491 free, high-quality past-paper-style questions on Statistics and probability for IB Maths AA HL, each with a full mark scheme. Covers: Sampling, bias and outliers; Measures of central tendency and dispersion; 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 Standard Level →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.4,143 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1AAHL4-q00024.1 Sampling, bias and outliersPaper 1Easy6 marks
A college has 750 students, numbered 001 to 750. A simple random sample of 5 students is chosen using the random three-digit numbers below, read from left to right. A number is used only if it corresponds to a student who has not already been selected.
512 093 781 512 000 347 906 228 650 119
(a)
Write down the numbers of the five students selected.
(2)
(b)
State what makes this a simple random sample.
(1)
(c)
The numbers of hours that the five selected students spent on voluntary work last term were 12, 7, 15, 0 and 9. Use the sample to estimate the total number of hours of voluntary work done by all 750 students last term.
(2)
(d)
Suggest one reason why this estimate may be unreliable.
(1)
Show mark scheme
Working
Mark
Notes
(a)
rejecting 781, 000 and 906 (no such student)
A1
rejecting the repeated 512: students 512, 093, 347, 228, 650
A1
(b)
every possible group of 5 students has the same probability of being chosen
R1
Accept “each student is equally likely to be chosen and choices are made at random without repetition”.
(c)
sample mean \(=\dfrac{43}{5}=8.6\) hours
(M1)
\(750\times8.6=6450\) hours
A1
(d)
the sample is very small (5 out of 750), so the sample mean may differ greatly from the population mean
R1
Accept other valid reasons, e.g. self-reported hours may be inaccurate.
2AAHL4-q00114.3 Measures of central tendency and dispersionPaper 1Easy16 marks
The marks, \(x\), of two classes in the same test are summarized below. For each class the standard deviation is that of the marks of all the students in the class.
Class
Number of students
Mean mark
Standard deviation
P
10
60
9
Q
8
69
6
For a set of \(n\) marks, the variance is \(\sigma^{2}=\dfrac{\sum x^{2}}{n}-\bar{x}^{2}\).
(a)
Show that, for class P, \(\sum x^{2}=36\,810\).
(2)
(b)
For class Q, find \(\sum x\) and \(\sum x^{2}\).
(2)
(c)
Find the mean mark of all 18 students.
(2)
(d)
Show that the standard deviation of the marks of all 18 students is 9.
(3)
(e)
The teacher scales every mark using \(y=1.2x-4\). Write down the mean and the standard deviation of the scaled marks of all 18 students.
(2)
(f)
Before scaling, it is discovered that one mark in class Q was recorded as 61 instead of 69. Find the correct mean and the correct standard deviation of the marks of class Q.
(3)
(g)
For class P, the lower quartile is 54 and the upper quartile is 62. The lowest mark in class P is 42. Determine whether this mark is an outlier.
3AAHL4-q00124.4 Correlation and regression of y on xPaper 1Easy8 marks
Different masses, \(x\) kg, were hung from a spring and its length, \(y\) cm, was measured. One length was not recorded.
\(x\) (kg)
2
4
5
7
8
10
\(y\) (cm)
13
18
22
\(w\)
31
37
The regression line of \(y\) on \(x\) for the complete data is \(y=3x+6.5\).
(a)
Write down the mean of the masses.
(1)
(b)
Using the fact that the regression line passes through the mean point, find the value of \(w\).
(3)
(c)
Interpret, in context, the value 3 and the value 6.5 in the equation of the line.
(2)
(d)
Use the regression line to estimate the length of the spring when a mass of 6 kg is hung from it.
(1)
(e)
Explain why the line should not be used to estimate the length when a mass of 25 kg is hung from it.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(\bar{x}=6\)
A1
(b)
\(\bar{y}=3(6)+6.5=24.5\)
M1
\(13+18+22+w+31+37=6\times24.5=147\)
M1
\(w=26\)
A1
(c)
3: the length increases by 3 cm for each extra kilogram
R1
6.5: the (estimated) length of the spring with no mass attached is 6.5 cm
R1
(d)
\(24.5\) cm
A1
(e)
25 kg is far outside the range of the data (extrapolation); the spring may be overstretched and stop behaving linearly
R1
Answers: (a) 6 · (b) \(w=26\) · (d) 24.5 cm
4AAHL4-q00164.5 Sample spaces and probabilityPaper 1Easy6 marks
The seven letters of the word PLANETS are arranged in a random order. All arrangements are equally likely.
(a)
Write down the number of possible arrangements.
(1)
(b)
Show that the probability that the two vowels, A and E, are next to each other is \(\dfrac{2}{7}\).
(3)
(c)
Find the probability that the first letter and the last letter are both consonants.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(7!=5040\)
A1
(b)
treating AE as one block: \(6!\) arrangements of 6 objects
M1
multiplying by 2 for the order AE or EA: \(2\times6!=1440\)
M1
\(\dfrac{1440}{5040}\)
A1
\(=\dfrac{2}{7}\)
AG
(c)
\(\dfrac{5}{7}\times\dfrac{4}{6}\)
(M1)
\(\dfrac{10}{21}\)
A1
Answers: (a) 5040 · (c) \(\tfrac{10}{21}\)
5AAHL4-q00224.6 Combined, conditional and independent eventsPaper 1Easy8 marks
Three students, Ali, Bo and Cy, each try to solve a puzzle, independently of each other. The probabilities that Ali, Bo and Cy solve it are \(\dfrac{1}{2}\), \(\dfrac{1}{3}\) and \(\dfrac{1}{4}\) respectively. The puzzle is solved if at least one of them solves it.
(a)
Find the probability that the puzzle is solved.
(2)
(b)
Find the probability that exactly one of the students solves the puzzle.
(2)
(c)
Given that the puzzle is solved, find the probability that exactly one of the students solved it.
(2)
(d)
A fourth student, who solves the puzzle with probability \(p\) independently of the others, joins them. Find the least value of \(p\) for which the probability that the puzzle is solved is at least 0.9.
7AAHL4-q00404.9 The normal distributionPaper 1Easy8 marks
A machine dispenses coffee into cups. The volume, \(X\) ml, of coffee in a cup is normally distributed with mean \(\mu\) ml and standard deviation \(\sigma\) ml. Approximately 16% of cups contain more than 215 ml and approximately 2.5% of cups contain less than 185 ml.
In this question use the facts that approximately 68%, 95% and 99.7% of a normal distribution lie within one, two and three standard deviations of the mean.
Normal curve; shaded tails are the proportions of cups above 215 ml and below 185 ml (ml). Diagram not to scale
(a)
Write down two equations in \(\mu\) and \(\sigma\).
(2)
(b)
Hence find the value of \(\mu\) and the value of \(\sigma\).
(2)
(c)
Estimate the percentage of cups that contain between 195 ml and 225 ml.
(2)
(d)
The machine fills 2000 cups. Estimate the number of these cups that contain more than 235 ml.
(2)
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Working
Mark
Notes
(a)
\(\mu+\sigma=215\)
A1
Since \(\tfrac{100-68}{2}=16\%\) lies above \(\mu+\sigma\).
\(\mu-2\sigma=185\)
A1
Since \(\tfrac{100-95}{2}=2.5\%\) lies below \(\mu-2\sigma\).
(b)
attempt to solve simultaneously, e.g. \(3\sigma=30\)
M1
\(\sigma=10\), \(\mu=205\)
A1
Both values required.
(c)
195 to 225 is \(\mu-\sigma\) to \(\mu+2\sigma\): \(34\%+47.5\%\)
(M1)
FT their \(\mu,\sigma\).
\(81.5\%\)
A1
(d)
\(235=\mu+3\sigma\), so proportion \(\approx\dfrac{100-99.7}{2}\%=0.15\%\)
8AAHL4-q00504.11 Conditional probability and independencePaper 1Easy8 marks
Events \(A\) and \(B\) are such that \(\mathrm{P}(A\mid B)=\dfrac{1}{3}\), \(\mathrm{P}(B\mid A)=\dfrac{1}{4}\) and \(\mathrm{P}(A\cap B)=\dfrac{1}{10}\).
(a)
Find \(\mathrm{P}(A)\) and \(\mathrm{P}(B)\).
(2)
(b)
Find \(\mathrm{P}(A\cup B)\).
(2)
(c)
Determine whether \(A\) and \(B\) are independent.
(2)
(d)
Find \(\mathrm{P}(A'\mid B')\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{P}(B)=\dfrac{\mathrm{P}(A\cap B)}{\mathrm{P}(A\mid B)}\) or \(\mathrm{P}(A)=\dfrac{\mathrm{P}(A\cap B)}{\mathrm{P}(B\mid A)}\)
9AAHL4-q00544.12 Standardization and unknown normal parametersPaper 1Easy6 marks
The masses, \(X\) grams, of eggs laid by a breed of hen are normally distributed with mean \(\mu\) and standard deviation \(\sigma\). Let \(Z\) be the standard normal variable, \(Z\sim\mathrm{N}(0,\,1)\). It is given that \(\mathrm{P}(X\lt 52)=\mathrm{P}(Z\lt 0.8)\) and \(\mathrm{P}(X\lt 37)=\mathrm{P}(Z\lt -1.2)\).
(a)
Find the value of \(\mu\) and the value of \(\sigma\).
(4)
(b)
Given that \(\mathrm{P}(Z\lt 1)=0.8413\), find \(\mathrm{P}(X\gt 53.5)\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac{52-\mu}{\sigma}=0.8\) or \(\dfrac{37-\mu}{\sigma}=-1.2\)
10AAHL4-q00604.13 Bayes’ theoremPaper 1Easy6 marks
In a large population, 5% of people have a certain condition. A screening test gives a positive result for 90% of people who have the condition and for 10% of people who do not have the condition. A person is chosen at random and tested.
Tree diagram with the probabilities given in the question (C: has the condition, +: positive test)
(a)
Find the probability that the test gives a positive result.
(2)
(b)
Given that the test gives a positive result, find the probability that the person has the condition.
(2)
(c)
Of 1000 people tested, find the expected number who test positive but do not have the condition, and hence comment on the result in part (b).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(0.05\times0.9+0.95\times0.1\)
(M1)
\(0.14\)
A1
(b)
\(\dfrac{0.045}{0.14}\)
(M1)
Bayes’ theorem or tree diagram; FT their (a).
\(\dfrac{9}{28}\)
A1
Accept \(0.321\).
(c)
\(1000\times0.95\times0.1=95\)
A1
only 45 positives are expected from people with the condition, so most positive results are false positives (because the condition is rare)
R1
Answers: (a) 0.14 · (b) \(\frac{9}{28}\) · (c) 95
11AAHL4-q00614.13 Bayes’ theoremPaper 1Easy6 marks
An insurance company classifies its drivers into three age groups. The tree diagram shows the proportion of drivers in each age group and the probability that a driver in that group makes a claim in a year.
Diagram not to scale
A driver is chosen at random.
(a)
Find the probability that the driver makes a claim in a year.
(2)
(b)
Given that the driver makes a claim, find the probability that the driver is under 25.
(2)
(c)
Given that the driver does not make a claim, find the probability that the driver is over 60.
12AAHL4-q00644.13 Bayes’ theoremPaper 1Easy7 marks
In a town, the probability that it rains on a given day is 0.3. When it rains, the evening forecast for that day had predicted rain with probability 0.8. When it does not rain, the forecast had predicted rain with probability 0.1.
Tree diagram with the probabilities given in the question (F: forecast predicts rain)
(a)
Find the probability that the forecast predicts rain.
(2)
(b)
Given that the forecast predicts rain, find the probability that it rains.
(2)
(c)
Given that the forecast does not predict rain, find the probability that it does not rain.
(2)
(d)
State, with a reason, whether a forecast of rain or a forecast of no rain is more reliable.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(0.3(0.8)+0.7(0.1)\)
(M1)
\(0.31\)
A1
(b)
\(\dfrac{0.24}{0.31}\)
(M1)
\(\dfrac{24}{31}\)
A1
Accept \(0.774\).
(c)
\(\dfrac{0.7(0.9)}{1-0.31}=\dfrac{0.63}{0.69}\)
(M1)
\(\dfrac{21}{23}\)
A1
Accept \(0.913\).
(d)
a forecast of no rain, since \(\dfrac{21}{23}\gt\dfrac{24}{31}\)
13AAHL4-q00734.13 Bayes’ theoremPaper 1Easy14 marks
Ana cycles to school each day by one of three routes, \(A\), \(B\) or \(C\). She chooses the route by rolling a fair six-sided die: a score of 1, 2 or 3 means route \(A\), a score of 4 or 5 means route \(B\), and a score of 6 means route \(C\).
The probability that she is late for school is \(\dfrac{1}{10}\) on route \(A\), \(\dfrac{1}{5}\) on route \(B\) and \(\dfrac{2}{5}\) on route \(C\). Let \(L\) be the event that Ana is late.
Tree diagram with the probabilities given in the question (L: Ana is late)
(a)
Show that \(\mathrm{P}(L)=\dfrac{11}{60}\).
(2)
(b)
Given that Ana is late, find the probability that she used route \(C\).
(2)
(c)
Given that Ana is not late, find the probability that she used route \(A\).
(3)
(d)
Find the expected number of days on which Ana is late in a period of 30 school days.
(1)
(e)
Find the probability that Ana is late on exactly one of two particular days. Assume that days are independent.
(2)
(f)
Determine whether the events \(L\) and “Ana uses route \(C\)” are independent. Justify your answer.
(2)
(g)
Ana improves her cycling on route \(C\) so that the probability of being late on route \(C\) becomes \(x\). The probabilities for routes \(A\) and \(B\) do not change, and \(\mathrm{P}(L)\) becomes \(\dfrac{1}{6}\). Find \(x\).
19AAHL4-q00904.14 Variance, continuous random variables and linear transformationsPaper 1Easy7 marks
The continuous random variable \(X\) has probability density function \(f(x)=k\sqrt{x}\) for \(0\le x\le4\), and \(f(x)=0\) otherwise, where \(k\) is a constant.
(a)
Find the value of \(k\).
(2)
(b)
Find the median of \(X\), giving your answer in the form \(p\sqrt[3]{q}\), where \(p,q\in\mathbb{Z}^{+}\) and \(q\) is prime.
21IBMA2H-051-01-024.6 Combined, conditional and independent eventsPaper 1Easy4 marks
No calculator
A running club has 240 members: 90 juniors, 110 seniors and 40 veterans. The club secretary wants to survey 48 members about a proposed change to training times.
(a)
The secretary uses a stratified sample by age category. Find the number of members to be chosen from each category.
(2)
(b)
State one advantage of this stratified sample over simply asking the first 48 members who arrive at training.
(1)
(c)
The secretary instead hands the survey out only at the Sunday 7 a.m. session. Suggest why this sample could be biased.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sampling fraction \(\frac{48}{240}=\frac{1}{5}\) applied to each group
✓ 1
A1 juniors 18, seniors 22, veterans 8
✓ 1
Part (b)
R1 each age category is represented in proportion to its size, so the sample is more representative of the whole club
✓ 1
Part (c)
R1 members who cannot or do not attend early Sunday sessions (for example juniors or those who work weekends) are excluded, so their views on training times are under-represented
✓ 1
Common error: A convenience sample is not 'random' just because the secretary did not choose the people deliberately. Every member must have a known chance of selection.
22IBMA2H-051-01-034.3 Measures of central tendency and dispersionPaper 1Easy5 marks
No calculator
A phone manufacturer records the battery life, in hours, of a sample of handsets under a standard video-streaming test. For this sample the lower quartile is 14.2 hours and the upper quartile is 18.6 hours. An outlier is defined as a value more than 1.5 × IQR below the lower quartile or above the upper quartile.
(a)
Find the interquartile range.
(1)
(b)
Find the boundaries beyond which a battery life would be classed as an outlier.
(2)
(c)
Two handsets recorded battery lives of 6.9 hours and 24.8 hours. Determine, with a reason, whether each value is an outlier.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 IQR = 18.6 − 14.2 = 4.4 hours
✓ 1
Part (b)
M1 1.5 × 4.4 = 6.6, subtracted from Q1 and added to Q3
R1 7.6 ≤ 24.8 ≤ 25.2, so 24.8 hours is not an outlier
✓ 1
Exam tip: State the comparison with the boundary explicitly, e.g. '6.9 < 7.6'. A bare 'yes' or 'no' does not earn the reasoning mark.
23IBMA2H-051-01-044.2 Presentation of dataPaper 1Easy4 marks
No calculator
A smart doorbell uses face recognition. Of the people who ring the doorbell, \(\frac{1}{5}\) are residents of the house. If the visitor is a resident, the doorbell recognises them with probability \(\frac{9}{10}\). If the visitor is not a resident, the doorbell wrongly 'recognises' them as a resident with probability \(\frac{1}{20}\). A visitor is recognised as a resident. Find the probability that the visitor really is a resident.
Tree diagram with the probabilities given in the question
(a)
Find P(resident | recognised).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt at total probability of being recognised: \(\frac{1}{5}\times\frac{9}{10}+\frac{4}{5}\times\frac{1}{20}\)
M1 use of conditional probability: \(\frac{9/50}{11/50}\)
✓ 1
A1 \(\frac{9}{11}\)
✓ 1
Common error: \(\frac{9}{10}\) is P(recognised | resident), not P(resident | recognised). Reversing the condition is the whole point of Bayes' theorem.
24IBMA2H-051-05-024.2 Presentation of dataPaper 1Easy6 marks
No calculator
The reaction time, \(T\) seconds, of an elite sprinter at the start of a race is modelled by a normal distribution with mean 0.160 s. It is known that \(P(T < 0.120) = 0.07\).
Normal curve of reaction time T (seconds) with mean 0.160; shaded region T < 0.120. Diagram not to scale
(a)
Write down \(P(T > 0.200)\), justifying your answer.
(2)
(b)
Find \(P(0.120 < T < 0.160)\).
(1)
(c)
Given that the sprinter's reaction time in a race is greater than 0.120 s, find the exact probability that it is less than 0.200 s.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 0.120 and 0.200 are equidistant from the mean 0.160 and the normal curve is symmetric about the mean
✓ 1
A1 \(P(T > 0.200) = 0.07\)
✓ 1
Part (b)
A1 0.5 − 0.07 = 0.43
✓ 1
Part (c)
M1 use of conditional probability \(\frac{P(0.120 < T < 0.200)}{P(T > 0.120)}\)
Common error: In the conditional part, the denominator is P(T > 0.120) = 0.93, not 1. Dividing by the probability of the given event is what turns an ordinary probability into a conditional one.
25IBMA2H-051-05-094.9 The normal distributionPaper 1Easy7 marks
No calculator
A bottling machine fills water bottles. The volume, \(V\) mL, in a bottle is normally distributed with mean \(\mu\) mL and standard deviation 4 mL. It is known that \(P(V < 494) = 0.0668\). You may use the following values, where \(Z \sim N(0, 1)\): \(P(Z < 0.5) = 0.6915\), \(P(Z < 1.5) = 0.9332\), \(P(Z < 2) = 0.9772\).
Normal curve of volume V (mL); shaded region V < 494. Diagram not to scale
(a)
Find the value of \(\mu\).
(2)
(b)
Write down \(P(V > 506)\).
(1)
(c)
Find \(P(494 < V < 502)\).
(2)
(d)
A new machine also has mean 500 mL, but only 2.28% of its bottles contain less than 494 mL. Find the standard deviation of the volume for the new machine.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{494 - \mu}{4} = -1.5\)
✓ 1
A1 \(\mu\) = 500
✓ 1
Part (b)
A1 0.0668
✓ 1
Part (c)
M1 502 corresponds to \(z = 0.5\), so the probability is \(0.6915 - 0.0668\)
✓ 1
A1 0.6247
✓ 1
Part (d)
M1 \(\frac{494 - 500}{\sigma} = -2\)
✓ 1
A1 \(\sigma\) = 3 mL
✓ 1
Common error: The z-value for a lower tail is negative. Writing (494 − μ)/4 = 1.5 gives μ = 488, which is below 494 and contradicts a lower-tail probability that is less than 0.5.
26IBMA2H-051-08-014.2 Presentation of dataPaper 1Easy5 marks
No calculator
A laboratory tests the battery life, in hours, of 60 smartphones of the same model under a standard video-streaming load. The results are summarised below. Minimum = 6.2, lower quartile = 9.8, median = 11.5, upper quartile = 13.4, maximum = 19.9. An outlier is defined as a value more than 1.5 × IQR below the lower quartile or more than 1.5 × IQR above the upper quartile.
Box plot of the five-number summary (whiskers drawn to the minimum and maximum)
(a)
Find the interquartile range.
(1)
(b)
Determine whether the minimum or the maximum battery life is an outlier. Justify your answer.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 IQR = 13.4 − 9.8 = 3.6 hours
✓ 1
Part (b)
M1 calculates 1.5 × IQR = 1.5 × 3.6 = 5.4
✓ 1
A1 upper boundary 13.4 + 5.4 = 18.8 hours
✓ 1
A1 lower boundary 9.8 − 5.4 = 4.4 hours
✓ 1
R1 19.9 > 18.8 so the maximum is an outlier; 6.2 > 4.4 so the minimum is not an outlier
✓ 1
Exam tip: Always check both ends of the data. A full justification compares each extreme value with its own boundary.
27IBMA2H-051-08-044.3 Measures of central tendency and dispersionPaper 1Easy6 marks
No calculator
The resting heart rates, in beats per minute, of 11 endurance athletes are listed in ascending order: 48, 51, 52, 55, 57, a, 61, 62, 64, 66, b The median heart rate is 59. Quartiles are found using the method in which the median is excluded from each half of the data.
(a)
Write down the value of a.
(1)
(b)
Find the lower quartile and the upper quartile.
(2)
(c)
Given that b is an integer and is an outlier, find the smallest possible value of b. (An outlier is more than 1.5 × IQR above the upper quartile.)
A1 upper quartile = 64 (median of 61, 62, 64, 66, b)
✓ 1
Part (c)
M1 IQR = 12 and boundary = 64 + 1.5 × 12
✓ 1
A1 boundary = 82
✓ 1
A1 b > 82 so smallest integer b = 83
✓ 1
Common error: A value equal to the boundary is not an outlier; it must be strictly greater than 82, so 82 itself is not the answer.
28IBMA2H-051-00-014.14 Variance, continuous random variables and linear transformationsPaper 1Easy5 marks
No calculator
A delivery drone touches down on a circular landing pad of radius 3 m. The distance, X metres, of the touchdown point from the centre of the pad is modelled by the probability density function \(f(x)=k(9-x^2)\) for \(0\le x\le 3\), and \(f(x)=0\) otherwise.
M1 \(E(X)=\frac{1}{18}\int_0^3 (9x-x^3)\,dx\) A1 \(\frac{1}{18}\left[\frac{9x^2}{2}-\frac{x^4}{4}\right]_0^3=\frac{1}{18}\left(\frac{81}{2}-\frac{81}{4}\right)=\frac{1}{18}\cdot\frac{81}{4}\) A1 \(E(X)=\frac{9}{8}\) m
✓ 3
Exam tip: For E(X) multiply the pdf by x before integrating: the integrand becomes \(9x-x^3\), not \(9-x^2\).
29IBMA2H-051-00-034.14 Variance, continuous random variables and linear transformationsPaper 1Easy6 marks
No calculator
A thermocouple on a space probe measures a temperature T °C. Over a long period, E(T) = 20 and Var(T) = 4. The probe transmits the reading in degrees Fahrenheit, F, where \(F=1.8T+32\). A separate calibration circuit outputs a voltage R millivolts, where \(R=50-2.5T\).
(a)
Find E(F).
(2)
(b)
Find the standard deviation of F.
(2)
(c)
Find Var(R).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(E(F)=1.8E(T)+32=1.8(20)+32\) A1 \(E(F)=68\)
✓ 2
Part (b)
M1 \(\text{Var}(F)=1.8^2\times 4=12.96\) or SD\((F)=1.8\times 2\) A1 SD\((F)=3.6\) °F
Common error: A negative multiplier does not make the variance negative: \(\text{Var}(aT+b)=a^2\text{Var}(T)\), and the constant b has no effect on spread.
30IBMA2H-051-04-014.11 Conditional probability and independencePaper 1Easy6 marks
No calculator
A smartphone app records two types of alert during a day. Let A be the event that a battery alert occurs and B be the event that a storage alert occurs. For a randomly chosen day, P(A) = 2/5, P(B) = 1/4 and P(A ∪ B) = 11/20.
(a)
Find P(A ∩ B).
(2)
(b)
Show that A and B are independent.
(2)
(c)
Find P(A | B′).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of P(A ∪ B) = P(A) + P(B) − P(A ∩ B), giving 2/5 + 1/4 − 11/20 ; A1 P(A ∩ B) = 1/10
✓ 2
Part (b)
M1 P(A) × P(B) = 2/5 × 1/4 = 1/10 ; R1 this equals P(A ∩ B), so A and B are independent
Exam tip: Once you have shown A and B are independent, A and B′ are also independent, so P(A | B′) = P(A) = 2/5 gives a quick check of your answer.
31IBMA2H-051-04-034.11 Conditional probability and independencePaper 1Easy6 marks
No calculator
During one pass of a space probe, two ground stations independently attempt to receive its signal. Station P receives the signal with probability 2/3 and station Q receives it with probability 3/5.
(a)
Find the probability that at least one station receives the signal.
(2)
(b)
Find the probability that exactly one station receives the signal.
(2)
(c)
Given that exactly one station receives the signal, find the probability that it is station P.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 1 − (1/3)(2/5) ; A1 13/15
✓ 2
Part (b)
M1 (2/3)(2/5) + (1/3)(3/5) ; A1 7/15
✓ 2
Part (c)
M1 (4/15)/(7/15) ; A1 4/7
✓ 2
Exam tip: 'At least one' is usually fastest as 1 − P(none). 'Exactly one' needs both cases added: P receives and Q misses, or P misses and Q receives.
32IBMA2H-051-04-044.6 Combined, conditional and independent eventsPaper 1Easy6 marks
No calculator
An engineering school has 240 students. Of these, 130 take Robotics (R), 95 take Materials (M) and 50 take neither subject.
Venn diagram (R: Robotics, M: Materials). Diagram not to scale
(a)
Find the number of students who take both subjects.
(2)
(b)
A student is chosen at random. Given that the student takes Materials, find the probability that the student also takes Robotics.
(2)
(c)
Determine whether the events R and M are independent. Justify your answer.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 130 + 95 − x + 50 = 240 ; A1 x = 35
✓ 2
Part (b)
M1 35/95 ; A1 7/19
✓ 2
Part (c)
M1 compares P(R) = 130/240 = 13/24 with P(R | M) = 7/19 (or compares P(R ∩ M) = 35/240 with P(R)P(M)) ; R1 these are not equal, so R and M are not independent
✓ 2
Common error: Adding 130 + 95 double-counts the students who take both subjects. Set up an equation for the overlap that includes the 50 who take neither.
33IBMA2H-051-07-024.2 Presentation of dataPaper 1Easy5 marks
No calculator
The finishing times, T minutes, of runners in a city marathon are modelled by a normal distribution with mean μ minutes and standard deviation 18 minutes. It is known that 6.68% of runners take longer than 270 minutes to finish. You may use the fact that if Z ~ N(0, 1) then P(Z < 1.5) = 0.9332.
Normal curve of finishing time T (minutes); shaded region T > 270. Diagram not to scale
(a)
Show that μ = 243.
(3)
(b)
The organisers award a gold medal to the fastest 6.68% of runners. Find the time, in minutes, that a runner must beat to receive a gold medal.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(T > 270) = 0.0668 = 1 − 0.9332, so the standardised value of 270 is z = 1.5 M1 forming (270 − μ)/18 = 1.5 A1 μ = 270 − 27 = 243
✓ 3
Part (b)
M1 by symmetry the required standardised value is z = −1.5, so t = 243 − 1.5 × 18 A1 t = 216 minutes
✓ 2
Exam tip: In a no-calculator question, the given value of P(Z < 1.5) can be used for both tails. Symmetry of the standard normal curve gives P(Z < −1.5) = 0.0668 with no further information needed.
34IBMA2H-051-07-104.9 The normal distributionPaper 1Easy7 marks
No calculator
A coffee machine dispenses a volume V ml of coffee into each cup, where V ~ N(μ, σ2). It is known that P(V < 180) = P(V > 220) = 0.0228. You may use the facts that if Z ~ N(0, 1) then P(Z < 2) = 0.9772 and P(Z < 1) = 0.8413.
(a)
Write down the value of μ, giving a reason for your answer.
(2)
(b)
Find the value of σ.
(2)
(c)
The machine is adjusted so that the mean changes but σ stays the same. After the adjustment, P(V < 190) = 0.0228. Find the new mean.
(2)
(d)
Find the probability that, after the adjustment, a cup contains more than 220 ml.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 the two tail probabilities are equal, so 180 and 220 are symmetric about the mean A1 μ = 200 ml
Exam tip: Equal tail probabilities at two values mean the mean lies exactly halfway between them. Spotting this saves solving simultaneous equations.
35IBMA2H-051-06-014.11 Conditional probability and independencePaper 1Easy6 marks
No calculator
A fitness app records two events for a randomly chosen user on a randomly chosen day. \(A\): the user walks more than 10 000 steps. \(B\): the user records more than 7 hours of sleep. It is known that \(P(A)=\frac{2}{5}\), \(P(B\mid A)=\frac{3}{4}\) and \(P(A\cup B)=\frac{4}{5}\).
(a)
Find \(P(A\cap B)\).
(2)
(b)
Find \(P(B)\).
(2)
(c)
Determine whether events \(A\) and \(B\) are independent. Justify your answer.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(P(A\cap B)=P(B\mid A)\,P(A)=\frac{3}{4}\times\frac{2}{5}\) ; A1 \(P(A\cap B)=\frac{3}{10}\)
✓ 2
Part (b)
M1 use of \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), i.e. \(\frac{4}{5}=\frac{2}{5}+P(B)-\frac{3}{10}\) ; A1 \(P(B)=\frac{7}{10}\)
✓ 2
Part (c)
M1 compares \(P(A)P(B)=\frac{2}{5}\times\frac{7}{10}=\frac{7}{25}\) with \(P(A\cap B)=\frac{3}{10}\) (or compares \(P(B\mid A)=\frac{3}{4}\) with \(P(B)=\frac{7}{10}\)) ; R1 the values are not equal, so \(A\) and \(B\) are not independent
✓ 2
Exam tip: For an independence test, state both values you are comparing and then write a clear conclusion; a bare 'not independent' with no numbers earns no reasoning mark.
36IBMA2H-051-06-034.14 Variance, continuous random variables and linear transformationsPaper 1Easy6 marks
No calculator
A deep-space probe sends a signal that may be received by two ground stations, P and Q, independently of each other. Station Q receives the signal with probability 0.6 and station P receives it with probability \(p\). The probability that at least one station receives the signal is 0.9.
(a)
Find the value of \(p\).
(3)
(b)
Given that at least one station receives the signal, find the probability that exactly one station receives it.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(P(\text{at least one})=1-P(\text{neither})\) ; M1 \(1-(1-p)(0.4)=0.9\) ; A1 \(p=0.75\)
Exam tip: 'At least one' is almost always fastest through the complement: 1 minus the probability that neither event happens.
37IBMA2H-051-06-044.6 Combined, conditional and independent eventsPaper 1Easy5 marks
No calculator
A workshop makes drone propellers on two machines. In one week, machine X made 240 propellers, of which 18 were cracked, and machine Y made 160 propellers, of which 6 were cracked. One propeller is chosen at random from the 400 made that week.
(a)
Find the probability that the propeller is cracked.
(1)
(b)
Given that the propeller is cracked, find the probability that it was made on machine X.
(2)
(c)
Two different cracked propellers are chosen at random for analysis. Find the probability that both were made on machine Y.
Common error: Choosing two different propellers is sampling without replacement: the second fraction must be \(\frac{5}{23}\), not \(\frac{6}{24}\) again.
38IBMA2H-051-03-024.8 Binomial distributionPaper 1Easy5 marks
No calculator
A delivery drone makes n deliveries in a week. Each delivery is late independently with probability p. Let X be the number of late deliveries in a week. It is known that E(X) = 6 and Var(X) = 4.2.
(a)
Find the value of p.
(2)
(b)
Find the value of n.
(1)
(c)
Find Var(5 − 2X).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 dividing np(1 − p) = 4.2 by np = 6 to get 1 − p = 0.7
✓ 1
A1 p = 0.3
✓ 1
Part (b)
A1 n = 6 ÷ 0.3 = 20
✓ 1
Part (c)
M1 use of Var(aX + b) = a2Var(X), giving (−2)2 × 4.2
✓ 1
A1 16.8
✓ 1
Common error: The constant 5 has no effect on the variance, and the factor −2 is squared. Var(5 − 2X) is 4Var(X). It is not −2Var(X) and it is not 5 − 2Var(X).
39IBMA2H-051-09-014.10 Regression line of x on yPaper 1Easy6 marks
No calculator
A ground station records the signal delay, in milliseconds (ms), of eight pings to a low-orbit satellite: 12, 15, 9, x, 14, 11, 16, 13 The mean delay is 13 ms. Quartiles are found as the medians of the lower and upper halves of the ordered data.
(a)
Find the value of x.
(2)
(b)
Find the median delay.
(2)
(c)
Find the interquartile range of the delays.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sets up \(\frac{90+x}{8}=13\) or states the total is 104 A1 x = 14
✓ 2
Part (b)
M1 orders the data: 9, 11, 12, 13, 14, 14, 15, 16 A1 median = \(\frac{13+14}{2}\) = 13.5 ms
✓ 2
Part (c)
M1 Q1 = 11.5 and Q3 = 14.5 A1 IQR = 3 ms
✓ 2
Exam tip: Always put the data in order, including the value you have just found, before reading off the median or quartiles.
40IBMA2H-051-09-024.10 Regression line of x on yPaper 1Easy5 marks
No calculator
A starting-block sensor records sprinters' reaction times in seconds. For one heat the mean reaction time is 0.162 s and the standard deviation is 0.021 s. It is then found that the sensor added a constant lag of 0.008 s to every reading. Each reading is corrected by removing the lag and is then converted to milliseconds.
(a)
Find the mean of the corrected reaction times, in milliseconds.
(2)
(b)
Find the standard deviation of the corrected reaction times, in milliseconds.
(2)
(c)
Write down the variance of the corrected reaction times, stating its units.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 subtracts 0.008 from the mean and multiplies by 1000 A1 154 ms
✓ 2
Part (b)
M1 recognises the standard deviation is unchanged by subtracting a constant but is multiplied by 1000 A1 21 ms
✓ 2
Part (c)
A1 441 ms2
✓ 1
Common error: Subtracting a constant shifts the mean but leaves the spread unchanged; do not subtract 0.008 from the standard deviation.
