Find this answer, and explain why the two estimates are so different for these data.
(3)
(f)
Write down the coordinates of the point where the two regression lines intersect.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(r=0.610\)
A1
\(0.609749\ldots\)
moderate positive (linear) correlation
A1
(b)
gradient \(0.626\)
A1
\(0.625738\ldots\)
\(y=0.626x+6.98\)
A1
\(6.97874\ldots\)
(c)
gradient \(0.594\)
A1
\(0.594170\ldots\)
\(x=0.594y+0.643\)
A1
\(0.643497\ldots\)
(d)
\(x=0.594170(15)+0.643497\)
(M1)
Must use the line of x on y.
\(9.56\) years
A1
\(9.55605\ldots\) Accept 9.55 to 9.56.
(e)
\(15=0.625738x+6.97874\)
(M1)
\(x=12.8\) years
A1
\(12.8188\ldots\)
the correlation is only moderate (\(|r|\) not close to 1), so the two regression lines are far apart; they coincide only when \(|r|=1\)
R1
(f)
\((7.625,\ 11.75)\)
A1
The mean point \((\bar{x},\,\bar{y})\). Accept \((7.63,\ 11.8)\).
Answers: (a) 0.610 · (b) \(y=0.626x+6.98\) · (c) \(x=0.594y+0.643\) · (d) 9.56 years · (e) 12.8 years · (f) \((7.625,\,11.75)\)
420AASL4-q01094.11 Conditional probability and independencePaper 2Hard7 marks
In a country, the heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm, and the heights of adult women are normally distributed with mean 165 cm and standard deviation 6.5 cm. A man is described as tall if his height is more than 185 cm, and a woman is described as tall if her height is more than 170 cm. One man and one woman are chosen independently at random.
(a)
Find the probability that the man is tall and the probability that the woman is tall.
(2)
(b)
Find the probability that exactly one of them is tall.
(3)
(c)
Given that exactly one of them is tall, find the probability that it is the man.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{P}(\text{man tall})=0.159\)
A1
\(0.158655\ldots\)
\(\mathrm{P}(\text{woman tall})=0.221\)
A1
\(0.220878\ldots\)
(b)
\(m(1-w)+w(1-m)\) with their values
M1
Two products, added.
\(0.123611\ldots+0.185833\ldots\)
(A1)
\(0.309\)
A1
\(0.309446\ldots\)
(c)
\(\dfrac{0.123611\ldots}{0.309446\ldots}\)
(M1)
\(0.399\)
A1
\(0.399460\ldots\)
Answers: (a) 0.159, 0.221 · (b) 0.309 · (c) 0.399
421AASL4-q01114.11 Conditional probability and independencePaper 2Hard15 marks
In a youth athletics squad, long-jump distances, \(L\) metres, are normally distributed with mean 5.6 m and standard deviation 0.32 m, and 100 m sprint times, \(T\) seconds, are normally distributed with mean 12.8 s and standard deviation 0.45 s. An athlete qualifies for a regional final if she jumps more than 6.0 m or runs 100 m in less than 12.2 s (or both). For any one athlete, \(L\) and \(T\) are independent.
(a)
Find \(\mathrm{P}(L\gt 6.0)\).
(2)
(b)
Find \(\mathrm{P}(T\lt 12.2)\).
(2)
(c)
Find the probability that a randomly chosen athlete qualifies.
(3)
(d)
Given that an athlete qualifies, find the probability that she achieves both standards.
(2)
(e)
Eight athletes from the squad are chosen at random. Find the probability that at least two of them qualify.
(3)
(f)
The organisers change the long-jump standard to \(d\) metres, keeping the sprint standard at 12.2 s, so that the probability that an athlete qualifies is 0.25. Find \(d\).
422AASL4-q01144.12 Standardization and inverse normalPaper 2Hard8 marks
The masses of bags of flour filled by a machine are normally distributed with mean \(\mu\) grams and standard deviation \(\sigma\) grams. It is known that 8% of bags have a mass less than 990 g and 15% of bags have a mass greater than 1030 g.
(a)
Find the value of \(\mu\) and the value of \(\sigma\).
(5)
(b)
The machine is adjusted so that the mean changes but the standard deviation does not. After the adjustment, only 2% of bags have a mass less than 990 g. Find the new mean.
(3)
Show mark scheme
Working
Mark
Notes
(a)
use of inverse normal on \(Z\): \(\mathrm{P}(Z\lt z_{1})=0.08\) or \(\mathrm{P}(Z\lt z_{2})=0.85\)
(M1)
\(z_{1}=-1.40507\ldots\) and \(z_{2}=1.03643\ldots\)
A1
\(\dfrac{990-\mu}{\sigma}=-1.40507\ldots\) and \(\dfrac{1030-\mu}{\sigma}=1.03643\ldots\)
Answers: (a) \(\mu=1013\) g, \(\sigma=16.4\) g · (b) 1024 g (1023.6)
423AASL4-q01174.12 Standardization and inverse normalPaper 2Hard8 marks
The masses, \(M\) grams, of fish in a fish farm are modelled by a normal distribution in which the standard deviation is 25% of the mean. It is known that 5% of the fish have a mass greater than 120 g.
(a)
Find the mean and the standard deviation of \(M\).
(4)
(b)
Find the probability that a fish has a mass less than 50 g.
(2)
(c)
Find \(\mathrm{P}(M\lt 0)\) and comment on what your answer shows about the suitability of the model.
424AASL4-q01184.12 Standardization and inverse normalPaper 2Hard8 marks
Metal rods are cut to a target length of 250 mm. A rod is accepted if its length is between 245 mm and 255 mm. The lengths of the rods are normally distributed with mean \(\mu\) mm and standard deviation \(\sigma\) mm.
(a)
For \(\mu=251\) and \(\sigma=1.8\), find the probability that a rod is accepted.
(2)
(b)
With \(\sigma=1.8\), the probability that a rod is accepted must be at least 0.99. Find the set of possible values of \(\mu\), giving the end points correct to two decimal places.
(3)
(c)
The mean is set to \(\mu=250\). Find the greatest value of \(\sigma\) for which at least 99.5% of rods are accepted.
(3)
Show mark scheme
Working
Mark
Notes
(a)
normalcdf(245, 255, 251, 1.8)
(M1)
\(0.986\)
A1
\(0.986436\ldots\)
(b)
\(\mathrm{P}(245\lt X\lt 255)=0.99\) solved for \(\mu\) with technology (graph or solver)
M1
\(\mu=249.23\) or \(\mu=250.77\)
A1
\(249.234\ldots,\ 250.765\ldots\), symmetric about 250.
Answers: (a) 0.986 · (b) \(249.23\le\mu\le250.77\) · (c) 1.78 mm
425IBMA2S-050-02-114.9 The normal distributionPaper 2Hard12 marks
Calculator
The lifetime, L hours, of an LED light bulb is modelled by a normal distribution. The median lifetime is 25 000 hours, and 5% of bulbs fail before 18 000 hours.
(a)
Write down the mean of L.
(1)
(b)
Find the standard deviation of L.
(2)
(c)
Find P(L > 30 000).
(2)
(d)
Find the least value of n.
(4)
(e)
Find P(L > 30 000 | L > 20 000).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 μ = 25 000 hours, since the median equals the mean for a normal distribution
Common error: In part (d), solving the inequality gives n > 54.03. Rounding this to 54 gives a probability just below 0.999. The least integer that works is 55.
426IBMA2S-050-02-124.9 The normal distributionPaper 2Hard10 marks
Calculator
The diameter, D mm, of ball bearings from a production line is normally distributed with mean μ and standard deviation σ, where σ = μ/50. It is found that 3% of bearings have a diameter greater than 8.2 mm.
(a)
Find μ.
(4)
(b)
Find P(D < 7.8).
(2)
(c)
Find the probability that a pack is returned.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitutes σ = μ/50 into a standardised equation ; A1 z = 1.88079 for an upper-tail area of 0.03 ; M1 (8.2 − μ)/(μ/50) = 1.88079, rearranged to μ(1 + 1.88079/50) = 8.2 ; A1 μ = 7.90 mm (7.9027…)
Exam tip: When σ is given in terms of μ, replace σ before standardising so the equation has only one unknown. Then factorise out μ to solve it.
427IBMA2S-050-07-124.7 Discrete random variablesPaper 2Hard14 marks
Calculator
In the prize round of a trivia app, a player chooses one of three options. • Option A: answer one question; win $1500 if correct, nothing otherwise. • Option B: answer two questions; win $3000 if both are correct, $500 if exactly one is correct, nothing otherwise. • Option C: take a guaranteed $400 without answering.
The player answers each question correctly with probability p, independently, where 0 < p ≤ 1. Let Y be the player's winnings, in dollars, under Option B.
(a)
For p = 0.6, find the probability distribution of Y.
(3)
(b)
For p = 0.6, find E(Y).
(2)
(c)
Show that, in general, \(E(Y)=2000p^2+1000p\).
(3)
(d)
Find the set of values of p for which Option B has a greater expected winnings than Option A.
(3)
(e)
Find the least value of p for which the expected winnings under Option B are at least the $400 guaranteed by Option C.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses independence to multiply probabilities
✓ 1
A1 \(P(Y=3000)=0.36\) and \(P(Y=0)=0.16\)
✓ 1
A1 \(P(Y=500)=2(0.6)(0.4)=0.48\)
✓ 1
Part (b)
M1 \(3000(0.36)+500(0.48)+0(0.16)\)
✓ 1
A1 \(E(Y)=1320\) dollars
✓ 1
Part (c)
M1 \(P(\text{exactly one correct})=2p(1-p)\)
✓ 1
M1 \(E(Y)=3000p^2+500\cdot2p(1-p)\)
✓ 1
A1 \(3000p^2+1000p-1000p^2=2000p^2+1000p\)
✓ 1
Part (d)
M1 \(2000p^2+1000p>1500p\)
✓ 1
A1 \(2000p^2-500p>0\), i.e. \(500p(4p-1)>0\)
✓ 1
A1 \(\frac14<p\le1\)
✓ 1
Part (e)
M1 \(2000p^2+1000p=400\)
✓ 1
A1 \(10p^2+5p-2=0\) or equivalent
✓ 1
A1 \(p=0.262\)
✓ 1
Common error: When solving the inequality in part (d), do not divide by p without noting p > 0, and remember to restrict the answer to p ≤ 1. The negative root in part (e) is not a valid probability.
428IBMA2S-050-01-114.6 Combined events and diagramsPaper 2Hard13 marks
Calculator
Of all emails arriving at an office account, 30% are spam. A filter flags 95% of spam emails and also wrongly flags 2% of legitimate emails. Emails arrive independently of one another, and what happens on one day is independent of every other day.
Diagram not to scale
(a)
Find the probability that a randomly chosen email is flagged.
(2)
(b)
An email is flagged. Find the probability that it is legitimate.
(3)
(c)
The account receives 120 emails each day. Find the expected number of legitimate emails that are wrongly flagged in one day.
(2)
(d)
Find the probability that on a given day at least 2 legitimate emails are wrongly flagged.
(3)
(e)
Using your unrounded answer to part (d), find the probability that, over a 5-day working week, at least 2 legitimate emails are wrongly flagged on exactly 4 of the days.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(0.3 \times 0.95 + 0.7 \times 0.02\)
✓ 1
A1 0.299
✓ 1
Part (b)
M1 uses conditional probability \(\frac{P(\text{legitimate and flagged})}{P(\text{flagged})}\)
✓ 1
M1 \(\frac{0.014}{0.299}\)
✓ 1
A1 0.0468
✓ 1
Part (c)
M1 P(legitimate and flagged) \(= 0.014\), multiplied by 120
Common error: In part (c), using 0.02 instead of \(0.7 \times 0.02\) applies the filter's error rate as if every email were legitimate. Only 70% of all emails are legitimate.
429IBMA2S-050-04-114.8 The binomial distributionPaper 2Hard15 marks
Calculator
A basketball player takes a sequence of free throws. She scores her first throw with probability \(0.8\). For each later throw, if she scored the previous throw she scores with probability \(0.85\); if she missed the previous throw she scores with probability \(0.6\). She takes exactly three throws.
(a)
Find the probability that she scores both of her first two throws.
(2)
(b)
Find the probability that she scores exactly one of her first two throws.
(3)
(c)
Find the probability that she scores her second throw.
(2)
(d)
Determine whether the events 'scores the first throw' and 'scores the second throw' are independent.
(2)
(e)
Find the probability that she scores all three throws.
(2)
(f)
Given that she scores exactly two of the three throws, find the probability that she missed her first throw.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(0.8\times0.85\)
✓ 1
A1 \(0.68\)
✓ 1
Part (b)
M1 score-then-miss \(0.8\times0.15=0.12\)
✓ 1
M1 miss-then-score \(0.2\times0.6=0.12\)
✓ 1
A1 \(0.24\)
✓ 1
Part (c)
M1 \(0.8\times0.85+0.2\times0.6\)
✓ 1
A1 \(0.8\)
✓ 1
Part (d)
M1 compares \(P(F\cap S)=0.68\) with \(P(F)P(S)=0.8\times0.8=0.64\)
✓ 1
R1 \(0.68\neq0.64\), so not independent
✓ 1
Part (e)
M1 \(0.8\times0.85\times0.85\)
✓ 1
A1 \(0.578\)
✓ 1
Part (f)
M1 identifies the three sequences SSM, SMS, MSS
✓ 1
A1 probabilities \(0.102\), \(0.072\), \(0.102\); total \(0.276\)
Common error: Equal marginal probabilities, \(P(F)=P(S)=0.8\), do not make the throws independent; the dependence shows up only in the joint probability.
430IBMA2S-050-05-094.2 Presentation of dataPaper 2Hard13 marks
Calculator
An engineer tested the tensile strength, s kN, of 80 steel bolts. 40 ≤ s < 44: 5 44 ≤ s < 48: 12 48 ≤ s < 52: p 52 ≤ s < 56: 21 56 ≤ s < 60: q 60 ≤ s < 64: 4 Exactly half of the bolts have a strength less than 52 kN.
(a)
Find the value of p and the value of q.
(3)
(b)
Find an estimate of the mean tensile strength.
(2)
(c)
Write down the cumulative frequencies at s = 44, 48, 52, 56, 60 and 64.
(2)
(d)
From a cumulative frequency curve, the lower quartile is 48.5 kN and the upper quartile is 55.8 kN. Find the interquartile range.
(1)
(e)
Bolts whose strength is more than 1.5 × IQR below the lower quartile are treated as outliers. Explain why none of these bolts is an outlier.
(2)
(f)
The bolts are acceptable if at least 75% have a strength of at least 48 kN. Determine whether this batch is acceptable.
M1 use of mid-interval values 42, 46, 50, 54, 58, 62 with frequencies A1 4164/80 = 52.1 kN (52.05)
✓ 2
Part (c)
M1 running totals A1 5, 17, 40, 61, 76, 80
✓ 2
Part (d)
A1 55.8 − 48.5 = 7.3 kN
✓ 1
Part (e)
M1 lower boundary = 48.5 − 1.5 × 7.3 = 37.55 kN R1 every bolt has s ≥ 40 > 37.55, so none is an outlier
✓ 2
Part (f)
M1 80 − 17 = 63 bolts with s ≥ 48 A1 63/80 = 78.75% R1 78.75% ≥ 75%, so the batch is acceptable
✓ 3
Exam tip: In part (e) the individual values are unknown, but the lowest class boundary of 40 kN is enough to rule out any low outlier.
431IBMA2S-050-00-084.12 Standardization and inverse normalPaper 2Hard13 marks
Calculator
A factory makes phone screens on two machines. Machine X makes 70% of the screens and machine Y makes the rest. The thickness of screens from machine X is normally distributed with mean 0.70 mm and standard deviation 0.02 mm. The thickness of screens from machine Y is normally distributed with mean 0.72 mm and standard deviation 0.015 mm. A screen is rejected if its thickness is greater than 0.74 mm.
(a)
Find the probability that a screen made by machine X is rejected.
(2)
(b)
Find the probability that a screen made by machine Y is rejected.
(2)
(c)
A screen is chosen at random from the factory's total output. Find the probability that it is rejected.
(3)
(d)
Given that a randomly chosen screen is rejected, find the probability that it was made by machine Y.
(3)
(e)
Two screens are chosen at random, independently, from the total output. Find the probability that exactly one of them is rejected.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(P(T_X>0.74)\) or \(P(Z>2)\)
✓ 1
A1 0.0228 (0.022750…)
✓ 1
Part (b)
M1 \(P(T_Y>0.74)\) or \(P(Z>1.333\ldots)\)
✓ 1
A1 0.0912 (0.091211…)
✓ 1
Part (c)
M1 one correct branch product, e.g. \(0.7\times0.02275\)
✓ 1
M1 sum of both products \(0.7\times0.02275+0.3\times0.09121\)
✓ 1
A1 0.0433 (0.043288…)
✓ 1
Part (d)
M1 numerator \(0.3\times0.09121=0.027363\)
✓ 1
M1 dividing by answer to (c)
✓ 1
A1 0.632
✓ 1
Part (e)
M1 \(p(1-p)\) with \(p=0.043288\)
✓ 1
M1 multiplying by 2 for the two orders
✓ 1
A1 0.0828
✓ 1
Common error: In (e) forgetting the factor 2 gives only the probability that the first is rejected and the second is not; both orders must be counted.
432IBMA2S-050-00-114.11 Conditional probability and independencePaper 2Hard12 marks
Calculator
The distances, \(L\) metres, achieved by a long jumper are normally distributed with mean 7.6 m and standard deviation \(\sigma\) m. 20% of her jumps exceed 7.9 m.
(a)
Find the value of \(\sigma\).
(3)
(b)
Find the probability that a jump is less than 7.2 m.
(2)
(c)
A jump counts as a qualifying jump if it exceeds 7.5 m. Given that a jump is a qualifying jump, find the probability that it exceeds 7.9 m.
(4)
(d)
In a competition she takes three independent jumps. Find the probability that at least one jump exceeds 7.9 m.
M1 use of complement, \(1-P(\text{no jump exceeds 7.9 m})\)
✓ 1
M1 \(0.8^3=0.512\)
✓ 1
A1 0.488
✓ 1
Common error: In (c) multiplying \(P(L>7.9)\) by \(P(L>7.5)\) treats nested events as independent; since exceeding 7.9 m implies exceeding 7.5 m, the intersection is just \(P(L>7.9)\).
433IBMA2S-050-06-074.12 Standardization and inverse normalPaper 2Hard11 marks
Calculator
A factory has two production lines making steel bolts. The standard deviations below are calculated with divisor n, treating each sample as the complete data set. Line A: 40 bolts sampled, mean length 50.2 mm, standard deviation 0.5 mm. Line B: n bolts sampled, mean length 49.6 mm, standard deviation 0.4 mm. The mean length of all the sampled bolts combined is 49.84 mm.
(a)
Find the value of n.
(3)
(b)
Find \(\sum x^2\) for the lengths of the bolts from Line A.
(2)
(c)
Find the standard deviation of the lengths of all the sampled bolts combined.
(4)
(d)
A bolt in the combined sample has length 50.9 mm. Find how many standard deviations this is above the combined mean.
A1 1.99 standard deviations (accept 2.00 from σ = 0.531)
✓ 1
Common error: Rounding the sums of squares early wipes out the variance, which is a tiny difference of two large numbers. Keep full values until the final step.
434IBMA2S-050-06-114.3 Measures of central tendency and dispersionPaper 2Hard12 marks
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A planetary science lab measures the masses, x grams, of 12 meteorite fragments. To simplify the arithmetic, each mass is coded as y = x − 50. For the coded values, \(\sum y = 30\) and \(\sum y^2 = 410\).
(a)
Find the mean mass of the fragments.
(2)
(b)
Find the standard deviation of the masses.
(3)
(c)
Show that \(\sum x^2 = 33410\).
(2)
(d)
A 13th fragment of mass m grams is added and the mean of all 13 masses becomes 53 g. Find m.
(2)
(e)
Find the standard deviation of the 13 masses.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\bar{y} = \frac{30}{12} = 2.5\)
✓ 1
A1 \(\bar{x} = 50 + 2.5 = 52.5\) g
✓ 1
Part (b)
M1 \(\sigma_y^2 = \frac{410}{12} - 2.5^2\)
✓ 1
A1 variance = 27.9 (27.916…)
✓ 1
R1 subtracting 50 does not change the spread, so σx = σy = 5.28 g
Common error: \(\sum x^2\) is not \(\sum y^2 + 12 \times 50^2\): the cross term \(100\sum y\) from expanding \((y + 50)^2\) is often forgotten.
435IBMA2S-050-03-044.10 Regression of x on yPaper 2Hard13 marks
Calculator
An aerospace team carries out static tests on small solid rocket motors. For each motor they record the mass of propellant, x kg, and the burn time, y seconds. One burn time was lost from the records and is denoted by k.
x: 1, 3, 5, 7, 9 y: 3, 10, 12, 14, k
The regression line of y on x for these data has equation \(y = 2x + 2\).
(a)
Show that \(k = 21\).
(3)
(b)
Using \(k = 21\), find the value of Pearson's product-moment correlation coefficient, r, and interpret it.
(3)
(c)
Use the regression line of y on x to estimate the burn time of a motor containing 6 kg of propellant.
(2)
(d)
A mission requires a burn time of 16 seconds. The team wants to estimate the propellant mass needed. (i) State, with a reason, which regression line should be used. (ii) Find the equation of this line. (iii) Hence estimate the propellant mass needed.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses the fact that the regression line passes through \((\bar{x}, \bar{y})\)
M1 (iii) substitutes \(y = 16\) into the x-on-y line
✓ 1
A1 6.88 kg (117/17 = 6.882…); also accept 6.89 kg from the 3 s.f. line
✓ 1
Common error: Rearranging the y-on-x line to make x the subject does not give the line of x on y. Unless r = ±1 the two lines are different, so use the x-on-y line from your GDC to predict x. Keep the unrounded coefficients in your calculator when you substitute.
436IBMA2S-050-03-094.4 Correlation and regression of y on xPaper 2Hard13 marks
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A sports scientist investigates how the length of a tennis player's warm-up, x minutes, relates to the speed of her first serve, y km/h, measured by a radar gun.