41IBMA2H-051-09-044.14 Variance, continuous random variables and linear transformationsPaper 1Easy6 marks
No calculator
The number of defective cells, X, on a randomly chosen solar panel has the following probability distribution: P(X = 0) = 0.55, P(X = 1) = 0.25, P(X = 2) = 0.12, P(X = 3) = k.
Common error: Var(X) = E(X²) − [E(X)]². Squaring E(X) is not the same as finding E(X²), so you must square each value of x first.
42IBMA2H-051-02-014.3 Measures of central tendency and dispersionPaper 1Easy4 marks
No calculator
A delivery drone logs \(X\), the number of GPS signal dropouts during a one-hour flight. The probability distribution of \(X\) is given by \(P(X=x)=k(x^2+1)\), for \(x=0,1,2,3\), where \(k\) is a constant.
Exam tip: Write out each probability as a multiple of k before adding. A quick check that the four probabilities sum to 1 catches arithmetic slips before they carry into E(X).
43IBMA2H-051-02-024.7 Discrete random variablesPaper 1Easy6 marks
No calculator
A board-game app simulates rolling a fair four-sided die, with faces numbered 1 to 4, and a fair six-sided die, with faces numbered 1 to 6. The two results are independent. Event \(A\): the sum of the two scores is 7. Event \(B\): the product of the two scores is even.
(a)
Find \(P(A)\).
(2)
(b)
Find \(P(B)\).
(2)
(c)
Determine whether events \(A\) and \(B\) are independent. Justify your answer.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 identifies 24 equally likely outcomes and lists the favourable pairs (1,6), (2,5), (3,4), (4,3)
✓ 1
A1 \(P(A)=\frac{4}{24}=\frac{1}{6}\)
✓ 1
Part (b)
M1 the product is odd only when both scores are odd: \(\frac{2}{4}\times\frac{3}{6}=\frac{1}{4}\), or counts 18 favourable outcomes out of 24
✓ 1
A1 \(P(B)=1-\frac{1}{4}=\frac{3}{4}\)
✓ 1
Part (c)
R1 a sum of 7 needs one odd and one even score, so the product is always even and \(P(A\cap B)=P(A)=\frac{1}{6}\)
✓ 1
A1 \(P(A)P(B)=\frac{1}{6}\times\frac{3}{4}=\frac{1}{8}\neq\frac{1}{6}\), so \(A\) and \(B\) are not independent
✓ 1
(Alternative: \(P(A\mid B)=\frac{1/6}{3/4}=\frac{2}{9}\neq P(A)\), so the events are not independent. This earns full credit.)
—
Common error: Independent is not the same as mutually exclusive. Here every outcome in A is also in B (A is a subset of B), so the events are certainly not mutually exclusive. To test independence, compare P(A∩B) with P(A)P(B), or compare P(A|B) with P(A).
44IBMA3H-4-19r6-014.11 Conditional probability and independencePaper 1Easy5 marks
No calculator
In a clinical trial, each patient takes two drugs. Let A be the event that a patient reports drowsiness and B the event that a patient reports nausea. The events A and B are independent, with P(A) = 0.4 and P(A ∪ B) = 0.7.
(a)
Find P(B).
(3)
(b)
Find the probability that a randomly chosen patient reports drowsiness but not nausea.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
✓ 1
M1 use of independence, P(A ∩ B) = P(A)P(B), giving 0.4 + p − 0.4p = 0.7
✓ 1
A1 P(B) = 0.5
✓ 1
Part (b)
M1 P(A ∩ B′) = P(A) × P(B′) = 0.4 × (1 − 0.5)
✓ 1
A1 0.2
✓ 1
Common error: Mutually exclusive and independent are different ideas: for independent events with non-zero probabilities, P(A ∩ B) is not zero.
45IBMA3H-4-19r7-014.1 Sampling, bias and outliersPaper 1Easy6 marks
No calculator
A team of archaeologists has 480 pottery sherds from a trench, numbered 1 to 480 in the order in which they were found. They choose a systematic sample of 24 sherds. The first sherd selected is number 7.
(a)
Write down the sampling interval.
(2)
(b)
Find the number of the 12th sherd selected.
(2)
(c)
Instead, the team chooses the 24 largest sherds from the whole collection. Explain why this sample may give a biased estimate of the mean thickness of all 480 sherds.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses 480 ÷ 24
✓ 1
A1 interval = 20
✓ 1
Part (b)
M1 attempt to use \(u_{12} = 7 + 11 \times 20\)
✓ 1
A1 227
✓ 1
Part (c)
R1 not every sherd has an equal chance of being selected (selection is not random)
✓ 1
R1 larger sherds may be systematically thicker, so the sample is not representative of the collection
✓ 1
Exam tip: For a systematic sample, the k-th item selected is first + (k − 1) × interval.
46IBMA3H-4-19r7-114.10 Regression line of x on yPaper 1Easy5 marks
No calculator
A data centre has two cooling lines, A and B. Line A produces 60% of the server units and line B produces the rest. 2% of units from A are defective and 5% of units from B are defective. A unit is chosen at random.
Tree diagram showing the probabilities given in the question; D = unit is defective
(a)
Find the probability that the unit is defective.
(2)
(b)
Given that the unit is defective, find the probability that it came from line A.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(D) = P(A)P(D | A) + P(B)P(D | B) = 0.6 × 0.02 + 0.4 × 0.05
✓ 1
A1 0.032
✓ 1
Part (b)
M1 P(A | D) = P(A ∩ D) ÷ P(D)
✓ 1
A1 0.012 ÷ 0.032
✓ 1
A1 3/8 (= 0.375)
✓ 1
Exam tip: Bayes' theorem is the conditional probability formula with the total probability of the evidence in the denominator.
47IBMA3H-4-19r9-074.14 Variance, continuous random variables and linear transformationsPaper 1Easy6 marks
No calculator
The duration, \(X\) hours, of a lava flow event has probability density function \(f(x)=kx(4-x^2)\) for \(0\le x\le2\), and \(f(x)=0\) otherwise, where \(k\) is a constant.
Graph of the probability density function, not to scale
Exam tip: In a 'show that' question, write the integral equation first and then the substitution of limits; every line carries a mark.
48IBMA3H-4-19r14-024.6 Combined, conditional and independent eventsPaper 1Easy5 marks
No calculator
In a study of patients at a clinic, A is the event "the patient has a fever" and B is the event "the patient has a cough". Events A and B are independent, with P(A) = 0.3 and P(A ∪ B) = 0.58.
(a)
Find P(B).
(3)
(b)
Find P(A ∩ B′).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use independence: P(A ∩ B) = P(A)P(B) = 0.3p
✓ 1
M1 P(A ∪ B) = P(A) + P(B) − P(A ∩ B), so 0.58 = 0.3 + p − 0.3p
✓ 1
A1 P(B) = 0.4
✓ 1
Part (b)
M1 P(A)P(B′) = 0.3 × 0.6 (or 0.3 − 0.12)
✓ 1
A1 0.18
✓ 1
Common error: Do not confuse independent with mutually exclusive: for independent events P(A ∩ B) is the product, not zero.
49IBMA3H-4-19r15-014.14 Variance, continuous random variables and linear transformationsPaper 1Easy6 marks
No calculator
The discrete random variable \(X\) is the number of members of a household who catch a seasonal virus after one member becomes ill at home. Its probability distribution is \(P(X=x)=k(5-x)\) for \(x=1,2,3,4\), where \(k\) is a constant.
(a)
Show that \(k=0.1\).
(2)
(b)
Find \(E(X)\).
(2)
(c)
Find \(P(X>E(X))\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sum of probabilities set equal to 1: \(4k+3k+2k+k=1\)
✓ 1
A1 \(10k=1\) so \(k=0.1\)
✓ 1
Part (b)
M1 attempt at \(\sum xP(X=x)\): \(1(0.4)+2(0.3)+3(0.2)+4(0.1)\)
✓ 1
A1 \(E(X)=2\)
✓ 1
Part (c)
M1 identifies the values \(x=3\) and \(x=4\)
✓ 1
A1 \(P=0.2+0.1=0.3\) (ECF from (b) if consistent)
✓ 1
Exam tip: In a 'show that' question, write the equation you start from; the answer given is not enough by itself.
50IBMA3H-4-19r15-044.7 Discrete random variablesPaper 1Easy6 marks
No calculator
A retail loyalty game uses two fair dice: a four-sided die numbered 1 to 4 and a six-sided die numbered 1 to 6. A customer rolls both dice once and adds the two scores.
(a)
Write down the number of equally likely outcomes in the sample space.
(1)
(b)
Find the probability that the sum is at least 8.
(2)
(c)
A customer wins a voucher if the sum is at least 8 or both dice show an even number. Find the probability of winning a voucher.
M1 identifies the 6 outcomes with both dice even: (2,2), (2,4), (2,6), (4,2), (4,4), (4,6)
✓ 1
M1 recognises the overlap (2,6), (4,4), (4,6) is counted once, so 9 outcomes
✓ 1
A1 \(\frac{9}{24}=\frac38\) (0.375)
✓ 1
Exam tip: List the sample space systematically (a table works well) so overlapping outcomes are not counted twice.
51IBMA3H-4-19r15-114.2 Presentation of dataPaper 1Easy5 marks
No calculator
A kiosk sells \(X\) items per hour, where \(E(X)=5\) and \(E(X^2)=29\). The kiosk's net income per hour is \(N=12X-20\) euros.
(a)
Find \(\mathrm{Var}(X)\).
(2)
(b)
Find \(E(N)\) and the standard deviation of \(N\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\mathrm{Var}(X)=E(X^2)-[E(X)]^2=29-25\)
✓ 1
A1 4
✓ 1
Part (b)
A1 \(E(N)=12(5)-20=40\)
✓ 1
M1 \(\mathrm{Var}(N)=12^2\times4=576\)
✓ 1
A1 standard deviation 24 (€)
✓ 1
Exam tip: For \(\mathrm{Var}(aX+b)\) the multiplier is squared; for the standard deviation it is \(|a|\).
52IBMA3H-4-19r17-064.11 Conditional probability and independencePaper 1Easy5 marks
No calculator
A swimming club has two coaches. Coach A trains \(\frac{2}{5}\) of the swimmers and Coach B trains the other \(\frac{3}{5}\). The probability that a swimmer trained by Coach A qualifies for the county meet is \(\frac{3}{4}\). The probability that a swimmer trained by Coach B qualifies is \(\frac{1}{3}\). A swimmer is chosen at random from the club.
Tree diagram showing the probabilities given in the question; Q = swimmer qualifies for the county meet
(a)
Find the probability that the swimmer qualifies for the county meet.
(2)
(b)
Given that the swimmer qualifies, find the probability that the swimmer is trained by Coach A.
M1 use of Bayes' theorem or \(\dfrac{\mathrm{P}(A\cap Q)}{\mathrm{P}(Q)}\)
✓ 1
A1 \(\dfrac{3/10}{1/2}\) (ECF from (a))
✓ 1
A1 \(\dfrac{3}{5}\)
✓ 1
Exam tip: The answer to a Bayes question should differ from the prior: here the prior for Coach A is 2/5 but the posterior is 3/5, because Coach A's swimmers qualify more often.
53IBMA3H-4-19r17-104.6 Combined, conditional and independent eventsPaper 1Easy7 marks
No calculator
A library trolley holds \(12\) returned books: \(7\) are fiction and \(5\) are non-fiction. Two books are selected at random from the trolley, one after the other, without replacement.
F = fiction, N = non-fiction; first book, then second book (no replacement)
(a)
Find the probability that both books are fiction.
(2)
(b)
Find the probability that exactly one of the two books is fiction.
(2)
(c)
Given that at least one of the books is fiction, find the probability that the second book selected is fiction.
M1 \(\mathrm{P}(\text{at least one fiction})=1-\dfrac{5}{12}\times\dfrac{4}{11}=\dfrac{28}{33}\)
✓ 1
M1 numerator: \(\mathrm{P}(\text{second is fiction})=\dfrac{7}{12}\) (the second being fiction already means at least one is fiction)
✓ 1
A1 \(\dfrac{7/12}{28/33}=\dfrac{11}{16}\)
✓ 1
Exam tip: When the second draw is without replacement, the second-draw probabilities use the reduced total (11) and the reduced count of the type already chosen.
54IBMA3H-4-19r19-034.3 Measures of central tendency and dispersionPaper 1Easy7 marks
No calculator
A band plays an encore after each concert. The number of encore songs, X, has probability distribution 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 E(X).
(2)
(c)
Find Var(X).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sum of probabilities = 5k + 4k + 3k + 2k = 14k = 1
Common error: Var(X) = E(X²) − [E(X)]²: squaring the mean after subtracting, rather than before, is a frequent slip.
55IBMA3H-4-18r20-044.2 Presentation of dataPaper 1Easy4 marks
No calculator
The delay times of trains on a line have mean 12.4 minutes and standard deviation 1.5 minutes. A compensation cost is calculated for each train as €3 for each minute of delay plus a fixed amount of €4. Let \(x\) be the delay in minutes, so that the cost in euros is \(y=3x+4\).
(a)
Find the mean cost per train.
(2)
(b)
Find the standard deviation of the cost.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(3\times12.4+4\)
✓ 1
A1 €41.20
✓ 1
Part (b)
M1 multiply by 3 only: \(3\times1.5\)
✓ 1
A1 €4.50
✓ 1
Common error: Adding a constant to every value changes the mean but not the standard deviation; multiplying by a constant changes both.
56IBMA3H-4-18r21-034.7 Discrete random variablesPaper 1Easy7 marks
No calculator
A community energy cooperative runs a prize draw. Two fair four-sided dice, each numbered 1, 2, 3, 4, are rolled together. Let \(D\) be the positive difference between the two scores.
(a)
Show that \(P(D=1)=\frac38\).
(2)
(b)
Find \(P(D\geq2)\).
(2)
(c)
Let \(A\) be the event \(D=1\) and \(B\) the event that the total score is 5. Find \(P(A\cup B)\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 identifies the 6 outcomes \((1,2),(2,1),(2,3),(3,2),(3,4),(4,3)\) out of 16
✓ 1
A1 \(\frac{6}{16}=\frac38\) (AG)
✓ 1
Part (b)
M1 \(P(D=0)=\frac{4}{16}\) and uses complement \(1-\frac14-\frac38\) (or counts 6 outcomes)
✓ 1
A1 \(\frac38\)
✓ 1
Part (c)
M1 \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
✓ 1
A1 \(P(B)=\frac{4}{16}\) and \(P(A\cap B)=\frac{2}{16}\) (outcomes \((2,3)\) and \((3,2)\))
✓ 1
A1 \(\frac{6+4-2}{16}=\frac12\) (= 0.5)
✓ 1
Common error: List the outcomes in an event before counting; outcomes in both events must be subtracted once so they are not counted twice.
57IBMA3H-4-19r23-034.14 Variance, continuous random variables and linear transformationsPaper 1Easy6 marks
No calculator
The waiting time, X minutes, of a patient in a hospital emergency department has E(X) = 24 and Var(X) = 36. The staffing cost of a patient, in euros, is Y = aX + b, where a > 0. It is known that E(Y) = 150 and Var(Y) = 144.
Common error: Adding a constant b shifts the mean but leaves the variance unchanged, and the multiplier a is squared in the variance.
58IBMA3H-4-19r23-084.2 Presentation of dataPaper 1Easy5 marks
No calculator
The waiting times, in minutes, of 21 patients at a hospital clinic have the following summary: minimum 6, lower quartile 22, median 33, upper quartile 48, maximum 100. The mean waiting time is 40 minutes. A value is classed as an outlier if it is more than 1.5 × IQR above the upper quartile or more than 1.5 × IQR below the lower quartile.
Box plot of the five-number summary (whiskers to the minimum and maximum)
(a)
Write down the interquartile range.
(1)
(b)
Show that the maximum waiting time is an outlier.
(2)
(c)
The outlier is removed from the data. Find the mean waiting time of the remaining patients.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 IQR = 48 − 22 = 26
✓ 1
Part (b)
M1 upper boundary = 48 + 1.5 × 26 = 87
✓ 1
R1 100 > 87, so 100 minutes is an outlier
✓ 1
Part (c)
M1 total = 21 × 40 = 840, remaining total = 840 − 100 = 740
✓ 1
A1 mean = 740 ÷ 20 = 37 minutes
✓ 1
Common error: When a value is removed, divide by the new number of data items (20), not the original 21.
59IBMA3H-4-19r24-014.11 Conditional probability and independencePaper 1Easy6 marks
No calculator
In a clinical trial of a new medicine, \(A\) is the event that a patient's symptoms improve and \(B\) is the event that the patient reports side effects. It is given that \(P(A)=0.64\), \(P(B)=0.25\) and \(P(A\cap B)=0.16\).
Venn diagram showing the information given in the question
(a)
Show that \(A\) and \(B\) are independent.
(2)
(b)
Find \(P(A\cup B)\).
(2)
(c)
Find \(P(A'\mid B)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to find \(P(A)\times P(B)=0.64\times0.25\)
✓ 1
R1 \(=0.16=P(A\cap B)\), hence independent
✓ 1
Part (b)
M1 \(P(A)+P(B)-P(A\cap B)=0.64+0.25-0.16\)
✓ 1
A1 \(0.73\)
✓ 1
Part (c)
M1 \(\dfrac{P(A'\cap B)}{P(B)}=\dfrac{0.25-0.16}{0.25}\), or use of independence so that \(P(A'\mid B)=P(A')\)
✓ 1
A1 \(0.36\)
✓ 1
Common error: Independent events are not the same as mutually exclusive events: for independent events with non-zero probabilities, \(P(A\cap B)\) is never zero.
60IBMA3H-4-18r25-024.11 Conditional probability and independencePaper 1Easy6 marks
No calculator
In a cycling sportive, A is the event that a randomly chosen rider has completed the event before, and B is the event that the rider uses a power meter. P(A) = 0.45, P(B) = 0.4 and P(A ∪ B) = 0.67.
(a)
Find P(A ∩ B).
(2)
(b)
Find P(A | B).
(2)
(c)
Determine whether A and B are independent, giving a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use P(A ∪ B) = P(A) + P(B) − P(A ∩ B): 0.67 = 0.45 + 0.4 − P(A ∩ B)
✓ 1
A1 P(A ∩ B) = 0.18
✓ 1
Part (b)
M1 use P(A | B) = \(\frac{P(A\cap B)}{P(B)}\) = \(\frac{0.18}{0.4}\)
✓ 1
A1 P(A | B) = 0.45 (ECF from (a))
✓ 1
Part (c)
M1 compare P(A | B) = 0.45 with P(A) = 0.45 (or P(A)P(B) = 0.18 with P(A ∩ B) = 0.18)
✓ 1
R1 equal, so A and B are independent
✓ 1
Exam tip: To test independence, show that P(A ∩ B) = P(A) × P(B) (or P(A | B) = P(A)) and state the conclusion in words.
61IBMA3H-4-18r25-044.7 Discrete random variablesPaper 1Easy5 marks
No calculator
The number of water bottles, X, that a rider takes from the support car during a stage has probability distribution P(X = x) = \(\frac{x}{10}\), for x = 1, 2, 3, 4.
(a)
Show that this is a valid probability distribution.
(2)
(b)
Find E(X).
(2)
(c)
Find P(X ≥ 3).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sum the probabilities: 0.1 + 0.2 + 0.3 + 0.4
✓ 1
R1 the sum is 1 (and every probability is between 0 and 1) (AG)
Exam tip: 'Show that' a distribution is valid: both conditions matter, namely every probability lies between 0 and 1 and they add to exactly 1.
62IBMA3H-4-20r28-034.1 Sampling, bias and outliersPaper 1Easy6 marks
No calculator
The manager of a theatre wants to know whether audience members think ticket prices are fair. She asks the first 40 people who leave after a Saturday matinee to complete a questionnaire.
(a)
Name the sampling technique used. [1]
(1)
(b)
Give two reasons why this sample may not be representative of all audience members. [2]
(2)
(c)
On one evening 600 tickets are sold: 360 for the stalls, 150 for the circle and 90 for the balcony. The manager decides to take a stratified sample of 60 people. Find the number to be sampled from each section. [3]
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 convenience sampling (accept opportunity or non-random sampling)
✓ 1
Part (b)
A1 any valid reason, e.g. only matinee attenders are included, so evening audiences are excluded
✓ 1
A1 a second valid reason, e.g. only Saturday attenders, or the first people to leave may share particular views, or those who stay behind are excluded
Exam tip: A stratified sample takes each group in proportion to its size, so every group uses the same sampling fraction. Check that your group sizes add up to the sample size.
63IBMA3H-4-19r32-014.2 Presentation of dataPaper 1Easy6 marks
No calculator
The times, in hours, taken by a large group of climbers to reach the summit of a mountain from the high camp have the following five-number summary: minimum 4.2, lower quartile 5.6, median 6.4, upper quartile 7.8, maximum 11.9. A value is classed as an outlier if it is more than 1.5 × IQR above the upper quartile or more than 1.5 × IQR below the lower quartile.
Box plot of the five-number summary (whiskers to the minimum and maximum)
(a)
Find the interquartile range.
(1)
(b)
Show that the maximum time of 11.9 hours is an outlier.
(3)
(c)
Determine whether the minimum time of 4.2 hours is an outlier. Justify your answer.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 IQR = 7.8 − 5.6 = 2.2 hours
✓ 1
Part (b)
M1 attempt to use 1.5 × IQR
✓ 1
A1 upper boundary = 7.8 + 1.5 × 2.2 = 11.1
✓ 1
R1 11.9 > 11.1, so it is an outlier
✓ 1
Part (c)
A1 lower boundary = 5.6 − 1.5 × 2.2 = 2.3
✓ 1
R1 4.2 > 2.3, so the minimum is not an outlier
✓ 1
Exam tip: Write the boundary value you are testing against, then compare it explicitly with the data value.
64IBMA3H-4-19r32-044.4 Correlation and regression of y on xPaper 1Easy5 marks
No calculator
A mobile phone engineer records the battery drain, x (% per hour), and the screen brightness setting, y (arbitrary units), for a sample of phones. The summary statistics are: mean of x = 30, mean of y = 50, standard deviation of x = 4.2, standard deviation of y = 6.0 and the product moment correlation coefficient r = 0.85.
(a)
Find the gradient of the regression line of x on y.
(2)
(b)
Use the regression line of x on y to estimate the battery drain for a phone with brightness setting 54. Give your answer to three significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 gradient = r × sx/sy = 0.85 × 4.2/6.0
✓ 1
A1 0.595
✓ 1
Part (b)
M1 line passes through (30, 50): x − 30 = 0.595(y − 50)
✓ 1
A1 x = 0.595 × 4 + 30 = 32.38
✓ 1
A1 32.4 (% per hour)
✓ 1
Common error: To estimate x from a given y use the regression line of x on y, not the line of y on x rearranged.
65IBMA3H-4-18r33-114.11 Conditional probability and independencePaper 1Easy6 marks
No calculator
A building design may include solar panels (event A) and a green roof (event B). For a randomly chosen design, P(A) = 0.3, P(B) = p and P(A ∪ B) = 0.72.
(a)
Given that A and B are mutually exclusive, find the value of p.
(2)
(b)
Given instead that A and B are independent, find the value of p.
(3)
(c)
In the case where A and B are independent, find the probability that a design has neither solar panels nor a green roof.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(A ∪ B) = P(A) + P(B), so 0.72 = 0.3 + p
✓ 1
A1 p = 0.42
✓ 1
Part (b)
M1 P(A ∩ B) = 0.3p
✓ 1
M1 0.72 = 0.3 + p − 0.3p, so 0.7p = 0.42
✓ 1
A1 p = 0.6
✓ 1
Part (c)
A1 1 − 0.72 = 0.28
✓ 1
Exam tip: Mutually exclusive events have P(A ∩ B) = 0; independent events have P(A ∩ B) = P(A)P(B). These are different conditions.
66IBMA3H-4-19r37-034.1 Sampling, bias and outliersPaper 1Easy7 marks
No calculator
A bridge has 600 rivets that must be checked for corrosion: 240 in the deck, 210 in the towers and 150 in the cables. The inspector will examine a sample of 40 rivets.
(a)
Explain why a stratified sample would be more suitable than a simple random sample for this inspection.
(2)
(b)
Instead, the inspector chooses the 40 rivets that are easiest to reach from the walkway. Name this type of sampling and give one reason why the sample may be biased.
(2)
(c)
Calculate how many rivets should be taken from each section for a stratified sample of size 40.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 stratified sampling ensures every section of the bridge is represented in proportion to its size
✓ 1
R1 a simple random sample could, by chance, include too few or no rivets from one section
✓ 1
Part (b)
A1 convenience sampling
✓ 1
R1 easy-to-reach rivets may be in less exposed positions, so they are not representative of all rivets
✓ 1
Part (c)
M1 40 × (section size ÷ 600)
✓ 1
A1 deck 16 and towers 14
✓ 1
A1 cables 10
✓ 1
Exam tip: In stratified sampling, each stratum's sample size is its share of the population multiplied by the total sample size; check that the sizes add to the total.
67IBMA3H-4-20r44-034.11 Conditional probability and independencePaper 1Easy6 marks
No calculator
A shipping company records, for each voyage, whether it is late (\(L\)) and whether any cargo is damaged (\(D\)). For a randomly chosen voyage, \(P(L)=0.4\), \(P(D)=0.3\) and \(P(L\cap D)=0.18\).
Venn diagram showing the information given in the question
(a)
Find \(P(L\cup D)\).
(2)
(b)
Show that \(L\) and \(D\) are not independent.
(2)
(c)
Find \(P(D\mid L)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(L) + P(D) − P(L ∩ D) = 0.4 + 0.3 − 0.18
✓ 1
A1 0.52
✓ 1
Part (b)
M1 P(L) × P(D) = 0.4 × 0.3 = 0.12
✓ 1
R1 0.12 ≠ 0.18 = P(L ∩ D), so the events are not independent
✓ 1
Part (c)
M1 P(D ∩ L) ÷ P(L) = 0.18 ÷ 0.4
✓ 1
A1 0.45
✓ 1
Exam tip: To test independence compare \(P(A\cap B)\) with \(P(A)P(B)\); comparing \(P(A\mid B)\) with \(P(A)\) is equivalent.
68IBMA3H-4-20r44-044.14 Variance, continuous random variables and linear transformationsPaper 1Easy4 marks
No calculator
A volcano observatory has four seismic sensors. On a day with a minor tremor, each sensor independently detects it with probability \(\frac{1}{3}\). Let \(X\) be the number of sensors that detect the tremor.
(a)
Find \(P(X=2)\), giving your answer as a fraction in its simplest form.
(3)
(b)
Find \(E(X)\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 X ~ B(4, 1/3) and use of C(4,2)
✓ 1
M1 6 × (1/3)2 × (2/3)2
✓ 1
A1 24/81 = 8/27
✓ 1
Part (b)
A1 4 × 1/3 = 4/3
✓ 1
Exam tip: In an exact binomial calculation, write the three factors (the coefficient, \(p^r\) and \(q^{n-r}\)) separately before multiplying.
69IBMA3H-4-20r46-014.11 Conditional probability and independencePaper 1Easy6 marks
No calculator
For a particular flight, event \(A\) is "the departure is delayed" and event \(B\) is "there is a strong headwind". \(P(A)=0.4\) and \(P(B)=0.35\). Events \(A\) and \(B\) are independent.
(a)
Find \(P(A\cap B)\).
(2)
(b)
Find \(P(A\cup B)\).
(2)
(c)
Find \(P(A\cap B')\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(P(A\cap B)=P(A)P(B)\)
✓ 1
A1 0.14
✓ 1
Part (b)
M1 \(0.4+0.35-0.14\) (ECF from (a))
✓ 1
A1 0.61
✓ 1
Part (c)
M1 \(P(A)-P(A\cap B)\) or \(0.4\times0.65\)
✓ 1
A1 0.26
✓ 1
Exam tip: Independence lets you multiply probabilities directly; do not also subtract an overlap as if the events were mutually exclusive.
70IBMA3H-4-20r46-034.14 Variance, continuous random variables and linear transformationsPaper 1Easy7 marks
No calculator
A violinist makes \(X\) mistakes in a short recital piece, where \(X\) takes the values 0, 1, 2 and 3 and \(P(X=x)=k(5-x)\).
Exam tip: Square each value of \(x\) before weighting by its probability when finding \(E(X^2)\); it is not \([E(X)]^2\).
71IBMA3H-4-20r46-044.7 Discrete random variablesPaper 1Easy6 marks
No calculator
Two aircraft are each assigned at random one of four gates, numbered 1 to 4, at different times of the day; both may be assigned the same gate. Let \(M\) be the larger of the two gate numbers, with \(M\) equal to the common number if they match.
(a)
Write down the number of equally likely outcomes in the sample space.
(1)
(b)
Find \(P(M=3)\).
(2)
(c)
Given that \(M=4\), find the probability that the two gate numbers add up to 5.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 16
✓ 1
Part (b)
M1 identify the outcomes (1,3), (2,3), (3,1), (3,2), (3,3)
Exam tip: For a conditional probability from a sample space, restrict to the outcomes that satisfy the given condition and count within that smaller set.
72AAHL4-q00014.1 Sampling, bias and outliersPaper 1Medium9 marks
A factory makes 600 items in a day, numbered 1 to 600 in the order they are made. A systematic sample of 30 items is taken as follows: an integer \(r\) is chosen at random from 1 to 20, and items \(r,\ r+20,\ r+40,\ \ldots,\ r+580\) are selected.
(a)
Find the probability that item 137 is selected.
(1)
(b)(i)
Find the probability that
items 137 and 157 are both selected;
(1)
(b)(ii)
items 137 and 140 are both selected.
(1)
(c)(i)
Instead, a simple random sample of 30 of the 600 items is taken.
Write down the probability that item 137 is selected.
(1)
(c)(ii)
Show that the probability that items 137 and 140 are both selected is \(\dfrac{29}{11\,980}\).
(2)
(d)
Hence explain why the systematic sample is not a simple random sample, even though every item has the same probability of being selected.
(1)
(e)(i)
Find the probability that at least one of items 137 and 140 is selected
in the systematic sample;
(1)
(e)(ii)
in the simple random sample.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(137=17+6(20)\), so item 137 is selected only when \(r=17\): probability \(\dfrac{1}{20}\)
A1
(b)(i)
\(\dfrac{1}{20}\)
A1
Both are selected exactly when \(r=17\).
(b)(ii)
\(0\)
A1
137 and 140 differ by 3, not a multiple of 20.
(c)(i)
\(\dfrac{30}{600}=\dfrac{1}{20}\)
A1
(c)(ii)
\(\dfrac{\binom{598}{28}}{\binom{600}{30}}\) or \(\dfrac{30}{600}\times\dfrac{29}{599}\)
M1
\(=\dfrac{870}{359\,400}\)
A1
\(=\dfrac{29}{11\,980}\)
AG
(d)
in a simple random sample every set of 30 items is equally likely, but in the systematic sample some pairs (e.g. 137 and 140) can never occur together, so not all samples are equally likely
R1
(e)(i)
\(\dfrac{1}{20}+\dfrac{1}{20}=\dfrac{1}{10}\)
A1
The two events are mutually exclusive (\(r=17\) or \(r=20\)).
Answers: (a) \(\tfrac{1}{20}\) · (b)(i) \(\tfrac{1}{20}\), (ii) 0 · (c)(i) \(\tfrac{1}{20}\) · (e)(i) \(\tfrac{1}{10}\), (ii) \(\tfrac{1169}{11\,980}\)
73AAHL4-q00084.3 Measures of central tendency and dispersionPaper 1Medium8 marks
A data set has mean 40, standard deviation 5, median 38 and interquartile range 6. Each value \(x\) in the data set is transformed to \(y=ax+b\), where \(a,b\in\mathbb{R}\) and \(a\neq0\). The transformed data have mean 70 and standard deviation 15.
(a)
Find the two possible pairs of values of \(a\) and \(b\).
(4)
(b)
It is also given that the median of the transformed data is greater than its mean. Find the value of \(a\) and the value of \(b\).
(2)
(c)
For the values in part (b), find the interquartile range of the transformed data, and state which quartile of the original data becomes the lower quartile of the transformed data.
(2)
Show mark scheme
Working
Mark
Notes
(a)
standard deviation of \(y\) is \(|a|\times5=15\)
M1
\(a=3\) or \(a=-3\)
A1
\(40a+b=70\)
M1
\(a=3,\ b=-50\) and \(a=-3,\ b=190\)
A1
(b)
medians: \(3(38)-50=64\) and \(-3(38)+190=76\)
M1
\(76\gt 70\), so \(a=-3,\ b=190\)
A1
(c)
IQR \(=|-3|\times6=18\)
A1
the upper quartile of \(x\) (multiplying by a negative number reverses the order)
A1
Answers: (a) \((3,-50)\) or \((-3,190)\) · (b) \(a=-3,\ b=190\) · (c) 18; the upper quartile
74AAHL4-q00094.3 Measures of central tendency and dispersionPaper 1Medium9 marks
The table shows the number of siblings, \(x\), of each student in a group, where \(k\in\mathbb{N}\).
Number of siblings, \(x\)
0
1
2
3
4
Frequency
6
9
\(k\)
7
5
(a)
Find the set of values of \(k\) for which the median number of siblings is 2.
(3)
(b)
Given also that 1 is the only mode, find the possible values of \(k\).
(2)
(c)
Show that the mean number of siblings is less than 2 for every value of \(k\).
(2)
(d)
Find the least value of \(k\) for which the median is 2 and the mean differs from the median by less than 0.1.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(6+9=15\) values are less than 2; total \(27+k\)
M1
the median is 2 exactly when the middle value(s) lie in positions 16 to \(15+k\) (e.g. \(k=3\): 30 values, 15th and 16th values 1 and 2, median 1.5; \(k=4\): 31 values, 16th value 2)
M1
\(k\ge4\)
A1
Accept \(k\in\{4,5,6,\ldots\}\).
(b)
need \(k\lt 9\) (\(k=9\) gives two modes)
M1
\(k=4,5,6,7,8\)
A1
(c)
mean \(=\dfrac{9+2k+21+20}{27+k}=\dfrac{50+2k}{27+k}\)
M1
\(\dfrac{50+2k}{27+k}\lt 2\iff 50+2k\lt 54+2k\), which is always true
R1
(d)
\(2-\dfrac{50+2k}{27+k}=\dfrac{4}{27+k}\lt 0.1\)
M1
\(27+k\gt 40\), so \(k=14\)
A1
Check: \(k=13\) gives a difference of exactly 0.1.
76AAHL4-q00204.6 Combined, conditional and independent eventsPaper 1Medium8 marks
A bag contains \(n\) red counters and 3 blue counters, where \(n\ge1\). Two counters are taken at random from the bag without replacement.
(a)
Show that the probability that the two counters are the same colour is \(\dfrac{n^{2}-n+6}{(n+3)(n+2)}\).
(3)
(b)
Given that this probability is \(\dfrac{1}{2}\), find the possible values of \(n\).