It is later found that the radar gun was wrongly positioned for the final reading.
(a)
Using all six data points, find (i) the value of Pearson's product-moment correlation coefficient, r; (ii) the equation of the regression line of y on x.
(4)
(b)
Explain the effect of the final reading on your answers to the first part.
(1)
(c)
The final reading is removed. For the remaining five points, find (i) the value of r; (ii) the equation of the regression line of y on x.
(4)
(d)
Using the more appropriate regression line, estimate the first-serve speed after a warm-up of 23 minutes.
(2)
(e)
Give two reasons why your estimate of the first-serve speed after a 23-minute warm-up is likely to be reliable.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 evidence of using a GDC (or summary statistics) with all six points
R1 the point (30, 160) is an outlier that reverses the trend: it makes the correlation weak and negative and the regression line slope downward, even though the other five points rise steadily
✓ 1
Part (c)
M1 evidence of using a GDC (or summary statistics) with the five remaining points
✓ 1
A1 (i) \(r = 0.997\) (0.997245…)
✓ 1
A1 (ii) gradient 1.9
✓ 1
A1 intercept 144, giving \(y = 1.9x + 144\)
✓ 1
Part (d)
M1 substitutes \(x = 23\) into \(y = 1.9x + 144\)
✓ 1
A1 188 km/h (187.7)
✓ 1
Part (e)
R1 23 minutes lies within the range of the remaining data (20 to 28 minutes), so this is interpolation
✓ 1
R1 \(r = 0.997\) shows a very strong positive linear correlation once the faulty reading is removed
✓ 1
Exam tip: A single faulty reading can change r dramatically, even its sign. If a data point is known to be an error, remove it and recalculate before making predictions.
437IBMA2S-050-09-074.3 Measures of central tendency and dispersionPaper 2Hard7 marks
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The flight times of 20 test flights of a delivery drone were recorded. The mean flight time was 15.5 minutes and the standard deviation was 6.00 minutes. It was later found that one flight time had been recorded as 41 minutes when it should have been 14 minutes.
(a)
Find the correct mean flight time.
(3)
(b)
Find the correct standard deviation of the flight times.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 original total Σx = 20 × 15.5 = 310
✓ 1
M1 corrected total = 310 − 41 + 14 = 283
✓ 1
A1 mean = 283 ÷ 20 = 14.15 minutes
✓ 1
Part (b)
M1 use of σ2 = Σx2/n − x̄2 to give Σx2 = 20(62 + 15.52) = 5525
✓ 1
M1 corrected Σx2 = 5525 − 412 + 142 = 4040
✓ 1
M1 new variance = 4040/20 − 14.152 = 1.7775
✓ 1
A1 σ = 1.33 minutes
✓ 1
Exam tip: You cannot correct a standard deviation directly. Work back to Σx and Σx², replace the wrong value in each, and then recalculate.
438IBMA2S-050-08-064.10 Regression of x on yPaper 2Hard15 marks
Calculator
The temperature sensor on a small satellite's solar panel has failed, so engineers plan to estimate the panel temperature from its power output. Before the failure, six readings of panel temperature x °C and power output y W were recorded. Temperature x (°C): 15, 25, 35, 45, 55, k Power y (W): 8.4, 8.1, 7.9, 7.5, 7.2, 6.9 The mean of the six temperatures is 40 °C.
(a)
Find the value of k.
(2)
(b)
Find the value of the correlation coefficient r and interpret it. Explain why the engineers should use the regression line of x on y.
(3)
(c)
Find the equation of the regression line of x on y. Hence estimate the panel temperature when the power output is 7.0 W.
(4)
(d)
Write down the coordinates of the point where the regression line of x on y meets the regression line of y on x, giving a reason. Comment on the reliability of using the regression line of x on y to estimate the temperature when the power output is 5.0 W.
(3)
(e)
Find the equation of the regression line of y on x. Hence estimate the power output when the panel temperature is 30 °C.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(\frac{15+25+35+45+55+k}{6} = 40\)
✓ 1
A1 \(k = 65\)
✓ 1
Part (b)
A1 \(r = -0.997\) (−0.99746…; accept −0.9975)
✓ 1
R1 very strong negative linear correlation
✓ 1
R1 temperature (x) is to be estimated from a known power output (y)
✓ 1
Part (c)
A1 gradient −32.9 (−32.851…)
✓ 1
A1 intercept 292 (291.86…), i.e. \(x = -32.9y + 292\)
✓ 1
M1 substitutes \(y = 7.0\)
✓ 1
A1 61.9 °C (61.90… from unrounded coefficients; accept 61.7 from rounded coefficients)
✓ 1
Part (d)
A1 \((40, 7.67)\)
✓ 1
R1 both regression lines pass through the mean point \((\bar{x},\bar{y})\)
✓ 1
R1 unreliable: 5.0 W is outside the range of the data (6.9–8.4 W), so it is extrapolation
✓ 1
Part (e)
A1 \(y = -0.0303x + 8.88\)
✓ 1
M1 substitutes \(x = 30\)
✓ 1
A1 7.97 W (7.9695…)
✓ 1
Common error: Rearranging the y on x line to make x the subject does not give the x on y line. Each line needs its own regression unless \(|r| = 1\).
439IBMA3S-4-16r5-094.6 Combined events and diagramsPaper 2Hard11 marks
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A group of 70 music students were asked whether they play guitar (G) or drums (D). 44 play guitar and 36 play drums. The number who play both instruments is twice the number who play neither.
(a)
Find the number of students who play both instruments.
(3)
(b)
Find the probability that a randomly chosen student plays guitar only.
(1)
(c)
Find P(D | G′).
(2)
(d)
Two students are chosen at random, without replacement. Find the probability that both play exactly one of the two instruments.
(3)
(e)
Determine whether G and D are independent.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 let the number playing neither be n and both be 2n; guitar only = 44 − 2n, drums only = 36 − 2n
✓ 1
M1 (44 − 2n) + (36 − 2n) + 2n + n = 70
✓ 1
A1 n = 10, so 20 play both
✓ 1
Part (b)
A1 24/70 = 12/35
✓ 1
Part (c)
M1 students who do not play guitar = 16 + 10 = 26, of whom 16 play drums
R1 the values differ so the events are not independent
✓ 1
Common error: Set up the Venn diagram with a single unknown first, then fill every region in terms of that unknown. Subtract the overlap when finding 'only' regions.
440IBMA3S-4-16r5-144.2 Presentation of dataPaper 2Hard10 marks
Calculator
The delays, in minutes, of 100 flights are shown in the table. Delay t: 0 ≤ t < 10: 18 flights; 10 ≤ t < 20: 24; 20 ≤ t < 40: 30; 40 ≤ t < 80: 28.
(a)
A histogram is drawn for these data. Find the frequency density for the class 40 ≤ t < 80.
(2)
(b)
Estimate the mean delay.
(3)
(c)
Assuming that delays in each class are evenly spread, estimate the probability that a flight is delayed by more than 30 minutes.
(2)
(d)
Given that a flight is delayed by more than 10 minutes, estimate the probability that the delay is more than 30 minutes.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 frequency density = frequency ÷ class width = 28/40
✓ 1
A1 0.7
✓ 1
Part (b)
M1 use of mid-interval values 5, 15, 30, 60
✓ 1
A1 (18 × 5 + 24 × 15 + 30 × 30 + 28 × 60)/100
✓ 1
A1 30.3 minutes
✓ 1
Part (c)
M1 half of the 20 ≤ t < 40 class: 15 flights
✓ 1
A1 (15 + 28)/100 = 0.43
✓ 1
Part (d)
M1 flights delayed by more than 10 minutes: 24 + 30 + 28 = 82
✓ 1
M1 use of conditional probability: 43/82
✓ 1
A1 0.524
✓ 1
Exam tip: State the 'evenly spread' assumption before splitting a class. For a conditional probability, count only the flights that satisfy the condition in the denominator.
441IBMA3S-4-16r6-114.9 The normal distributionPaper 2Hard12 marks
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The thickness, T mm, of pottery sherds from an excavation is normally distributed with mean 8.4 and standard deviation 1.2.
Normal distribution curve, not to scale
(a)
Find the probability that a sherd is thicker than 10 mm.
(2)
(b)
The thickest 5% of sherds are classed as 'very thick'. Find the least thickness of a very thick sherd.
(2)
(c)
A sherd thicker than 10 mm is classed as 'thick'. A sample of 20 sherds is chosen at random. Let X be the number of thick sherds. Use the unrounded probability from part (a) in the rest of this question. (c) Find P(X ≥ 3).
(3)
(d)
Given that the sample contains at least one thick sherd, find the probability that it contains exactly two thick sherds.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to find P(T > 10)
✓ 1
A1 0.0912
✓ 1
Part (b)
M1 attempt to use inverse normal with P(T < t) = 0.95
Common error: For a conditional probability, divide by the probability of the condition, not by 1.
442IBMA3S-4-16r6-144.10 Regression of x on yPaper 2Hard14 marks
Calculator
Archaeologists measure the femur length, x cm, and the estimated stature, y cm, of eight skeletons.
x
40.2
42.5
43.1
44.8
45.9
47.0
48.3
49.6
y
156
161
163
167
170
172
176
178
Scatter diagram of the data
(a)
Find the correlation coefficient and the equation of the regression line of x on y. Give the coefficients correct to 4 significant figures.
(4)
(b)
A skeleton has stature 165 cm. Estimate its femur length, and comment on the reliability of your estimate.
(3)
(c)
At another site the statures of adults are normally distributed with mean 168 cm and standard deviation 7.5 cm. Find the probability that a randomly chosen adult is taller than 175 cm.
(2)
(d)
The tallest 10% of adults at this site are taller than h cm. Find h to 4 significant figures.
(2)
(e)
Use the equation from part (a) to estimate the femur length of an adult whose stature is h cm, using h = 177.6, and state whether this estimate is likely to be reliable.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 r = 0.999
✓ 1
M1 attempt to find regression of x on y
✓ 1
A1 gradient 0.4150
✓ 1
A1 x = 0.4150y − 24.49
✓ 1
Part (b)
M1 substitute y = 165
✓ 1
A1 44.0 cm
✓ 1
R1 reliable: 165 is within the data range 156 to 178 and r is very close to 1
✓ 1
Part (c)
M1 attempt to find P(Y > 175) with Y ~ N(168, 7.52)
✓ 1
A1 0.175
✓ 1
Part (d)
M1 attempt to use inverse normal with P(Y < h) = 0.9
✓ 1
A1 h = 177.6
✓ 1
Part (e)
M1 substitute y = 177.6
✓ 1
A1 49.2 cm
✓ 1
R1 177.6 is within the data range (below 178), so interpolation; likely to be reliable
✓ 1
Common error: Choose the regression line by the variable you want to estimate: to estimate femur length from stature, use x on y.
443IBMA3S-4-16r8-134.12 Standardization and inverse normalPaper 2Hard11 marks
Calculator
A machine fills pots of yoghurt. The volume V mL in a pot is normally distributed with mean μ and standard deviation 4.0 mL. The machine is set so that exactly 1% of pots contain less than 500 mL.
Normal distribution curve, not to scale
(a)
Find the value of μ. Give your answer to 4 significant figures.
(3)
(b)
A pot is overfilled if it contains more than 515 mL. Using the unrounded value of μ, find the probability that a pot is overfilled. Give your answer to 3 significant figures.
(2)
(c)
A box contains 12 pots, filled independently. Find the probability that at least one pot in the box is overfilled. Give your answer to 3 significant figures.
(3)
(d)
A pot is rejected if it contains less than 500 mL or more than 515 mL. Given that a pot is rejected, find the probability that it was underfilled. Give your answer to 3 significant figures.
Common error: Keep the unrounded μ in later parts. Using 509.3 instead of 509.305... changes the tail probability in the third significant figure.
444IBMA3S-4-16r8-144.7 Discrete random variablesPaper 2Hard10 marks
Calculator
The number of dropped calls, X, in a block of n calls on a telecoms network is modelled by the binomial distribution B(n, p). It is known that E(X) = 6 and Var(X) = 4.2.
(a)
Find the values of n and p.
(3)
(b)
Find P(X = 6). Give your answer to 3 significant figures.
(2)
(c)
Find the smallest integer k such that P(X ≤ k) ≥ 0.95.
(2)
(d)
Find P(X > 6 | X ≥ 4). Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 np = 6 and np(1 − p) = 4.2
✓ 1
M1 divide to get 1 − p = 0.7
✓ 1
A1 p = 0.3 and n = 20
✓ 1
Part (b)
M1 P(X = 6) for B(20, 0.3)
✓ 1
A1 0.192
✓ 1
Part (c)
M1 P(X ≤ 8) = 0.887 and P(X ≤ 9) = 0.952
✓ 1
A1 k = 9
✓ 1
Part (d)
M1 P(X ≥ 7) / P(X ≥ 4), since X ≥ 7 is contained in X ≥ 4
✓ 1
A1 P(X ≥ 7) = 0.3920 and P(X ≥ 4) = 0.8929
✓ 1
A1 0.439
✓ 1
Exam tip: Dividing Var(X) by E(X) gives 1 − p directly, which is quicker than solving simultaneous equations.
445IBMA3S-4-16r9-124.7 Discrete random variablesPaper 2Hard11 marks
Calculator
An online shop sells a dress in sizes S, M and L. Of all orders, 25% are for S, 45% for M and 30% for L. The probability that an item is returned is 0.12 for S, 0.08 for M and 0.20 for L.
Tree diagram
(a)
Find the probability that a randomly chosen order is returned.
(3)
(b)
Given that an order is returned, find the probability that it was for size L, to 3 significant figures.
(3)
(c)
In a month there are 5000 orders. Estimate the number of size L orders that are returned and the total number of returned orders.
(2)
(d)
The shop wants the overall return probability to fall to 0.100 by changing only the return probability r for size L. Find r to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sum of products of size probability and return probability
✓ 1
A1 0.25 × 0.12 + 0.45 × 0.08 + 0.30 × 0.20
✓ 1
A1 0.126
✓ 1
Part (b)
M1 P(L ∩ returned) = 0.30 × 0.20 = 0.06
✓ 1
M1 0.06 / 0.126
✓ 1
A1 0.476
✓ 1
Part (c)
A1 5000 × 0.06 = 300 size L returns
✓ 1
A1 5000 × 0.126 = 630 returns in total
✓ 1
Part (d)
M1 0.03 + 0.036 + 0.30r = 0.100
✓ 1
A1 0.30r = 0.034
✓ 1
A1 r = 0.113
✓ 1
Exam tip: Build the table of products (size probability times return probability) once and reuse it for every part.
446IBMA3S-4-16r9-134.9 The normal distributionPaper 2Hard10 marks
Calculator
A robotics team tests two batches of batteries. Batch A has 30 batteries with a mean life of 4.2 hours and a standard deviation of 0.8 hours. Batch B has 20 batteries with a mean life of 5.1 hours and a standard deviation of 1.1 hours. Use the population formula σ² = Σx²/n − mean².
(a)
Find the mean life of all 50 batteries.
(2)
(b)
Show that for batch A, Σx² = 548.4.
(2)
(c)
Find the standard deviation of the lives of all 50 batteries, to 3 significant figures.
(4)
(d)
A 51st battery is tested. The mean life of all 51 batteries is now 4.58 hours. Find the life of the 51st battery.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (30 × 4.2 + 20 × 5.1) / 50
✓ 1
A1 4.56 hours
✓ 1
Part (b)
M1 Σx² = n(σ² + mean²) = 30(0.8² + 4.2²)
✓ 1
A1 30 × 18.28 = 548.4
✓ 1
Part (c)
M1 Σx² for batch B = 20(1.1² + 5.1²) = 544.4
✓ 1
M1 variance = (548.4 + 544.4)/50 − 4.56²
✓ 1
A1 1.0624
✓ 1
A1 1.03 hours
✓ 1
Part (d)
M1 (228 + m)/51 = 4.58
✓ 1
A1 m = 5.58 hours
✓ 1
Common error: Standard deviations cannot be averaged; combine the sums of squares and then subtract the square of the combined mean.
447IBMA3S-4-17r12-114.9 The normal distributionPaper 2Hard10 marks
Calculator
The sugar content, \(X\) grams per 100 ml, of a fruit drink is normally distributed with mean \(\mu\) and standard deviation \(\sigma\). It is known that \(\mathrm{P}(X < 11.0) = 0.15\) and \(\mathrm{P}(X > 16.0) = 0.05\).
Normal distribution curve, not to scale
(a)
Write down the \(z\)-values corresponding to \(\mathrm{P}(X<11.0)\) and to \(\mathrm{P}(X<16.0)\), to 3 decimal places.
(2)
(b)
Find \(\mu\) and \(\sigma\), each to 3 significant figures.
(5)
(c)
Using \(\mu = 12.9\) and \(\sigma = 1.86\), find \(\mathrm{P}(12.0 < X < 14.0)\) to 3 significant figures.
(2)
(d)
Out of 1000 bottles, estimate how many have a sugar content between 12.0 and 14.0 grams per 100 ml. Use the 3 significant figure probability from (c).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(z = -1.036\) for the cumulative probability 0.15
✓ 1
A1 \(z = 1.645\) for the cumulative probability 0.95
✓ 1
Part (b)
M1 \(\frac{11.0-\mu}{\sigma} = -1.036\)
✓ 1
M1 \(\frac{16.0-\mu}{\sigma} = 1.645\)
✓ 1
M1 eliminate one unknown, e.g. subtract: \(\frac{5.0}{\sigma} = 2.681\)
✓ 1
A1 \(\sigma = 1.86\)
✓ 1
A1 \(\mu = 12.9\)
✓ 1
Part (c)
M1 normal cdf (or standardise) with the correct parameters and limits
✓ 1
A1 0.409
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Part (d)
A1 \(1000 \times 0.409 = 409\) bottles (ECF from (c))
✓ 1
Common error: A lower-tail probability below 0.5 gives a negative \(z\)-value; check the sign of each \(z\) before forming the two equations.
448IBMA3S-4-17r12-134.9 The normal distributionPaper 2Hard8 marks
Calculator
The volume, \(V\) ml, of juice in a carton is normally distributed with mean 1005 and standard deviation 6. Each carton is labelled '1000 ml'.
Normal distribution curve, not to scale
(a)
Find the probability that a carton contains less than 1000 ml. Give your answer to 3 significant figures.
(2)
(b)
The standard deviation stays at 6. The producer changes the mean so that only 1% of cartons contain less than 1000 ml. Find the new mean, to the nearest millilitre.
(3)
(c)
The producer decides not to change the mean. A pack contains 6 cartons chosen at random. Using the probability from (a), find the probability that at least one carton in the pack contains less than 1000 ml. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(z = \frac{1000-1005}{6} = -0.833\) or normal cdf with correct parameters
✓ 1
A1 0.202
✓ 1
Part (b)
M1 \(\mathrm{P}(V<1000) = 0.01\) so \(z = -2.326\)
✓ 1
M1 \(\frac{1000-\mu}{6} = -2.326\)
✓ 1
A1 \(\mu = 1014\) ml
✓ 1
Part (c)
M1 cartons are independent: \(\mathrm{P}(\text{none under 1000 ml}) = (1-0.202)^6\)
✓ 1
M1 complement: \(1 - (1-0.202)^6\)
✓ 1
A1 0.742
✓ 1
Common error: 'At least one' is best found as \(1 - \mathrm{P}(\text{none})\), not by adding several cases.
449IBMA3S-4-16r21-064.9 The normal distributionPaper 2Hard7 marks
Calculator
The time T minutes taken by players to complete a level of a game is normally distributed. 20% of players take less than 12 minutes and 10% of players take more than 20 minutes.
Normal distribution curve, not to scale
(a)
Find the mean and the standard deviation of T, each to 3 significant figures. [5]
(5)
(b)
Use the unrounded values from (a) to find the probability that a randomly chosen player takes between 14 and 18 minutes. Give your answer to 3 significant figures. [2]
A1 correct second equation \(\frac{20-\mu}{\sigma}=1.2816\) from \(P(T<20)=0.9\)
✓ 1
M1 attempt to solve the two equations simultaneously
✓ 1
A1 σ = 3.77 (3.7679...)
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A1 μ = 15.2 (15.171...)
✓ 1
Part (b)
M1 attempt to find P(14 < T < 18)
✓ 1
A1 0.396 (0.39564...)
✓ 1
Common error: Both unknowns need two separate z-equations; the 10% 'more than' condition must be converted to a cumulative probability of 0.9 before using inverse normal.
450IBMA3S-4-16r21-114.9 The normal distributionPaper 2Hard12 marks
Calculator
The plasma concentration C, in ng/mL, of a drug two hours after a standard dose is normally distributed with mean 84 and standard deviation 12.
Normal distribution curve, not to scale
(a)
Find P(C > 100), giving your answer to 3 significant figures. [2]
(2)
(b)
A concentration is therapeutic if 70 ≤ C ≤ 100. Find the probability that a patient's concentration is therapeutic, to 3 significant figures. [2]
(2)
(c)
A patient is known to have a concentration above 70. Find the probability that it is above 100. Give your answer to 3 significant figures. [3]
(3)
(d)
Find the concentration that is exceeded by 10% of patients, to 3 significant figures. [2]
(2)
(e)
The dose is changed so that the mean becomes μ' with the standard deviation unchanged, and P(C > 100) = 0.02. Find μ', to 3 significant figures. [3]
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to find P(C > 100)
✓ 1
A1 0.0912 (0.091211...)
✓ 1
Part (b)
M1 attempt to find P(70 ≤ C ≤ 100)
✓ 1
A1 0.787 (0.78712...)
✓ 1
Part (c)
M1 conditional probability P(C > 100 | C > 70) = P(C > 100)/P(C > 70)
✓ 1
A1 P(C > 70) = 0.8783 (0.87833...)
✓ 1
A1 0.104 (0.10385...)
✓ 1
Part (d)
M1 attempt to find k with P(C < k) = 0.9
✓ 1
A1 99.4 (99.379...)
✓ 1
Part (e)
M1 z = 2.054 (2.0537...) for P(Z < z) = 0.98
✓ 1
M1 (100 − μ')/12 = 2.0537...