(3)
(c)
Using the larger value of \(n\), three counters are taken at random from the bag without replacement. Find the probability that all three are the same colour.
or \(\dfrac{\binom{6}{3}+\binom{3}{3}}{\binom{9}{3}}\)
\(\dfrac{21}{84}=\dfrac{1}{4}\)
A1
Answers: (b) \(n=1\) or \(n=6\) · (c) \(\tfrac{1}{4}\)
77AAHL4-q00264.7 Discrete random variablesPaper 1Medium10 marks
The probability distribution of a discrete random variable \(X\) is given by \(\mathrm{P}(X=x)=a+bx\), for \(x\in\{1,\,2,\,3,\,4\}\), where \(a\) and \(b\) are constants.
(a)
Show that \(\mathrm{E}(X)=2.5+5b\).
(3)
(b)
By considering the conditions on a probability distribution, find the set of possible values of \(b\).
(3)
(c)
Hence write down the least possible value and the greatest possible value of \(\mathrm{E}(X)\).
(1)
(d)
Given that \(\mathrm{E}(X)=3\), find \(\mathrm{P}(X=4\mid X\ge2)\).
78AAHL4-q00274.7 Discrete random variablesPaper 1Medium9 marks
A board is divided into an \(n\times n\) grid of squares, where \(n\ge2\). A counter is placed on a square chosen at random. Let \(X\) be the number of squares that share an edge with the square on which the counter is placed.
(a)
For a \(4\times4\) grid, find the probability distribution of \(X\).
(3)
(b)
For a \(4\times4\) grid, find \(\mathrm{E}(X)\).
(1)
(c)
Show that, for an \(n\times n\) grid, \(\mathrm{E}(X)=\dfrac{4(n-1)}{n}\).
(3)
(d)
A second counter is then placed on a different square, chosen at random. Hence, or otherwise, show that the probability that the two counters are on squares that share an edge is \(\dfrac{4}{n(n+1)}\).
80AAHL4-q00314.7 Discrete random variablesPaper 1Medium15 marks
A fair coin is tossed four times. Let \(X\) be the length of the longest run of consecutive heads; for example, the outcome HHTH gives \(X=2\), and TTTT gives \(X=0\). The table shows part of the probability distribution of \(X\).
\(x\)
0
1
2
3
4
\(\mathrm{P}(X=x)\)
\(\tfrac{1}{16}\)
\(\tfrac{1}{16}\)
(a)
Show that \(\mathrm{P}(X=3)=\dfrac{1}{8}\).
(2)
(b)
Complete the table of the probability distribution of \(X\) shown above.
(3)
(c)
Show that \(\mathrm{E}(X)=\dfrac{27}{16}\).
(2)
(d)
Find \(\mathrm{Var}(X)\).
(3)
(e)
In a game, a player pays \(c\) euros to play and receives \(4X\) euros. Find the value of \(c\) for which the game is fair.
(2)
(f)
Find the variance of the amount received.
(1)
(g)
Given that \(X\ge2\), find the probability that the first toss was a head.
(2)
Show mark scheme
Working
Mark
Notes
(a)
outcomes HHHT and THHH only
M1
Both outcomes listed, and no others.
\(\dfrac{2}{16}\)
A1
\(=\dfrac{1}{8}\)
AG
(b)
listing outcomes with no two consecutive heads: TTTH, TTHT, THTT, HTTT, HTHT, THTH, HTTH
M1
Or listing the five outcomes with X = 2: HHTT, THHT, TTHH, HHTH, HTHH.
81AAHL4-q00334.8 Binomial distributionPaper 1Medium9 marks
A random variable \(X\) follows a binomial distribution \(\mathrm{B}(n,\,p)\), where \(0\lt p\lt 1\). It is given that \(\mathrm{P}(X=1)=3\,\mathrm{P}(X=0)\) and \(4\,\mathrm{P}(X=2)=5\,\mathrm{P}(X=1)\).
(a)
Show that \(np=3(1-p)\).
(2)
(b)
Show that \((n-1)p=\dfrac{5}{2}(1-p)\).
(2)
(c)
Hence find the value of \(p\) and the value of \(n\).
(3)
(d)
Find \(\mathrm{P}(X\le1)\), giving your answer as a fraction.
82AAHL4-q00484.11 Conditional probability and independencePaper 1Medium8 marks
A card is drawn at random from a standard pack of 52 playing cards (13 cards in each of the four suits hearts, diamonds, clubs and spades; each suit contains one king, one queen and one jack). Let \(H\) be the event “the card is a heart”, \(K\) the event “the card is a king” and \(C\) the event “the card is a king, a queen or a jack”.
(a)
Show that \(H\) and \(K\) are independent.
(2)
(b)
By finding \(\mathrm{P}(K\mid C)\), show that \(K\) and \(C\) are not independent.
(2)
(c)
Events \(A\) and \(B\) are mutually exclusive, with \(\mathrm{P}(A)\gt 0\) and \(\mathrm{P}(B)\gt 0\). Prove that \(A\) and \(B\) are not independent.
(2)
(d)
The two of clubs is removed from the pack and a card is drawn at random from the remaining 51 cards. Determine whether \(H\) and \(K\) are now independent.
\(\dfrac{13}{51}\times\dfrac{4}{51}=\dfrac{52}{2601}\neq\dfrac{51}{2601}=\dfrac{1}{51}\), so not independent
R1
Or \(\mathrm{P}(K\mid H)=\dfrac{1}{13}\neq\dfrac{4}{51}\).
Answers: (b) \(\mathrm{P}(K\mid C)=\frac13\) · (d) not independent
83AAHL4-q00534.11 Conditional probability and independencePaper 1Medium16 marks
In a town, if it rains on one day, the probability that it rains on the next day is \(\dfrac{3}{5}\). If it does not rain on one day, the probability that it rains on the next day is \(\dfrac{1}{5}\). It rains on day 1. Let \(p_{n}\) be the probability that it rains on day \(n\), so \(p_{1}=1\).
Tree for days 2 and 3 given rain on day 1 (R = rains, N = does not rain)
(a)
Find \(p_{2}\) and \(p_{3}\).
(3)
(b)
Show that \(p_{n+1}=\dfrac{1}{5}+\dfrac{2}{5}p_{n}\).
(2)
(c)
By considering \(p_{n}-\dfrac{1}{3}\), show that \(p_{n}=\dfrac{1}{3}+\dfrac{2}{3}\left(\dfrac{2}{5}\right)^{n-1}\).
(4)
(d)
Write down the value of \(p_{n}\) in the long run.
(1)
(e)
Find the least value of \(n\) for which \(p_{n}-\dfrac{1}{3}\lt\dfrac{1}{150}\).
(3)
(f)
Given that it rains on day 3, find the probability that it did not rain on day 2.
84AAHL4-q00624.13 Bayes’ theoremPaper 1Medium7 marks
A veterinary test is used on cattle in a region where 10% of cattle have a disease. The test is positive for 90% of cattle with the disease. For cattle without the disease, the test is positive with probability \(q\), where \(0\lt q\lt 1\). Let \(D\) be the event that an animal has the disease and \(+\) the event that its test is positive.
Tree diagram with the probabilities given in the question
(a)
Show that \(\mathrm{P}(D\mid +)=\dfrac{1}{1+10q}\).
(2)
(b)
The test is acceptable if \(\mathrm{P}(D\mid +)\ge0.8\). Find the largest value of \(q\) for which the test is acceptable.
(2)
(c)
For this value of \(q\), find the probability that an animal whose test is negative does not have the disease.
85AAHL4-q00634.13 Bayes’ theoremPaper 1Medium10 marks
A survey estimates the proportion, \(\pi\), of drivers who have driven through a red traffic light. Each driver privately rolls a fair six-sided die. If the score is 1 or 2, the driver answers “Yes” whatever the truth; otherwise the driver answers the question “Have you ever driven through a red light?” truthfully. Let \(T\) be the event that the driver has driven through a red light and \(Y\) the event that the driver answers “Yes”.
(a)
Show that \(\mathrm{P}(Y)=\dfrac{1+2\pi}{3}\).
(2)
(b)
In the survey, 40% of drivers answer “Yes”. Find the value of \(\pi\).
(2)
(c)
Find \(\mathrm{P}(T\mid Y)\).
(2)
(d)
The procedure is changed: a score of 1 means answer “Yes”, a score of 2 means answer “No”, and otherwise the driver answers truthfully. Taking \(\pi=0.1\), find \(\mathrm{P}(T\mid Y)\) and \(\mathrm{P}(T\mid Y')\).
(4)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{P}(Y)=\dfrac13(1)+\dfrac23\,\pi\)
M1
Two branches of the die outcome.
\(=\dfrac{1}{3}+\dfrac{2\pi}{3}\)
A1
\(=\dfrac{1+2\pi}{3}\)
AG
(b)
\(\dfrac{1+2\pi}{3}=0.4\)
M1
\(\pi=0.1\)
A1
(c)
\(\mathrm{P}(Y\mid T)=1\), so \(\mathrm{P}(T\mid Y)=\dfrac{0.1\times1}{0.4}\)
86AAHL4-q00664.13 Bayes’ theoremPaper 1Medium8 marks
In a game show a prize is placed behind one of three closed doors, 1, 2 and 3, each door being equally likely. A contestant chooses door 1. The host, who knows where the prize is, then opens one of doors 2 and 3, always choosing a door that does not hide the prize. If the prize is behind door 1, the host opens door 3 with probability \(q\) and door 2 with probability \(1-q\), where \(0\le q\le1\). The host opens door 3.
(a)
In the case \(q=\dfrac12\), show that the probability that the prize is behind door 2 is \(\dfrac23\).
(3)
(b)
For a general value of \(q\), find, in terms of \(q\), the probability that the prize is behind door 1.
(3)
(c)
The contestant may switch to door 2. Show that switching never decreases the contestant’s probability of winning.
\(\dfrac{1}{1+q}\ge\dfrac{q}{1+q}\) because \(q\le1\), so switching is never worse (equal only when \(q=1\))
R1
Answers: (b) \(\frac{q}{1+q}\) · (c) \(\mathrm{P}(\text{win by switching})=\frac{1}{1+q}\)
87AAHL4-q00684.13 Bayes’ theoremPaper 1Medium7 marks
A shop buys all its light bulbs from three suppliers, P, Q and R. Supplier P provides half of the bulbs, supplier Q provides a proportion \(x\) and supplier R provides the rest. The probability that a bulb is faulty is 0.02 for supplier P, 0.04 for supplier Q and 0.08 for supplier R. Half of all faulty bulbs come from supplier R.
(a)
Find the value of \(x\).
(4)
(b)
A bulb is chosen at random and is not faulty. Find the probability that it came from supplier P.
88AAHL4-q00694.13 Bayes’ theoremPaper 1Medium8 marks
Let \(A\) and \(B\) be events with \(0\lt\mathrm{P}(A)\lt1\) and \(0\lt\mathrm{P}(B)\lt1\).
(a)
Show that \(\mathrm{P}(B\mid A)-\mathrm{P}(B)=\dfrac{\mathrm{P}(B)\,\mathrm{P}(B')\big(\mathrm{P}(A\mid B)-\mathrm{P}(A\mid B')\big)}{\mathrm{P}(A)}\).
(4)
(b)
Hence state a condition on \(\mathrm{P}(A\mid B)\) and \(\mathrm{P}(A\mid B')\) for which knowing that \(A\) has occurred increases the probability of \(B\).
(1)
(c)
Hence show that if \(\mathrm{P}(A\mid B)=\mathrm{P}(A\mid B')\) then \(A\) and \(B\) are independent.
(1)
(d)
Given that \(\mathrm{P}(B)=0.4\), \(\mathrm{P}(A\mid B)=0.3\) and \(\mathrm{P}(A\mid B')=0.2\), use the result in part (a) to find \(\mathrm{P}(B\mid A)\).
89AAHL4-q00704.13 Bayes’ theoremPaper 1Medium9 marks
A transmitter sends the digit 1 with probability \(\dfrac35\) and the digit 0 with probability \(\dfrac25\). To reduce errors, each digit is sent three times (as 111 or 000). Each of the three received digits is independently changed (1 to 0 or 0 to 1) with probability \(\dfrac{1}{10}\).
(a)
Find the probability that the receiver gets 110, in that order.
(3)
(b)
Given that the receiver gets 110, find the probability that the digit 1 was sent.
(2)
(c)
The receiver decodes each block as the digit that appears at least twice. Find the probability that a block is decoded incorrectly.
(2)
(d)
Given that the receiver gets 100, find the probability that the digit 0 was sent.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{P}(110\mid1)=\left(\dfrac{9}{10}\right)^{2}\dfrac{1}{10}=\dfrac{81}{1000}\) and \(\mathrm{P}(110\mid0)=\left(\dfrac{1}{10}\right)^{2}\dfrac{9}{10}=\dfrac{9}{1000}\)
Reject \(a=-\tfrac{13}{4}\) since \(a\gt1\). Award A0 if both values are given as the answer.
(c)
\(\mathrm{E}(X)=1.1\), \(\sigma=\sqrt{2.29}\) with \(1.5\lt\sigma\lt1.6\), so the interval is approximately \(]-0.4,\ 2.6[\), containing \(x=0\) and \(x=1\) only
(M1)
Since \(1.5^{2}=2.25\lt2.29\lt2.56=1.6^{2}\).
\(0.5+0.3=0.8\)
A1
Answers: (b) \(a=4\) · (c) \(0.8\)
91AAHL4-q00854.14 Variance, continuous random variables and linear transformationsPaper 1Medium8 marks
The continuous random variable \(X\) has probability density function \(f(x)=k(9-x^{2})\) for \(0\le x\le3\), and \(f(x)=0\) otherwise.
(a)
Find the value of \(k\).
(2)
(b)
Find \(\mathrm{E}(X)\).
(2)
(c)
Find \(\mathrm{P}\big(X\lt\mathrm{E}(X)\big)\).
(2)
(d)
Hence state, with a reason, whether the median of \(X\) is greater than or less than \(\mathrm{E}(X)\).
\(\mathrm{P}\big(X\lt\mathrm{E}(X)\big)=\dfrac{549}{1024}\gt\dfrac{512}{1024}=\dfrac12\), so half of the probability lies below \(\mathrm{E}(X)\) already
93AAHL4-q00884.14 Variance, continuous random variables and linear transformationsPaper 1Medium8 marks
The continuous random variable \(X\) has the probability density function \(f\) shown. The graph consists of line segments joining \((0,0)\), \((1,h)\), \((3,h)\) and \((4,0)\), and \(f(x)=0\) for \(x\lt 0\) and \(x\gt 4\).
94AAHL4-q00894.14 Variance, continuous random variables and linear transformationsPaper 1Medium9 marks
The continuous random variable \(X\) has probability density function \(f(x)=\dfrac{1}{b-a}\) for \(a\le x\le b\), and \(f(x)=0\) otherwise, where \(a\lt b\).
(a)
Show that \(\mathrm{E}(X)=\dfrac{a+b}{2}\).
(2)
(b)
Show that \(\mathrm{Var}(X)=\dfrac{(b-a)^{2}}{12}\).
(3)
(c)
Given that \(\mathrm{E}(X)=7\) and \(\mathrm{Var}(X)=3\), find the value of \(a\) and the value of \(b\).
95AAHL4-q00924.14 Variance, continuous random variables and linear transformationsPaper 1Medium8 marks
The continuous random variable \(X\) has probability density function \(f(x)=k\mathrm{e}^{-x}\) for \(0\le x\le\ln4\), and \(f(x)=0\) otherwise, where \(k\) is a constant.
(a)
Show that \(k=\dfrac43\).
(2)
(b)
Find the median of \(X\), giving your answer in the form \(\ln\dfrac{p}{q}\), where \(p,q\in\mathbb{Z}^{+}\).
(3)
(c)
Find \(\mathrm{E}(X)\), giving your answer in the form \(a+b\ln2\), where \(a,b\in\mathbb{Q}\).
98AAHL4-q00954.14 Variance, continuous random variables and linear transformationsPaper 1Medium9 marks
A machine cuts square tiles. The side length, \(X\) cm, of a randomly chosen tile is modelled by the probability density function \(f(x)=\dfrac{x-1}{2}\) for \(1\le x\le3\), and \(f(x)=0\) otherwise.
Graph of the probability density function f
(a)
Find the expected side length.
(2)
(b)
Find the expected area of a tile.
(2)
(c)
Find the standard deviation of the perimeter of a tile.
(3)
(d)
Find the probability that the area of a randomly chosen tile is greater than \(4\ \text{cm}^{2}\).
101IBMA2H-051-01-074.13 Bayes’ theoremPaper 1Medium6 marks
No calculator
An email provider's spam filter flags 92% of spam emails and 3% of legitimate emails. A proportion p of all incoming emails are spam. Given that an email has been flagged, the probability that it is spam is 0.8.
(a)
Find the exact value of p.
(4)
(b)
Find the exact proportion of all incoming emails that are flagged.
Exam tip: When the prior is the unknown, write P(flagged) in terms of p first, then form a single equation. Multiply out before collecting terms.
102IBMA2H-051-01-084.1 Sampling, bias and outliersPaper 1Medium6 marks
No calculator
A machine makes bolts using five cutting heads that operate in strict rotation: head 1, head 2, head 3, head 4, head 5, head 1, and so on. In one day it makes 6000 bolts, numbered in order of production. A quality inspector selects bolt number 3 and then every 50th bolt after it (bolts 3, 53, 103, …).
(a)
Name this sampling method.
(1)
(b)
Explain why this sample would give a biased estimate of the proportion of faulty bolts produced by the machine.
(2)
(c)
Suggest an interval that would avoid this problem while keeping a systematic method.
(1)
(d)
Find the number of bolts in the inspector's sample.
(1)
(e)
Describe how a simple random sample of 120 bolts could be chosen instead.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 systematic sampling
✓ 1
Part (b)
R1 50 is a multiple of 5, so every sampled bolt comes from the same cutting head (head 3)
✓ 1
R1 faults caused by the other four heads cannot be detected, so the sample does not represent the whole output
✓ 1
Part (c)
R1 any interval not a multiple of 5, e.g. every 49th or 51st bolt, so all heads are sampled
✓ 1
Part (d)
A1 \(\frac{5953-3}{50}+1\) = 120 bolts
✓ 1
Part (e)
R1 use a random number generator to select 120 different numbers from 1 to 6000 and inspect those bolts
✓ 1
Common error: Systematic sampling is not automatically unbiased. Check whether the sampling interval lines up with any repeating pattern in the population.
103IBMA2H-051-01-104.11 Conditional probability and independencePaper 1Medium7 marks
No calculator
A goalkeeper studies a penalty taker. The taker shoots to the goalkeeper's left 60% of the time and to the right otherwise. When shooting left, the taker uses a long run-up with probability 0.7. When shooting right, the taker uses a long run-up with probability 0.2.
Tree diagram with the probabilities given in the question (run-up)
(a)
Given that the taker uses a long run-up, find the probability that the shot goes left.
(3)
(b)
The goalkeeper dives left after every long run-up and right after every short run-up. Find the probability that the goalkeeper dives the correct way.
(2)
(c)
Given that the goalkeeper dives the wrong way, find the probability that the shot went right.
Common error: 'Diving the wrong way' has two branches (long run-up and shot right, or short run-up and shot left). Include both in the denominator.
104IBMA2H-051-01-114.2 Presentation of dataPaper 1Medium7 marks
No calculator
A food-delivery app records the delivery times, in minutes, of 25 orders. The mean time is 32.4 minutes, the standard deviation is 8.5 minutes, the lower quartile is 27 minutes and the upper quartile is 35 minutes. One order is later removed from the data, and the mean of the remaining 24 times is 31.5 minutes. An outlier is a value more than 1.5 × IQR beyond the nearer quartile.
(a)
Find the value above which a delivery time is an outlier.
(2)
(b)
Find the delivery time that was removed and determine whether it was an outlier.
(3)
(c)
State, with a reason, whether the standard deviation of the remaining 24 times is greater or smaller than 8.5 minutes.
(1)
(d)
The app collects customer feedback only from users who have just given the delivery a 5-star rating. Explain why this gives a biased picture of customer satisfaction.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 IQR = 35 − 27 = 8 and 35 + 1.5 × 8
✓ 1
A1 47 minutes
✓ 1
Part (b)
M1 total of 25 times = 25 × 32.4 = 810 and total of 24 times = 24 × 31.5 = 756
✓ 1
A1 removed time = 810 − 756 = 54 minutes
✓ 1
R1 54 > 47, so it was an outlier
✓ 1
Part (c)
R1 smaller, because the removed value is 54 − 32.4 = 21.6 minutes from the mean, which is far more than one standard deviation (8.5 minutes), so removing it reduces the spread
✓ 1
Part (d)
R1 only satisfied customers are sampled, so dissatisfied views are excluded and satisfaction is overestimated
✓ 1
Exam tip: Multiplying the mean by n gives the total of the data. Comparing the totals before and after a value is removed gives the removed value directly.
105IBMA2H-051-05-104.2 Presentation of dataPaper 1Medium7 marks
No calculator
Raw scores, \(X\), on an aptitude test are modelled by \(X \sim N(60, 10^2)\). Scores are reported on a scaled form \(S = aX + b\), where \(a > 0\), chosen so that \(E(S) = 100\) and the standard deviation of \(S\) is 15.
(a)
Find the value of \(a\) and the value of \(b\).
(4)
(b)
Given that \(P(X > 75) = 0.0668\), find \(P(77.5 < S < 122.5)\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\text{Var}(S) = a^2 \times 10^2 = 15^2\)
✓ 1
A1 \(a = 1.5\), rejecting the negative root since \(a > 0\)
✓ 1
M1 \(E(S) = 1.5 \times 60 + b = 100\)
✓ 1
A1 \(b = 10\)
✓ 1
Part (b)
M1 converting the bounds: \(S = 77.5 \Rightarrow X = 45\) and \(S = 122.5 \Rightarrow X = 75\)
✓ 1
R1 45 and 75 are symmetric about the mean 60, so \(P(X < 45) = 0.0668\)
✓ 1
A1 \(1 - 2(0.0668) = 0.8664\)
✓ 1
Exam tip: Because S is an increasing linear function of X, an inequality on S converts directly into one on X. Solving 1.5X + 10 = 122.5 avoids any need for new z-values.
106IBMA2H-051-08-064.2 Presentation of dataPaper 1Medium8 marks
No calculator
The temperature, in °C, of a satellite component was logged over one orbit. The summary statistics are: minimum = −40, lower quartile = −10, median = 5, upper quartile = 20, maximum = 95. An outlier is a value more than 1.5 × IQR beyond the nearer quartile.
Box plot of the five-number summary (whiskers drawn to the minimum and maximum)
(a)
Show that the maximum temperature is an outlier.
(3)
(b)
The temperatures are converted to degrees Fahrenheit using F = 1.8C + 32. Find the median and the IQR of the temperatures in °F.
(3)
(c)
By finding the upper outlier boundary in °F, verify that the maximum temperature is still an outlier after conversion.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 IQR = 20 − (−10) = 30
✓ 1
A1 upper boundary = 20 + 1.5 × 30 = 65
✓ 1
R1 95 > 65, so 95 is an outlier
✓ 1
Part (b)
A1 median = 1.8 × 5 + 32 = 41 °F
✓ 1
M1 the IQR is multiplied by 1.8 and is unaffected by adding 32
A1 boundary = 149 °F; maximum = 1.8 × 95 + 32 = 203 °F > 149, so still an outlier
✓ 1
Common error: Adding a constant shifts the quartiles but not the IQR. Do not add 32 to the IQR.
107IBMA2H-051-08-084.2 Presentation of dataPaper 1Medium9 marks
No calculator
The number of goals scored by a football team in each of its 40 matches last season is summarised below. 0 goals: p matches 1 goal: 9 matches 2 goals: q matches 3 goals: 7 matches 4 goals: 4 matches 5 goals: 2 matches The mean number of goals per match is 2.
(a)
Find the value of p and the value of q.
(4)
(b)
Find the median number of goals.
(2)
(c)
The data are to be displayed on a box-and-whisker diagram. Determine whether the match in which 5 goals were scored should be shown as an outlier.
R1 upper boundary = 3 + 1.5 × 2 = 6; since 5 < 6 it is not an outlier
✓ 1
Common error: Matches with 0 goals contribute nothing to the total of goals but do count towards the 40 matches; both equations are needed.
108IBMA2H-051-08-124.2 Presentation of dataPaper 1Medium11 marks
No calculator
A histogram of the journey times, X minutes, of a courier's drone flights is used to build a piecewise-constant probability density function: f(x) = 0.05 for 0 ≤ x < 4 f(x) = 0.15 for 4 ≤ x < 8 f(x) = k for 8 ≤ x ≤ 10 f(x) = 0 otherwise.
(a)
Find the value of k.
(2)
(b)
Find the median journey time.
(2)
(c)
Find the exact interquartile range.
(3)
(d)
Find E(X) and use it with the median to comment on the skewness of the distribution.
Common error: Each integral for E(X) must use the density for its own interval. Applying a single density across 0 to 10 gives the wrong mean.
109IBMA2H-051-00-054.14 Variance, continuous random variables and linear transformationsPaper 1Medium10 marks
No calculator
At a hospital triage desk, the waiting time, X minutes, before a patient is first assessed is modelled by the probability density function \(f(x)=\frac{x}{6}\) for \(0\le x\le 2\), \(f(x)=a(6-x)\) for \(2<x\le 6\), \(f(x)=0\) otherwise, where a is a positive constant.
(a)
Show that \(a=\frac{1}{12}\).
(2)
(b)
Write down the mode of X.
(1)
(c)
Find the exact value of E(X).
(3)
(d)
Find the exact value of the median waiting time.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\int_0^2\frac{x}{6}dx+\int_2^6 a(6-x)dx=1\), giving \(\frac{1}{3}+8a=1\) A1 \(8a=\frac{2}{3}\Rightarrow a=\frac{1}{12}\) AG
✓ 2
Part (b)
A1 2 minutes (f increases on [0, 2] and decreases on [2, 6])
✓ 1
Part (c)
M1 \(E(X)=\int_0^2\frac{x^2}{6}dx+\int_2^6\frac{6x-x^2}{12}dx\) A1 \(\frac{4}{9}\) and \(\frac{1}{12}\left[3x^2-\frac{x^3}{3}\right]_2^6=\frac{20}{9}\) A1 \(E(X)=\frac{8}{3}\) minutes
✓ 3
Part (d)
R1 \(P(X\le2)=\frac{1}{3}<\frac{1}{2}\), so the median m satisfies \(2<m<6\) M1 \(\int_m^6\frac{6-x}{12}dx=\frac{1}{2}\) (or equivalent from 0) A1 \(\frac{(6-m)^2}{24}=\frac{1}{2}\Rightarrow(6-m)^2=12\) A1 \(m=6-2\sqrt{3}\) minutes, rejecting \(6+2\sqrt{3}\) as outside the domain
✓ 4
Exam tip: With a piecewise pdf, find the probability in the first piece before setting up the median equation; it tells you which piece contains the median.
110IBMA2H-051-00-084.14 Variance, continuous random variables and linear transformationsPaper 1Medium10 marks
No calculator
The time, X seconds, that a smartphone takes to recognise its owner's face is modelled by the probability density function \(f(x)=ax+b\) for \(0\le x\le 2\), and \(f(x)=0\) otherwise, where a and b are constants. It is known that \(E(X)=\frac{7}{6}\).
(a)
Find the value of a and the value of b.
(5)
(b)
Find the exact value of Var(X).
(3)
(c)
The software reports a latency index \(L=6X-1\). Find Var(L).
Exam tip: Two unknowns need two equations: one from total probability equal to 1 and one from the given mean. Solve them simultaneously before working out any further moments.
111IBMA2H-051-00-094.10 Regression line of x on yPaper 1Medium7 marks
No calculator
A smartwatch company records the ambient temperature, x °C, and the battery drain rate, y % per hour, on ten days. The temperatures range from −10 °C to 25 °C. The regression line of y on x is \(y=-0.18x+9.6\) and Pearson's product-moment correlation coefficient is \(r=-0.87\).
(a)
Interpret the gradient of the regression line in context.
(1)
(b)
Estimate the drain rate when the temperature is 15 °C.
(2)
(c)
The temperatures are converted to degrees Fahrenheit, t, using \(t=1.8x+32\). State the value of the correlation coefficient between t and y, giving a reason.
(2)
(d)
Find the equation of the regression line of y on t.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 for each 1 °C rise in temperature, the drain rate decreases by about 0.18 percentage points per hour
✓ 1
Part (b)
M1 \(-0.18(15)+9.6\) A1 6.9 % per hour
✓ 2
Part (c)
R1 a linear transformation with a positive multiplier does not change the strength or direction of linear correlation A1 \(r=-0.87\)
Common error: Substituting \(x=1.8t+32\) instead of rearranging to make x the subject. Express the old variable in terms of the new one before substituting.
112IBMA2H-051-04-064.13 Bayes’ theoremPaper 1Medium7 marks
No calculator
An electrician's box contains 8 fuses, of which 5 are working and 3 are faulty. Fuses are taken out one at a time at random, without replacement, and tested.
Tree for the first two fuses tested (F = faulty, W = working; no replacement)
(a)
Find the probability that the first two fuses tested are both faulty.
(2)
(b)
Find the probability that the first working fuse is found on the third test.
(2)
(c)
Given that at least one of the first two fuses tested is faulty, find the probability that both of them are faulty.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 3/8 × 2/7 ; A1 3/28
✓ 2
Part (b)
M1 3/8 × 2/7 × 5/6 ; A1 5/56
✓ 2
Part (c)
M1 P(at least one faulty) = 1 − 5/8 × 4/7 = 9/14 ; M1 (3/28)/(9/14) ; A1 1/6
✓ 3
Common error: Without replacement the second probability changes: after one faulty fuse is removed there are 2 faulty fuses out of 7, not 3 out of 8.
113IBMA2H-051-04-084.6 Combined, conditional and independent eventsPaper 1Medium11 marks
No calculator
Events A and B are independent, with P(A) = p, P(B) = 2p and P(A ∪ B) = 7/9.
(a)
Show that 18p2 − 27p + 7 = 0.
(3)
(b)
Hence find the value of p.
(3)
(c)
Prove that, for any independent events A and B, the events A′ and B are also independent.
(3)
(d)
Find P(A | A ∪ B).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(A ∪ B) = p + 2p − P(A ∩ B) ; M1 P(A ∩ B) = p × 2p = 2p2 by independence ; A1 3p − 2p2 = 7/9 rearranged to 18p2 − 27p + 7 = 0
✓ 3
Part (b)
M1 factorises (3p − 1)(6p − 7) = 0 or uses the quadratic formula ; A1 p = 1/3 or p = 7/6 ; R1 rejects p = 7/6 since a probability cannot exceed 1, so p = 1/3
✓ 3
Part (c)
M1 P(A′ ∩ B) = P(B) − P(A ∩ B) ; M1 substitutes P(A ∩ B) = P(A)P(B) ; R1 = P(B)(1 − P(A)) = P(A′)P(B), hence A′ and B are independent
Common error: Do not accept both roots of the quadratic without checking: any value giving a probability greater than 1 must be rejected with a reason.
114IBMA2H-051-04-094.14 Variance, continuous random variables and linear transformationsPaper 1Medium11 marks
No calculator
The lifetime, T years, of a smartwatch battery is modelled by the probability density function \(f(t)=kt(4-t)\) for \(0 \le t \le 4\), and \(f(t)=0\) otherwise.
(a)
Show that \(k=\frac{3}{32}\).
(2)
(b)
Find P(T > 3).
(3)
(c)
Given that a battery has lasted more than 2 years, find the probability that it lasts more than 3 years.
(3)
(d)
Two such watches have batteries that last independently. Given that at least one of the batteries lasts more than 2 years, find the probability that both batteries last more than 3 years.
(3)
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Marking point
Mark
Notes
Part (a)
M1 sets k∫04 (4t − t2) dt = 1 and integrates correctly to k[2t2 − t3/3]04 ; A1 evaluates k(32 − 64/3) = k × 32/3 = 1, so k = 3/32
M1 P(T > 3 | T > 2) = P(T > 3)/P(T > 2) ; A1 P(T > 2) = 1/2 by symmetry about t = 2 or by integration ; A1 (5/32)/(1/2) = 5/16
✓ 3
Part (d)
M1 P(at least one > 2) = 1 − (1/2)2 = 3/4 ; M1 recognises that 'both > 3' is contained in 'at least one > 2', giving (5/32)2/(3/4) ; A1 25/768
✓ 3
Exam tip: When one event lies entirely inside the conditioning event, the intersection is just the smaller event itself. This means you do not need to find the intersection separately.
115IBMA2H-051-07-054.10 Regression line of x on yPaper 1Medium7 marks
No calculator
The time delay, D milliseconds, for a signal to travel between a ground station and a communications satellite is modelled by a normal distribution with mean μ and standard deviation σ. It is found that 10% of delays are longer than 290 ms and 20% of delays are shorter than 237 ms. You may use the facts that if Z ~ N(0, 1) then P(Z < 0.84) = 0.80 and P(Z < 1.28) = 0.90.
(a)
Show that σ = 25 and find the value of μ.
(5)
(b)
A signal is selected at random and its delay is known to be longer than 237 ms. Find the exact probability that its delay is longer than 290 ms.
(2)
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Mark
Notes
Part (a)
M1 standardised value of 290 is z = 1.28 M1 standardised value of 237 is z = −0.84 M1 forming 290 − μ = 1.28σ and 237 − μ = −0.84σ A1 subtracting gives 53 = 2.12σ, so σ = 25 A1 μ = 258 ms
Exam tip: Subtracting the two equations removes μ straight away. When the z-values are given to two decimal places, the arithmetic is designed to come out exactly.
116IBMA2H-051-06-054.13 Bayes’ theoremPaper 1Medium7 marks
No calculator
A cloud service routes each request to one of three data centres: 50% go to centre A, 30% to centre B and 20% to centre C. The probability that a request times out is 0.02 at A, 0.05 at B and \(p\) at C. Let \(T\) be the event that a randomly chosen request times out.
Tree diagram with the probabilities given in the question (T: times out)
(a)
Show that \(P(T)=0.025+0.2p\).
(1)
(b)
Given that \(P(C\mid T)=0.4\), find the value of \(p\).
(3)
(c)
Given that a request did not time out, find the probability that it was routed to centre A.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(P(T)=0.5\times0.02+0.3\times0.05+0.2p=0.01+0.015+0.2p\) leading to \(0.025+0.2p\)
A1 \(P(T)=0.025+\frac{1}{60}=\frac{1}{24}\), so \(P(T')=\frac{23}{24}\) ; M1 \(P(A\cap T')=0.5\times0.98=\frac{49}{100}\) divided by \(\frac{23}{24}\) ; A1 \(\frac{294}{575}\) (≈0.511)
✓ 3
Exam tip: When an unknown appears in Bayes' theorem, clear the fraction first and collect terms in \(p\); the resulting equation is linear.