✓ 1
A1 μ' = 75.4 (75.355...)
✓ 1
Exam tip: In a conditional probability the denominator is the probability of the condition, here P(C > 70), not 1.
451IBMA3S-4-16r21-134.7 Discrete random variablesPaper 2Hard13 marks
Calculator
A game has two modes, A and B. Scores in mode A are normally distributed with mean 560 and standard deviation 80. Scores in mode B are normally distributed with mean 60 and standard deviation 8. A player's scores in the two modes are independent. A player scores 660 in mode A and 72 in mode B.
(a)
Find the z-score for each of the player's scores. [2]
(2)
(b)
State, with a reason, in which mode the player performed better relative to the other players. [1]
(1)
(c)
Find the score in mode A that is exceeded by the top 5% of players. Give your answer to the nearest whole number. [2]
(2)
(d)
Find the score in mode B that has the same z-score as a score of 660 in mode A. [2]
(2)
(e)
A player plays each mode once. Find the probability that the player scores more than 660 in mode A and more than 75 in mode B. Give your answer to 3 significant figures. [3]
(3)
(f)
A player scored more than 660 in mode A. Find the probability that the player scored more than 700 in mode A. Give your answer to 3 significant figures. [3]
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 mode A: z = (660 − 560)/80 = 1.25
✓ 1
A1 mode B: z = (72 − 60)/8 = 1.5
✓ 1
Part (b)
R1 mode B, because the z-score is higher (1.5 > 1.25)
✓ 1
Part (c)
M1 z = 1.645 (1.6449...) for P(Z < z) = 0.95, so 560 + 1.6449 × 80
Common error: To compare scores from different distributions, compare z-scores, not raw scores or raw differences from the mean.
452IBMA3S-4-16r22-074.12 Standardization and inverse normalPaper 2Hard6 marks
Calculator
The incubation period T days of a disease is normally distributed with mean μ and standard deviation σ. It is known that P(T < 3.0) = 0.10 and P(T < 8.0) = 0.85.
Normal distribution curve, not to scale
(a)
Write down two equations connecting μ and σ.
(3)
(b)
Hence find the values of μ and σ, correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to find the z-values using inverse normal
✓ 1
A1 (3.0 − μ) ÷ σ = −1.28
✓ 1
A1 (8.0 − μ) ÷ σ = 1.04
✓ 1
Part (b)
M1 attempt to solve the two equations simultaneously
✓ 1
A1 σ = 2.16
✓ 1
A1 μ = 5.76 (accept 5.77 from a rounded σ)
✓ 1
Exam tip: Keep the z-values at full calculator precision when solving, and round only at the end.
453IBMA3S-4-16r22-134.10 Regression of x on yPaper 2Hard12 marks
Calculator
The wind speed W m s−1 at a site is normally distributed with mean 7.5 and standard deviation σ. It is known that P(W > 10) = 0.12.
Normal distribution curve, not to scale
(a)
Find the value of σ.
(3)
(b)
A turbine operates when 3.5 ≤ W ≤ 25. Find the probability that it operates on a given day. Use the unrounded value of σ.
(2)
(c)
Find the expected number of days, out of 30, on which the turbine operates.
(2)
(d)
Given that the turbine is operating, find the probability that the wind speed is greater than 10 m s−1.
(3)
(e)
The wind speed exceeds w m s−1 on 90% of days. Find the value of w.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(W < 10) = 0.88, so use inverse normal to find z = 1.175
✓ 1
M1 (10 − 7.5) ÷ σ = 1.175
✓ 1
A1 σ = 2.13
✓ 1
Part (b)
M1 attempt to use the normal distribution with mean 7.5 and the unrounded σ
✓ 1
A1 0.970
✓ 1
Part (c)
M1 30 × 0.970
✓ 1
A1 29.1
✓ 1
Part (d)
M1 P(10 < W ≤ 25) ÷ P(3.5 ≤ W ≤ 25)
✓ 1
A1 numerator ≈ 0.12, because P(W > 25) is negligible
✓ 1
A1 0.124
✓ 1
Part (e)
M1 use of inverse normal with P(W < w) = 0.10
✓ 1
A1 4.77
✓ 1
Exam tip: For 'given that' questions, the condition forms the denominator, and the numerator is the overlap of the two events.
454IBMA3S-4-16r22-144.12 Standardization and inverse normalPaper 2Hard11 marks
Calculator
A coffee roaster fills bags that are labelled 250 g. The mass of coffee in a bag is normally distributed with mean μ g and standard deviation 4.0 g. The roaster wants 2% of bags to contain less than 250 g.
Normal distribution curve, not to scale
(a)
Find the value of μ, correct to 1 decimal place.
(3)
(b)
A new machine has a mean of 255 g. The standard deviation is adjusted so that 2% of bags still contain less than 250 g. Find the new standard deviation, correct to 3 significant figures.
(3)
(c)
Coffee costs €0.0095 per gram. Using μ = 258.2 for the first machine and 255 for the new machine, find the saving in coffee cost for 10 000 bags.
(2)
(d)
With the new machine, 10 bags are chosen at random. Find the probability that at most one bag contains less than 250 g.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of inverse normal: z = −2.054 for probability 0.02
✓ 1
M1 (250 − μ) ÷ 4.0 = −2.054
✓ 1
A1 μ = 258.2
✓ 1
Part (b)
M1 (250 − 255) ÷ σ = −2.054
✓ 1
M1 σ = 5 ÷ 2.054
✓ 1
A1 2.43
✓ 1
Part (c)
M1 (258.2 − 255) × 10 000 × 0.0095
✓ 1
A1 €304
✓ 1
Part (d)
M1 identify the binomial model B(10, 0.02)
✓ 1
M1 P(X = 0) + P(X = 1)
✓ 1
A1 0.984
✓ 1
Exam tip: A 'proportion of bags' is a probability: once you know it is 0.02, you can switch from the normal model to the binomial for a sample of bags.
455IBMA3S-4-16r24-124.4 Correlation and regression of y on xPaper 2Hard12 marks
Calculator
A hospital screens patients for a condition. 4% of patients have the condition. The test is positive for 92% of patients who have the condition and positive for 7% of patients who do not have it. Let D be the event that a patient has the condition and T the event that the test is positive.
Tree diagram
(a)
Find P(D ∩ T).
(2)
(b)
Show that P(T) = 0.104.
(3)
(c)
Find P(D | T). Give your answer to 3 significant figures.
(3)
(d)
Interpret your answer to (c) in context.
(1)
(e)
Two patients are tested independently. Find the probability that exactly one test is positive. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.04 × 0.92
✓ 1
A1 0.0368
✓ 1
Part (b)
M1 P(T) = P(D∩T) + P(D′∩T)
✓ 1
A1 0.96 × 0.07 = 0.0672
✓ 1
A1 0.0368 + 0.0672 = 0.104
✓ 1
Part (c)
M1 P(D|T) = P(D∩T) ÷ P(T)
✓ 1
A1 0.0368 ÷ 0.104
✓ 1
A1 0.354
✓ 1
Part (d)
R1 Only about 35.4% of patients with a positive test actually have the condition (so most positive results are false positives)
✓ 1
Part (e)
M1 P(T′) = 1 − 0.104 = 0.896
✓ 1
M1 2 × 0.104 × 0.896
✓ 1
A1 0.186
✓ 1
Common error: P(D | T) and P(T | D) are different: 0.92 is the chance of a positive test given the condition, not the chance of the condition given a positive test.
456IBMA3S-4-16r24-134.4 Correlation and regression of y on xPaper 2Hard12 marks
Calculator
A beekeeper records, for eight apiaries, the number of hives x and the mean mass of honey collected per hive y (kg). x: 8, 12, 15, 20, 24, 30, 36, 40 y: 9.8, 9.1, 8.7, 8.0, 7.5, 6.4, 5.9, 5.0
Scatter diagram of the data
(a)
Find the Pearson correlation coefficient r. Give your answer to 3 significant figures.
(2)
(b)
Describe the correlation between x and y, in context.
(2)
(c)
Find the equation of the regression line of y on x. Give the coefficients to 3 significant figures.
(2)
(d)
Interpret the gradient and the intercept of the line in context.
(2)
(e)
Use the unrounded line to estimate the mean mass of honey per hive for an apiary with 28 hives. Give your answer to 3 significant figures.
(2)
(f)
Explain why the line should not be used for an apiary with 80 hives.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of calculator for r
✓ 1
A1 r = −0.998
✓ 1
Part (b)
A1 Very strong negative linear correlation
✓ 1
A1 Apiaries with more hives tend to collect less honey per hive
✓ 1
Part (c)
A1 gradient −0.145 (−0.144842…)
✓ 1
A1 intercept 10.9 (10.8995…), so y = −0.145x + 10.9
✓ 1
Part (d)
A1 Each additional hive is associated with a decrease of 0.145 kg in the mean mass per hive
✓ 1
A1 The intercept, 10.9 kg, is the predicted mass per hive when there are zero hives, which has no practical meaning
✓ 1
Part (e)
M1 −0.144842 × 28 + 10.8995
✓ 1
A1 6.84 kg
✓ 1
Part (f)
R1 80 is far outside the range of the data (8 to 40), so this is extrapolation
✓ 1
R1 The line gives −0.144842 × 80 + 10.8995 = −0.688, a negative mass, which is impossible
✓ 1
Exam tip: When asked to interpret an intercept, check whether x = 0 is realistic; if not, say the value has no practical meaning.
457IBMA3S-4-16r24-144.6 Combined events and diagramsPaper 2Hard12 marks
Calculator
In a Formula 1 season, 120 races are analysed. Let A be the event that it rains during a race and B the event that a safety car is used. Rain occurred in 48 races and a safety car was used in 60 races. Events A and B are independent.
(a)
Show that the number of races in which it rained and a safety car was used is 24.
(3)
(b)
Find the number of races in which neither rain nor a safety car occurred.
(2)
(c)
Find P(A | B′).
(2)
(d)
Two of the 120 races are chosen at random, without replacement. Find the probability that a safety car was used in both. Give your answer to 3 significant figures.
(2)
(e)
Given that it rained in the first race chosen, find the probability that it rained in the second race. Give your answer to 3 significant figures.
(2)
(f)
Explain why your answer to (c) equals P(A).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(A) = 48/120 = 0.4 and P(B) = 60/120 = 0.5
✓ 1
M1 P(A∩B) = P(A) × P(B) = 0.2
✓ 1
A1 0.2 × 120 = 24
✓ 1
Part (b)
M1 n(A∪B) = 48 + 60 − 24 = 84
✓ 1
A1 120 − 84 = 36
✓ 1
Part (c)
M1 n(A∩B′) = 48 − 24 = 24 and n(B′) = 60
✓ 1
A1 24 ÷ 60 = 0.4
✓ 1
Part (d)
M1 60/120 × 59/119
✓ 1
A1 0.248
✓ 1
Part (e)
A1 47/119
✓ 1
A1 0.395
✓ 1
Part (f)
R1 A and B are independent, so A and B′ are also independent; knowing B′ occurred does not change the probability of A
✓ 1
Common error: When choosing without replacement from a fixed group of races, reduce both the numerator and the denominator for the second choice when the first race has the same property.
458IBMA3S-4-17r33-124.9 The normal distributionPaper 2Hard7 marks
Calculator
The number of laps, L, that a set of tyres lasts in a Formula 1 race is normally distributed with mean 38. It is known that 10% of sets of tyres last for more than 45 laps.
Normal distribution curve, not to scale
(a)
Find the standard deviation of L, to three significant figures.
(3)
(b)
A set of tyres has already completed 35 laps. Using the unrounded value of σ, find the probability that it lasts for more than 42 laps. Give your answer to three significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 z = 1.28… for an upper-tail probability of 0.10
✓ 1
M1 (45 − 38)/σ = 1.28…
✓ 1
A1 σ = 5.46
✓ 1
Part (b)
M1 P(L > 42 | L > 35) = P(L > 42) / P(L > 35)
✓ 1
A1 P(L > 42) = 0.232
✓ 1
A1 P(L > 35) = 0.709
✓ 1
A1 0.327
✓ 1
Common error: When the tyres have already lasted 35 laps, divide by P(L > 35); P(L > 42) on its own ignores the information given.
459IBMA3S-4-17r33-134.3 Measures of central tendency and dispersionPaper 2Hard8 marks
Calculator
The lengths of stay, in whole days, of 12 patients on a hospital ward are listed in increasing order. The lower quartile is 4 days, the upper quartile is 10 days, the 6th value is 7 and the 7th value is 9. The mean length of stay is 8.5 days. The longest stay, x days, is classed as an outlier, where an outlier is any value greater than Q3 + 1.5 × IQR.
(a)
Find the least possible value of x.
(3)
(b)
Find the median length of stay for the 12 patients, and the median once the outlier is removed.
(2)
(c)
It is found that x = 29. Find the mean length of stay of the other 11 patients, to three significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 IQR = 10 − 4 = 6
✓ 1
M1 Q3 + 1.5 × IQR = 10 + 9 = 19
✓ 1
A1 least whole-number value is x = 20
✓ 1
Part (b)
A1 12 patients: (7 + 9)/2 = 8 days
✓ 1
A1 11 patients: the 6th value, 7 days
✓ 1
Part (c)
M1 total of all 12 values = 12 × 8.5 = 102
✓ 1
M1 (102 − 29)/11 (or 73/11)
✓ 1
A1 6.64 days
✓ 1
Exam tip: To recover a total from a mean, multiply the mean by the number of values; then remove or add values before dividing again.
460IBMA3S-4-18r36-114.2 Presentation of dataPaper 2Hard15 marks
Calculator
The waiting times, t seconds, of 200 people for a lift in a tall building are recorded. The cumulative frequency is shown below.
t ≤ 10: 14 t ≤ 20: 46 t ≤ 30: 104 t ≤ 40: 158 t ≤ 50: 188 t ≤ 60: 200
Assume the times are spread evenly within each class. Give answers to 3 significant figures.
(a)
Estimate the median waiting time.
(3)
(b)
Estimate the lower quartile and the upper quartile.
(4)
(c)
Hence find the interquartile range.
(1)
(d)
A person waits for 70 seconds. Using the unrounded value of the IQR, show that this waiting time is an outlier. (An outlier is more than 1.5 × IQR above the upper quartile.)
(3)
(e)
Estimate the mean waiting time.
(2)
(f)
Estimate the number of people who waited more than 45 seconds.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 median is the 100th value, in the class 20 < t ≤ 30
✓ 1
M1 20 + (100 − 46)/(104 − 46) × 10
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A1 29.3 s
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Part (b)
M1 lower quartile is the 50th value: 20 + (50 − 46)/58 × 10
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A1 20.7 s
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M1 upper quartile is the 150th value: 30 + (150 − 104)/54 × 10
M1 cumulative frequency at 45 s: 158 + (5/10) × 30 = 173
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A1 200 − 173 = 27
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Common error: Cumulative frequency tables give the number of values up to each upper bound, so subtract to find class frequencies before calculating the mean.
461IBMA3S-4-18r36-134.2 Presentation of dataPaper 2Hard13 marks
Calculator
A logistics firm records the distance, d km, of each of 280 deliveries made by a van.
0 ≤ d < 5: 30 deliveries 5 ≤ d < 10: 55 deliveries 10 ≤ d < 20: 90 deliveries 20 ≤ d < 40: 60 deliveries 40 ≤ d < 70: 45 deliveries
The data are to be shown on a histogram. Assume distances are spread evenly within each class. Give non-exact answers to 3 significant figures.
(a)
Find the frequency density for the class 10 ≤ d < 20 and for the class 40 ≤ d < 70.
(2)
(b)
Estimate the mean distance.
(3)
(c)
Estimate the median distance.
(3)
(d)
Estimate the proportion of deliveries with a distance between 12 km and 25 km.
(3)
(e)
The van must recharge after any delivery of more than 30 km. Estimate the number of deliveries that need a recharge.
M1 median is the 140th value, in the class 10 ≤ d < 20 (cumulative frequency 85 before it)
✓ 1
M1 10 + (140 − 85)/90 × 10
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A1 16.1 km
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Part (d)
M1 12 to 20 km: 8/10 × 90 = 72
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M1 20 to 25 km: 5/20 × 60 = 15, total 87
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A1 87/280 = 0.311
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Part (e)
M1 30 to 40 km: 10/20 × 60 = 30
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A1 30 + 45 = 75
✓ 1
Common error: On a histogram with unequal class widths, the bar height is frequency density, not frequency; always convert before estimating a count.
462IBMA3S-4-18r37-044.6 Combined events and diagramsPaper 2Hard10 marks
Calculator
A regional airline's morning flight is either on time or late. Whether it is on time depends on the previous day: if the previous day's flight was on time, the probability that the next is on time is 0.8; if the previous flight was late, the probability that the next is on time is 0.55. Let \(p\) be the probability that Monday's flight is on time.
Tree diagram: Monday, then Tuesday
(a)
Show that the probability that Tuesday's flight is on time is \(0.55+0.25p\).
(3)
(b)
The probability that Tuesday's flight is on time is 0.7. Find \(p\).
(2)
(c)
Given that Tuesday's flight is on time, find the probability that Monday's flight was on time.
(3)
(d)
Assume the probabilities are the same every week and weeks are independent. Find the expected number of weeks, out of 50, in which both Monday's and Tuesday's flights are on time.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 attempt to use a tree diagram or total probability
✓ 1
A1 \(P(\text{Tue on time})=0.8p+0.55(1-p)\)
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A1 \(=0.55+0.25p\) (AG)
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Part (b)
M1 \(0.55+0.25p=0.7\)
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A1 \(p=0.6\)
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Part (c)
M1 \(P(\text{both on time})=0.8\times0.6=0.48\)
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M1 \(\dfrac{0.48}{0.7}\)
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A1 0.686 (accept \(\dfrac{24}{35}\))
✓ 1
Part (d)
M1 \(50\times0.48\)
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A1 24
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Common error: The probability that two flights are both on time uses the first-day probability multiplied by the conditional probability, not 0.8 alone.
463IBMA3S-4-18r37-114.3 Measures of central tendency and dispersionPaper 2Hard14 marks
Calculator
The turnaround times, in minutes, of ten aircraft at a regional airport are 28, 31, 33, 35, 36, 38, 40, 41, 42, 73. A value is an outlier if it is more than \(1.5\times\text{IQR}\) above the upper quartile or below the lower quartile.
(a)
Find \(Q_1\), \(Q_3\) and the interquartile range.
(3)
(b)
Show that 73 is an outlier.
(3)
(c)
The value 73 is removed. Find the mean and the standard deviation of the remaining nine values.
(3)
(d)
A tenth turnaround time \(t\) is then added to the nine values so that the mean of the ten values is 37. Find \(t\).
(2)
(e)
Find the standard deviation of the ten values including \(t\).
(3)
Show mark scheme
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Mark
Notes
Part (a)
A1 \(Q_1=33\)
✓ 1
A1 \(Q_3=41\)
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A1 IQR = 8
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Part (b)
M1 \(1.5\times8=12\)
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M1 \(41+12=53\)
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R1 \(73>53\), so 73 is an outlier
✓ 1
Part (c)
A1 mean 36 minutes
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M1 attempt to find the standard deviation, using GDC or \(\sigma^2=\dfrac{\sum(x-\mu)^2}{9}\)
✓ 1
A1 4.47 minutes (accept \(\sqrt{20}\))
✓ 1
Part (d)
M1 \(10\times37-9\times36=370-324\)
✓ 1
A1 46
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Part (e)
M1 \(\sum x^2\) for the nine values \(=9(20+36^2)=11\,844\), or list the ten values in the GDC
✓ 1
M1 \(\sigma^2=\dfrac{11\,844+46^2}{10}-37^2=27\)
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A1 5.20 minutes (accept \(\sqrt{27}\))
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Exam tip: Use the same quartile method throughout: with ten values, split the data into two halves of five and take the median of each half.
464IBMA3S-4-18r37-144.4 Correlation and regression of y on xPaper 2Hard12 marks
Calculator
A grower records the greenhouse temperature, \(x\) °C, and the number of days, \(y\), that tomatoes take to ripen. \(x\): 18, 20, 22, 24, 26, 28, 30 \(y\): 41, 38, 35, 34, 29, 27, 24
Scatter diagram of the data
(a)
Find the Pearson product-moment correlation coefficient, \(r\).
(2)
(b)
Find the equation of the regression line of \(y\) on \(x\).
(2)
(c)
Interpret the gradient of this line in context.
(1)
(d)
Find the equation of the regression line of \(x\) on \(y\).
(2)
(e)
Both regression lines pass through \((\bar{x},\bar{y})\). Find \(\bar{x}\) and \(\bar{y}\), and show, using the unrounded value of \(\bar{y}\), that \((\bar{x},\bar{y})\) lies on the line in part (d).
(2)
(f)
The grower wants the tomatoes to ripen in 20 days. Estimate the temperature needed, and comment on the reliability of the estimate.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to find \(r\) using the GDC
✓ 1
A1 \(-0.994\)
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Part (b)
A1 \(y=-1.41x+\dots\)
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A1 \(y=-1.41x+66.4\)
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Part (c)
A1 each 1 °C rise in temperature is associated with tomatoes ripening about 1.41 days sooner
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Part (d)
A1 \(x=-0.700y+\dots\)
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A1 \(x=-0.700y+46.8\)
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Part (e)
A1 \(\bar{x}=24\) and \(\bar{y}=\dfrac{228}{7}\approx32.6\)
✓ 1
M1 \(-0.7\times\dfrac{228}{7}+46.8=24=\bar{x}\)
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Part (f)
M1 substitute \(y=20\) into \(x=-0.7y+46.8\)
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A1 32.8 °C
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R1 20 days is outside the data range 24 to 41 days, so this extrapolation is unreliable
✓ 1
Common error: When the aim is to estimate \(x\) from \(y\), use the regression line of \(x\) on \(y\) and do not extrapolate far beyond the data.
465IBMA3S-4-17r38-114.4 Correlation and regression of y on xPaper 2Hard9 marks
Calculator
The table shows the time in orbit, x months, and the total radiation dose, y units, absorbed by a sensor on a space probe. x: 2, 4, 6, 8, 10, 12, 14, 16 y: 11.2, 15.7, 18.5, 22.7, 24.6, 38.6, 31.6, 36.6
Scatter diagram of the data
(a)
Find the correlation coefficient r, and the equation of the regression line of y on x. Give all values to 3 significant figures.