117IBMA2H-051-06-064.14 Variance, continuous random variables and linear transformationsPaper 1Medium11 marks
No calculator
The lifetime, \(T\) years, of a type of drone battery is modelled by the probability density function \(f(t)=k(4t-t^2)\) for \(0\le t\le4\), and \(f(t)=0\) otherwise.
(a)
Show that \(k=\frac{3}{32}\).
(2)
(b)
Find \(P(T>3\mid T>2)\).
(4)
(c)
Determine whether the events \(T>2\) and \(T<3\) are independent.
(3)
(d)
Three such batteries operate independently. Each has already lasted 2 years. Find the probability that at least one of them lasts more than 3 years in total.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sets \(\int_0^4 k(4t-t^2)\,dt=1\) ; A1 \(k\left[2t^2-\frac{t^3}{3}\right]_0^4=\frac{32k}{3}=1\), so \(k=\frac{3}{32}\)
✓ 2
Part (b)
M1 uses \(\frac{P(T>3)}{P(T>2)}\) with integrals of \(f\) ; A1 \(P(T>3)=\frac{3}{32}\left(\frac{32}{3}-9\right)=\frac{5}{32}\) ; A1 \(P(T>2)=\frac{3}{32}\left(\frac{32}{3}-\frac{16}{3}\right)=\frac{1}{2}\) ; A1 \(\frac{5}{16}\)
✓ 4
Part (c)
M1 \(P(2<T<3)=\frac{1}{2}-\frac{5}{32}=\frac{11}{32}\) ; A1 \(P(T<3)=\frac{27}{32}\), so \(P(T>2)P(T<3)=\frac{27}{64}\) ; R1 \(\frac{27}{64}\ne\frac{22}{64}\), so not independent
Common error: For \(P(T>3\mid T>2)\) the intersection of the events is simply \(T>3\); integrating over \(2<t<3\) for the numerator finds the wrong event.
118IBMA2H-051-06-084.6 Combined, conditional and independent eventsPaper 1Medium7 marks
No calculator
A drawer contains 8 USB drives, of which 3 are faulty and 5 work. Two drives are taken at random without replacement.
Tree for the first two drives taken (F = faulty, W = works; no replacement)
(a)
Find the probability that both drives are faulty.
(2)
(b)
Given that at least one of the two drives is faulty, find the probability that both are faulty.
(3)
(c)
All 8 drives are returned. Drives are now tested one at a time, without replacement, until a working drive is found. Find the probability that the third drive tested is the first working one.
Common error: The condition 'at least one faulty' is not the same as 'the first drive is faulty'; using \(\frac{3}{8}\) as the denominator gives \(\frac{2}{7}\), which answers a different question.
119IBMA2H-051-06-124.6 Combined, conditional and independent eventsPaper 1Medium11 marks
No calculator
A voice assistant receives a sequence of spoken commands. The first command is recognised with probability 0.7. For each later command, if the previous command was recognised, it is recognised with probability 0.8; if the previous command was not recognised, it is recognised with probability 0.6. Let \(R_n\) be the event that the \(n\)th command is recognised.
Tree diagram with the probabilities given in the question (R: recognised, N: not recognised; commands 1, 2, 3)
(a)
Find \(P(R_2)\).
(2)
(b)
Find \(P(R_1\mid R_2)\).
(2)
(c)
Find \(P(R_3)\).
(3)
(d)
Show that \(R_1\) and \(R_3\) are not independent.
Common error: \(P(R_3\mid R_1)\) is not 0.8: the second command sits between them, so both of its outcomes must be included before reaching the third command.
120IBMA2H-051-03-054.6 Combined, conditional and independent eventsPaper 1Medium7 marks
No calculator
A space probe has four independent manoeuvring thrusters. When commanded, each thruster fires correctly with probability \(\frac{2}{3}\). The probe can complete a course correction only if at least three thrusters fire correctly. Let X be the number of thrusters that fire correctly.
(a)
Find P(X = 3).
(2)
(b)
Find the probability that the probe can complete a course correction.
(2)
(c)
Given that the probe completes a course correction, find the probability that all four thrusters fired correctly.
M1 use of conditional probability P(X = 4 | X ≥ 3) = P(X = 4) ÷ P(X ≥ 3)
✓ 1
M1 (16/81) ÷ (48/81)
✓ 1
A1 1/3
✓ 1
Exam tip: In part (c) the event X = 4 is already contained in X ≥ 3, so the intersection is simply P(X = 4). Do not multiply the two probabilities together.
121IBMA2H-051-03-064.4 Correlation and regression of y on xPaper 1Medium7 marks
No calculator
A sports scientist records x, the hours of training per week, and y, the increase in power output in watts, for a group of cyclists. For these data: the regression line of y on x is y = 0.8x + 2.4 the regression line of x on y is x = 1.1y − 1.5
Let Sxx = Σ(x − \(\bar{x}\))2, Syy = Σ(y − \(\bar{y}\))2 and Sxy = Σ(x − \(\bar{x}\))(y − \(\bar{y}\)). You are given that the gradient of the regression line of y on x is \(\frac{S_{xy}}{S_{xx}}\). You are also given that Pearson's product-moment correlation coefficient is \(r = \frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}\).
(a)
Find the mean of x and the mean of y.
(3)
(b)
By interchanging the roles of x and y, write down the gradient of the regression line of x on y in terms of Sxy and Syy. Hence show that the product of the two gradients equals r2.
(2)
(c)
Hence find the exact value of r.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 both lines pass through (\(\bar{x}\), \(\bar{y}\)), so solve simultaneously, e.g. x = 1.1(0.8x + 2.4) − 1.5
✓ 1
A1 \(\bar{x}\) = 9.5
✓ 1
A1 \(\bar{y}\) = 10
✓ 1
Part (b)
M1 gradient of x on y = Sxy/Syy
✓ 1
A1 product = Sxy2/(SxxSyy) = (Sxy/√(SxxSyy))2 = r2, as required
✓ 1
Part (c)
M1 r2 = 0.8 × 1.1 = 0.88 = 22/25
✓ 1
A1 r = √22/5, taking the positive root because both gradients are positive
✓ 1
Common error: Use the gradient of the x on y line exactly as written, 1.1. Do not take its reciprocal, 1/1.1. The reciprocal is the slope of that line drawn on y-against-x axes. Using it gives 0.8/1.1 instead of r².
122IBMA2H-051-03-094.8 Binomial distributionPaper 1Medium6 marks
No calculator
A 3D printer produces parts in batches of 10. Each part is defective independently with probability p, where 0 < p < 1. Let X be the number of defective parts in a batch. It is known that P(X = 1) = P(X = 2).
M1 equating and dividing by p(1 − p)8, valid since 0 < p < 1
✓ 1
A1 10(1 − p) = 45p
✓ 1
A1 p = 2/11
✓ 1
Part (b)
M1 np(1 − p) = 10 × (2/11) × (9/11)
✓ 1
A1 180/121
✓ 1
Common error: Do not expand the powers of (1 − p). Divide both sides by the common factor p(1 − p)⁸ instead, which leaves a linear equation.
123IBMA2H-051-09-064.3 Measures of central tendency and dispersionPaper 1Medium7 marks
No calculator
Five segments of fibre-optic cable have lengths, in metres, 4, 7, 9, p and q, where p < q. The mean length is 6 m and the variance of the lengths is 5.2 m2, where variance is \(\frac{\sum x^2}{n}-\bar{x}^2\).
Exam tip: When the mean and variance are both given, turn each into an equation in the unknowns: one for Σx and one for Σx².
124IBMA2H-051-09-074.3 Measures of central tendency and dispersionPaper 1Medium6 marks
No calculator
The response times, in ms, of a web server during one hour have this five-number summary: minimum 18, Q1 = 31, median 37, Q3 = 44, maximum 79. An outlier is a value more than 1.5 × IQR above Q3 or below Q1.
Box plot of the five-number summary (whiskers drawn to the minimum and maximum)
(a)
Write down the interquartile range.
(1)
(b)
Determine whether the maximum response time is an outlier.
(2)
(c)
After a software update, Q1 remains 31 ms and the maximum remains 79 ms. Find the least value of Q3 for which 79 ms would not be an outlier.
(3)
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Marking point
Mark
Notes
Part (a)
A1 IQR = 13 ms
✓ 1
Part (b)
M1 upper fence = 44 + 1.5 × 13 = 63.5 R1 79 > 63.5 so it is an outlier
Common error: Changing Q₃ also changes the IQR, so the fence depends on Q₃ twice; do not keep the old IQR of 13.
125IBMA2H-051-09-094.14 Variance, continuous random variables and linear transformationsPaper 1Medium13 marks
No calculator
The waiting time, X minutes, for a ride-share car to arrive at a pickup point is modelled by the probability density function \(f(x)=\frac{1}{5}e^{-x/5}\) for x ≥ 0, and f(x) = 0 otherwise.
Graph of the probability density function f
(a)
Show that the cumulative distribution function is \(F(x)=1-e^{-x/5}\) for x ≥ 0.
(2)
(b)
Find the exact median waiting time.
(2)
(c)
Show that the interquartile range is 5 ln 3 minutes.
(3)
(d)
Use integration by parts to find E(X).
(4)
(e)
Without using a calculator, explain why the mean waiting time is greater than the median, and state what this indicates about the distribution.
M1 \(E(X)=\int_0^{\infty}\frac{x}{5}e^{-x/5}\,dx\) M1 parts with u = x, dv = \(\frac{1}{5}e^{-x/5}\)dx A1 \(\left[-xe^{-x/5}\right]_0^{\infty}+\int_0^{\infty}e^{-x/5}\,dx\) A1 E(X) = 5 minutes
✓ 4
Part (e)
R1 ln 2 < 1 because 2 < e, so 5 ln 2 < 5 R1 the distribution is positively skewed (long tail of long waits pulls the mean up)
✓ 2
Exam tip: For an improper integral, evaluate the limit of the boundary term explicitly: xe^(−x/5) → 0 as x → ∞.
126IBMA2H-051-02-054.14 Variance, continuous random variables and linear transformationsPaper 1Medium7 marks
No calculator
A satellite operator records \(X\), the number of ground stations in contact with a particular satellite at a randomly chosen moment. The probability distribution of \(X\) is shown below. \(x\): 1, 2, 3, 4 \(P(X=x)\): \(p\), 0.3, \(q\), 0.2 It is known that \(E(X)=2.5\).
(a)
Find the value of \(p\) and the value of \(q\).
(4)
(b)
Find \(\mathrm{Var}(5-2X)\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses the sum of the probabilities: \(p+q+0.5=1\), so \(p+q=0.5\)
✓ 1
M1 uses the expectation: \(p+0.6+3q+0.8=2.5\), so \(p+3q=1.1\)
✓ 1
A1 \(q=0.3\)
✓ 1
A1 \(p=0.2\)
✓ 1
Part (b)
M1 \(E(X^2)=1(0.2)+4(0.3)+9(0.3)+16(0.2)=7.3\)
✓ 1
A1 \(\mathrm{Var}(X)=7.3-2.5^2=1.05\)
✓ 1
A1 \(\mathrm{Var}(5-2X)=(-2)^2(1.05)=4.2\)
✓ 1
Common error: Var(5−2X) is neither −2Var(X) nor 5+4Var(X). Adding a constant has no effect on the variance, and the coefficient of X is squared.
127IBMA2H-051-02-074.6 Combined, conditional and independent eventsPaper 1Medium7 marks
No calculator
When serving in tennis, Mira's first serve lands in with probability 0.6. If it lands in, she wins the point with probability 0.75. If her first serve is out, she takes a second serve, which lands in with probability 0.9. If her second serve lands in, she wins the point with probability 0.5. If both serves are out (a double fault), she loses the point.
Tree diagram with the probabilities given in the question (first serve, then second serve)
(a)
Find the probability that Mira wins the point.
(3)
(b)
Given that Mira wins the point, find the probability that her first serve landed in.
(2)
(c)
Given that Mira loses the point, find the probability that she served a double fault.
Common error: A double fault has probability 0.4×0.1, not 0.1. The second serve only happens after the first serve has missed.
128IBMA2H-051-02-104.7 Discrete random variablesPaper 1Medium7 marks
No calculator
A wireless sensor repeatedly tries to connect to a hub until it succeeds. The number of attempts needed, \(X\), has probability distribution \(P(X=x)=k\left(\frac{1}{3}\right)^x\), for \(x=1,2,3,\ldots\)
(a)
Show that \(k=2\).
(2)
(b)
Find the probability that an even number of attempts is needed.
(2)
(c)
By differentiating \(\sum_{x=0}^{\infty}r^x=\frac{1}{1-r}\) for \(|r|<1\), find an expression for \(\sum_{x=1}^{\infty}xr^{x-1}\). Hence find \(E(X)\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sums the geometric series: \(k\cdot\frac{1/3}{1-1/3}=1\)
Exam tip: The even outcomes 2, 4, 6, … form a geometric series with common ratio 1/9, not 1/3. Find the first term and the ratio before you apply the sum to infinity.
129IBMA3H-4-19r6-074.2 Presentation of dataPaper 1Medium7 marks
No calculator
The masses, x mg, of the active ingredient in 20 tablets are summarised by Σx = 300 and Σx2 = 4700.
(a)
Find the mean and the variance of the masses.
(3)
(b)
Each mass is multiplied by 3 and then 4 is subtracted, giving new values y = 3x − 4. Find the mean and the variance of y.
(2)
(c)
A twenty-first tablet is found to have mass exactly 15 mg. Find the variance of the 21 masses, giving your answer as a fraction.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 mean = 300/20 = 15
✓ 1
M1 variance = Σx2/n − mean2 = 4700/20 − 152
✓ 1
A1 variance = 10
✓ 1
Part (b)
A1 mean = 3 × 15 − 4 = 41
✓ 1
A1 variance = 32 × 10 = 90
✓ 1
Part (c)
M1 Σx2 = 4700 + 225 = 4925, mean still 15, variance = 4925/21 − 152
✓ 1
A1 200/21
✓ 1
Common error: Subtracting a constant shifts the mean but leaves the variance unchanged; multiplying by 3 multiplies the variance by 9, not 3.
130IBMA3H-4-19r7-054.9 The normal distributionPaper 1Medium6 marks
No calculator
The number of disk failures, X, in a data centre in one week has probability distribution \(\mathrm{P}(X = x) = k(x + 1)\) for \(x = 0, 1, 2, 3\), where k is a constant.
(a)
Show that \(k = 0.1\).
(2)
(b)
Find \(\mathrm{E}(X)\).
(2)
(c)
Each failure costs £250 to repair and there is a fixed weekly monitoring cost of £40. Find the expected weekly cost.
Exam tip: The probabilities of a discrete distribution must sum to 1: use this to find an unknown constant first.
131IBMA3H-4-19r7-094.14 Variance, continuous random variables and linear transformationsPaper 1Medium10 marks
No calculator
The depth, X metres, below the surface at which a coin is found in a 4 m deep trench has probability density function \(f(x) = kx(4 - x)\) for \(0 \le x \le 4\), and \(f(x) = 0\) otherwise.
Common error: Var(X) = E(X²) − [E(X)]²; do not forget to subtract the square of the mean.
132IBMA3H-4-19r9-114.7 Discrete random variablesPaper 1Medium11 marks
No calculator
At a theatre fair, a player spins wheel P, which shows 1, 2 or 3 with equal probability, and independently spins wheel Q, which shows 0 or 2 with equal probability. The score \(S\) is the sum of the two numbers shown.
(a)
Draw up the probability distribution of \(S\) (state each value and its probability).
(3)
(b)
Find \(E(S)\).
(2)
(c)
Find \(\operatorname{Var}(S)\).
(3)
(d)
Given that wheel P shows at least 2, find the probability that \(S\ge4\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sample space of 6 equally likely outcomes considered
✓ 1
A1 values 1, 2, 3, 4, 5 with correct sums
✓ 1
A1 \(P(S=1)=P(S=2)=P(S=4)=P(S=5)=\frac16\) and \(P(S=3)=\frac13\)
Exam tip: List the sample space systematically so that repeated sums, such as the two ways of scoring 3, are not missed.
133IBMA3H-4-19r14-074.14 Variance, continuous random variables and linear transformationsPaper 1Medium10 marks
No calculator
A home battery charged by solar panels supplies power for X hours after sunset, where X has probability density function \(f(x) = kx(4 - x)\) for \(0 \le x \le 4\), and \(f(x) = 0\) otherwise.
(a)
Show that \(k = \frac{3}{32}\).
(3)
(b)
Write down E(X), giving a reason.
(1)
(c)
Find Var(X).
(4)
(d)
The battery's running cost, in £, is \(C = 5 - 3X\). Find Var(C).
Exam tip: For a pdf that is symmetric about a line, the mean lies on that line, which can save an integration.
134IBMA3H-4-19r14-104.10 Regression line of x on yPaper 1Medium7 marks
No calculator
A retailer records whether customers buy online (event A) or buy in store (event B) in one week; a customer can do both. P(A) = 0.55, P(B) = 0.40 and P(A′ ∩ B′) = 0.15.
Venn diagram showing the information given in the question
(a)
Find P(A ∩ B).
(3)
(b)
Determine whether A and B are independent.
(2)
(c)
Find P(A | B).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(A ∪ B) = 1 − 0.15 = 0.85
✓ 1
M1 P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.55 + 0.40 − 0.85
✓ 1
A1 0.10
✓ 1
Part (b)
M1 P(A)P(B) = 0.55 × 0.40 = 0.22
✓ 1
R1 0.22 ≠ 0.10, so A and B are not independent
✓ 1
Part (c)
M1 P(A ∩ B) / P(B) = 0.10 / 0.40
✓ 1
A1 0.25
✓ 1
Exam tip: To test independence, compare P(A ∩ B) with P(A)P(B) and state a conclusion in words.
135IBMA3H-4-19r17-074.14 Variance, continuous random variables and linear transformationsPaper 1Medium9 marks
No calculator
The time, \(X\) hours, that a visitor spends in a library on a given day has probability density function \(f(x)=kx(3-x)\) for \(0\le x\le3\), and \(f(x)=0\) otherwise, where \(k\) is a constant.
(a)
Show that \(k=\dfrac{2}{9}\).
(3)
(b)
Write down \(\mathrm{E}(X)\), giving a reason for your answer.
Common error: Var(X) = E(X²) − [E(X)]²: forgetting to square the mean, or integrating x f(x) when E(X²) is needed, are the usual slips.
136IBMA3H-4-19r19-094.4 Correlation and regression of y on xPaper 1Medium8 marks
No calculator
A youth orchestra records the weekly rehearsal time x (hours) and a performance score y for each member. The regression line of y on x is y = 0.5x + 12 and the regression line of x on y is x = 1.5y − 14.
(a)
Find the mean rehearsal time and the mean score.
(3)
(b)
Find the exact value of the product moment correlation coefficient r.
(3)
(c)
The standard deviation of x is 6. Find the exact value of the standard deviation of y.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 both lines pass through (x̄, ȳ), so solve simultaneously, e.g. x = 1.5(0.5x + 12) − 14
✓ 1
A1 x̄ = 16
✓ 1
A1 ȳ = 20
✓ 1
Part (b)
M1 r2 = product of the two gradients = 0.5 × 1.5
✓ 1
A1 r = √3/2
✓ 1
R1 positive root since both gradients are positive
✓ 1
Part (c)
M1 gradient of y on x = r × sy/sx, so 0.5 = (√3/2) × sy/6
✓ 1
A1 sy = 6/√3 = 2√3
✓ 1
Common error: The gradient of the line of y on x is r sy/sx; the gradient of the line of x on y is r sx/sy. Mixing these up inverts the ratio.
137IBMA3H-4-18r20-084.2 Presentation of dataPaper 1Medium6 marks
No calculator
The five-number summaries of the daily distances, in km, cycled by two athletes, A and B, over 60 days of training are shown.
Minimum
\(Q_1\)
Median
\(Q_3\)
Maximum
A
12
18
24
27
45
B
14
20
22
24
30
Box plots of the five-number summaries
(a)
Find the interquartile range of each athlete's distances.
(2)
(b)
Show that the maximum for athlete A is an outlier, using the rule that an outlier is greater than \(Q_3+1.5\times\text{IQR}\).
(3)
(c)
State, with a reason, which athlete's daily distances are more consistent.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 athlete A: 9 km
✓ 1
A1 athlete B: 4 km
✓ 1
Part (b)
M1 \(Q_3+1.5\times\text{IQR}=27+1.5\times9\)
✓ 1
A1 40.5
✓ 1
R1 \(45>40.5\), so 45 is an outlier
✓ 1
Part (c)
R1 athlete B, because the interquartile range is smaller
✓ 1
Exam tip: When comparing two data sets, quote a measure of centre and a measure of spread, and relate each to the context.
138IBMA3H-4-19r24-064.11 Conditional probability and independencePaper 1Medium7 marks
No calculator
In a trial of 60 patients, \(A\) is the event that a patient reports nausea and \(B\) is the event that a patient reports dizziness. A Venn diagram of the numbers of patients gives: \(A\) only, \(x+4\); both \(A\) and \(B\), \(x\); \(B\) only, \(2x\); neither, \(56-4x\). At least one patient reports both symptoms, and \(A\) and \(B\) are independent.
Venn diagram of the numbers of patients (60 in total)
(a)
Find the value of \(x\).
(5)
(b)
Find \(P(B\mid A')\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(P(A)=\dfrac{2x+4}{60}\) and \(P(B)=\dfrac{3x}{60}\)
✓ 1
M1 use of \(P(A)\times P(B)=P(A\cap B)\), with \(P(A\cap B)=\dfrac{x}{60}\)
✓ 1
A1 \((2x+4)(3x)=60x\)
✓ 1
M1 simplifies and solves, e.g. \(6x(x-8)=0\)
✓ 1
R1 \(x=8\), with \(x=0\) rejected because at least one patient reports both
Exam tip: With a Venn diagram of counts, write \(P(A)\) and \(P(B)\) as totals over the whole circle, not just the part labelled 'only', before testing independence.
139IBMA3H-4-20r28-074.11 Conditional probability and independencePaper 1Medium7 marks
No calculator
At a theatre, A is the event that a randomly chosen audience member bought a drink in the interval and B is the event that the same person bought a programme. It is given that \(\mathrm{P}(A)=0.5\), \(\mathrm{P}(B)=0.4\) and \(\mathrm{P}(A\cup B)=0.65\).
(a)
Find \(\mathrm{P}(A\cap B)\). [2]
(2)
(b)
Find \(\mathrm{P}(A\mid B)\). [2]
(2)
(c)
Determine whether A and B are independent. Justify your answer. [2]
R1 \(0.2\neq0.25\), so the events are not independent (ECF from (a))
✓ 1
Part (d)
A1 \(1-0.65=0.35\)
✓ 1
Exam tip: To test independence compare \(\mathrm{P}(A\cap B)\) with \(\mathrm{P}(A)\mathrm{P}(B)\), or compare \(\mathrm{P}(A\mid B)\) with \(\mathrm{P}(A)\), and state the conclusion in words.
140IBMA3H-4-20r28-104.14 Variance, continuous random variables and linear transformationsPaper 1Medium8 marks
No calculator
The time, X hours, for which a volcanic tremor lasts is modelled by the probability density function \(f(x)=kx(4-x)\) for \(0\leq x\leq4\), and \(f(x)=0\) otherwise, where k is a constant.
Exam tip: For a continuous variable use \(\mathrm{Var}(X)=\mathrm{E}(X^2)-[\mathrm{E}(X)]^2\) with \(\mathrm{E}(X^2)=\int x^2f(x)\,dx\). A negative variance signals an arithmetic slip.
141IBMA3H-4-20r28-114.14 Variance, continuous random variables and linear transformationsPaper 1Medium8 marks
No calculator
The time, X minutes, at which a lighting cue is triggered during a festival performance has probability density function \(f(x)=k\sin\!\left(\frac{\pi x}{60}\right)\) for \(0\leq x\leq60\), and \(f(x)=0\) otherwise.
(a)
Show that \(k=\frac{\pi}{120}\). [3]
(3)
(b)
Find \(\mathrm{P}(X<20)\). [3]
(3)
(c)
Write down the mode and the median of X, giving a reason for each. [2]
A1 mode is 30, where \(\sin\!\left(\frac{\pi x}{60}\right)\) is greatest
✓ 1
A1 median is 30, because the graph of \(f\) is symmetric about \(x=30\)
✓ 1
Exam tip: The mode is where the density is greatest. If the graph of the density is symmetric, the mean and the median both lie on the line of symmetry.
142IBMA3H-4-19r32-104.4 Correlation and regression of y on xPaper 1Medium8 marks
No calculator
A mountaineering club records the mass of the pack, x kg, carried by each member and the time, y minutes, taken to climb a fixed 300 m section of a route. The regression line of y on x is y = 0.8x + 4 and the regression line of x on y is x = 0.45y + 4.6.
(a)
Find the mean pack mass and the mean climbing time.
(3)
(b)
Find the value of the product moment correlation coefficient r, justifying its sign.
(3)
(c)
The standard deviation of the pack masses is 7.5 kg. Find the standard deviation of the climbing times.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 both regression lines pass through the point (x̄, ȳ)
✓ 1
M1 solve simultaneously, e.g. x = 0.45(0.8x + 4) + 4.6, so 0.64x = 6.4
✓ 1
A1 x̄ = 10 kg and ȳ = 12 minutes
✓ 1
Part (b)
M1 r² = 0.8 × 0.45 (product of the two gradients)
✓ 1
A1 r² = 0.36, so |r| = 0.6
✓ 1
R1 both gradients are positive, so r = +0.6
✓ 1
Part (c)
M1 gradient of y on x = r × sy/sx, so 0.8 = 0.6 × sy/7.5
✓ 1
A1 sy = 10 minutes
✓ 1
Exam tip: The product of the two regression gradients is r², so r has the same sign as both gradients.
143IBMA3H-4-18r33-074.6 Combined, conditional and independent eventsPaper 1Medium6 marks
No calculator
A planning authority considers buildings in a city. Let F be the event that a building passes a fire inspection and S the event that it has a steel frame. It is known that P(F) = 0.4, P(S | F) = 0.45 and P(S | F′) = 0.2.
(a)
Find P(S).
(3)
(b)
Use Bayes' theorem to find P(F | S).
(2)
(c)
State, with a reason, whether F and S are independent.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(F ∩ S) = 0.4 × 0.45 = 0.18
✓ 1
M1 P(F′ ∩ S) = 0.6 × 0.2 = 0.12
✓ 1
A1 0.3
✓ 1
Part (b)
M1 P(F | S) = 0.18 / 0.3 (ECF from (a))
✓ 1
A1 0.6
✓ 1
Part (c)
R1 not independent, since P(F | S) = 0.6 ≠ P(F) = 0.4
✓ 1
Exam tip: Independence needs P(A | B) = P(A); a single mismatch in the probabilities is enough to show events are not independent.
144IBMA3H-4-18r33-124.1 Sampling, bias and outliersPaper 1Medium6 marks
No calculator
A logistics company wants to check the labelling of parcels leaving a sorting hub. A conveyor belt carries 1000 parcels in a day, numbered 1 to 1000 in the order they pass a scanner. An inspector checks parcel number 3 and then every 8th parcel after that.
(a)
State the name of this sampling method and find the number of parcels checked.
(2)
(b)
The parcels are loaded onto the belt from eight docks in a repeating cycle, one parcel from each dock in turn. Explain why the sample may be biased.
(1)
(c)
The company switches to quota sampling, with an inspector choosing any 20 parcels from the day shift and 20 from the night shift. Give one limitation of quota sampling in this situation.
(1)
(d)
Find the number of the last parcel checked in the systematic sample.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 systematic sampling
✓ 1
A1 125 parcels
✓ 1
Part (b)
R1 every 8th parcel comes from the same dock, so only one dock is represented
✓ 1
Part (c)
R1 any valid limitation, e.g. the inspector chooses the parcels, not randomly, so selection bias is possible
✓ 1
Part (d)
M1 3 + 8 × 124 (the 125th parcel in the sample)
✓ 1
A1 995
✓ 1
Common error: A systematic sample is only unbiased if the pattern in the list does not match the sampling interval; check the structure of the data first.
145IBMA3H-4-19r37-054.11 Conditional probability and independencePaper 1Medium9 marks
No calculator
A probe carries several instruments. In each part, the events refer to separate pairs of instruments transmitting during a given hour.
(a)
Events A and B are independent, with P(A) = 3/5 and P(B) = 1/4. Find P(A ∩ B) and P(A ∪ B).
(3)
(b)
Events C and D are mutually exclusive, with P(C) = 0.35 and P(C ∪ D) = 0.8. Find P(D).
(2)
(c)
Events E and F satisfy P(E) = 0.5, P(F) = 0.4 and P(E ∪ F) = 0.8. Find P(E | F), and hence state with a reason whether E and F are independent.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(A ∩ B) = P(A)P(B)
✓ 1
A1 P(A ∩ B) = 3/20
✓ 1
A1 P(A ∪ B) = 7/10
✓ 1
Part (b)
M1 P(C ∪ D) = P(C) + P(D)
✓ 1
A1 P(D) = 0.45
✓ 1
Part (c)
M1 P(E ∩ F) = 0.5 + 0.4 − 0.8 = 0.1
✓ 1
M1 P(E | F) = 0.1 ÷ 0.4
✓ 1
A1 P(E | F) = 0.25
✓ 1
R1 0.25 ≠ P(E) = 0.5, so E and F are not independent
✓ 1
Exam tip: To test independence, compare P(E | F) with P(E), or P(E ∩ F) with P(E)P(F); a statement without a comparison earns no reasoning mark.
146IBMA3H-4-20r44-144.14 Variance, continuous random variables and linear transformationsPaper 1Medium6 marks
No calculator
A volcanic gas sensor is triggered (\(S\)) with probability 0.9 when an eruption (\(E\)) is imminent and with probability 0.1 when no eruption is imminent. The probability that an eruption is imminent is \(p\). Given that the sensor is triggered, the probability that an eruption is imminent is 0.75.
(a)
Find the value of \(p\).
(4)
(b)
Hence find the probability that the sensor is triggered.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 Bayes: P(E|S) = 0.9p ÷ (0.9p + 0.1(1 − p))
✓ 1
M1 set equal to 0.75, so 0.9p = 0.75(0.8p + 0.1)
✓ 1
A1 0.3p = 0.075
✓ 1
A1 p = 0.25
✓ 1
Part (b)
M1 0.25 × 0.9 + 0.75 × 0.1
✓ 1
A1 0.3
✓ 1
Exam tip: When the posterior is given and a prior is unknown, write Bayes' theorem with the unknown and solve the resulting linear equation.
147IBMA3H-4-20r46-114.13 Bayes’ theoremPaper 1Medium5 marks
No calculator
On a farm, 10% of plants have a disease. A test gives a positive result for 90% of diseased plants and for 20% of healthy plants. Let \(D\) be the event that a plant has the disease and \(+\) the event that the test is positive.
(a)
Find \(P(+)\).
(2)
(b)
Find the probability that a plant has the disease, given that its test is positive.
(2)
(c)
Explain why your answer to (b) is much smaller than 0.9.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(0.1\times0.9+0.9\times0.2\)
✓ 1
A1 0.27
✓ 1
Part (b)
M1 \(P(D|+)=\frac{0.09}{0.27}\)
✓ 1
A1 \(\frac{1}{3}\)
✓ 1
Part (c)
R1 healthy plants are far more common, so the false positives (0.18) outnumber the true positives (0.09)
✓ 1
Exam tip: A test's detection rate and the probability of disease given a positive test are different quantities; Bayes' theorem links them.
148AAHL4-q00044.2 Presentation of dataPaper 1Hard9 marks
A data set consists of seven integers \(x_{1}\le x_{2}\le\cdots\le x_{7}\). The quartiles are found by taking the median of the lower three values and the median of the upper three values. The box-and-whisker diagram summarizes the data.
(a)
Write down the values of \(x_{1},\ x_{2},\ x_{4},\ x_{6}\) and \(x_{7}\).
(2)
(b)
Determine whether the largest value is an outlier.
(2)
(c)
Find the least possible value and the greatest possible value of the mean of the data.
(3)
(d)
Given that the mean of the data is equal to the median, find the values of \(x_{3}\) and \(x_{5}\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
recognizing \(Q_{1}=x_{2}\), median \(=x_{4}\), \(Q_{3}=x_{6}\)
only possible with the least values: \(x_{3}=6,\ x_{5}=10\)
A1
Answers: (a) 3, 6, 10, 14, 21 · (b) not an outlier · (c) 10 and \(\tfrac{78}{7}\) · (d) \(x_{3}=6,\ x_{5}=10\)
149AAHL4-q00054.2 Presentation of dataPaper 1Hard10 marks
The time, \(t\) minutes, that each of 200 visitors spent in a gallery was recorded. For \(0\le t\le60\), the cumulative frequency curve is modelled by \(F(t)=\dfrac{200}{1+\mathrm{e}^{-(t-30)/5}}\), as shown.
(a)
Write down the median time.
(1)
(b)
Show that the lower quartile is \(30-5\ln3\).
(2)
(c)
Write down the upper quartile.
(1)
(d)
Show that a visitor’s time is an upper outlier if it is greater than \(30+20\ln3\).
(2)
(e)
Use the model to estimate the number of visitors whose times are upper outliers.
(2)
(f)
Find the 90th percentile, giving your answer in the form \(a+b\ln3\), where \(a,b\in\mathbb{Z}\).
Answers: (a) 30 min · (c) \(30+5\ln3\) · (e) about 2 · (f) \(30+10\ln3\)
150AAHL4-q00174.5 Sample spaces and probabilityPaper 1Hard11 marks
Two integers \(a\) and \(b\) are chosen independently and at random from \(\{1,\,2,\,\ldots,\,n\}\), where \(n\ge2\). Every ordered pair \((a,\ b)\) is equally likely.
(a)
Write down \(\mathrm{P}(a=b)\).
(1)
(b)
Hence show that \(\mathrm{P}(a\lt b)=\dfrac{n-1}{2n}\).
(2)
(c)
Show that \(\mathrm{P}(|a-b|\le1)=\dfrac{3n-2}{n^{2}}\).
(3)
(d)
Given that \(\mathrm{P}(|a-b|\le1)=\dfrac{7}{25}\), find the value of \(n\).
(3)
(e)
For this value of \(n\), find \(\mathrm{P}(a\lt b\mid |a-b|\le1)\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac{n}{n^{2}}=\dfrac{1}{n}\)
A1
(b)
by symmetry \(\mathrm{P}(a\lt b)=\mathrm{P}(a\gt b)=\dfrac{1}{2}\left(1-\dfrac{1}{n}\right)\)
M1
\(=\dfrac{1}{2}\cdot\dfrac{n-1}{n}\)
A1
\(=\dfrac{n-1}{2n}\)
AG
(c)
counting pairs with \(b=a+1\): \(n-1\) pairs; similarly \(n-1\) pairs with \(a=b+1\)
M1
favourable pairs \(=n+2(n-1)=3n-2\)
A1
out of \(n^{2}\) equally likely pairs
A1
\(\dfrac{3n-2}{n^{2}}\)
AG
(d)
\(7n^{2}=25(3n-2)\Rightarrow 7n^{2}-75n+50=0\)
M1
\((n-10)(7n-5)=0\)
A1
\(n=10\) (rejecting \(n=\tfrac{5}{7}\), not an integer)
151AAHL4-q00214.6 Combined, conditional and independent eventsPaper 1Hard10 marks
In a game of tennis, when the score reaches “deuce” the game continues until one player has won two more points than the other. Ana wins each point against Bea with probability \(p\), independently of all other points, where \(0\lt p\lt 1\). The game is at deuce.
(a)(i)
Let \(p=\dfrac{2}{3}\).
Write down the probability that Ana wins the next two points.
(1)
(a)(ii)
Find the probability that the score returns to deuce after the next two points.
(2)
(b)
Hence show that, when \(p=\dfrac{2}{3}\), the probability that Ana wins the game is \(\dfrac{4}{5}\).
(3)
(c)
Show that, for general \(p\), the probability that Ana wins the game is \(\dfrac{p^{2}}{1-2p+2p^{2}}\).
(2)
(d)
Find the value of \(p\) for which the probability that Ana wins the game is 0.9.
(2)
Show mark scheme
Working
Mark
Notes
(a)(i)
\(\dfrac{4}{9}\)
A1
(a)(ii)
Ana wins exactly one of the two points: \(2\times\dfrac{2}{3}\times\dfrac{1}{3}\)
Answers: (a)(i) \(\tfrac{4}{9}\), (ii) \(\tfrac{4}{9}\) · (d) \(p=\tfrac{3}{4}\)
152AAHL4-q00254.7 Discrete random variablesPaper 1Hard10 marks
The probability distribution of a discrete random variable \(X\) is given by \(\mathrm{P}(X=x)=\dfrac{k}{x(x+1)}\), for \(x\in\{1,\,2,\,\ldots,\,9\}\), where \(k\) is a constant.
(a)
By writing \(\dfrac{1}{x(x+1)}=\dfrac{1}{x}-\dfrac{1}{x+1}\), show that \(k=\dfrac{10}{9}\).
(3)
(b)
Find \(\mathrm{P}(X\le3)\).
(2)
(c)
Find \(\mathrm{P}(X\gt 4\mid X\gt 1)\).
(3)
(d)
Find the least integer \(m\) for which \(\mathrm{P}(X\ge m)\lt 0.1\).
153AAHL4-q00294.7 Discrete random variablesPaper 1Hard15 marks
A town has \(n\) cafés, where \(n\ge3\). Each of three friends chooses one of the cafés at random, independently of the others, to visit on Saturday. Let \(X\) be the number of different cafés chosen by the three friends.
(a)(i)
In this part, \(n=4\).
Find the probability distribution of \(X\).
(3)
(a)(ii)
Hence find \(\mathrm{E}(X)\).
(1)
(b)
Show that, for general \(n\), \(\mathrm{P}(X=1)=\dfrac{1}{n^{2}}\) and \(\mathrm{P}(X=3)=\dfrac{(n-1)(n-2)}{n^{2}}\).
(3)
(c)
Hence show that \(\mathrm{P}(X=2)=\dfrac{3(n-1)}{n^{2}}\).
(2)
(d)
Show that \(\mathrm{E}(X)=3-\dfrac{3}{n}+\dfrac{1}{n^{2}}\).
(2)
(e)
Find the least value of \(n\) for which the probability that the three friends choose three different cafés is greater than 0.9.
(3)
(f)
Describe what happens to \(\mathrm{E}(X)\) as \(n\) increases, and interpret this in context.
(1)
Show mark scheme
Working
Mark
Notes
(a)(i)
e.g. \(\mathrm{P}(X=3)=1\times\dfrac{3}{4}\times\dfrac{2}{4}\)
155AAHL4-q00654.13 Bayes’ theoremPaper 1Hard9 marks
A fair six-sided die is rolled to choose a box. If the score is 1, box X is chosen; if the score is 2 or 3, box Y is chosen; otherwise box Z is chosen. Box X contains 3 red and 1 blue ball, box Y contains 2 red and 2 blue balls and box Z contains 2 red and 4 blue balls. Two balls are then taken at random from the chosen box, without replacement.
(a)
Find the probability that both balls are red.
(3)
(b)
Given that both balls are red, find the probability that box Z was chosen.
(2)
(c)
Given that exactly one of the balls is red, find the probability that box Y was chosen.
156AAHL4-q00674.13 Bayes’ theoremPaper 1Hard11 marks
A multiple-choice question has four options, exactly one of which is correct. A student knows the answer with probability \(p\), where \(0\lt p\lt 1\). If the student knows the answer, the student answers correctly; otherwise the student guesses one of the four options at random. Let \(K\) be the event that the student knows the answer and \(C\) the event that the student answers correctly.
(a)
Show that \(\mathrm{P}(C)=\dfrac{1+3p}{4}\).
(2)
(b)
Show that \(\mathrm{P}(K\mid C)=\dfrac{4p}{1+3p}\).
(2)
(c)
Find the value of \(p\) for which \(\mathrm{P}(K\mid C)=0.9\).
(2)
(d)(i)
Let \(g(p)=\mathrm{P}(K\mid C)-p\), the increase in the probability that the student knows the answer once a correct answer is seen.
Find the value of \(p\) for which \(g(p)\) is a maximum. Justify that it is a maximum.
Or completing the square, or vertex of the parabola.
\(p=\tfrac{5}{16}\), which lies in \(\left]0,\tfrac12\right]\), and \(\dfrac{\mathrm{d}^{2}}{\mathrm{d}p^{2}}=-32\lt 0\) so a maximum
A1
Must check that the value lies in the domain from (a) and justify the maximum.
maximum \(\mathrm{Var}(X)=\dfrac{50}{16}-\dfrac{25}{16}=\dfrac{25}{16}\)
A1
Answers: (a) \(0\lt p\le\tfrac12\) · (c) \(\tfrac{25}{16}\) when \(p=\tfrac5{16}\)
158AAHL4-q00824.14 Variance, continuous random variables and linear transformationsPaper 1Hard10 marks
The discrete random variable \(X\) takes only the values \(0\) and \(c\), where \(c\gt 0\), with \(\mathrm{P}(X=c)=p\) and \(\mathrm{P}(X=0)=1-p\), where \(0\lt p\lt 1\).
(a)
Show that \(\mathrm{Var}(X)=\mathrm{E}(X)\big(c-\mathrm{E}(X)\big)\).
(3)
(b)
Given that \(\mathrm{E}(X)=2\) and \(\mathrm{Var}(X)=6\), find the value of \(c\) and the value of \(p\).
(3)
(c)
For the values in part (b), the random variable \(Y=aX+b\), where \(a\gt0\), takes only the values \(-1\) and \(1\). Find the value of \(a\) and the value of \(b\), and hence find \(\mathrm{E}(Y)\) and \(\mathrm{Var}(Y)\).
(4)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{E}(X)=pc\) and \(\mathrm{E}(X^{2})=pc^{2}\)
A1
\(\mathrm{Var}(X)=pc^{2}-p^{2}c^{2}\)
M1
Use of \(\mathrm{E}(X^{2})-[\mathrm{E}(X)]^{2}\).
\(=pc\,(c-pc)\)
A1
Factorizing.
\(=\mathrm{E}(X)\big(c-\mathrm{E}(X)\big)\)
AG
(b)
\(6=2(c-2)\)
M1
Using the result of (a).
\(c=5\)
A1
\(p=\dfrac{2}{5}\)
A1
From \(pc=2\).
(c)
\(b=-1\) and \(a=\dfrac25\)
A1
From \(a(0)+b=-1\) and \(5a+b=1\).
\(\mathrm{E}(Y)=\tfrac25(2)-1\) and \(\mathrm{Var}(Y)=\left(\tfrac25\right)^{2}(6)\)
M1
Use of \(\mathrm{E}(aX+b)=a\mathrm{E}(X)+b\) and \(\mathrm{Var}(aX+b)=a^{2}\mathrm{Var}(X)\). FT their \(a\), \(b\).
Write down \(\mathrm{E}(X)\), giving a reason for your answer.
(1)
(c)
Given that \(\mathrm{Var}(X)=\dfrac{13}{45}\), find the value of \(a\) and the value of \(b\).
(4)
(d)
For these values, find \(\mathrm{P}\left(|X|\lt\dfrac{1}{2}\right)\).
(3)
(e)
For these values, write down the mode of \(X\).
(1)
(f)
Now suppose that \(a\) and \(b\) can take any values for which \(f\) is a probability density function. Find the set of possible values of \(\mathrm{Var}(X)\).
162IBMA2H-051-08-094.2 Presentation of dataPaper 1Hard10 marks
No calculator
A network engineer records the download times, t seconds, of 500 files. The cumulative frequency curve is modelled by C(t) = 500(1 − e−0.25t), t ≥ 0, where C(t) is the number of files with download time at most t seconds. An outlier is a value more than 1.5 × IQR beyond the nearer quartile.
(a)
Find the exact median download time.
(2)
(b)
Show that the interquartile range is 4 ln 3 seconds.
(3)
(c)
Find the exact number of files predicted by the model to be outliers, and hence state this number to the nearest integer. You may use √3 ≈ 1.73.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sets 500(1 − e−0.25t) = 250
✓ 1
A1 t = 4 ln 2 seconds
✓ 1
Part (b)
M1 sets C(t) = 125 and C(t) = 375
✓ 1
A1 Q1 = 4 ln(4/3) and Q3 = 4 ln 4
✓ 1
A1 IQR = 4 ln 4 − 4 ln(4/3) = 4 ln 3
✓ 1
Part (c)
R1 lower boundary 4 ln(4/3) − 6 ln 3 < 0, so no outliers at the lower end
A1 \(\frac{125\sqrt{3}}{9}\), approximately 24 files
✓ 1
Exam tip: Keep logarithms in exact form: e to the power of minus (ln 4 + 1.5 ln 3) simplifies to 1/(4 × 3√3) without any decimal working.
163IBMA2H-051-00-114.14 Variance, continuous random variables and linear transformationsPaper 1Hard13 marks
No calculator
A basketball analytics app models the number of points, X, scored by a team on a single possession by the distribution below, where p and q are constants. x: 0, 1, 2, 3 P(X = x): p, 0.1, q, 0.2 It is given that E(X) = 1.5.
(a)
Find the value of p and the value of q.
(4)
(b)
Find Var(X).
(3)
(c)
The app converts each possession into a rating \(S=aX+b\), where \(a>0\), chosen so that E(S) = 10 and the standard deviation of S is 5. Find the exact values of a and b.
M1 \(S>10\Leftrightarrow X>\frac{10-b}{a}=\frac{3\sqrt{5}}{2\sqrt{5}}=1.5\), so X = 2 or 3 A1 \(0.4+0.2=0.6\)
✓ 2
Common error: Setting \(a\times1.25=5\), which confuses variance with standard deviation. The standard deviation of S is a times the standard deviation of X, not a times the variance.
164IBMA2H-051-04-124.6 Combined, conditional and independent eventsPaper 1Hard14 marks
No calculator
In a robotics contest, two drones, Ana and Ben, take turns trying to land on a moving platform, with Ana going first. On each of her attempts Ana lands successfully with probability 1/3, and on each of his attempts Ben lands successfully with probability 1/4, independently of all other attempts. The first drone to land wins, and the contest then stops.
(a)
Find the probability that Ana wins on her second attempt.
(2)
(b)
Show that the probability that Ana wins is 2/3.
(4)
(c)
Given that Ana wins, find the probability that she wins on her first attempt.
(2)
(d)
Given that Ben wins, find the probability that the contest ends within the first four attempts in total (that is, two attempts by each drone).
(3)
(e)
Ben's drone is upgraded so that his probability of landing on each attempt is q, where 0 < q ≤ 1. Ana's probability is unchanged. Find the value of q for which each drone is equally likely to win.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (2/3)(3/4)(1/3) ; A1 1/6
✓ 2
Part (b)
M1 recognises a geometric series ; M1 first term 1/3 and common ratio (2/3)(3/4) = 1/2 ; M1 uses S∞ = a/(1 − r) ; A1 (1/3)/(1/2) = 2/3
M1 P(Ana wins) = (1/3)/(1 − (2/3)(1 − q)) ; M1 sets this equal to 1/2 (the contest ends with probability 1, so P(Ben wins) = 1 − P(Ana wins)) ; A1 q = 1/2
✓ 3
Exam tip: Each complete round of two misses returns the contest to its starting state, so the common ratio of the geometric series is the probability that both drones miss in one round.
165IBMA2H-051-06-114.11 Conditional probability and independencePaper 1Hard12 marks
No calculator
\(A\) and \(B\) are independent events with \(P(A)>P(B)\).
(a)
Prove that \(A'\) and \(B'\) are independent.
(3)
(b)
It is given that \(P(A\cap B)=\frac{1}{3}\) and \(P(A'\cap B')=\frac{1}{6}\). Find \(P(A)\) and \(P(B)\).
(5)
(c)
Determine whether \(A\) and \(A\cup B\) are independent.
M1 \(P(A\cap(A\cup B))=P(A)=\frac{2}{3}\) compared with \(P(A)P(A\cup B)=\frac{2}{3}\times\frac{5}{6}=\frac{5}{9}\) ; R1 \(\frac{2}{3}\ne\frac{5}{9}\), so not independent
Common error: Writing \(P(A'\cap B')=1-P(A\cap B)\) is wrong: the complement of 'neither' is 'at least one', so \(P(A'\cap B')=1-P(A\cup B)\).
166IBMA2H-051-03-124.13 Bayes’ theoremPaper 1Hard10 marks
No calculator
An automated telescope searches n independent regions of sky each night. In each region it detects a transient event with probability p, where 0 < p < 1. Let X ~ B(n, p) be the number of regions in which an event is detected.
(a)
Show that, for r = 0, 1, …, n − 1, \(\frac{P(X = r+1)}{P(X = r)} = \frac{(n-r)p}{(r+1)(1-p)}\).
(3)
(b)
For n = 20 and p = 0.3, use the ratio result shown above to find the most likely number of regions in which an event is detected.
(4)
(c)
For n = 20, find the value of p for which P(X = 4) = P(X = 5).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writing P(X = r + 1) = nCr+1pr+1(1 − p)n−r−1 and P(X = r) = nCrpr(1 − p)n−r
✓ 1
M1 simplifying the ratio of binomial coefficients: [n!/((r+1)!(n−r−1)!)] ÷ [n!/(r!(n−r)!)] = (n − r)/(r + 1)
✓ 1
A1 the powers give p/(1 − p), leading to the given result
✓ 1
Part (b)
M1 ratio > 1 gives (20 − r)(0.3) > (r + 1)(0.7)
✓ 1
A1 6 − 0.3r > 0.7r + 0.7, so r < 5.3
✓ 1
R1 P(X = r + 1) > P(X = r) for r ≤ 5 and P(X = r + 1) < P(X = r) for r ≥ 6, so the probabilities increase up to r = 6 and then decrease
✓ 1
A1 mode = 6
✓ 1
Part (c)
M1 setting the ratio with r = 4 equal to 1: 16p/(5(1 − p)) = 1
✓ 1
A1 16p = 5 − 5p
✓ 1
A1 p = 5/21
✓ 1
Exam tip: A ratio of consecutive probabilities is the cleanest way to locate a mode. Successive probabilities increase while the ratio is greater than 1 and decrease once it falls below 1.
167IBMA2H-051-09-124.14 Variance, continuous random variables and linear transformationsPaper 1Hard14 marks
No calculator
The time, X seconds, taken by a smartphone's face-recognition system to unlock is modelled by the probability density function f(x) = ax + b for 0 ≤ x ≤ 2, and f(x) = 0 otherwise, where a and b are constants. It is given that E(X) = \(\frac{7}{6}\).
(a)
Find the value of a and the value of b.
(5)
(b)
Find Var(X).
(3)
(c)
The phone's display reports a score Y = 6X − 1. Find E(Y) and Var(Y).
M1 \(\int_0^m \frac{x+1}{4}\,dx=\frac{1}{2}\) A1 \(\frac{m^2}{2}+m=2\), i.e. m2 + 2m − 4 = 0 M1 uses quadratic formula and rejects the negative root A1 median = √5 − 1 seconds
✓ 4
Common error: Var(aX + b) = a²Var(X): the constant b has no effect and the multiplier must be squared, so Var(Y) is not 6 × 11/36 − 1.
168IBMA2H-051-02-114.6 Combined, conditional and independent eventsPaper 1Hard16 marks
No calculator
Two esports teams, Aurora and Borealis, play a best-of-five series. The first team to win three games wins the series, and no further games are played. Aurora wins each game with probability \(\frac{2}{3}\), independently of other games. Let \(X\) be the number of games played in the series.
(a)
Show that \(P(X=3)=\frac{1}{3}\).
(2)
(b)
Find \(P(X=4)\) and \(P(X=5)\).
(4)
(c)
Find \(E(X)\).
(2)
(d)
Given that Aurora wins the series, find the probability that all five games were played.
(4)
(e)
Suppose instead that Aurora wins each game with probability \(p\), where \(0<p<1\). Show that \(P(X=3)=3p^2-3p+1\), and hence find the minimum possible value of \(P(X=3)\).
M1 completes the square, \(3\left(p-\frac{1}{2}\right)^2+\frac{1}{4}\), or solves \(6p-3=0\)
✓ 1
A1 minimum value \(\frac{1}{4}\), when \(p=\frac{1}{2}\)
✓ 1
Exam tip: For a series that is won in 4 games, the deciding game must be won by the series winner. Count arrangements of the earlier games only: C(3,2), not C(4,3).
169IBMA3H-4-19r6-144.14 Variance, continuous random variables and linear transformationsPaper 1Hard13 marks
No calculator
The time, X hours, for which a dose of a drug keeps a patient's blood concentration above a threshold has probability density function \(f(x) = ax\) for \(0 \le x \le 2\), \(f(x) = b(6-x)\) for \(2 < x \le 6\), and \(f(x) = 0\) otherwise, where a and b are positive constants and f is continuous.
(a)
Find the value of a and the value of b.
(4)
(b)
Write down the mode of X.
(1)
(c)
Find E(X).
(4)
(d)
Find the exact value of the median of X.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 continuity at x = 2: 2a = 4b
✓ 1
M1 total area equals 1: ½ × 2 × 2a + ½ × 4 × 4b = 1, i.e. 2a + 8b = 1
✓ 1
A1 b = 1/12
✓ 1
A1 a = 1/6
✓ 1
Part (b)
A1 2 hours (maximum of f)
✓ 1
Part (c)
M1 E(X) = ∫02x × x/6 dx + ∫26x(6 − x)/12 dx
✓ 1
A1 first integral 4/9
✓ 1
A1 second integral 20/9
✓ 1
A1 E(X) = 8/3
✓ 1
Part (d)
M1 P(X < 2) = 1/3 < 1/2, so the median m lies in (2, 6]
✓ 1
M1 1/3 + ∫2m (6 − x)/12 dx = 1/2
✓ 1
A1 m2 − 12m + 24 = 0
✓ 1
A1 m = 6 − 2√3 (reject 6 + 2√3, which exceeds 6)
✓ 1
Exam tip: Check which piece of a piecewise density the median falls in by comparing the cumulative probability at the boundary with 1/2 before integrating.
170IBMA3H-4-19r7-144.14 Variance, continuous random variables and linear transformationsPaper 1Hard13 marks
No calculator
A pulse generator in a data centre produces a signal whose phase, X radians, is a continuous random variable with probability density function \(f(x) = \frac{1}{2}\sin x\) for \(0 \le x \le \pi\), and \(f(x) = 0\) otherwise.
(a)
Show that f is a valid probability density function.
(2)
(b)
Use integration by parts to show that \(\mathrm{E}(X) = \frac{\pi}{2}\).
Exam tip: For E(X²) with a trigonometric pdf, expect to integrate by parts twice; check your answer is positive and less than E(X²).
171IBMA3H-4-19r9-124.14 Variance, continuous random variables and linear transformationsPaper 1Hard13 marks
No calculator
The profit, \(X\) thousand euros, from a theatre production has probability density function \(f(x)=kx(2-x)\) for \(0\le x\le2\), and \(f(x)=0\) otherwise.
(a)
Show that \(k=\frac34\).
(3)
(b)
Find \(E(X)\).
(2)
(c)
Show that \(\operatorname{Var}(X)=\frac15\).
(4)
(d)
The company's bonus pool is \(Y=aX+b\) thousand euros, with \(a>0\). Given that \(E(Y)=50\) and \(\operatorname{Var}(Y)=45\), find \(a\) and \(b\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\int_0^2 kx(2-x)\,dx=1\)
✓ 1
A1 \(k\left[x^2-\frac{x^3}{3}\right]_0^2\)
✓ 1
A1 \(\frac{4k}{3}=1\) so \(k=\frac34\)
✓ 1
Part (b)
M1 \(\int_0^2 x\cdot\frac34x(2-x)\,dx\) (or symmetry about \(x=1\) stated)
Common error: Use \(\operatorname{Var}(X)=E(X^2)-[E(X)]^2\); forgetting to square \(E(X)\), or integrating \(x\,f(x)\) instead of \(x^2f(x)\), are frequent slips.
172IBMA3H-4-19r9-144.14 Variance, continuous random variables and linear transformationsPaper 1Hard11 marks
No calculator
The strength of a seismic tremor burst at time \(x\) hours after it begins, for \(0\le x\le\pi\), is modelled by a random variable \(X\) with probability density function \(f(x)=k\sin x\), and \(f(x)=0\) otherwise.
(a)
Show that \(k=\frac12\).
(2)
(b)
State the value of \(E(X)\), giving a reason.
(2)
(c)
Find \(P\left(X<\frac\pi3\right)\).
(2)
(d)
Show that \(\operatorname{Var}(X)=\frac{\pi^2}{4}-2\).
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\int_0^\pi k\sin x\,dx=1\)
✓ 1
A1 \(2k=1\) so \(k=\frac12\)
✓ 1
Part (b)
A1 \(E(X)=\frac\pi2\)
✓ 1
R1 the density is symmetric about \(x=\frac\pi2\)
✓ 1
Part (c)
M1 \(\frac12\left[-\cos x\right]_0^{\pi/3}\)
✓ 1
A1 \(\frac12\left(1-\frac12\right)=\frac14\)
✓ 1
Part (d)
M1 \(E(X^2)=\frac12\int_0^\pi x^2\sin x\,dx\)
✓ 1
M1 integration by parts at least once
✓ 1
A1 antiderivative \(-x^2\cos x+2x\sin x+2\cos x\)
✓ 1
A1 \(\int_0^\pi x^2\sin x\,dx=\pi^2-4\), so \(E(X^2)=\frac{\pi^2-4}{2}\)
Exam tip: When a variance question asks you to show a result, state \(E(X^2)\) and \([E(X)]^2\) separately before subtracting.
173IBMA3H-4-19r17-134.14 Variance, continuous random variables and linear transformationsPaper 1Hard12 marks
No calculator
The time, \(X\) hours, that a swimmer spends in the pool on a training day has probability density function \(f(x)=\begin{cases} kx, & 0\le x\le 2\\ \dfrac{k(6-x)}{2}, & 2<x\le 6\\ 0, & \text{otherwise}\end{cases}\) where \(k\) is a constant.
Graph of the probability density function, not to scale
M1 \(\int_0^2\dfrac{x}{6}\,dx=\dfrac{1}{3}<\dfrac{1}{2}\) so the median \(m\) lies in \(2<m\le6\)
✓ 1
M1 \(\int_m^6\dfrac{6-x}{12}\,dx=\dfrac{1}{2}\)
✓ 1
A1 \(\dfrac{(6-m)^2}{24}=\dfrac{1}{2}\)
✓ 1
A1 \((6-m)^2=12\)
✓ 1
R1 \(m=6-2\sqrt{3}\) since \(m<6\) (AG)
✓ 1
Part (c)
M1 split at \(x=2\): \(\int_1^2\dfrac{x}{6}\,dx+\int_2^4\dfrac{6-x}{12}\,dx\)
✓ 1
A1 \(\dfrac{1}{4}\) and \(\dfrac{1}{2}\) respectively
✓ 1
A1 \(\dfrac{3}{4}\)
✓ 1
Exam tip: With a piecewise density, find which piece contains the median by checking the accumulated probability at the break point first.
174IBMA3H-4-19r19-134.14 Variance, continuous random variables and linear transformationsPaper 1Hard10 marks
No calculator
An engineer checks an aircraft component for a fault. Each check independently detects the fault with probability p, where 0 < p ≤ 1. The inspection stops as soon as the fault is detected, or after the third check, whichever comes first. Let X be the number of checks carried out.
(a)
Show that E(X) = p2 − 3p + 3.
(4)
(b)
Given that E(X) = 1.75, find the value of p.
(3)
(c)
Find Var(X).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 P(X = 2) = p(1 − p)
✓ 1
A1 P(X = 3) = (1 − p)2, since a third check is carried out whenever the first two fail
✓ 1
M1 E(X) = p + 2p(1 − p) + 3(1 − p)2
✓ 1
A1 = p + 2p − 2p2 + 3 − 6p + 3p2 = p2 − 3p + 3
✓ 1
Part (b)
M1 p2 − 3p + 1.25 = 0
✓ 1
A1 (p − 0.5)(p − 2.5) = 0 so p = 0.5 or 2.5
✓ 1
R1 reject 2.5 because a probability cannot exceed 1, so p = 0.5
Common error: P(X = 3) is the probability that the first two checks both fail; the third check is then carried out whatever its result.
175IBMA3H-4-18r21-144.6 Combined, conditional and independent eventsPaper 1Hard9 marks
No calculator
A shop's loyalty draw uses a bag containing \(n\) red tokens and 4 blue tokens, where \(n\geq2\). Two tokens are drawn at random without replacement. The probability that the two tokens are different colours is \(\frac{8}{15}\).
(a)
Show that \(n^2-8n+12=0\).
(4)
(b)
Find the possible values of \(n\).
(2)
(c)
There are more red tokens than blue tokens. Find the probability that both tokens drawn are blue.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(P(\text{red then blue})=\frac{n}{n+4}\times\frac{4}{n+3}\)
M1 equates to \(\frac{8}{15}\) and cross-multiplies: \(15n=n^2+7n+12\)
✓ 1
A1 \(n^2-8n+12=0\) (AG)
✓ 1
Part (b)
M1 factorises \((n-2)(n-6)=0\)
✓ 1
A1 \(n=2\) or \(n=6\)
✓ 1
Part (c)
R1 \(n=6\), since \(n=2\) gives fewer red than blue tokens
✓ 1
M1 \(\frac{4}{10}\times\frac{3}{9}\)
✓ 1
A1 \(\frac{2}{15}\) (= 0.1333)
✓ 1
Exam tip: When a condition produces two solutions, check each against any extra information in the question before using it.
176IBMA3H-4-20r28-124.13 Bayes’ theoremPaper 1Hard8 marks
No calculator
A port authority flags some incoming ships for inspection. Thirty percent of incoming ships have travelled on route R. Ninety percent of ships from route R are flagged. A proportion p of the ships not from route R are flagged. Given that a ship is flagged, the probability that it came from route R is 0.6.
(a)
Use Bayes' theorem to show that \(\frac{0.27}{0.27+0.7p}=0.6\). [3]
(3)
(b)
Find the value of p. [3]
(3)
(c)
Find the probability that a randomly chosen ship is flagged. [2]
A1 denominator \(0.27+0.7p\), giving the equation (AG)
✓ 1
Part (b)
M1 \(0.27=0.6(0.27+0.7p)\)
✓ 1
A1 \(0.108=0.42p\)
✓ 1
A1 \(p=\frac{9}{35}\)
✓ 1
Part (c)
M1 \(0.3\times0.9+0.7\times\frac{9}{35}\) (ECF from (b))
✓ 1
A1 0.45
✓ 1
Exam tip: The denominator in Bayes' theorem is the total probability of the given event, found by adding over every branch of the tree that leads to it.
177IBMA3H-4-18r33-044.13 Bayes’ theoremPaper 1Hard8 marks
No calculator
A logistics company uses courier A for 60% of its deliveries and courier B for the other 40%. A delivery by courier A is late with probability p, and a delivery by courier B is late with probability 0.05. Let L be the event that a delivery is late.
Tree diagram showing the probabilities given in the question
(a)
Show that P(L) = 0.6p + 0.02.
(2)
(b)
Given that P(courier A | L) = 0.8, find the value of p.
(4)
(c)
Hence find the probability that a delivery chosen at random is late.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(L) = 0.6 × p + 0.4 × 0.05
✓ 1
A1 0.6p + 0.02 (AG)
✓ 1
Part (b)
M1 P(A | L) = 0.6p / (0.6p + 0.02)
✓ 1
A1 0.6p = 0.8(0.6p + 0.02)
✓ 1
M1 attempt to solve, e.g. 0.12p = 0.016
✓ 1
A1 p = 2/15
✓ 1
Part (c)
M1 substitute p = 2/15 into 0.6p + 0.02
✓ 1
A1 0.1
✓ 1
Exam tip: Write Bayes' theorem as an equation in the unknown first, then clear the fraction by multiplying through by the denominator.
178IBMA3H-4-19r37-134.3 Measures of central tendency and dispersionPaper 1Hard11 marks
No calculator
A probe has five instrument modules. The number of modules, X, switched on during a given hour has probability distribution P(X = x) = kx for x = 1, 2, 3, 4, 5, where k is a constant.
(a)
Find the value of k.
(2)
(b)
Show that E(X) = 11/3.
(2)
(c)
Show that Var(X) = 14/9.
(3)
(d)
The mean squared deviation of X from a value m is E[(X − m)2]. Given that m is an integer and that E[(X − m)2] = 2, find m.
Exam tip: E[(X − m)²] equals Var(X) + (E(X) − m)², which saves expanding the whole sum again.
179IBMA3H-4-20r46-124.14 Variance, continuous random variables and linear transformationsPaper 1Hard11 marks
No calculator
The discrete random variable \(X\) has the following distribution.
x: 1, 2, 3, 4 P(X = x): 0.2, \(a\), \(b\), 0.3
It is given that \(E(X)=2.7\).
(a)
Find the value of \(a\) and the value of \(b\).
(4)
(b)
Find \(\mathrm{Var}(X)\).
(3)
(c)
Find \(\mathrm{Var}(3-2X)\).
(2)
(d)
Two independent values of \(X\) are observed. Find the probability that both are greater than \(E(X)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(0.2+a+b+0.3=1\), so \(a+b=0.5\)
✓ 1
M1 \(0.2+2a+3b+1.2=2.7\), so \(2a+3b=1.3\)
✓ 1
A1 \(b=0.3\)
✓ 1
A1 \(a=0.2\)
✓ 1
Part (b)
M1 \(E(X^2)=0.2+0.8+2.7+4.8=8.5\)
✓ 1
M1 \(\mathrm{Var}(X)=8.5-2.7^2\)
✓ 1
A1 1.21
✓ 1
Part (c)
M1 \((-2)^2\mathrm{Var}(X)\)
✓ 1
A1 4.84 (ECF from (b))
✓ 1
Part (d)
M1 \(P(X>2.7)=P(X=3)+P(X=4)=0.6\)
✓ 1
A1 \(0.6^2=0.36\)
✓ 1
Common error: \(\mathrm{Var}(3-2X)\) is \((-2)^2\mathrm{Var}(X)\), which is positive; the constant 3 has no effect.
180AAHL4-q00064.2 Presentation of dataPaper 2Easy8 marks
The times, \(t\) minutes, that 40 people spent in a waiting room are summarized in the table, where \(p\) and \(q\) are positive integers.
Time, \(t\) (min)
\(0\le t\lt 10\)
\(10\le t\lt 20\)
\(20\le t\lt 30\)
\(30\le t\lt 40\)
\(40\le t\lt 50\)
Frequency
5
\(p\)
12
\(q\)
4
Using mid-interval values, the estimated mean time is 25.25 minutes.
(a)
Show that \(p+q=19\).
(1)
(b)
Find the value of \(p\) and the value of \(q\).
(3)
(c)
Write down the modal class and the class that contains the median.
(2)
(d)
Find an estimate of the standard deviation of the times.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(5+p+12+q+4=40\)
A1
\(p+q=19\)
AG
(b)
\(\dfrac{5(5)+15p+25(12)+35q+45(4)}{40}=25.25\)
M1
\(15p+35q=505\)
A1
\(p=8,\ q=11\)
A1
(c)
modal class \(20\le t\lt 30\)
A1
median class \(20\le t\lt 30\)
A1
Cumulative frequencies 5, 13, 25: the 20th and 21st values lie in this class.
(d)
GDC with mid-interval values 5, 15, 25, 35, 45 and frequencies 5, 8, 12, 11, 4
(M1)
FT from their \(p,\ q\).
\(11.7\) min
A1
\(11.7233\ldots\)
Answers: (b) \(p=8,\ q=11\) · (c) both \(20\le t\lt 30\) · (d) 11.7 min
181AAHL4-q00104.3 Measures of central tendency and dispersionPaper 2Easy8 marks
The lengths, \(t\) minutes, of 100 phone calls made from an office are summarized below.
Length, \(t\) (min)
\(0\le t\lt 2\)
\(2\le t\lt 4\)
\(4\le t\lt 6\)
\(6\le t\lt 8\)
\(8\le t\lt 10\)
Number of calls
18
34
27
15
6
Histogram of the data in the table (equal class widths)
(a)
Write down the modal class.
(1)
(b)
Find an estimate of the mean and an estimate of the standard deviation of the call lengths.
(3)
(c)
The office’s phone records give the exact lengths of the 100 calls, which satisfy \(\sum t=421.6\) and \(\sum t^{2}=2310\). Using \(\sigma^{2}=\dfrac{\sum t^{2}}{n}-\bar{t}^{\,2}\), find the exact mean and the exact standard deviation of the call lengths.
(3)
(d)
Explain why the values found in part (b) differ from those found in part (c).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(2\le t\lt 4\)
A1
(b)
mid-interval values 1, 3, 5, 7, 9 with the frequencies
the grouped estimates assume every call in a class lasts the mid-interval value; the actual lengths are spread within each class
R1
Answers: (a) \(2\le t\lt 4\) · (b) 4.14 min, 2.25 min · (c) 4.216 min, 2.31 min
182AAHL4-q00184.5 Sample spaces and probabilityPaper 2Easy8 marks
A computer chooses points \((x,\ y)\) uniformly at random in the unit square \(0\le x\le1,\ 0\le y\le1\).
Unit square; the region x² + y² < 1 is shaded
(a)
Show that the probability that a point satisfies \(x^{2}+y^{2}\lt 1\) is \(\dfrac{\pi}{4}\).