(3)
(b)
The reading at x = 12 was caused by a solar flare and is removed. For the remaining seven points, find r and the equation of the regression line of y on x, to 3 significant figures.
(3)
(c)
Explain why the line from (b) is a better model of the normal dose rate than the line from (a).
(1)
(d)
Use your equation from (b), with the coefficients to 3 significant figures, to estimate the dose at 18 months. Give your answer to 1 decimal place.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 r = 0.940
✓ 1
A1 gradient 1.90
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A1 intercept 7.82
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Part (b)
A1 r = 0.997
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A1 gradient 1.73
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A1 intercept 8.17
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Part (c)
R1 r is closer to 1 without the flare reading, so the points lie closer to a straight line (the flare reading was far from the trend)
✓ 1
Part (d)
M1 substitute x = 18
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A1 1.73×18 + 8.17 = 39.3 (ECF from (b))
✓ 1
Common error: Removing a point because it is inconvenient is not justified; give a reason that is physical, here the solar flare.
466IBMA3S-4-17r38-144.3 Measures of central tendency and dispersionPaper 2Hard8 marks
Calculator
A technician records 25 strain readings from a bridge. The mean is 38.4 and the population variance is 83.44. She then finds that one reading was recorded as 52 but should have been 25.
(a)
Find the correct mean.
(2)
(b)
Find the correct population standard deviation, to 3 significant figures.
(4)
(c)
A 26th reading of 30 is added to the corrected data. Find the new mean, to 3 significant figures.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 sum 25×38.4 = 960, then 960 − 52 + 25
✓ 1
A1 933/25 = 37.32
✓ 1
Part (b)
M1 Σx2 = 25(83.44 + 38.42) = 38 950
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A1 corrected Σx2 = 38 950 − 522 + 252 = 36 871
✓ 1
M1 variance = 36 871/25 − 37.322 (ECF from (a))
✓ 1
A1 9.06
✓ 1
Part (c)
M1 (933 + 30)/26 (ECF from (a))
✓ 1
A1 37.0
✓ 1
Exam tip: To correct a standard deviation, work with Σx and Σx² rather than trying to adjust the standard deviation directly.
467IBMA3S-4-18r39-124.8 The binomial distributionPaper 2Hard10 marks
Calculator
The random variable \(X\) is the number of delayed flights among the three connections on a passenger's journey. Its distribution is \(P(X=x)=k(x+1)^2\) for \(x\in\{0,1,2,3\}\), where \(k\) is a constant.
(a)
Show that \(k=\frac1{30}\).
(2)
(b)
Find \(E(X)\).
(2)
(c)
Find \(\operatorname{Var}(X)\), giving your answer to 3 significant figures.
(3)
(d)
Two passengers travel independently and each has this distribution. Find the probability that they have the same number of delayed flights, giving your answer to 3 significant figures.
Common error: For two independent copies of \(X\), the probability of equal values is the sum of the squared probabilities, not the square of a single probability.
468IBMA3S-4-19r41-114.9 The normal distributionPaper 2Hard10 marks
Calculator
The length, L metres, of a water pipe from a factory is normally distributed with standard deviation 0.12 m. It is known that 10% of the pipes are shorter than 2.90 m.
Normal distribution curve, not to scale
(a)
Find the mean length of the pipes, giving your answer to 4 significant figures.
(3)
(b)
Use the unrounded mean to find P(L > 3.20 | L > 3.10). Give your answer to 3 significant figures.
(4)
(c)
Eight pipes are chosen at random. Find the probability that exactly two of them are shorter than 2.90 m. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 z = invNorm(0.1) = −1.2816...
✓ 1
M1 (2.90 − μ)/0.12 = −1.2816...
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A1 μ = 3.054 m
✓ 1
Part (b)
M1 P(L > 3.20) = 0.1115... and P(L > 3.10) = 0.3500... (accept 3 s.f. or better)
✓ 1
M1 recognising that P(L > 3.20 | L > 3.10) = P(L > 3.20)/P(L > 3.10)
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M1 correct substitution into the ratio
✓ 1
A1 probability = 0.319
✓ 1
Part (c)
M1 recognising binomial with n = 8, p = 0.1
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M1 8C2 × 0.12 × 0.96
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A1 probability = 0.149
✓ 1
Common error: In a conditional probability the denominator is the probability of the condition, here P(L > 3.10), not 1.
469IBMA3S-4-19r41-124.11 Conditional probability and independencePaper 2Hard10 marks
Calculator
The ping times, in milliseconds, to two game servers are recorded. For server A there are 20 measurements with mean 15.2 and population standard deviation 2.1. For server B there are 30 measurements with mean 17.0 and population standard deviation 3.0.
(a)
Show that, for server A, Σx = 304 and Σx2 = 4709.
(3)
(b)
Find Σx and Σx2 for server B.
(2)
(c)
Find the mean of all 50 measurements.
(2)
(d)
Find the population standard deviation of all 50 measurements, giving your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 Σx = 20 × 15.2 = 304
✓ 1
M1 using σ2 = Σx2/n − mean2, so Σx2 = 20 × (2.12 + 15.22)
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A1 20 × 235.45 = 4709
✓ 1
Part (b)
A1 Σx = 30 × 17.0 = 510
✓ 1
A1 Σx2 = 30 × (3.02 + 17.02) = 8940
✓ 1
Part (c)
M1 (304 + 510)/50
✓ 1
A1 16.28
✓ 1
Part (d)
M1 combined Σx2 = 4709 + 8940 = 13 649
✓ 1
M1 variance = 13 649/50 − 16.282
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A1 σ = 2.82 ms
✓ 1
Exam tip: Standard deviations cannot be averaged; convert each group back to Σx and Σx², combine, then recompute.
470IBMA3S-4-19r41-144.4 Correlation and regression of y on xPaper 2Hard9 marks
Calculator
A water company studies x, the hours per day a pump runs, and y, the volume delivered in thousands of litres. The regression line of y on x is y = 1.2x + 4 and the regression line of x on y is x = 0.5y − 1. The values of y in the data range from 2 to 12.
(a)
Find the mean of x and the mean of y.
(3)
(b)
Find the value of the Pearson correlation coefficient, r, giving your answer to 3 significant figures.
(3)
(c)
Estimate the number of hours the pump must run to deliver 10 thousand litres.
(2)
(d)
Comment on the reliability of your estimate in part (c).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 both regression lines pass through the point (mean of x, mean of y)
✓ 1
M1 substitute one equation into the other, e.g. x = 0.5(1.2x + 4) − 1
✓ 1
A1 mean of x = 2.5 and mean of y = 7
✓ 1
Part (b)
M1 r2 = 1.2 × 0.5 = 0.6
✓ 1
A1 r = √0.6 = 0.775
✓ 1
R1 r is positive because both gradients are positive
✓ 1
Part (c)
M1 use the line of x on y with y = 10
✓ 1
A1 x = 4 hours
✓ 1
Part (d)
R1 reliable: y = 10 lies within the range of the data (interpolation) and r shows a fairly strong positive correlation
✓ 1
Exam tip: The product of the two regression gradients equals r², and the sign of r matches the sign of the gradients.
471IBMA3S-4-18r45-104.8 The binomial distributionPaper 2Hard12 marks
Calculator
The number of customers, X, who are connected to a mobile mast at a particular moment follows a binomial distribution X ~ B(n, p) with mean 12 and variance 7.2.
(a)
Find the value of n and the value of p.
(4)
(b)
Find P(X = 12).
(2)
(c)
Find P(X ≤ 14 | X ≥ 10).
(3)
(d)
Over five independent moments, find the probability that X ≥ 15 at least once. Use the unrounded value of P(X ≥ 15). Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 np = 12 and np(1 − p) = 7.2
✓ 1
M1 divide: 1 − p = 7.2 ÷ 12 = 0.6
✓ 1
A1 p = 0.4
✓ 1
A1 n = 30
✓ 1
Part (b)
M1 use of the binomial probability with n = 30, p = 0.4, k = 12
✓ 1
A1 0.147
✓ 1
Part (c)
M1 P(10 ≤ X ≤ 14) ÷ P(X ≥ 10)
✓ 1
A1 numerator 0.648 and denominator 0.824 (or unrounded values)
✓ 1
A1 0.787
✓ 1
Part (d)
A1 P(X ≥ 15) = 0.17537…
✓ 1
M1 1 − (1 − P(X ≥ 15))5
✓ 1
A1 0.619
✓ 1
Exam tip: Use the mean and variance of a binomial distribution (np and np(1 − p)) together: dividing the second by the first gives 1 − p directly.
472IBMA3S-4-18r45-114.8 The binomial distributionPaper 2Hard10 marks
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A telecoms company offers fibre broadband. Each customer in a large town independently has fibre with probability 0.35. A random sample of 20 customers is taken, and X is the number of them who have fibre.
(a)
Find E(X) and Var(X).
(2)
(b)
Find P(X = 8).
(2)
(c)
Find P(X > E(X)).
(3)
(d)
Find P(X ≤ 9 | X ≥ 5). Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 E(X) = 20 × 0.35 = 7
✓ 1
A1 Var(X) = 20 × 0.35 × 0.65 = 4.55
✓ 1
Part (b)
M1 use of the binomial probability with n = 20, p = 0.35, k = 8
✓ 1
A1 0.161
✓ 1
Part (c)
M1 E(X) = 7 so require P(X > 7)
✓ 1
M1 1 − P(X ≤ 7)
✓ 1
A1 0.399
✓ 1
Part (d)
M1 P(5 ≤ X ≤ 9) ÷ P(X ≥ 5)
✓ 1
A1 numerator 0.760 and denominator 0.882 (or unrounded values)
✓ 1
A1 0.862
✓ 1
Common error: 'Greater than 7' means 8 or more, so use 1 − P(X ≤ 7), not 1 − P(X ≤ 8).
473IBMA3S-4-18r45-134.3 Measures of central tendency and dispersionPaper 2Hard12 marks
Calculator
Climbers on a mountain choose either the east ridge route or another route. Let A be the event that a climber takes the east ridge and B the event that a climber reaches the summit. It is given that P(A) = 0.55, P(B | A) = 0.4 and P(B | A′) = 0.2.
Tree diagram
(a)
Find P(A ∩ B).
(2)
(b)
Find P(B).
(3)
(c)
Given that a climber reached the summit, find the probability that the climber took the east ridge. Give your answer to 3 significant figures.
(2)
(d)
Show that A and B are not independent.
(2)
(e)
Five climbers are chosen at random, independently. Find the probability that at least two of them reach the summit. Give your answer to 3 significant figures.
Exam tip: Build the probability of the second event from both branches (route taken and not taken), then use the total to get the reverse conditional probability.
474IBMA3S-4-19r45-124.6 Combined events and diagramsPaper 2Hard12 marks
Calculator
Each student in a large music class independently passes a grade exam with probability p. For a group of n students, the number X who pass satisfies X ~ B(n, p), with E(X) = 6.4 and Var(X) = 1.28.
(a)
Find the value of p and the value of n.