(2)
(b)
In a simulation of 5000 points, 3921 satisfy \(x^{2}+y^{2}\lt 1\). Use this result to find an estimate of \(\pi\).
(2)
(c)
By finding the area of the region \(y\gt x^{2}\) inside the square, find the expected number of the 5000 points that satisfy \(y\gt x^{2}\).
(3)
(d)
Explain how the estimate in part (b) could be made more reliable.
(1)
Show mark scheme
Working
Mark
Notes
(a)
region is a quarter of a circle of radius 1, area \(\dfrac{\pi}{4}\); square has area 1
M1
probability \(=\dfrac{\pi/4}{1}\)
A1
\(=\dfrac{\pi}{4}\)
AG
(b)
\(\dfrac{\pi}{4}\approx\dfrac{3921}{5000}\)
(M1)
\(\pi\approx3.14\)
A1
\(3.1368\)
(c)
\(1-\displaystyle\int_{0}^{1}x^{2}\,\mathrm{d}x\)
(M1)
\(\dfrac{2}{3}\)
A1
\(5000\times\dfrac{2}{3}=3333.3\), about 3333
A1
(d)
use many more points: relative frequency gets closer to the theoretical probability as the number of trials increases
R1
Answers: (b) 3.14 · (c) about 3333
183AAHL4-q00244.6 Combined, conditional and independent eventsPaper 2Easy16 marks
The 400 students in Years 11 and 12 of a school were asked how they usually travel to school. The results are shown in the table.
Walk
Bus
Car
Total
Year 11
70
88
62
220
Year 12
50
72
58
180
Total
120
160
120
400
(a)
A stratified sample of 40 of these students, stratified by year group, is to be chosen. Find the number of Year 11 students and the number of Year 12 students in the sample.
(2)
(b)
A student is chosen at random from the 400. Write down the probability that the student travels by bus.
(1)
(c)(i)
Determine whether the events “travels by bus” and “is in Year 12” are independent.
(2)
(c)(ii)
Determine whether the events “walks” and “is in Year 12” are independent.
(2)
(d)
A student chosen at random does not walk. Find the probability that the student is in Year 11.
(2)
(e)
Three students are chosen at random from the 400, without replacement. Find the probability that all three walk.
(2)
(f)
Two Year 12 students are chosen at random, without replacement. Find the probability that they travel by different methods.
(3)
(g)
In another survey, the first 40 students to arrive at school one morning are questioned. Name this sampling method and give a reason why it may give a biased estimate of the proportion of students who travel by bus.
eg students arriving first may mostly come on the same bus (or are dropped off by car), so the sample is not representative of how all students travel
R1
Accept any reason linking arrival time to the method of travel.
Answers: (a) 22 and 18 · (b) 0.4 · (c)(i) independent (ii) not independent · (d) \(\tfrac{15}{28}\) · (e) 0.0265 · (f) 0.663 · (g) convenience
184AAHL4-q00364.8 Binomial distributionPaper 2Easy7 marks
Mira claims that she can predict the suit (hearts, diamonds, clubs or spades) of a playing card before it is turned over. To test her, a card is drawn at random from a full pack of 52 cards (13 of each suit) and she predicts its suit; the card is then replaced and the pack shuffled. This is done 48 times. Assume that Mira is only guessing, so that each prediction is correct with probability 0.25, independently. Let \(X\) be the number of correct predictions.
(a)
Find the mean and the standard deviation of \(X\).
(2)
(b)
Find \(\mathrm{P}(X\ge16)\).
(2)
(c)
The tester will accept Mira’s claim if she makes at least \(k\) correct predictions. Find the least value of \(k\) for which the probability that a person who is only guessing has her claim accepted is less than 0.01.
(2)
(d)
Explain why the card is replaced and the pack shuffled after each prediction.
so that the probability of a correct guess stays at 0.25 for every prediction and the predictions are independent (binomial conditions)
R1
Answers: (a) 12, 3 · (b) 0.123 · (c) \(k=20\)
185AAHL4-q00464.10 Regression line of x on yPaper 2Easy7 marks
A biologist records the air temperature, \(x\) °C, and the number of chirps, \(y\), made by a cricket in 15 seconds on eight evenings.
\(x\) (°C)
15
17
19
20
22
24
26
28
\(y\) (chirps)
22
27
29
33
35
40
42
47
Scatter diagram of the data in the table
The biologist wants to use these data to estimate the temperature from a count of chirps.
(a)
Find the equation of the regression line of \(x\) on \(y\).
(2)
(b)
Use your equation to estimate the temperature on an evening when a cricket makes 36 chirps in 15 seconds.
(2)
(c)
Interpret the gradient of your line in the context of the question.
(1)
(d)
On a cold evening a cricket makes 5 chirps in 15 seconds. Explain why the line should not be used to estimate the temperature on this evening.
(1)
(e)
Explain why the line found in part (a) should not be used to predict the number of chirps when the temperature is 21 °C.
(1)
Show mark scheme
Working
Mark
Notes
(a)
gradient \(0.533\)
A1
\(0.532667\ldots\)
intercept \(3.06\), giving \(x=0.533y+3.06\)
A1
\(3.06456\ldots\) Award A0A0 for \(y=1.86x-5.34\) (line of \(y\) on \(x\)).
(b)
\(x=0.532667(36)+3.06457\)
(M1)
\(22.2\) °C
A1
\(22.2405\ldots\) Accept 22.2–22.3 from rounded coefficients.
(c)
each extra chirp in 15 seconds corresponds to an increase in temperature of about 0.533 °C
A1
Must be in context and refer to the change in x per unit change in y.
(d)
5 is outside the range of the \(y\)-data (22 to 47), so it would be extrapolation
R1
(e)
to predict \(y\) from \(x\) the regression line of \(y\) on \(x\) must be used
R1
Answers: (a) \(x=0.533y+3.06\) · (b) 22.2 °C
186AAHL4-q00514.11 Conditional probability and independencePaper 2Easy7 marks
An archer shoots 8 arrows at a target. For each arrow, the probability that it hits the centre is 0.3, independently of all other arrows. Let \(X\) be the number of arrows that hit the centre.
(a)
Find \(\mathrm{P}(X=2)\).
(2)
(b)
Given that at least one arrow hits the centre, find the probability that exactly two arrows hit the centre.
(3)
(c)
Given that at least one arrow hits the centre, find the probability that the first arrow hits the centre.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(X\sim\mathrm{B}(8,\,0.3)\)
(M1)
\(0.296\)
A1
\(0.296475\ldots\)
(b)
\(\dfrac{\mathrm{P}(X=2)}{\mathrm{P}(X\ge1)}\)
(M1)
\(\mathrm{P}(X\ge1)=1-0.7^{8}=0.942352\)
(A1)
\(0.315\)
A1
\(0.314612\ldots\)
(c)
\(\dfrac{0.3}{0.942352}\)
(M1)
Recognizing that “first arrow hits” is a subset of “at least one hit”.
\(0.318\)
A1
\(0.318352\ldots\)
Answers: (a) 0.296 · (b) 0.315 · (c) 0.318
187AAHL4-q00594.12 Standardization and unknown normal parametersPaper 2Easy12 marks
The time, \(T\) minutes, for a ferry crossing is normally distributed with mean 48 minutes. 20% of crossings take more than 52 minutes.
Normal curve of crossing time T (minutes) with mean 48; shaded region T > 52. Diagram not to scale
(a)
Find the standard deviation of \(T\).
(3)
(b)
Find the probability that a crossing takes more than 55 minutes.
(1)
(c)
The ferry makes 6 crossings each day, with times independent of each other. Find the probability that on a particular day at least 2 crossings take more than 52 minutes.
(2)
(d)
Find the expected number of crossings that take more than 52 minutes in a period of 5 days.
(1)
(e)
Find the least number of crossings that must be made for the probability that at least one of them takes more than 55 minutes to exceed 0.95.
(3)
(f)
The company publishes a scheduled crossing time of \(s\) minutes, chosen so that only 5% of crossings take longer than \(s\) minutes. Find \(s\).
188AAHL4-q00724.13 Bayes’ theoremPaper 2Easy5 marks
In a climbing club, 60% of the members are experienced and 40% are novices. On each attempt at a difficult route, an experienced climber succeeds with probability 0.8 and a novice succeeds with probability 0.3, independently of other attempts. A member is chosen at random and attempts the route repeatedly until the first success.
(a)
Find the probability that an experienced climber succeeds for the first time on the third attempt.
(1)
(b)
Find the probability that the chosen member succeeds for the first time on the third attempt.
(2)
(c)
Given that the chosen member succeeds for the first time on the third attempt, find the probability that the member is experienced.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(0.2^{2}\times0.8=0.032\)
A1
(b)
\(0.6(0.032)+0.4(0.7^{2}\times0.3)\)
(M1)
\(0.078\)
A1
(c)
\(\dfrac{0.0192}{0.078}\)
(M1)
\(0.246\)
A1
\(\tfrac{16}{65}=0.246153\ldots\)
Answers: (a) 0.032 · (b) 0.078 · (c) 0.246
189AAHL4-q00984.14 Variance, continuous random variables and linear transformationsPaper 2Easy4 marks
The maximum daily temperature in a city in April, \(T\) °C, has mean \(15\) and standard deviation \(4\). The temperature in degrees Fahrenheit is \(F=1.8T+32\).
(a)
Find \(\mathrm{E}(F)\).
(2)
(b)
Find the standard deviation of \(F\) and \(\mathrm{Var}(F)\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(1.8(15)+32\)
(M1)
\(59\) °F
A1
(b)
\(\sigma_{F}=1.8\times4=7.2\)
A1
Award A0 for \(1.8\times4+32\).
\(\mathrm{Var}(F)=1.8^{2}\times16=51.84\)
A1
Answers: (a) 59 °F · (b) 7.2 °F, 51.84
190AAHL4-q01004.14 Variance, continuous random variables and linear transformationsPaper 2Easy6 marks
A bakery bakes 35 loaves of a speciality bread each day. The number of these loaves sold on a randomly chosen day, \(X\), has the distribution shown.
\(x\)
\(20\)
\(25\)
\(30\)
\(35\)
\(\mathrm{P}(X=x)\)
\(0.1\)
\(0.3\)
\(0.4\)
\(0.2\)
The daily profit, in euros, from this bread is \(D=3.2X-60\).
(a)
Find \(\mathrm{E}(X)\) and the standard deviation of \(X\).
(3)
(b)
Hence find the expected daily profit and the standard deviation of the daily profit.
(2)
(c)
Find the probability that the profit on a randomly chosen day exceeds €30.
191AAHL4-q01134.14 Variance, continuous random variables and linear transformationsPaper 2Easy15 marks
Three fair six-sided dice are rolled. Let \(D\) be the number of different scores shown. For example, the scores 4, 1, 4 give \(D=2\).
(a)
Show that \(\mathrm{P}(D=3)=\dfrac{5}{9}\).
(2)
(b)
Find \(\mathrm{P}(D=1)\) and \(\mathrm{P}(D=2)\).
(2)
(c)
Find \(\mathrm{E}(D)\) and \(\mathrm{Var}(D)\).
(3)
(d)
In a game, a player pays €2 to roll the three dice and receives \(12-4D\) euros. Find the mean and the standard deviation of the amount received, and hence state the expected gain of the player per game.
(3)
(e)
The game is played 30 times. Find the probability that all three scores are different in at least 20 of these games.
(3)
(f)
Given that at least one of the dice shows a 6, find the probability that \(D=1\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac{6\times5\times4}{6^{3}}\) or \(1\times\dfrac{5}{6}\times\dfrac{4}{6}\)
192IBMA2H-051-01-014.13 Bayes’ theoremPaper 2Easy4 marks
Calculator
A satellite manufacturer buys star-tracker sensors from three suppliers. Supplier A provides 50% of the sensors, supplier B provides 30% and supplier C provides 20%. The probabilities that a sensor from A, B and C is defective are 0.02, 0.03 and 0.05 respectively. A sensor is chosen at random.
Tree diagram with the probabilities given in the question (D: defective)
(a)
Find the probability that the sensor is defective.
(2)
(b)
Given that the sensor is defective, find the probability that it came from supplier C.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of total probability: 0.5×0.02 + 0.3×0.03 + 0.2×0.05
✓ 1
A1 P(D) = 0.029
✓ 1
Part (b)
M1 use of Bayes' theorem: P(C | D) = \(\frac{0.2\times 0.05}{0.029}\)
✓ 1
A1 P(C | D) = 0.345
✓ 1
Exam tip: The denominator in Bayes' theorem is the total probability you found in part (a). Reuse it rather than recalculating it.
193IBMA2H-051-05-014.12 Standardization and unknown normal parametersPaper 2Easy5 marks
Calculator
The charge capacity, \(X\) mAh, of a lithium-ion battery made for a particular smartphone is normally distributed with mean 4500 mAh and standard deviation 60 mAh.
Normal curve for X, with the region 4440 < X < 4590 shaded
(a)
Find the probability that a randomly chosen battery has a capacity between 4440 mAh and 4590 mAh.
(2)
(b)
A batch contains 2000 of these batteries. Find the expected number of batteries in the batch with capacity between 4440 mAh and 4590 mAh.
(1)
(c)
Batteries with capacity below \(k\) mAh are recycled. 10% of batteries are recycled. Find the value of \(k\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 correct use of normal cdf with lower bound 4440, upper bound 4590, μ = 4500, σ = 60 (or z-values −1 and 1.5)
✓ 1
A1 0.775 (0.774538…)
✓ 1
Part (b)
A1 2000 × 0.774538… ≈ 1549 (accept 1550)
✓ 1
Part (c)
M1 use of inverse normal with area 0.1, or \(\frac{k-4500}{60} = -1.28155…\)
✓ 1
A1 \(k\) = 4423.1…, so \(k\) ≈ 4420 (3 s.f.)
✓ 1
Exam tip: Write down the distribution and the bounds you enter into the calculator, e.g. P(4440 < X < 4590) with X ~ N(4500, 60²). If the final value is wrong, the method mark can still be earned from what is written.
194IBMA2H-051-05-034.9 The normal distributionPaper 2Easy6 marks
Calculator
The mass, \(M\) mg, of active ingredient in a tablet produced by a pharmaceutical machine is normally distributed with mean 250 mg and standard deviation \(\sigma\) mg. It is found that 2% of tablets contain more than 256 mg of active ingredient.
Normal curve of mass M (mg) with mean 250; shaded region M > 256. Diagram not to scale
(a)
Find the value of \(\sigma\).
(3)
(b)
Tablets containing less than 244 mg or more than 256 mg of active ingredient are rejected. Write down the proportion of tablets that are rejected.
(1)
(c)
A pack contains 30 tablets, taken at random from production before any are rejected. Find the probability that none of the tablets in the pack would be rejected.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(P(M > 256) = 0.02\) leading to \(P(M < 256) = 0.98\)
✓ 1
M1 \(\frac{256-250}{\sigma} = 2.05374…\)
✓ 1
A1 \(\sigma\) = 2.92 mg (2.92149…)
✓ 1
Part (b)
A1 244 and 256 are both 6 mg from the mean, so by symmetry the proportion is 0.02 + 0.02 = 0.04
✓ 1
Part (c)
M1 recognising \(B(30, 0.04)\) with zero successes, or \(0.96^{30}\)
✓ 1
A1 0.294 (0.293858…)
✓ 1
Exam tip: When σ is unknown, standardise: write (256 − 250)/σ = z and find z from the inverse normal of N(0, 1). You cannot enter an unknown σ directly into the calculator's inverse normal function.
195IBMA2H-051-08-024.2 Presentation of dataPaper 2Easy6 marks
Calculator
A hospital records the waiting time, t minutes, of 200 patients arriving at its emergency department. The cumulative frequencies are: t ≤ 20: 14 t ≤ 40: 47 t ≤ 60: 102 t ≤ 80: 158 t ≤ 100: 186 t ≤ 120: 200 Assume the times are spread evenly within each class.
(a)
Write down the number of patients who waited more than 40 minutes but no more than 60 minutes.
(1)
(b)
Estimate the median waiting time.
(2)
(c)
Estimate the percentage of patients who waited more than 90 minutes.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 102 − 47 = 55 patients
✓ 1
Part (b)
M1 locates the 100th value in the class 40 < t ≤ 60 and interpolates: 40 + (100 − 47)/55 × 20
✓ 1
A1 median ≈ 59.3 minutes
✓ 1
Part (c)
M1 interpolates between (80, 158) and (100, 186)
✓ 1
A1 cumulative frequency at 90 minutes ≈ 172
✓ 1
A1 (200 − 172)/200 × 100 = 14%
✓ 1
Common error: Reading the cumulative frequency at 90 gives the number who waited at most 90 minutes; subtract from the total to get those who waited longer.
196IBMA2H-051-08-034.2 Presentation of dataPaper 2Easy5 marks
Calculator
A delivery company records the flight time, m minutes, of 60 drone deliveries. The data are grouped into classes of equal width, with the following frequencies: 10 ≤ m < 15: 8 15 ≤ m < 20: 15 20 ≤ m < 25: k 25 ≤ m < 30: 12 30 ≤ m < 35: 5
(a)
Find the value of k.
(2)
(b)
Write down the modal class.
(1)
(c)
Use the mid-interval values to estimate the mean flight time.
Exam tip: With grouped data the mean is only an estimate, because the midpoint stands in for every value in its class.
197IBMA2H-051-00-024.4 Correlation and regression of y on xPaper 2Easy7 marks
Calculator
A sports scientist sets a treadmill at a fixed speed and records an athlete's steady heart rate, y beats per minute (bpm), at six different inclines, x%. Incline x (%): 0, 2, 4, 6, 8, 10 Heart rate y (bpm): 112, 118, 127, 133, 141, 146
Scatter diagram of the data in the table
(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, in the form \(y=ax+b\).
(2)
(c)
Use your equation to estimate the athlete's heart rate at an incline of 5.5%.
(2)
(d)
Explain why it would not be appropriate to use the equation to estimate the heart rate at an incline of 25%.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A2 \(r=0.998\) (0.99767…); a correct value from the GDC earns both marks
R1 25% lies well outside the range of inclines recorded (0–10%), so this is extrapolation and the linear relationship may not continue
✓ 1
Exam tip: Give r to 3 significant figures but keep the full GDC values of a and b when making a prediction, rounding only the final answer.
198IBMA2H-051-00-044.8 Binomial distributionPaper 2Easy6 marks
Calculator
The number of faulty chips, X, on a randomly chosen silicon wafer has the following probability distribution. x: 0, 1, 2, 3 P(X = x): 0.5, 0.3, 0.15, 0.05 Each wafer costs the manufacturer C dollars to process, where \(C=40X+15\).
M1 \(E(X^2)=0.3+0.6+0.45=1.35\) and uses \(E(X^2)-[E(X)]^2\) A1 \(\text{Var}(X)=1.35-0.5625=0.7875\)
✓ 2
Part (c)
M1 SD\((C)=40\sqrt{0.7875}\) or \(\text{Var}(C)=1600\times0.7875=1260\) A1 SD\((C)=35.5\) dollars
✓ 2
Common error: Using \(40\,\text{Var}(X)\) instead of \(40^2\,\text{Var}(X)\), or adding 15 to the variance. Square the multiplier and ignore the constant.
199IBMA2H-051-04-024.13 Bayes’ theoremPaper 2Easy5 marks
Calculator
At a wind farm, 4% of turbines have a developing bearing fault. A vibration sensor flags a turbine with a fault with probability 0.92 and flags a turbine without a fault with probability 0.07. A turbine is chosen at random.
Tree diagram with the probabilities given in the question
(a)
Find the probability that the sensor flags the turbine.
(2)
(b)
Given that the sensor flags the turbine, find the probability that it has a bearing fault.
(2)
(c)
Comment on the reliability of a flag from this sensor.
M1 0.0368/0.104, or equivalently 0.0368/(0.0368 + 0.0672) ; A1 0.354
✓ 2
Part (c)
R1 the conditional probability found in the previous part is less than 0.5, so most flagged turbines do not actually have a fault and a flag alone is not reliable evidence of a fault
✓ 1
Common error: Do not confuse P(flag | fault) = 0.92 with P(fault | flag). Faults are rare, so false flags from the many healthy turbines outnumber the true flags.
200IBMA2H-051-07-014.8 Binomial distributionPaper 2Easy5 marks
Calculator
A manufacturer tests the lithium-ion batteries used in a smartphone. The number of full charge cycles, L, that a battery completes before reaching the end of its useful life is modelled by a normal distribution with mean 820 cycles and standard deviation σ cycles. Testing shows that 6% of batteries reach the end of their useful life before completing 700 cycles.
(a)
Find the value of σ.
(3)
(b)
The manufacturer advertises that 90% of batteries complete more than c cycles. Find the value of c.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of inverse normal with probability 0.06, giving z = −1.5548 M1 setting up (700 − 820)/σ = −1.5548 A1 σ = 77.2 cycles
✓ 3
Part (b)
M1 P(L > c) = 0.9 so P(L < c) = 0.1, giving c = 820 − 1.2816 × 77.18 or equivalent inverse normal method A1 c = 721 cycles
✓ 2
Common error: "90% exceed c" means c is the 10th percentile, so it lies below the mean. Entering 0.9 directly as a left-tail area gives a value above 820, which should be an obvious warning sign.
201IBMA2H-051-07-044.11 Conditional probability and independencePaper 2Easy5 marks
Calculator
At a maternity unit, the birth mass, M kg, of full-term babies is modelled by a normal distribution with mean μ kg and standard deviation 0.45 kg. The 30th percentile of birth mass is 3.12 kg.
(a)
Find the value of μ.
(3)
(b)
Find the interquartile range of the birth masses.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 z = −0.5244 for a lower-tail probability of 0.3 M1 (3.12 − μ)/0.45 = −0.5244 A1 μ = 3.36 kg
✓ 3
Part (b)
M1 IQR = 2 × 0.6745 × 0.45, or Q1 = 3.052 and Q3 = 3.660 found A1 IQR = 0.607 kg
✓ 2
Exam tip: The interquartile range of a normal distribution depends only on σ. It is 2 × 0.6745σ, so the mean is not needed for part (b). This also gives a quick check on the answer.
202IBMA2H-051-06-024.6 Combined, conditional and independent eventsPaper 2Easy5 marks
Calculator
A tennis player's first serve lands in with probability 0.62. If the first serve lands in, she wins the point with probability 0.74. If the first serve does not land in, she takes a second serve, which lands in with probability 0.91; if the second serve lands in, she wins the point with probability 0.52. If both serves fail to land in, she loses the point.
Tree diagram with the probabilities given in the question (first serve, then second serve)
(a)
Find the probability that she wins a given point on her serve.
(3)
(b)
Given that she won the point, find the probability that her first serve landed in.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 product for first-serve route \(0.62\times0.74=0.4588\) ; M1 product for second-serve route \(0.38\times0.91\times0.52=0.179816\) ; A1 \(P(\text{win})=0.639\) (0.638616)
✓ 3
Part (b)
M1 \(\frac{0.4588}{0.638616}\) ; A1 0.718
✓ 2
Common error: In part (b) the denominator is the probability of winning found in part (a), not 0.62; dividing by the wrong event gives a probability for the wrong condition.
203IBMA2H-051-03-014.6 Combined, conditional and independent eventsPaper 2Easy5 marks
Calculator
A small satellite carries 12 identical solar cells. During its first year in orbit, each cell fails independently with probability 0.08. Let X be the number of cells that fail during the first year.
(a)
Find the probability that exactly 2 cells fail during the first year.
(2)
(b)
The satellite meets its power target if at least 10 of its cells are still working at the end of the first year. Find the probability that the satellite meets its power target.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 recognising X ~ B(12, 0.08) and using P(X = 2)
✓ 1
A1 0.183
✓ 1
Part (b)
M1 at least 10 working is equivalent to X ≤ 2
✓ 1
M1 P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2), or use of binomial cdf
✓ 1
A1 0.935
✓ 1
Common error: Read carefully whether the question counts failures or successes. 'At least 10 working' means 'at most 2 failed'. It does not mean P(X ≥ 10).
204IBMA2H-051-03-034.10 Regression line of x on yPaper 2Easy5 marks
Calculator
An engineer tests a steel cable. She applies a load of x kN and measures the resulting extension y mm. The results are shown below. x (kN): 10, 14, 18, 25, 29, 36 y (mm): 2.1, 3.0, 3.8, 5.2, 6.1, 7.4 On a later occasion the extension of a cable is measured, and the engineer wishes to estimate the load that was applied.
Scatter diagram of the data in the table
(a)
Find the equation of the regression line of x on y, giving your answer in the form x = ay + b.
(2)
(b)
Explain why the regression line of x on y is the appropriate line to use here.
(1)
(c)
A cable has an extension of 4.5 mm. Use your equation to estimate the load applied to it.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 a = 4.90 (4.89949…)
✓ 1
A1 b = −0.538 (−0.53769…)
✓ 1
Part (b)
R1 the value of y (extension) is known and the value of x (load) is to be estimated, so x must be regressed on y
✓ 1
Part (c)
M1 substituting y = 4.5 into the x on y line
✓ 1
A1 21.5 kN
✓ 1
Exam tip: Choose the regression line by asking which variable you are given. When y is given and x is to be estimated, use the x on y line. Do not rearrange the y on x line, because the two lines are different unless r = ±1.
205IBMA2H-051-03-044.14 Variance, continuous random variables and linear transformationsPaper 2Easy5 marks
Calculator
A smartphone's face-recognition unlock fails on any attempt with probability 0.04, independently of all other attempts.
(a)
Find the least number of attempts, n, such that the probability of at least one failure in n attempts is greater than 0.5.
(3)
(b)
On one day, 25 face-recognition unlock attempts are made. Find the probability that exactly one of these 25 attempts fails.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 forming the inequality 1 − 0.96n > 0.5
✓ 1
M1 attempt to solve, e.g. n > ln 0.5 ÷ ln 0.96 = 16.98…, or a table of values showing 0.9616 = 0.5204 and 0.9617 = 0.4996
✓ 1
A1 n = 17
✓ 1
Part (b)
M1 25 × 0.04 × 0.9624, or binomial pdf with n = 25, p = 0.04, x = 1
✓ 1
A1 0.375
✓ 1
Common error: Dividing by ln 0.96 reverses the inequality, because ln 0.96 is negative. Always check the integer answer and the integer just below it in the original inequality, using enough significant figures to see which side of 0.5 each value falls.
206IBMA2H-051-09-034.2 Presentation of dataPaper 2Easy5 marks
Calculator
The screen-on battery life, t hours, of 50 smartphones of the same model was tested. The results are: 4 ≤ t < 6: 5 phones 6 ≤ t < 8: 12 phones 8 ≤ t < 10: 18 phones 10 ≤ t < 12: 11 phones 12 ≤ t < 14: 4 phones
Histogram of the data in the question (equal class widths)
(a)
Write down the modal class.
(1)
(b)
Find an estimate of the mean battery life.
(2)
(c)
Find an estimate of the standard deviation of the battery life.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 8 ≤ t < 10
✓ 1
Part (b)
M1 uses midpoints 5, 7, 9, 11, 13 with frequencies, Σfx = 444 A1 mean ≈ 8.88 hours
✓ 2
Part (c)
M1 uses midpoints in GDC or \(\sqrt{\frac{4178}{50}-8.88^2}\) A1 σ ≈ 2.17 hours
✓ 2
Exam tip: Enter class midpoints, not class boundaries, as the data values, and give the population standard deviation σ from your GDC.
207IBMA2H-051-02-034.13 Bayes’ theoremPaper 2Easy5 marks
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In a climbing-wall speed challenge, a climber is allowed up to three attempts at a route. She stops as soon as she completes the route. On each attempt the probability that she completes it is 0.82, independently of other attempts. If she fails her first two attempts, she takes a third attempt, whatever its outcome. Let \(X\) be the number of attempts she makes.
(a)
Find the probability distribution of \(X\).
(3)
(b)
Find the expected number of attempts she makes.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(P(X=2)=0.18\times0.82\)
✓ 1
A1 \(P(X=1)=0.82\), \(P(X=2)=0.1476\)
✓ 1
A1 \(P(X=3)=0.18^2=0.0324\)
✓ 1
Part (b)
M1 \(1(0.82)+2(0.1476)+3(0.0324)\)
✓ 1
A1 \(E(X)=1.2124\approx1.21\)
✓ 1
Common error: P(X=3) is not 0.18²×0.82. She takes a third attempt whenever the first two fail, whether or not the third succeeds, so P(X=3)=0.18².
208IBMA2H-051-02-044.13 Bayes’ theoremPaper 2Easy5 marks
Calculator
A phone uses face recognition to unlock. Of all unlock attempts, 4% are made by an unauthorised person. The system rejects an unauthorised person with probability 0.95. It wrongly rejects the owner with probability 0.02.
Tree diagram with the probabilities given in the question (Rej.: rejected, Acc.: accepted)
(a)
Find the probability that a randomly chosen unlock attempt is rejected.
(2)
(b)
An unlock attempt is rejected. Find the probability that it was made by an unauthorised person.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(0.04\times0.95+0.96\times0.02\)
✓ 1
A1 \(0.0572\)
✓ 1
Part (b)
M1 uses conditional probability \(\frac{P(U\cap R)}{P(R)}\)
Exam tip: Draw a two-stage tree: first the type of user, then reject or accept. The required denominator is the total of every branch that ends in 'reject'.
209IBMA3H-4-19r6-024.2 Presentation of dataPaper 2Easy7 marks
Calculator
In an online game, 20 + k players each record the number of levels, x, completed in one session. The results are shown in the table.
x
1
2
3
4
5
Frequency
3
k
8
6
3
The mean number of levels completed is 3.04.
(a)
Find the value of k.
(3)
(b)
Find the median number of levels completed.
(2)
(c)
Find the population standard deviation of the number of levels completed, correct to 3 significant figures.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 total frequency 20 + k and Σfx = 66 + 2k
✓ 1
M1 (66 + 2k)/(20 + k) = 3.04
✓ 1
A1 k = 5
✓ 1
Part (b)
M1 there are 25 players, so the median is the 13th value; cumulative frequencies 3, 8, 16
✓ 1
A1 median = 3
✓ 1
Part (c)
M1 attempt at σ using Σfx2 = 266, i.e. √(266/25 − 3.042)
✓ 1
A1 σ = 1.18
✓ 1
Exam tip: Once the frequency table is in your calculator's list editor, mean, median and standard deviation all come from one output screen; check that you read σn and not sn−₁.
210IBMA3H-4-19r6-034.12 Standardization and unknown normal parametersPaper 2Easy6 marks
Calculator
The water pressure, X bar, at a hydrant in a town's supply network is modelled by a normal distribution with mean 3.8 and standard deviation 0.25. Give all answers to 3 significant figures.
Normal curve for X, with the region X < 3.4 shaded
(a)
Find the probability that the pressure is below 3.4 bar.
(2)
(b)
Find the probability that the pressure is between 3.5 bar and 4.2 bar.
(2)
(c)
The pressure is exceeded on 5% of days. Find this pressure.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to standardise or use normal cdf, e.g. z = (3.4 − 3.8)/0.25 = −1.6
✓ 1
A1 0.0548
✓ 1
Part (b)
M1 P(3.5 < X < 4.2) = P(−1.2 < Z < 1.6)
✓ 1
A1 0.830
✓ 1
Part (c)
M1 use of inverse normal with P(X > k) = 0.05, i.e. z = 1.645 (area 0.95 to the left)
✓ 1
A1 k = 4.21
✓ 1
Common error: For an upper-tail probability such as P(X > k) = 0.05, the inverse normal command needs the area to the left, 0.95, not 0.05.
211IBMA3H-4-19r6-044.10 Regression line of x on yPaper 2Easy5 marks
Calculator
A pharmacologist records the daily dose, x mg, given to seven patients and the blood concentration, y mg/L, measured four hours later.
x
20
30
40
50
60
70
80
y
1.9
2.8
3.1
4.6
5.2
5.9
7.4
Scatter diagram of the data in the table
(a)
Find the equation of the regression line of x on y, giving the coefficients to 4 significant figures.
(2)
(b)
Use your equation to estimate the dose that would give a blood concentration of 4.0 mg/L. Give your answer to 3 significant figures.
(2)
(c)
Explain why the regression line of y on x should not be used for the estimate in part (b).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt at linear regression with y as the independent variable
✓ 1
A1 x = 11.06y + 1.190
✓ 1
Part (b)
M1 substitute y = 4.0
✓ 1
A1 45.4 mg
✓ 1
Part (c)
R1 the line of y on x minimises vertical distances (errors in y), so it is the line for estimating y from x, not x from y
✓ 1
Common error: Rearranging the y on x line to make x the subject gives a different line from the x on y regression whenever the correlation is not perfect.
212IBMA3H-4-19r7-024.3 Measures of central tendency and dispersionPaper 2Easy7 marks
Calculator
The times, in minutes, taken by ten cyclists to complete a hill climb are 31, 34, 35, 36, 38, 39, 41, 42, 44, 70. A value is an outlier if it is more than 1.5 × IQR above the upper quartile or below the lower quartile.
(a)
Find the lower quartile, the upper quartile and the interquartile range.
(3)
(b)
Show that 70 is an outlier, and state whether any other time is an outlier.
(2)
(c)
The outlier is removed. Calculate the mean time of the remaining nine cyclists, to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 Q1 = 35
✓ 1
A1 Q3 = 42
✓ 1
A1 IQR = 7
✓ 1
Part (b)
M1 upper fence = 42 + 1.5 × 7 = 52.5
✓ 1
R1 70 > 52.5 so 70 is an outlier; lower fence is 24.5 and no other value lies outside the fences
Common error: Quartiles are found from the data, not from the fences; use the fences only to test individual values.
213IBMA3H-4-19r7-034.2 Presentation of dataPaper 2Easy7 marks
Calculator
The response time, X milliseconds, of a data-centre server is normally distributed with mean 120 and standard deviation 15.
(a)
Find \(\mathrm{P}(X > 150)\).
(2)
(b)
Find \(\mathrm{P}(100 < X < 130)\).
(2)
(c)
Five per cent of responses take longer than t milliseconds. Find t, to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use normal distribution with mean 120 and standard deviation 15
✓ 1
A1 0.0228
✓ 1
Part (b)
M1 correct bounds 100 and 130 used
✓ 1
A1 0.656
✓ 1
Part (c)
M1 uses inverse normal with probability 0.95 (or 0.05 in the upper tail), or z = 1.6449
✓ 1
M1 t = 120 + 1.6449 × 15
✓ 1
A1 145 ms (144.67)
✓ 1
Exam tip: Sketch the normal curve and shade the required area before entering anything into the calculator.
214IBMA3H-4-19r9-014.2 Presentation of dataPaper 2Easy7 marks
Calculator
A volcano observatory records the height reached by the ash column on each of 11 eruptions, in km, listed in increasing order: 2.1, 2.4, 2.8, 3.0, 3.3, 3.5, 3.9, 4.2, 4.6, 5.1, 9.4 The lower quartile is \(Q_1 = 2.8\) and the upper quartile is \(Q_3 = 4.6\). A reading is classed as an outlier if it is more than \(1.5\times\)IQR above \(Q_3\) or more than \(1.5\times\)IQR below \(Q_1\).
(a)
Show that exactly one reading is an outlier, and identify it.
(3)
(b)
The outlier is removed. Find the mean height of the remaining 10 eruptions.
(2)
(c)
State, with a reason, whether the median of the 11 readings or the mean of the 11 readings changes more when the outlier is removed. (The median of the remaining 10 readings is 3.4 km.)
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 IQR \(= 4.6-2.8 = 1.8\)
✓ 1
M1 upper fence \(= 4.6 + 1.5\times 1.8 = 7.3\)
✓ 1
R1 9.4 > 7.3 so 9.4 is an outlier, and lower fence \(2.8-2.7=0.1\) means no reading is below it, so it is the only outlier
✓ 1
Part (b)
M1 sum of remaining readings \(= 34.9\) (or \(44.3-9.4\))
✓ 1
A1 mean \(= 3.49\) km
✓ 1
Part (c)
R1 the mean changes more (from about 4.03 to 3.49)
✓ 1
R1 median changes only from 3.5 to 3.4, because the median depends on the middle positions, not on how extreme the largest value is
✓ 1
Exam tip: Always show the fence calculation for an outlier test; a bare statement that a value 'looks too big' earns no marks.
215IBMA3H-4-19r9-024.6 Combined, conditional and independent eventsPaper 2Easy6 marks
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A theatre has 600 seats: 360 in the stalls, 150 in the circle and 90 in the balcony. The manager wants the opinions of 40 audience members about the new seating. An assistant stands in the bar during the interval and questions the first 40 people who agree to speak to her.
(a)
Name this sampling method and give one reason why the sample may be biased.
(2)
(b)
Explain why a stratified sample would be more representative of the whole audience.
(1)
(c)
The manager decides to use a stratified sample of 40 people. Find how many should come from the circle and how many from the balcony.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 convenience sampling
✓ 1
R1 e.g. only people who visit the bar are included, so those who stayed in their seats or whose seats are in one section may be under-represented
✓ 1
Part (b)
R1 each seating area is represented in proportion to its size, so every group is included in the correct proportion
✓ 1
Part (c)
M1 circle: \(40\times\frac{150}{600}\)
✓ 1
A1 circle \(=10\)
✓ 1
A1 balcony: \(40\times\frac{90}{600}=6\)
✓ 1
Common error: Stratified sample sizes are proportional to the size of each group in the population, not equal across the groups.
216IBMA3H-4-19r9-034.14 Variance, continuous random variables and linear transformationsPaper 2Easy6 marks
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A shipping company records the number of days, \(X\), by which a cargo ship is delayed leaving port. The probability distribution is given by \(P(X=x)=k(x+2)\) for \(x=0,1,2,3\), where \(k\) is a constant.
(a)
Find the value of \(k\).
(2)
(b)
Find \(E(X)\).
(2)
(c)
Each day of delay costs the company €1200. Find the expected delay cost for one sailing, to the nearest euro. Use the exact value of \(E(X)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(2k+3k+4k+5k=1\)
✓ 1
A1 \(k=\frac{1}{14}\)
✓ 1
Part (b)
M1 \(E(X)=\sum xP(X=x)\) with attempt at substitution, e.g. \(\frac{0\times2+1\times3+2\times4+3\times5}{14}\)
Exam tip: Probabilities in a distribution must sum to 1; use that to find the unknown constant before anything else.
217IBMA3H-4-19r14-014.13 Bayes’ theoremPaper 2Easy6 marks
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A clinic uses a rapid test that gives a false positive for 8% of healthy people. Fifteen healthy people are tested. Let X be the number of false positives.
(a)
State two assumptions needed for X to be modelled by a binomial distribution.
(2)
(b)
Find P(X = 2).
(2)
(c)
Find the probability that at least one of the fifteen people receives a false positive.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 the tests are independent of each other (the result for one person does not affect another)
✓ 1
R1 the probability of a false positive is constant (0.08) for every person
✓ 1
Part (b)
M1 use of X ~ B(15, 0.08) with P(X = 2) = C(15,2)(0.08)2(0.92)13
✓ 1
A1 0.227
✓ 1
Part (c)
M1 P(X ≥ 1) = 1 − P(X = 0) = 1 − 0.9215
✓ 1
A1 0.714
✓ 1
Exam tip: "At least one" is almost always easiest as 1 minus the probability of none.
218IBMA3H-4-19r14-034.10 Regression line of x on yPaper 2Easy6 marks
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A wind turbine's electrical output P (kW) was recorded at six different wind speeds v (m s−1). v: 4, 6, 8, 10, 12, 14 P: 38, 70, 112, 168, 198, 262
Scatter diagram of the data in the table
(a)
Find the Pearson correlation coefficient r.
(1)
(b)
Find the equation of the regression line of P on v.
(2)
(c)
Interpret the value of the gradient in this context.
(1)
(d)
Use the regression line, with the unrounded values from your calculator, to estimate the output at a wind speed of 9 m s−1. Give your answer to the nearest kW.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 r = 0.995
✓ 1
Part (b)
A1 gradient 22.3
✓ 1
A1 P = 22.3v − 59.2 (accept 22.29v − 59.24)
✓ 1
Part (c)
R1 each extra 1 m s−1 of wind speed is associated with about 22.3 kW more output
✓ 1
Part (d)
M1 substitute v = 9 into the regression line
✓ 1
A1 141 kW (accept 142 if rounded coefficients used)
✓ 1
Exam tip: Store the regression coefficients in your calculator rather than retyping rounded values, so that later estimates are accurate.
219IBMA3H-4-19r15-024.1 Sampling, bias and outliersPaper 2Easy6 marks
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A renewable energy company manages 480 wind turbines: 200 onshore and 280 offshore. An engineer wants to inspect a stratified sample of 36 turbines, with the strata being onshore and offshore.
(a)
Find the number of offshore turbines in the sample.
(2)
(b)
State one advantage of a stratified sample over a simple random sample in this situation.
(1)
(c)
A second engineer lists the turbines in order of serial number, picks a random starting point among the first ten and inspects every tenth turbine after it. Name this sampling method.
(1)
(d)
A third engineer inspects only the turbines closest to the access road. Explain why this sample may be biased.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(36\times\frac{280}{480}\)
✓ 1
A1 21
✓ 1
Part (b)
R1 guarantees both onshore and offshore turbines are represented, in proportion to their numbers (accept: reduces the chance of an unrepresentative sample)
✓ 1
Part (c)
A1 systematic sampling
✓ 1
Part (d)
R1 convenience sample: turbines near the road may differ in age, maintenance or exposure from those far away
✓ 1
R1 so the sample is not representative of all turbines and conclusions may be biased
✓ 1
Common error: Do not round each stratum separately without checking the total; here the sample sizes add to the required total.
220IBMA3H-4-19r17-014.11 Conditional probability and independencePaper 2Easy6 marks
Calculator
At a town library, \(A\) is the event that a randomly chosen member borrowed a fiction book last month and \(B\) is the event that the member borrowed an audiobook last month. It is known that \(\mathrm{P}(A)=0.45\), \(\mathrm{P}(B)=0.30\) and \(\mathrm{P}(A\cap B)=0.18\).
Venn diagram showing the information given in the question
(a)
Find \(\mathrm{P}(A\cup B)\).
(2)
(b)
Find \(\mathrm{P}(B\mid A)\).
(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 \(\mathrm{P}(A)+\mathrm{P}(B)-\mathrm{P}(A\cap B)\)
✓ 1
A1 \(0.45+0.30-0.18=0.57\)
✓ 1
Part (b)
M1 use of \(\dfrac{\mathrm{P}(A\cap B)}{\mathrm{P}(A)}\)
✓ 1
A1 \(\dfrac{0.18}{0.45}=0.4\)
✓ 1
Part (c)
M1 compare \(\mathrm{P}(A)\mathrm{P}(B)=0.45\times0.30=0.135\) with \(\mathrm{P}(A\cap B)\), or compare \(\mathrm{P}(B\mid A)\) with \(\mathrm{P}(B)\)
✓ 1
R1 \(0.135\neq0.18\) (or \(0.4\neq0.30\)), so \(A\) and \(B\) are not independent
✓ 1
Exam tip: To test independence, compare P(A∩B) with P(A)P(B); stating only that the events overlap is not a justification.
221IBMA3H-4-19r17-024.2 Presentation of dataPaper 2Easy7 marks
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The 100 m freestyle times, in seconds, of the swimmers in a club are modelled by a normal distribution with mean \(62.4\) and standard deviation \(3.2\). Let \(T\) denote the time of a randomly chosen swimmer.
Sketch of the normal curve, not to scale
(a)
Find \(\mathrm{P}(T<58)\).
(2)
(b)
Find the probability that a swimmer's time is between \(60\) s and \(65\) s.
(2)
(c)
The fastest \(10\%\) of swimmers in the club are selected for a regional squad. Find the slowest time, to 3 significant figures, that qualifies a swimmer for the squad.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use the normal distribution with mean \(62.4\) and standard deviation \(3.2\) (or \(z=\frac{58-62.4}{3.2}=-1.375\))
✓ 1
A1 \(0.0846\)
✓ 1
Part (b)
M1 attempt to find \(\mathrm{P}(60<T<65)\) using the normal distribution
✓ 1
A1 \(0.565\)
✓ 1
Part (c)
M1 \(\mathrm{P}(T<t)=0.1\) or use of inverse normal with \(z=-1.2816\)
✓ 1
M1 \(t=62.4-1.2816\times3.2\) (or inverse normal with mean \(62.4\) and standard deviation \(3.2\))
✓ 1
A1 \(t=58.3\) s
✓ 1
Common error: Selecting the fastest 10% means the lower tail: the required probability to the left of t is 0.1, not 0.9.
222IBMA3H-4-19r19-014.6 Combined, conditional and independent eventsPaper 2Easy5 marks
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An airline carried 1200 passengers on one route last week: 840 in economy, 240 in premium economy and 120 in business class. The airline wants a sample of 60 passengers to answer a satisfaction survey. An interviewer proposes to stand at gate 12 and ask the first 60 passengers who walk past.
(a)
State the name of this sampling method and give one reason why it may produce a biased sample.
(2)
(b)
The airline decides instead to use a stratified sample, with the cabins as the strata. Find the number of passengers sampled from each cabin.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
convenience sampling
✓ 1
only passengers using gate 12 at that time are available, so other flights, times or cabins may be under-represented (accept any valid reason)
✓ 1
Part (b)
M1 sample fraction 60/1200 (= 1/20) applied to each cabin
✓ 1
A1 economy 42 and premium economy 12
✓ 1
A1 business 6
✓ 1
Common error: In a stratified sample the number taken from each stratum is proportional to the stratum size; do not take equal numbers from each.
223IBMA3H-4-19r19-024.2 Presentation of dataPaper 2Easy7 marks
Calculator
A farmer records the wheat yield, in tonnes per hectare, of 12 fields: 5.8, 6.1, 6.4, 6.4, 6.7, 6.9, 7.0, 7.2, 7.3, 7.6, 8.0, 10.2. Find the quartiles as the medians of the lower six and upper six values.
(a)
Find the median, the lower quartile Q1 and the upper quartile Q3.
(3)
(b)
A value is classed as an outlier if it is more than 1.5 × IQR above Q3 or below Q1. Show that 10.2 is an outlier and that there is no outlier at the lower end.
(2)
(c)
The 10.2 is found to come from a field with a different crop, and is removed. Find the mean yield of the remaining 11 fields, to 3 significant figures.
A1 10.2 > 9.025, and lower boundary 6.4 − 1.575 = 4.825 is below 5.8
✓ 1
Part (c)
M1 sum of the 11 values = 75.4 divided by 11
✓ 1
A1 6.85 tonnes per hectare
✓ 1
Exam tip: Write the boundary values in full before comparing them with the data; the comparison is where the marks are.
224IBMA3H-4-19r19-044.3 Measures of central tendency and dispersionPaper 2Easy7 marks
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The number of lambs born to each of the 100 ewes on a farm last spring is summarised below. Lambs per ewe, x: 0, 1, 2, 3, 4 Number of ewes: 4, 31, 48, 14, 3 The 100 ewes form the whole flock.
(a)
Find the mean number of lambs per ewe.
(2)
(b)
Find the standard deviation of the number of lambs per ewe, to 3 significant figures.
(2)
(c)
The farmer is paid €40 for each lamb sold plus €25 for each ewe, so the payment for a ewe is y = 40x + 25 euros. Find the mean and the standard deviation of y.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (0×4 + 1×31 + 2×48 + 3×14 + 4×3)/100
✓ 1
A1 1.81
✓ 1
Part (b)
M1 use of the population standard deviation formula or GDC
✓ 1
A1 0.833
✓ 1
Part (c)
A1 mean = 40 × 1.81 + 25 = €97.40
✓ 1
M1 standard deviation of y = 40 × standard deviation of x (the 25 has no effect)
✓ 1
A1 €33.3
✓ 1
Exam tip: Adding a constant moves the mean but leaves the standard deviation unchanged; multiplying by a constant scales both.
225IBMA3H-4-18r20-014.10 Regression line of x on yPaper 2Easy7 marks
Calculator
A sports scientist records the weekly training time, \(x\) hours, and the 5 km run time, \(y\) minutes, of six club runners.
\(x\) (hours)
4
6
7
9
10
12
\(y\) (minutes)
28.4
27.1
26.8
25.0
24.9
23.2
Scatter diagram of the data in the table
(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 \(x\), with coefficients to 3 significant figures.
(2)
(c)
Interpret the gradient of the line in this context.
(1)
(d)
Estimate the 5 km time of a runner who trains for 8 hours per week.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of technology to find \(r\)
✓ 1
A1 \(r=-0.993\)
✓ 1
Part (b)
M1 use of technology to find the regression line
✓ 1
A1 \(y=-0.643x+31.0\)
✓ 1
Part (c)
R1 each extra hour of weekly training is associated with a decrease of about 0.643 minutes in 5 km time
✓ 1
Part (d)
M1 substitute \(x=8\) into the regression line
✓ 1
A1 25.9 minutes
✓ 1
Exam tip: Interpret a gradient with its units and the words 'on average' or 'associated with'; a strong correlation does not by itself prove that training causes the faster times.
226IBMA3H-4-18r20-024.2 Presentation of dataPaper 2Easy7 marks
Calculator
A rail operator records the number of intermediate stops, \(x\), made by each of 30 services on a route. The frequency of each value is shown in the table, where \(k\) is a positive integer.
\(x\)
0
1
2
3
4
Frequency
2
5
\(k\)
8
5
The mean number of intermediate stops is 2.3.
(a)
Find the value of \(k\).
(3)
(b)
Find the median number of stops.
(2)
(c)
A service is chosen at random. Find the probability that it makes more than 2 intermediate stops.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 total frequency \(20+k\) and \(\sum xf=49+2k\)
✓ 1
M1 \(\dfrac{49+2k}{20+k}=2.3\)
✓ 1
A1 \(k=10\)
✓ 1
Part (b)
M1 30 services, so median lies between the 15th and 16th values; cumulative frequencies 2, 7, 17
✓ 1
A1 median \(=2\)
✓ 1
Part (c)
M1 \(\dfrac{8+5}{30}\)
✓ 1
A1 \(\dfrac{13}{30}\)
✓ 1
Exam tip: With an unknown frequency, remember it appears in both the total frequency and in \(\sum xf\).
227IBMA3H-4-18r20-034.14 Variance, continuous random variables and linear transformationsPaper 2Easy6 marks
Calculator
A telescope detects a short-lived transient event on 35% of the nights it is used. Observations on different nights are independent. The telescope is used on 12 nights. Let \(X\) be the number of nights on which a transient event is detected.
(a)
State the distribution of \(X\).
(1)
(b)
Find \(P(X=5)\), to 3 significant figures.
(2)
(c)
Find \(P(X\le4)\), to 3 significant figures.
(1)
(d)
Find the probability that an event is detected on at least 6 of the nights, to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(X\sim B(12,\,0.35)\)
✓ 1
Part (b)
M1 use of binomial probability, for example \(\binom{12}{5}(0.35)^5(0.65)^7\)
✓ 1
A1 0.204
✓ 1
Part (c)
A1 0.583
✓ 1
Part (d)
M1 \(1-P(X\le5)\)
✓ 1
A1 0.213
✓ 1
Common error: 'At least 6' means \(X\ge6\), so use \(1-P(X\le5)\), not \(1-P(X\le6)\).
228IBMA3H-4-18r21-014.2 Presentation of dataPaper 2Easy6 marks
Calculator
The daily energy output, in MWh, of a wind farm on 11 days was recorded as: 24, 19, 28, 22, 31, 14, 25, 52, 18, 27, 21 Quartiles are to be found as the medians of the lower and upper halves of the ordered data, excluding the median itself.
(a)
Find the median, the lower quartile and the upper quartile.
(3)
(b)
Find the interquartile range.
(1)
(c)
A value is an outlier if it is more than \(1.5\times\) IQR above the upper quartile. Show that 52 MWh is an outlier.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 median \(=24\)
✓ 1
A1 lower quartile \(=19\)
✓ 1
A1 upper quartile \(=28\)
✓ 1
Part (b)
A1 \(28-19=9\) MWh (ECF from (a))
✓ 1
Part (c)
M1 \(28+1.5\times9=41.5\)
✓ 1
A1 \(52>41.5\), so 52 is an outlier
✓ 1
Exam tip: Put the data in order before reading any quartile, and show the outlier boundary as a number before comparing.
229IBMA3H-4-18r21-024.12 Standardization and unknown normal parametersPaper 2Easy7 marks
Calculator
In a health screening programme, the systolic blood pressure \(X\) mmHg of adults is normally distributed with mean 125 and standard deviation 12.
(a)
Find the standardized value (\(z\)-value) for a reading of 143 mmHg.
(2)
(b)
Find the probability that an adult has a reading above 143 mmHg.
(2)
(c)
The highest 10% of readings are referred for further tests. Find the lowest reading that is referred. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{143-125}{12}\)
✓ 1
A1 \(z=1.5\)
✓ 1
Part (b)
M1 attempt to find \(P(X>143)\) using the normal distribution (or \(P(Z>1.5)\))
✓ 1
A1 0.0668
✓ 1
Part (c)
M1 \(z=1.2816\) for an upper tail of 0.1
✓ 1
M1 \(125+12\times1.2816\)
✓ 1
A1 140 mmHg
✓ 1
Common error: For an upper-tail percentage, use the inverse normal with the area to the left, here 0.9, or the negative of the lower-tail \(z\)-value.
230IBMA3H-4-18r21-044.2 Presentation of dataPaper 2Easy7 marks
Calculator
The time \(t\) days taken to recover from an infection was recorded for 40 patients. \(0\leq t<4\): 6 patients \(4\leq t<8\): 18 patients \(8\leq t<12\): 11 patients \(12\leq t<16\): 4 patients \(16\leq t<20\): 1 patient
Histogram of the recovery times of 40 patients
(a)
Write down the cumulative frequencies for the five classes.
(2)
(b)
Write down the class that contains the median.
(2)
(c)
Estimate the mean recovery time.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 6, 24, 35
✓ 1
A1 39, 40
✓ 1
Part (b)
M1 identifies the 20th and 21st values, using \(\frac{40}{2}=20\)
✓ 1
A1 \(4\leq t<8\)
✓ 1
Part (c)
M1 uses class midpoints 2, 6, 10, 14, 18 with frequencies
✓ 1
A1 \(\Sigma ft=12+108+110+56+18=304\)
✓ 1
A1 \(\frac{304}{40}=7.6\) days
✓ 1
Exam tip: Multiply each frequency by the midpoint of its class, not by the class boundary, and divide by the total frequency.
231IBMA3H-4-19r23-064.2 Presentation of dataPaper 2Easy6 marks
Calculator
The time, X seconds, taken by a Formula 1 team to complete a pit stop is normally distributed with mean 2.62 s and standard deviation 0.18 s.
Sketch of the normal curve, not to scale
(a)
Find the probability that a pit stop takes less than 2.4 s.
(2)
(b)
The slowest 2% of pit stops take longer than t seconds. Find the value of t.
(2)
(c)
In a season the team completes 200 pit stops. Estimate the number that take less than 2.4 s, giving your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempts a normal probability calculation, e.g. P(Z < (2.4 − 2.62)/0.18)
✓ 1
A1 0.111
✓ 1
Part (b)
M1 uses the inverse normal with an upper-tail area of 0.02, giving z = 2.054
✓ 1
A1 t = 2.62 + 2.054 × 0.18 = 2.99
✓ 1
Part (c)
M1 200 × P(X < 2.4)
✓ 1
A1 22.2
✓ 1
Common error: For the slowest 2%, the area to the left of t is 0.98, not 0.02; using 0.02 gives a time below the mean.
232IBMA3H-4-19r23-114.10 Regression line of x on yPaper 2Easy6 marks
Calculator
The body temperature, T °C, of healthy adults is normally distributed with mean 36.8 and standard deviation 0.4. A hospital monitor displays the temperature in degrees Fahrenheit, F = 1.8T + 32.
(a)
Find the mean and the standard deviation of F.
(4)
(b)
Find the probability that the monitor displays a temperature above 99.5 °F. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 E(F) = 1.8 × 36.8 + 32
✓ 1
A1 mean = 98.24
✓ 1
M1 standard deviation of F = 1.8 × 0.4 (the +32 does not change the spread)
✓ 1
A1 standard deviation = 0.72
✓ 1
Part (b)
M1 uses F ~ N(98.24, 0.72²) with P(F > 99.5), i.e. z = 1.75
✓ 1
A1 0.0401
✓ 1
Exam tip: A linear transformation of a normal variable is again normal; find its new mean and standard deviation first, then use them directly.
233IBMA3H-4-19r24-024.6 Combined, conditional and independent eventsPaper 2Easy6 marks
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A water company supplies 9000 households in four districts: A (2400 households), B (3600), C (1800) and D (1200). It wants a stratified sample of 150 households to test water pressure at the tap.
(a)
Calculate the number of households to be sampled from each district.
(3)
(b)
State one advantage of stratified sampling over simple random sampling in this situation.
(1)
(c)
Instead, the company tests the first 150 households that phone to report low pressure. Name this type of sampling and give one reason why it may be biased.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sampling fraction \(\frac{150}{9000}\) (or \(\frac{1}{60}\)) used on the district sizes
✓ 1
A1 A: 40 and B: 60
✓ 1
A1 C: 30 and D: 20
✓ 1
Part (b)
R1 every district is represented in proportion to its size, so a small district cannot be missed by chance
✓ 1
Part (c)
A1 convenience sampling
✓ 1
R1 only households already experiencing problems are included, so the sample does not represent all households
✓ 1
Exam tip: In sampling questions, say what the sample is compared with: a sample is biased when some members of the population are more likely to be chosen than others.
234IBMA3H-4-19r24-034.2 Presentation of dataPaper 2Easy7 marks
Calculator
The times, in minutes, taken by 12 players to complete a game level are, in order: 14, 17, 18, 19, 21, 22, 24, 25, 27, 30, 31, 58.
(a)
Find the median, the lower quartile \(Q_1\) and the upper quartile \(Q_3\). (Take \(Q_1\) and \(Q_3\) as the medians of the lower and upper six values.)
(3)
(b)
Show that 58 is an outlier, using the rule that an outlier is greater than \(Q_3+1.5\times\text{IQR}\).
(2)
(c)
The outlier is removed. Find the mean of the remaining 11 times, to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 median \(=23\)
✓ 1
A1 \(Q_1=18.5\)
✓ 1
A1 \(Q_3=28.5\)
✓ 1
Part (b)
M1 \(\text{IQR}=28.5-18.5=10\) and \(Q_3+1.5\times10\)
✓ 1
R1 \(43.5<58\), so 58 is an outlier
✓ 1
Part (c)
M1 \(\dfrac{306-58}{11}\) (or sum of remaining times \(=248\))
✓ 1
A1 \(22.5\) minutes
✓ 1
Common error: Quartiles from a calculator can differ slightly from hand methods for small data sets; state the method you use, and apply the 1.5 × IQR rule with the same quartiles throughout.
235IBMA3H-4-19r24-084.1 Sampling, bias and outliersPaper 2Easy6 marks
Calculator
A game studio has 5000 registered players with IDs numbered 1 to 5000. It selects 200 players by systematic sampling: the first player selected has ID 7, and every \(k\)th ID after that is selected.
(a)
Find the value of \(k\).
(2)
(b)
Premium accounts are exactly those whose IDs are multiples of 25. Explain why the sample is biased.
(2)
(c)
Find the ID of the 200th player selected.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\dfrac{5000}{200}\)
✓ 1
A1 \(k=25\)
✓ 1
Part (b)
A1 every selected ID has the form \(25n+7\), so no premium account can be selected
✓ 1
R1 so the sample is not representative of all players
✓ 1
Part (c)
M1 \(7+25\times199\)
✓ 1
A1 \(4982\)
✓ 1
Exam tip: In systematic sampling, the \(n\)th selected item is the first plus \((n-1)\) intervals; check that the last value does not exceed the population size.
236IBMA3H-4-18r25-014.2 Presentation of dataPaper 2Easy5 marks
Calculator
An archaeologist records the depth, in cm, below the surface at which eight pottery shards were found: 14, 17, 18, 21, 23, 26, 29, 32
(a)
Calculate the mean depth.
(2)
(b)
Use your calculator to find the population standard deviation of the depths, to 3 significant figures.
M1 use of the statistics function of the calculator (population standard deviation σ)
✓ 1
A1 σ = 5.81 cm
✓ 1
Part (c)
A1 median = \(\frac{21+23}{2}\) = 22 cm
✓ 1
Common error: Calculators show two standard deviations: use the population one (σn) when the data are the whole group described, and the sample one only when asked for an unbiased estimate.
237IBMA3H-4-18r25-034.4 Correlation and regression of y on xPaper 2Easy5 marks
Calculator
A data centre records the number of active servers, x, in a rack and the power, y kW, used by the rack's cooling system. For 10 ≤ x ≤ 40, the regression line of y on x is y = 0.35x + 1.8 and the correlation coefficient is r = 0.96.
(a)
Describe the correlation between x and y.
(1)
(b)
Interpret the value 0.35 in this context.
(1)
(c)
Use the regression line to estimate the cooling power for a rack with 26 active servers.
(2)
(d)
A manager uses the line to estimate the cooling power for 80 active servers. Comment on the reliability of this estimate.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 strong positive (linear) correlation
✓ 1
Part (b)
A1 each additional active server increases the cooling power by 0.35 kW, on average
✓ 1
Part (c)
M1 substitute x = 26: 0.35(26) + 1.8
✓ 1
A1 y = 10.9 kW
✓ 1
Part (d)
R1 unreliable: 80 lies outside the range of the data (10 to 40), so this is extrapolation
✓ 1
Common error: When interpreting a gradient, always say 'on average' and give the change in y for each one-unit increase in x, with units.
238IBMA3H-4-18r26-014.6 Combined, conditional and independent eventsPaper 2Easy6 marks
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A library has 2400 registered members: 540 juniors, 1260 adults and 600 seniors. The librarian wants a sample of 120 members to answer a survey about opening hours.
(a)
Calculate how many members of each age group should be in a stratified sample of 120.
(3)
(b)
Give one reason why stratified sampling is more suitable here than simple random sampling.
(1)
(c)
Instead, the librarian asks the first 50 people who enter on a Tuesday between 10:00 and 12:00. Name this sampling method and state one source of bias.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 proportion, e.g. 540/2400 × 120
✓ 1
A1 juniors 27 and adults 63
✓ 1
A1 seniors 30
✓ 1
Part (b)
R1 e.g. guarantees every age group is represented in proportion to its size, because age groups may have different opening-hours needs
✓ 1
Part (c)
A1 convenience (or opportunity) sampling
✓ 1
R1 e.g. people who visit on weekday mornings are not representative of members who can only visit at other times
✓ 1
Exam tip: Proportional allocation: sample in each stratum = (stratum size ÷ population size) × total sample size.
239IBMA3H-4-18r26-024.12 Standardization and unknown normal parametersPaper 2Easy7 marks
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The length L metres of rotor blades made at a factory is normally distributed with mean 62.4 and standard deviation 0.8.
Sketch of the normal curve, not to scale
(a)
Find the probability that a blade is longer than 63.5 m.
(2)
(b)
Find the probability that a blade is between 61 m and 63.5 m long.
(2)
(c)
The longest 10% of blades are sent to an offshore site. Find the minimum length of a blade sent offshore, giving your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 standardise: z = (63.5 − 62.4)/0.8 = 1.375, or use GDC normal CDF
✓ 1
A1 0.0846
✓ 1
Part (b)
M1 correct normal probability setup for 61 < L < 63.5
✓ 1
A1 0.875
✓ 1
Part (c)
M1 P(L < c) = 0.9
✓ 1
A1 z = 1.2816
✓ 1
A1 63.4 m
✓ 1
Common error: Using the inverse normal function, convert a 'top 10%' statement to a lower-tail probability of 0.9 before you start.
240IBMA3H-4-18r26-054.4 Correlation and regression of y on xPaper 2Easy10 marks
Calculator
The table shows the maximum daily temperature x (°C) and the number of people y who used an outdoor swimming pool on seven days. x: 12, 15, 18, 21, 24, 27, 30 y: 38, 45, 61, 66, 84, 92, 103
(a)
Find the Pearson's product-moment correlation coefficient, r, to 3 significant figures.
(1)
(b)
Describe the correlation between x and y.
(1)
(c)
Find the equation of the regression line of y on x, in the form y = ax + b, with a and b to 3 significant figures.
(2)
(d)
Interpret the value of a in this context.
(1)
(e)
Use the regression line to estimate the number of people on a day when the maximum temperature is 20 °C. Give your answer to the nearest whole number.
(2)
(f)
A manager uses the line to predict the number of users when the maximum temperature is 45 °C. Find this prediction and comment on its reliability.
(2)
(g)
A colleague says the high value of r proves that warm weather causes more people to use the pool. Comment on this claim.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 r = 0.995
✓ 1
Part (b)
A1 very strong positive (linear) correlation
✓ 1
Part (c)
A1 a = 3.71
✓ 1
A1 b = −8.14
✓ 1
Part (d)
R1 for each 1 °C rise in maximum temperature, about 3.71 more people use the pool
✓ 1
Part (e)
M1 substitute x = 20
✓ 1
A1 66
✓ 1
Part (f)
A1 159
✓ 1
R1 45 °C is outside the range of the data (extrapolation), so the prediction is unreliable
✓ 1
Part (g)
R1 correlation does not by itself prove causation; other factors (e.g. school holidays) could also be involved
✓ 1
Exam tip: Always state which variable is which when interpreting a gradient: 'for each extra unit of x, y changes by a'.
241IBMA3H-4-18r26-084.8 Binomial distributionPaper 2Easy6 marks
Calculator
Gearboxes in a wind farm are checked by a vibration test. 3% of gearboxes have a fault. If a gearbox has a fault, the test shows 'positive' with probability 0.96. If a gearbox has no fault, the test shows 'positive' with probability 0.05.
Tree diagram showing the probabilities given in the question
(a)
Find the probability that a randomly chosen gearbox tests positive.
(3)
(b)
Given that a gearbox tests positive, find the probability that it has a fault.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of a tree or the law of total probability
✓ 1
A1 0.03 × 0.96 = 0.0288 and 0.97 × 0.05 = 0.0485
✓ 1
A1 P(positive) = 0.0773
✓ 1
Part (b)
M1 conditional probability formula or Bayes' theorem
✓ 1
A1 0.0288/0.0773
✓ 1
A1 0.373
✓ 1
Common error: Write 'given' probabilities as P(A | B) so you can see which event goes in the denominator.
242IBMA3H-4-20r28-014.2 Presentation of dataPaper 2Easy7 marks
Calculator
The heights, in km, of the ash columns produced by eight eruptions at a volcano are, in increasing order: 2.1, 2.8, 3.0, 3.4, 3.9, 4.6, 5.2, 12.1 For this question, take the lower quartile \(Q_1\) and upper quartile \(Q_3\) to be the medians of the lower four and upper four values respectively.
(a)
Calculate the mean height. Give your answer to 3 significant figures. [2]
(2)
(b)
Find the median and the interquartile range. [3]
(3)
(c)
A value is classed as an outlier if it is greater than \(Q_3+1.5\times\text{IQR}\). Show that 12.1 km is an outlier. [2]
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{37.1}{8}\)
✓ 1
A1 4.64 km
✓ 1
Part (b)
A1 median \(=\frac{3.4+3.9}{2}=3.65\) km
✓ 1
A1 \(Q_1=2.9\) and \(Q_3=4.9\)
✓ 1
A1 IQR \(=2.0\) km
✓ 1
Part (c)
M1 \(4.9+1.5\times2.0\) (ECF from (b))
✓ 1
R1 threshold is 7.9 km and \(12.1>7.9\), so 12.1 is an outlier
✓ 1
Exam tip: Show the numerical threshold and then compare it with the data value in a sentence. A bare statement that the value is an outlier gains nothing.
243IBMA3H-4-20r28-024.2 Presentation of dataPaper 2Easy6 marks
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The running time, X minutes, of the evening performances at a theatre is normally distributed with mean 142 and standard deviation 6.
(a)
Find \(\mathrm{P}(X>150)\). [2]
(2)
(b)
Find \(\mathrm{P}(138<X<145)\). [2]
(2)
(c)
Ten percent of performances are longer than t minutes. Find t, correct to 1 decimal place. [2]
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 standardise, \(z=\frac{150-142}{6}=1.33\), or use normal cdf
✓ 1
A1 0.0912
✓ 1
Part (b)
M1 normal cdf with limits 138 and 145
✓ 1
A1 0.439
✓ 1
Part (c)
M1 \(z=1.2816\) for the upper 10% and \(\frac{t-142}{6}=1.2816\)
✓ 1
A1 \(t=149.7\)
✓ 1
Common error: For \"the top 10%\" the area to the left of t is 0.9. Using 0.1 in the inverse normal function gives a value below the mean.
244IBMA3H-4-20r28-044.11 Conditional probability and independencePaper 2Easy6 marks
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A shipping company knows that, on a particular route, the probability of a storm delay is 0.3 and the probability of port congestion is 0.4. The two events are independent. Let S be the event that there is a storm delay and C the event that there is port congestion.
(a)
Find \(\mathrm{P}(S\cap C)\). [1]
(1)
(b)
Find the probability that at least one of the two problems occurs. [2]
(2)
(c)
Given that at least one of the two problems occurs, find the probability that there is a storm delay. Give your answer to 3 significant figures. [3]
A1 \(\frac{0.3}{0.58}\) (since \(S\) lies within \(S\cup C\))
✓ 1
A1 0.517
✓ 1
Common error: For independent events the union is not \(\mathrm{P}(S)+\mathrm{P}(C)\). Subtract the intersection to avoid counting the overlap twice.