(4)
(b)
Find P(X = 7), giving your answer to 3 significant figures.
(2)
(c)
A different group of m students each pass with probability 0.8, independently. Find the smallest value of m for which the probability that at least one student fails is greater than 0.99.
(3)
(d)
For the group of 8 students, find P(X = 8 | X ≥ 7). Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 np = 6.4
✓ 1
M1 np(1 − p) = 1.28
✓ 1
A1 1 − p = 0.2 so p = 0.8
✓ 1
A1 n = 8
✓ 1
Part (b)
M1 8 × 0.8^7 × 0.2
✓ 1
A1 0.336
✓ 1
Part (c)
M1 1 − 0.8^m > 0.99 (or 0.8^m < 0.01)
✓ 1
M1 m > ln 0.01 / ln 0.8 = 20.6 or table of values showing 0.8^20 = 0.0115 and 0.8^21 = 0.0092
Exam tip: Dividing Var(X) by E(X) gives 1 − p directly, which is a quick way to separate p from n.
475IBMA3S-4-19r45-144.3 Measures of central tendency and dispersionPaper 2Hard12 marks
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Two groups of music students record their daily practice times, in minutes. Group A has 20 students with mean 45 and population standard deviation 8. Group B has 10 students with mean 54 and population standard deviation 6.
(a)
Find Σx2 for Group A and for Group B.
(3)
(b)
Find the mean practice time of all 30 students.
(2)
(c)
Find the population standard deviation of the practice times of all 30 students, giving your answer to 3 significant figures.
(3)
(d)
It is later found that one value in Group A was recorded as 33 but should have been 43. Find the corrected population standard deviation of Group A, giving your answer to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of variance = Σx²/n − mean², e.g. 8² = Σx²/20 − 45²
✓ 1
A1 Group A: Σx² = 41 780
✓ 1
A1 Group B: Σx² = 10 × (36 + 54²) = 29 520
✓ 1
Part (b)
M1 (20 × 45 + 10 × 54)/30
✓ 1
A1 48 minutes
✓ 1
Part (c)
M1 Σx² for all students = 41 780 + 29 520 = 71 300
✓ 1
M1 variance = 71 300/30 − 48²
✓ 1
A1 8.52 minutes
✓ 1
Part (d)
M1 corrected Σx = 910 so mean = 45.5
✓ 1
M1 corrected Σx² = 41 780 − 33² + 43² = 42 540
✓ 1
M1 variance = 42 540/20 − 45.5²
✓ 1
A1 7.53 minutes
✓ 1
Exam tip: To combine groups, add their sums and sums of squares; you cannot average the standard deviations.
476IBMA3S-4-18r46-104.12 Standardization and inverse normalPaper 2Hard10 marks
Calculator
The height \(H\) km of the ash plume from a volcanic eruption is normally distributed with mean \(\mu\) and standard deviation \(\sigma\). It is known that \(P(H<8)=0.20\) and \(P(H>14)=0.10\).
Normal distribution curve, not to scale
(a)
Write down two equations connecting \(\mu\) and \(\sigma\).
(3)
(b)
Hence find \(\mu\) and \(\sigma\), correct to 3 significant figures.
(3)
(c)
Using \(\mu=10.4\) and \(\sigma=2.83\), find the probability that a plume is higher than \(12\) km.
(2)
(d)
Using \(\sigma=2.83\), find the interquartile range of the plume heights, correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(z=-0.8416\ldots\) for \(P(Z<z)=0.20\)
✓ 1
M1 \(z=1.2815\ldots\) for \(P(Z<z)=0.90\)
✓ 1
A1 \(\frac{8-\mu}{\sigma}=-0.8416\) and \(\frac{14-\mu}{\sigma}=1.2816\)
✓ 1
Part (b)
M1 subtracting the equations or substituting to eliminate one variable, e.g. \(6=2.1232\sigma\)
✓ 1
A1 \(\sigma=2.83\)
✓ 1
A1 \(\mu=10.4\)
✓ 1
Part (c)
M1 attempt to use \(H\sim N(10.4,2.83^2)\) with \(P(H>12)\)
✓ 1
A1 \(0.286\)
✓ 1
Part (d)
M1 \(Q_3-Q_1=2\times0.6745\times\sigma\), or finding both quartiles
✓ 1
A1 \(3.82\) km
✓ 1
Exam tip: For two unknowns, convert each probability to a \(z\)-value first. This gives two linear equations in \(\mu\) and \(\sigma\) that you can solve simultaneously.
477IBMA3S-4-18r46-114.6 Combined events and diagramsPaper 2Hard7 marks
Calculator
Each member of a school group of \(12\) people attending a play independently buys a programme with probability \(0.35\). Let \(X\) be the number of people in the group who buy a programme.
(a)
Find the standard deviation of \(X\).
(2)
(b)
Find the probability that at least \(5\) people in the group buy a programme.
(2)
(c)
Given that at least \(5\) people in the group buy a programme, find the probability that exactly \(6\) people buy one. Use the unrounded probabilities.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\sqrt{12\times0.35\times0.65}\)
✓ 1
A1 \(1.65\)
✓ 1
Part (b)
M1 \(1-P(X\leq4)\) or \(P(X\geq5)\) with \(X\sim B(12,0.35)\)
✓ 1
A1 \(0.417\)
✓ 1
Part (c)
M1 \(P(X=6\mid X\geq5)=\frac{P(X=6)}{P(X\geq5)}\)
✓ 1
M1 \(\frac{0.12810\ldots}{0.41665\ldots}\)
✓ 1
A1 \(0.307\)
✓ 1
Common error: In \(P(X=6\mid X\geq5)\) the numerator is \(P(X=6)\) only, because \(X=6\) already lies inside \(X\geq5\). Do not use \(P(X\geq6)\).
478IBMA3S-4-18r46-124.9 The normal distributionPaper 2Hard12 marks
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The mass \(M\) tonnes of a container loaded onto a cargo ship is normally distributed with mean \(18\) and standard deviation \(2.5\). A container is called heavy if its mass is greater than \(21\) tonnes.
Normal distribution curve, not to scale
(a)
Find the probability that a randomly chosen container is heavy.
(2)
(b)
Ten containers are chosen at random. Find the probability that exactly \(3\) of them are heavy. Use the unrounded value from (a).
(3)
(c)
Find the probability that at least one of the ten containers is heavy.
(2)
(d)
The heaviest \(10\%\) of containers are loaded last. Find the least mass of a container loaded last, correct to 3 significant figures.
(2)
(e)
Given that a container is heavy, find the probability that its mass is greater than \(23\) tonnes. Use unrounded probabilities.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use \(M\sim N(18,2.5^2)\) with \(P(M>21)\)
✓ 1
A1 \(0.115\)
✓ 1
Part (b)
M1 \(Y\sim B(10,p)\) with \(p=0.11507\ldots\)
✓ 1
M1 \(\binom{10}{3}p^3(1-p)^7\)
✓ 1
A1 \(0.0777\)
✓ 1
Part (c)
M1 \(1-(1-p)^{10}\) or \(1-P(Y=0)\)
✓ 1
A1 \(0.705\)
✓ 1
Part (d)
M1 attempt to find \(m\) with \(P(M<m)=0.90\)
✓ 1
A1 \(21.2\) tonnes
✓ 1
Part (e)
M1 \(P(M>23\mid M>21)=\frac{P(M>23)}{P(M>21)}\)
✓ 1
M1 \(\frac{0.02275\ldots}{0.11507\ldots}\)
✓ 1
A1 \(0.198\)
✓ 1
Exam tip: In a long question like this, label each probability you find, such as \(P(M>21)=p\), so that you can reuse it accurately in the later parts.
479IBMA3S-4-18r48-064.8 The binomial distributionPaper 2Hard10 marks
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In a Formula 1 team, each tyre change in a race is independently completed in under 2.5 seconds with probability 0.35. Let \(X\) be the number of tyre changes completed in under 2.5 seconds in \(n\) tyre changes.
(a)
For \(n = 6\), find the probability that at least one tyre change is completed in under 2.5 seconds, to 3 significant figures.
(2)
(b)
Find the smallest value of \(n\) for which the probability of at least one tyre change under 2.5 seconds is greater than 0.99.
(4)
(c)
For \(n = 8\), find the probability that exactly 2 tyre changes are under 2.5 seconds, given that at least one is. Use unrounded values and give your answer to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use \(1 - 0.65^6\)
✓ 1
A1 0.925
✓ 1
Part (b)
M1 set up \(1 - 0.65^n > 0.99\)
✓ 1
A1 \(0.65^n < 0.01\)
✓ 1
M1 attempt to solve, e.g. \(n > \frac{\ln 0.01}{\ln 0.65}\) (\(= 10.69\)) or by trial
✓ 1
A1 \(n = 11\)
✓ 1
Part (c)
M1 attempt to use conditional probability \(\frac{P(X = 2)}{P(X \ge 1)}\) (since \(X = 2\) implies \(X \ge 1\))
✓ 1
A1 \(P(X = 2) = 0.2587\)
✓ 1
A1 \(P(X \ge 1) = 1 - 0.65^8 = 0.9681\)
✓ 1
A1 0.267
✓ 1
Common error: Dividing by \(\ln 0.65\) (a negative number) flips the inequality sign, and the answer must be rounded up to a whole number of tyre changes.
480IBMA3S-4-18r48-134.4 Correlation and regression of y on xPaper 2Hard10 marks
Calculator
A beekeeper records the mean daily temperature \(x\) °C in July and the honey yield \(y\) kg per hive for eight apiaries: \(x\): 18, 20, 22, 24, 26, 28, 30, 32 \(y\): 9.8, 11.5, 12.1, 14.2, 15.0, 17.1, 17.4, 8.0
Scatter diagram of the data
(a)
Find \(r\) and the equation of the regression line of \(y\) on \(x\) for all eight apiaries, with all values to 3 significant figures.
(3)
(b)
The apiary with \(x = 32\) had a heatwave. Find \(r\) and the equation of the regression line of \(y\) on \(x\) for the other seven apiaries, with all values to 3 significant figures.
(3)
(c)
Use your answer to part (b), unrounded, to estimate the yield at a mean temperature of 25 °C, to 3 significant figures.
(2)
(d)
Explain why the equation from part (b) is more suitable than the one from part (a) for temperatures between 18 °C and 30 °C, and give one reason why it should not be used for 32 °C.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use a calculator for regression with all eight points
✓ 1
A1 \(r = 0.282\)
✓ 1
A1 \(y = 0.195x + 8.27\)
✓ 1
Part (b)
M1 attempt to use a calculator for regression with the seven points \(x = 18\) to \(30\)
✓ 1
A1 \(r = 0.990\)
✓ 1
A1 \(y = 0.659x - 1.94\)
✓ 1
Part (c)
M1 substitute \(x = 25\) into the regression line from (b)
✓ 1
A1 14.5 kg
✓ 1
Part (d)
R1 the heatwave value distorts the line and weakens the correlation in (a); the seven points are strongly linear
✓ 1
R1 32 °C is outside the range used and yield appears to drop in extreme heat, so the linear trend does not continue
✓ 1
Exam tip: An unusual data point can change both \(r\) and the gradient a lot; always say whether a regression line is being used inside the range of the data.
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