245IBMA3H-4-19r32-024.8 Binomial distributionPaper 2Easy6 marks
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A mobile network sends a large file as separate data packets. Each packet independently fails to arrive with probability 0.08. A file consists of 25 packets. Let X be the number of packets that fail to arrive.
(a)
Write down the distribution of X, and state one assumption needed for this model to be valid.
(2)
(b)
Find the probability that exactly two packets fail to arrive.
(2)
(c)
Find the probability that at least one packet fails to arrive.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 X ~ B(25, 0.08)
✓ 1
R1 packets fail independently of each other (or constant probability of failure for each packet)
✓ 1
Part (b)
M1 attempt to use binomial probability, e.g. C(25,2) × 0.08² × 0.9223
✓ 1
A1 P(X = 2) = 0.282
✓ 1
Part (c)
M1 P(X ≥ 1) = 1 − P(X = 0) = 1 − 0.9225
✓ 1
A1 = 0.876
✓ 1
Common error: 'At least one' is the complement of 'none': calculate 1 − P(X = 0) rather than adding many terms.
246IBMA3H-4-19r32-034.9 The normal distributionPaper 2Easy7 marks
Calculator
A food scientist is testing fruit bars. The sugar content, S grams, of a bar is normally distributed with mean 32.0 g and standard deviation 1.5 g.
(a)
Find P(S < 30).
(2)
(b)
Find the probability that a randomly chosen bar has a sugar content between 31 g and 34 g.
(2)
(c)
A batch contains 3000 bars. Find the expected number of bars in the batch with a sugar content greater than 35 g. Use the probability of 0.0228 (3 significant figures) in your calculation.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use the normal distribution, e.g. normalCdf with mean 32 and standard deviation 1.5
✓ 1
A1 0.0912
✓ 1
Part (b)
M1 P(31 < S < 34)
✓ 1
A1 0.656
✓ 1
Part (c)
M1 attempt to find P(S > 35)
✓ 1
A1 P(S > 35) = 0.0228
✓ 1
A1 expected number = 3000 × 0.0228 = 68.4
✓ 1
Exam tip: Sketch the normal curve and shade the required region before using your calculator, to check the probability is sensible.
247IBMA3H-4-18r33-034.13 Bayes’ theoremPaper 2Easy5 marks
Calculator
An architect arranges a survey of a building's foundations. The probability that a building has a foundation fault is 0.08. If a fault is present, the survey detects it with probability 0.95. If there is no fault, the survey wrongly reports a fault with probability 0.10.
(a)
Find the probability that the survey does not report a fault.
(2)
(b)
The survey does not report a fault. Find the probability that the building does have a fault. Give your answer to three significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.08 × 0.05 + 0.92 × 0.90 (both branches)
✓ 1
A1 0.832
✓ 1
Part (b)
M1 P(fault | no report) = P(fault ∩ no report) / P(no report)
✓ 1
A1 0.004 / 0.832 (ECF from (a))
✓ 1
A1 0.00481
✓ 1
Common error: 'Fault not detected' has probability 0.05 when a fault exists; do not confuse it with the 0.10 false-alarm probability.
248IBMA3H-4-18r33-084.9 The normal distributionPaper 2Easy5 marks
Calculator
Steel beams for a footbridge are cut to a nominal length of 3.50 m. The actual length of a beam, in metres, is normally distributed with mean 3.50 and standard deviation 0.02. A beam is accepted if its length is between 3.47 m and 3.53 m.
Sketch of the normal curve, not to scale
(a)
Write down the standardised value z corresponding to a length of 3.53 m.
(1)
(b)
Find the probability that a beam is accepted, to three significant figures.
(2)
(c)
A batch contains 500 beams. Using the unrounded probability from (b), estimate the number that are rejected, to the nearest whole number.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 z = 1.5
✓ 1
Part (b)
M1 P(3.47 < L < 3.53) = P(−1.5 < Z < 1.5)
✓ 1
A1 0.866
✓ 1
Part (c)
M1 500 × (1 − P(accepted))
✓ 1
A1 67
✓ 1
Common error: Remember that a symmetric interval about the mean has the same probability in each tail; rejected beams come from both ends.
249IBMA3H-4-18r33-094.2 Presentation of dataPaper 2Easy7 marks
Calculator
A logistics firm records the delivery times, in minutes, of eight parcels to a remote area. 22, 25, 27, 30, 31, 34, 36, 71
(a)
Find the lower quartile and the upper quartile.
(2)
(b)
A delivery time is an outlier if it is greater than Q3 + 1.5 × IQR. Find the greatest whole number of minutes that is not an outlier.
(3)
(c)
The delivery time of 71 minutes is removed from the data. Find the mean of the remaining times, to three significant figures.
Exam tip: When the fence is not a whole number, the greatest whole-number value that is not an outlier is the whole number just below it.
250IBMA3H-4-19r37-014.3 Measures of central tendency and dispersionPaper 2Easy9 marks
Calculator
A space probe records the temperature of a sensor, in °C, at eight times during its cruise phase: −41, −38, −36, −36, −33, −30, −27, −25.
(a)
Find the mean, the median and the mode of the eight readings.
(4)
(b)
A ninth reading x is recorded. The mean of all nine readings is −34. Find x.
(2)
(c)
Each of the original eight readings is converted to kelvin by adding 273. Find the new mean of these eight readings, and state what happens to their standard deviation.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to divide the total (−266) by 8
✓ 1
A1 mean = −33.25
✓ 1
A1 median = −34.5
✓ 1
A1 mode = −36
✓ 1
Part (b)
M1 9 × (−34) − (−266) or equivalent equation
✓ 1
A1 x = −40
✓ 1
Part (c)
M1 −33.25 + 273
✓ 1
A1 mean = 239.75 K
✓ 1
A1 the standard deviation is unchanged
✓ 1
Exam tip: Adding a constant to every value shifts the mean, median and mode by that constant but leaves every measure of spread unchanged.
251IBMA3H-4-19r37-024.2 Presentation of dataPaper 2Easy7 marks
Calculator
A coffee roaster records the roasting time, in minutes, of 30 batches of beans.
Time (minutes): 10, 11, 12, 13, 14 Number of batches: 3, 8, 12, 5, 2
(a)
Find the mean roasting time, correct to 3 significant figures.
(2)
(b)
Find the population standard deviation of the roasting times, correct to 3 significant figures.
(2)
(c)
Two batches are chosen at random, without replacement, from the 30. Find the probability that both batches took more than 12 minutes. Give your answer as an exact fraction or correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to find Σfx ÷ 30
✓ 1
A1 mean = 11.8 minutes
✓ 1
Part (b)
M1 use of GDC statistics or correct formula
✓ 1
A1 standard deviation = 1.04 minutes
✓ 1
Part (c)
M1 P(>12 minutes) = 7/30
✓ 1
M1 multiplying 7/30 by 6/29
✓ 1
A1 7/145 = 0.0483
✓ 1
Common error: When sampling without replacement, the second probability must use the reduced counts (6 out of 29), not the original 7 out of 30.
252IBMA3H-4-19r37-044.2 Presentation of dataPaper 2Easy8 marks
Calculator
The masses, in grams, of 40 bags of roasted coffee are summarised by a lower quartile of 248 g and an upper quartile of 256 g. The mean mass of the 40 bags is 252.05 g. A bag is classed as an outlier if its mass is more than 1.5 × IQR below the lower quartile or above the upper quartile.
(a)
Find the interquartile range.
(1)
(b)
The four most extreme masses in the sample are 233 g, 241 g, 262 g and 273 g. Determine which of these are outliers.
(4)
(c)
The outliers are removed from the sample. Find the mean mass of the remaining bags.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 IQR = 8 g
✓ 1
Part (b)
M1 1.5 × IQR = 12
✓ 1
A1 lower fence 236 and upper fence 268
✓ 1
A1 233 g is an outlier
✓ 1
A1 273 g is an outlier, and 241 g and 262 g are not
✓ 1
Part (c)
M1 total mass of all bags = 40 × 252.05 = 10082
✓ 1
M1 subtract 233 and 273 and divide by 38
✓ 1
A1 mean = 252 g
✓ 1
Common error: Remember to subtract the removed values from the total before dividing by the new number of values, 38 and not 40.
253IBMA3H-4-20r44-014.8 Binomial distributionPaper 2Easy5 marks
Calculator
A theatre sells 25 seats for a performance. Each ticket holder independently attends with probability 0.9. Let \(X\) be the number of ticket holders who attend.
(a)
State the distribution of \(X\) and find the probability that all 25 ticket holders attend. Give your answer to 3 significant figures.
(2)
(b)
Find the probability that at least 23 ticket holders attend. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 X ~ B(25, 0.9)
✓ 1
A1 0.925 = 0.0718
✓ 1
Part (b)
M1 P(X ≥ 23) = P(X = 23) + P(X = 24) + P(X = 25)
✓ 1
M1 correct use of the binomial pdf or cdf (e.g. 1 − P(X ≤ 22))
✓ 1
A1 0.537
✓ 1
Common error: \(P(X\ge 23)\) is \(1-P(X\le 22)\), not \(1-P(X\le 23)\).
254IBMA3H-4-20r44-024.10 Regression line of x on yPaper 2Easy5 marks
Calculator
A volcano observatory records the tremor amplitude \(x\) (in arbitrary units) and the sulfur dioxide flux \(y\) (in tonnes per day) on six days.
x
2.1
3.4
4.0
5.2
6.3
7.5
y
160
190
265
285
390
405
(a)
Find the Pearson correlation coefficient \(r\), to 3 significant figures, and describe the correlation.
(2)
(b)
Find the equation of the regression line of \(y\) on \(x\), giving the coefficients to 3 significant figures.
(2)
(c)
Use the unrounded regression line to estimate the flux when the amplitude is 4.5. Give your answer to the nearest tonne per day.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 r = 0.973
✓ 1
A1 strong positive linear correlation
✓ 1
Part (b)
A1 gradient 49.4
✓ 1
A1 y = 49.4x + 47.8
✓ 1
Part (c)
A1 270 tonnes per day
✓ 1
Exam tip: Name the regression line 'of \(y\) on \(x\)' when you quote it: the line of \(x\) on \(y\) is a different line and should not be used to predict \(y\).
255IBMA3H-4-20r46-024.10 Regression line of x on yPaper 2Easy6 marks
Calculator
A farmer records the rainfall during the growing season, \(x\) mm, and the wheat yield, \(y\) tonnes per hectare, on six fields.
Find the equation of the regression line of \(y\) on \(x\). Give the coefficients to 3 significant figures.
(3)
(b)
Interpret the gradient of the line in context.
(1)
(c)
Use your line to estimate the yield for a field with 75 mm of rainfall. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of a GDC or formulae for regression of \(y\) on \(x\)
✓ 1
A1 gradient 0.0554
✓ 1
A1 intercept 0.858, so \(y=0.0554x+0.858\)
✓ 1
Part (b)
A1 each extra millimetre of rainfall is associated with an increase of about 0.0554 tonnes per hectare in yield
✓ 1
Part (c)
M1 substitute \(x=75\) (unrounded or rounded coefficients both accepted)
✓ 1
A1 5.01 tonnes per hectare
✓ 1
Common error: Describe the gradient as a change in \(y\) per unit of \(x\), in context, rather than just quoting 'the gradient is 0.0554'.
256AAHL4-q00034.1 Sampling, bias and outliersPaper 2Medium10 marks
The times, in minutes, of 14 bus journeys on the same route were recorded.
Journey
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Time (min)
12
14
15
15
16
17
17
18
19
20
21
23
29
52
Use the quartiles given by your GDC. An outlier is a value more than \(1.5\times\) IQR from the nearest quartile.
(a)
Find the lower quartile and the upper quartile of the times.
(2)
(b)
Show that 52 is an outlier but 29 is not.
(2)
(c)
The time of 52 minutes is removed from the data. Show that 29 is now an outlier.
(2)
(d)
Find the mean of the times before and after 52 is removed, and state, with a reason, whether the mean or the median is more affected by its removal.
(3)
(e)
The 52-minute journey took place on a day when a road was closed. Comment on whether the analyst should have removed it, and on whether 29 should then also be removed.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(Q_{1}=15\)
A1
\(Q_{3}=21\)
A1
(b)
\(\text{IQR}=6\); upper boundary \(21+1.5(6)=30\)
M1
\(52\gt 30\) so 52 is an outlier; \(29\lt 30\) so 29 is not
A1
(c)
new quartiles \(Q_{1}=15,\ Q_{3}=20.5\), IQR \(=5.5\)
M1
upper boundary \(20.5+1.5(5.5)=28.75\lt 29\), so 29 is now an outlier
A1
(d)
mean before \(=\dfrac{288}{14}=20.6\) min
A1
\(20.571\ldots\)
mean after \(=\dfrac{236}{13}=18.2\) min
A1
\(18.153\ldots\)
the mean is more affected (it falls by about 2.4 min, whereas the median only changes from 17.5 to 17)
R1
(e)
52 is a genuine value (a real journey) so it may be kept, or analysed separately with the reason stated; repeatedly removing the new largest value is not valid, so 29 should not be removed
R1
Accept any reasoned comment that distinguishes a genuine outlier from an error and rejects automatic repeated removal.
Answers: (a) \(Q_{1}=15,\ Q_{3}=21\) · (d) 20.6 min and 18.2 min
257AAHL4-q00074.2 Presentation of dataPaper 2Medium15 marks
The finishing times, \(t\) minutes, of 120 runners in a 5 km run are summarized below.
Time, \(t\) (min)
\(15\le t\lt 20\)
\(20\le t\lt 25\)
\(25\le t\lt 30\)
\(30\le t\lt 35\)
\(35\le t\lt 40\)
\(40\le t\lt 45\)
Number of runners
8
22
35
28
17
10
Histogram of the data in the table (equal class widths)
(a)
Write down the modal class.
(1)
(b)
Find an estimate of the mean and an estimate of the standard deviation of the times.
(3)
(c)
Construct a cumulative frequency table and use it, assuming the times are evenly spread within each class, to estimate the median time.
(3)
(d)
Estimate the number of runners who finished in less than 33 minutes.
(2)
(e)
From the raw data, the fastest time was 15.8 minutes, the slowest time 44.9 minutes, the lower quartile 24.2 minutes, the median 29.4 minutes and the upper quartile 34.6 minutes. Show that there are no outliers.
(3)
(f)
Use the five-number summary in part (e) to comment on whether the times could be normally distributed.
(1)
(g)
It is later found that every recorded time was 0.5 minutes too long. Write down the corrected estimates of the mean and the standard deviation.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(25\le t\lt 30\)
A1
(b)
mid-interval values 17.5, 22.5, …, 42.5 with the frequencies
60th value lies in \(25\le t\lt 30\): \(25+\dfrac{60-30}{35}\times5\)
(M1)
\(29.3\) min
A1
\(29.2857\ldots\) Accept from 29.2 to 29.4 (e.g. 60.5th value).
(d)
\(65+\dfrac{3}{5}\times28\)
(M1)
\(82\) runners
A1
\(81.8\)
(e)
IQR \(=34.6-24.2=10.4\); \(1.5\times10.4=15.6\)
M1
boundaries \(24.2-15.6=8.6\) and \(34.6+15.6=50.2\)
A1
\(15.8\gt 8.6\) and \(44.9\lt 50.2\), so there are no outliers
R1
(f)
the box is symmetric about the median (5.2 each side) and the whiskers are similar in length (8.4 and 10.3), so a normal distribution is plausible
R1
(g)
mean \(=29.25\) min
A1
FT from (b).
standard deviation unchanged, \(6.70\) min
A1
Answers: (a) \(25\le t\lt 30\) · (b) 29.75 min, 6.70 min · (c) 29.3 min · (d) 82 · (g) 29.25 min, 6.70 min
258AAHL4-q00134.4 Correlation and regression of y on xPaper 2Medium10 marks
The table shows the number of public electric-vehicle charging points, \(y\) thousand, in a region in the year \(t\). An analyst codes the years using \(u=\dfrac{t-2008}{2}\).
Year, \(t\)
2008
2010
2012
2014
2016
2018
2020
2022
\(y\) (thousands)
3.1
4.0
5.2
5.9
7.3
8.1
9.0
10.4
Scatter diagram of the data in the table
(a)
Find the equation of the regression line of \(y\) on \(u\).
(2)
(b)
Write down the value of Pearson’s product-moment correlation coefficient, \(r\), between \(y\) and \(t\).
(1)
(c)
Hence find the equation of the regression line of \(y\) on \(t\), in the form \(y=at+b\).
(3)
(d)
Interpret the value of \(a\) in context, and explain why \(b\) has no practical meaning.
(2)
(e)
Use your regression line to predict the number of charging points in 2030, and comment on the reliability of this prediction.
(2)
Show mark scheme
Working
Mark
Notes
(a)
gradient \(1.03\) (\(1.02619\ldots\))
A1
\(y=1.03u+3.03\)
A1
Intercept \(3.03333\ldots\)
(b)
\(r=0.998\)
A1
\(0.998126\ldots\); coding \(t\) linearly with a positive factor does not change \(r\).
\(-1027.26\ldots\) Accept \(-1027\) or \(-1028\); \(b\) is very sensitive to rounding of \(a\), so full accuracy must be carried. Also accept direct GDC regression on \(t\).
(d)
the number of charging points increases by about 513 per year
R1
\(b\) is the predicted value in year 0, far outside the data (and negative), so it is meaningless
R1
(e)
\(u=11\): \(y=14.3\), about 14 300 charging points
A1
\(14.32\ldots\)
unreliable: 2030 is outside the range of the data (extrapolation); growth may not stay linear
259AAHL4-q00154.4 Correlation and regression of y on xPaper 2Medium15 marks
A grower investigates the effect of a fertilizer on tomato plants. Eight plots are treated with different amounts of fertilizer, \(x\) grams per square metre, and the mean yield, \(y\) kg per plant, is recorded.
\(x\) (g m−2)
10
20
30
40
50
60
70
80
\(y\) (kg)
2.6
3.1
3.0
3.7
3.9
4.5
4.4
5.0
Scatter diagram of the data in the table
(a)
Find Pearson’s product-moment correlation coefficient, \(r\), and describe the correlation between \(x\) and \(y\).
(2)
(b)
Find the equation of the regression line of \(y\) on \(x\), in the form \(y=ax+b\).
(2)
(c)
Interpret the value of \(a\) in context.
(1)
(d)
Use your regression line to estimate the mean yield per plant when 35 g m−2 of fertilizer is used.
(2)
(e)(i)
Each square metre contains 3 plants. Tomatoes are sold for €1.20 per kg and fertilizer costs €0.08 per gram. Let \(V(x)\) be the value of the predicted yield of one square metre minus the cost of the fertilizer used on it, in euros.
Show that \(V(x)=0.04x+8.19\).
(2)
(e)(ii)
The grower decides to use the amount of fertilizer, within the range of the data, that maximizes \(V(x)\). State this amount, and explain why the grower should not conclude that using even more fertilizer would be better.
(2)
(f)
When the amount found in part (e)(ii) is used, the yield of a plant is normally distributed with mean equal to the value predicted by the regression line and standard deviation 0.6 kg. Find the probability that a randomly chosen plant yields less than 4 kg.
(2)
(g)
Ten such plants are chosen at random. Find the probability that at most one of them yields less than 4 kg.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(r=0.978\)
A1
\(0.978399\ldots\)
strong positive (linear) correlation
A1
(b)
\(a=0.0333\)
A1
\(0.03333\ldots\)
\(b=2.28\); \(y=0.0333x+2.28\)
A1
\(2.275\) Award A1A0 if not written as an equation.
(c)
for each extra gram per square metre of fertilizer, the mean yield increases by about 0.0333 kg per plant
A1
Must refer to an increase of 1 unit in x.
(d)
\(0.033333(35)+2.275\)
(M1)
\(3.44\) kg
A1
\(3.44166\ldots\)
(e)(i)
\(V(x)=3\times1.20\times(0.033333x+2.275)-0.08x\)
M1
\(=0.12x+8.19-0.08x\)
A1
\(3.6\times\tfrac{1}{30}=0.12\) and \(3.6\times2.275=8.19\).
\(V(x)=0.04x+8.19\)
AG
(e)(ii)
\(80\) g m−2 (the largest amount in the data, since \(V\) is increasing)
A1
the linear model is only supported for 10 ≤ x ≤ 80; beyond this it is extrapolation and the yield may level off (or fall), so V may not keep increasing
R1
(f)
\(Y\sim\mathrm{N}(4.94167,\,0.6^{2})\), where \(\tfrac{80}{30}+2.275=4.94166\ldots\); \(\mathrm{P}(Y\lt4)\)
(M1)
\(0.0583\)
A1
\(0.0582722\ldots\) Accept 0.0582–0.0586 from a mean rounded to 4.94.
260AAHL4-q00234.6 Combined, conditional and independent eventsPaper 2Medium7 marks
A contestant answers three questions in a quiz. The probability that she answers the first question correctly is 0.7. For each later question, the probability of a correct answer is 0.8 if the previous answer was correct and 0.5 if the previous answer was wrong.
Tree diagram with the probabilities given in the question (C: correct, W: wrong)
(a)
Find the probability that she answers all three questions correctly.
(2)
(b)
Find the probability that she answers exactly two of the questions correctly.
(3)
(c)
Given that she answers exactly two questions correctly, find the probability that she answered the first question correctly.
261AAHL4-q00344.8 Binomial distributionPaper 2Medium8 marks
Each day a machine makes 20 components. Each component is defective with probability 0.06, independently of all others. A day is classed as “bad” if 3 or more of that day’s components are defective.
(a)
Find the probability that a given day is bad.
(2)
(b)
Find the probability that, in a five-day week, there are at least two bad days.
(3)
(c)
Find the expected number of bad days in a year of 250 working days.
(2)
(d)
State one assumption needed for your answer to part (b).
whether a day is bad is independent of other days (and the defect rate stays at 0.06 every day)
R1
Answers: (a) 0.115 · (b) 0.104 · (c) 28.7
262AAHL4-q00354.8 Binomial distributionPaper 2Medium7 marks
An archer shoots 10 arrows. Each arrow hits the target with probability \(p\), independently. Let \(X\) be the number of arrows that hit the target. The variance of \(X\) is 2.1.
(a)
Show that \(p^{2}-p+0.21=0\).
(2)
(b)
Find the two possible values of \(p\).
(1)
(c)
It is known that \(\mathrm{P}(X\ge6)\gt 0.5\). Determine the value of \(p\).
(2)
(d)
For this value of \(p\), find the probability that the number of hits equals its expected value.
263AAHL4-q00384.8 Binomial distributionPaper 2Medium16 marks
A laboratory tests blood samples for a rare condition. Each person has the condition with probability 0.04, independently of all other people. To save tests, the laboratory uses pooled testing.
The samples of a group of 8 people are mixed and the mixture is tested once. The mixture tests positive if and only if at least one person in the group has the condition.
If the mixture tests negative, no further tests are needed for the group.
If the mixture tests positive, each of the 8 samples is then tested individually.
Let \(N\) be the total number of tests carried out for a group of 8 people.
(a)
Find the probability that the mixture tests positive.
(2)
(b)
Show that \(\mathrm{E}(N)=3.23\), correct to three significant figures.
(3)
(c)
Find \(\mathrm{Var}(N)\).
(3)
(d)
Each test costs €12 and the laboratory also pays €30 to process each group. Find the mean and the standard deviation of the total cost, in euros, of testing a group.
(3)
(e)
Given that the mixture for a group tests positive, find the probability that exactly one person in the group has the condition.
(2)
(f)(i)
Find the expected number of tests per person under this scheme, and hence the percentage reduction in the expected number of tests compared with testing every person individually.
(2)
(f)(ii)
Explain why pooled testing would save fewer tests if the condition were much more common.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(1-\mathrm{P}(\text{nobody has the condition})=1-0.96^{8}\)
(M1)
Or \(X\sim\mathrm{B}(8,\,0.04)\), \(\mathrm{P}(X\ge1)\).
\(0.279\)
A1
\(0.278610\ldots\)
(b)
\(N=1\) with probability \(0.721389\ldots\) and \(N=9\) with probability \(0.278610\ldots\)
264AAHL4-q00394.8 Binomial distributionPaper 2Medium16 marks
A test has 40 multiple-choice questions, each with four options of which exactly one is correct. For each question, independently, Sami knows the answer with probability 0.6; otherwise he chooses one of the four options at random.
(a)
Show that the probability that Sami answers a question correctly is 0.7.
(2)
(b)
Given that Sami answers two particular questions correctly, find the probability that he knew the answers to both of them.
(2)
(c)
Let \(X\) be the number of questions Sami answers correctly. Find \(\mathrm{E}(X)\) and \(\mathrm{Var}(X)\).
(2)
(d)
Each correct answer scores 4 marks and each wrong answer scores \(-1\) mark. Let \(S\) be Sami’s total score. Show that \(S=5X-40\), and hence find the mean and the standard deviation of \(S\).
(3)
(e)
Find \(\mathrm{P}(S\ge120)\).
(3)
(f)
Rhea knows none of the answers and guesses every question at random. Find the probability that her total score is positive.
(2)
(g)
The test designer wants the expected score for a random guess on a single question to be zero. Keeping 4 marks for a correct answer, find the number of marks that should be deducted for a wrong answer.
265AAHL4-q00414.9 The normal distributionPaper 2Medium7 marks
At an airport, the time, \(W\) minutes, that a passenger waits for the shuttle bus is modelled by a normal distribution with mean 6.5 minutes and standard deviation 4 minutes.
Normal curve for W, with the region W > 10 shaded
(a)
Use this model to find the probability that a passenger waits more than 10 minutes.
(2)
(b)(i)
Use this model to find \(\mathrm{P}(W\lt 0)\).
(1)
(b)(ii)
Hence comment on the suitability of this model.
(1)
(c)
The bus company replaces the model by \(W\sim\mathrm{N}(6.5,\,\sigma^{2})\), choosing \(\sigma\) so that \(\mathrm{P}(W\lt 0)=0.001\). Find the value of \(\sigma\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{P}(W\gt 10)\) with \(W\sim\mathrm{N}(6.5,\,4^{2})\), e.g. sketch or GDC input
(M1)
\(0.191\)
A1
\(0.190786\ldots\)
(b)(i)
\(0.0521\)
A1
\(0.0520812\ldots\)
(b)(ii)
a waiting time cannot be negative, but the model gives about 5% of passengers a negative waiting time, so the model is not suitable (for short waiting times)
R1
Must refer to the size of the probability from (b)(i) and to waiting times not being negative.
(c)
\(\dfrac{0-6.5}{\sigma}=z\) where \(\mathrm{P}(Z\lt z)=0.001\)
266AAHL4-q00434.9 The normal distributionPaper 2Medium8 marks
The volume, \(V\) ml, of juice in a carton is normally distributed with mean 502 ml and standard deviation 1.8 ml. Each carton is labelled “500 ml”. Cartons are sold in packs of five, and the volumes in different cartons are independent.
Normal curve for V, with the region V < 500 shaded
(a)
Find the probability that a randomly chosen carton contains less than 500 ml.
(2)
(b)
Find the probability that none of the cartons in a pack contains less than 500 ml.
(2)
(c)
The manufacturer adjusts the mean volume to \(\mu\) ml, keeping the standard deviation at 1.8 ml, so that the probability that none of the cartons in a pack contains less than 500 ml is 0.9. Find the value of \(\mu\). Give your answer correct to two decimal places.
(4)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{P}(V\lt 500)\) with \(V\sim\mathrm{N}(502,\,1.8^{2})\)
(M1)
\(0.133\)
A1
\(0.133260\ldots\)
(b)
\((1-0.133260\ldots)^{5}\)
(M1)
Or binomial \(\mathrm{P}(Y=0)\) with \(Y\sim\mathrm{B}(5,\,0.13326)\).
\(0.489\)
A1
\(0.489151\ldots\)
(c)
\(\big(\mathrm{P}(V\ge 500)\big)^{5}=0.9\)
M1
\(\mathrm{P}(V\ge 500)=0.9^{1/5}=0.979148\ldots\)
A1
\(\dfrac{500-\mu}{1.8}=-2.03647\ldots\)
M1
Standardizing with the \(z\)-value for a lower tail of \(0.020851\ldots\)
\(\mu=503.67\) ml
A1
\(503.6656\ldots\) Award A0 for 503.66 (truncation) or 503.7 (not 2 d.p.).
Answers: (a) 0.133 · (b) 0.489 · (c) \(\mu=503.67\) ml
267AAHL4-q00454.9 The normal distributionPaper 2Medium16 marks
The error, \(X\) mm, in a length measured by a laser device is modelled by the probability density function
Write down \(\mathrm{E}(X)\), giving a reason for your answer.
(1)
(c)
Show that \(\mathrm{Var}(X)=\dfrac{1}{5}\).
(3)
(d)
Find \(\mathrm{P}(|X|\lt0.5)\).
(2)
(e)
An engineer instead models the error by \(Y\sim\mathrm{N}\left(0,\,\tfrac{1}{5}\right)\), which has the same mean and variance as \(X\). Find \(\mathrm{P}(|Y|\lt0.5)\).
(2)
(f)
Find the value of \(t\) such that \(\mathrm{P}(X\gt t)=0.05\), and the value of \(s\) such that \(\mathrm{P}(Y\gt s)=0.05\).
(4)
(g)
Find \(\mathrm{P}(|Y|\gt1)\), and hence comment on the suitability of the normal model.
Or \(\dfrac{3}{4}\left(\dfrac{2}{3}-t+\dfrac{t^{3}}{3}\right)=0.05\).
solving with GDC
(M1)
\(t=0.729\)
A1
\(0.729299\ldots\)
\(s=0.736\)
A1
\(0.735600\ldots\) (inverse normal).
(g)
\(\mathrm{P}(|Y|\gt1)=0.0253\)
A1
\(0.0253473\ldots\)
under the original model errors larger than 1 mm are impossible, but the normal model gives them a probability of about 2.5%, so it overestimates the tails (although it is close for central probabilities)
R1
Accept any comment comparing the bounded and unbounded models.
268AAHL4-q00474.10 Regression line of x on yPaper 2Medium15 marks
In a road-safety study, nine drivers were tested at different times after waking. For each driver the number of hours, \(x\), since waking and the reaction time, \(y\) milliseconds (ms), were recorded.
\(x\) (hours)
2
4
6
8
10
12
14
16
18
\(y\) (ms)
246
251
260
258
274
279
291
298
311
Scatter diagram of the data in the table
(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 \(x\), and use it to estimate the reaction time of a driver who has been awake for 11 hours.
(3)
(c)
A driver involved in an accident has a reaction time of 295 ms. Find the equation of the regression line of \(x\) on \(y\), and use it to estimate the number of hours since this driver woke.
(3)
(d)
Explain why the line of \(x\) on \(y\), rather than a rearrangement of the line in part (b), is used in part (c).
(1)
(e)
The reaction times of well-rested drivers are normally distributed with mean 255 ms and standard deviation 28 ms. Find the probability that a well-rested driver has a reaction time greater than 320 ms.
(2)
(f)
Find the reaction time, \(t\) ms, such that only 2% of well-rested drivers have a reaction time greater than \(t\).
(2)
(g)
Fifteen well-rested drivers are chosen at random. Find the probability that at least one of them has a reaction time greater than \(t\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(r=0.985\)
A1
\(0.985064\ldots\)
strong positive (linear) correlation
A1
FT their \(r\).
(b)
\(y=4.03x+234\)
A1
\(4.03333\ldots,\ 233.888\ldots\)
\(y=4.03333(11)+233.889\)
(M1)
\(278\) ms
A1
\(278.255\ldots\)
(c)
\(x=0.241y-56.0\)
A1
\(0.240583\ldots,\ -55.9732\ldots\)
\(x=0.240583(295)-55.9733\)
(M1)
\(15.0\) hours
A1
\(14.9987\ldots\) Accept 14.9–15.1 from rounded coefficients. Award (M1)A0 for 15.2 from rearranging the \(y\) on \(x\) line.
(d)
\(x\) is being estimated from a known value of \(y\), and the line of \(y\) on \(x\) minimizes errors in \(y\), not in \(x\)
R1
(e)
\(\mathrm{P}(Y\gt 320)\) with \(Y\sim\mathrm{N}(255,\,28^{2})\)
(M1)
\(0.0101\)
A1
\(0.0101318\ldots\)
(f)
\(\mathrm{P}(Y\gt t)=0.02\), e.g. invNorm(0.98, 255, 28)
(M1)
\(t=313\)
A1
\(312.504\ldots\) Accept 312.5.
(g)
\(1-0.98^{15}\)
(M1)
Or \(1-\mathrm{P}(N=0)\) with \(N\sim\mathrm{B}(15,\,0.02)\).
269AAHL4-q00554.12 Standardization and unknown normal parametersPaper 2Medium7 marks
The time, \(T\) minutes, that a worker takes to assemble a flat-pack desk is normally distributed with mean \(\mu\) and standard deviation \(\sigma\). It is known that 30% of workers take less than 20 minutes and 50% of workers take between 20 and 30 minutes.
Normal curve of assembly time T (minutes); shaded regions T < 20 and 20 < T < 30. Diagram not to scale
(a)
Find the value of \(\mu\) and the value of \(\sigma\).
(4)
(b)
After a training course the mean time is unchanged, but the standard deviation changes so that 60% of workers take between 20 and 30 minutes. Find the new standard deviation.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{P}(T\lt 30)=0.3+0.5=0.8\)
(M1)
\(z\)-values \(-0.524400\ldots\) and \(0.841621\ldots\)
(A1)
\(\dfrac{20-\mu}{\sigma}=-0.524400\) and \(\dfrac{30-\mu}{\sigma}=0.841621\)
M1
Award M0 if 0.5 is used as the cumulative probability for 30.
\(\mu=23.8,\ \sigma=7.32\)
A1
\(23.8388\ldots,\ 7.32052\ldots\) Both values required.
(b)
\(\mathrm{P}(20\lt T\lt 30)=0.6\) with \(T\sim\mathrm{N}(23.8388,\,s^{2})\)
M1
FT their \(\mu\).
use of GDC to solve, e.g. graph of \(y=\mathrm{normalcdf}(20,\,30,\,23.8388,\,s)\) intersecting \(y=0.6\)
(M1)
The interval is not symmetric about the mean, so a single \(z\)-value cannot be used.
\(s=5.82\)
A1
\(5.82458\ldots\) Accept 5.81–5.83 (5.816 from \(\mu=23.8\)).
270AAHL4-q00574.12 Standardization and unknown normal parametersPaper 2Medium8 marks
The diameters, \(D\) mm, of bolts made by a machine are normally distributed with mean 12 mm and standard deviation \(\sigma\) mm. A bolt is rejected if its diameter differs from 12 mm by more than 0.05 mm. Exactly 4% of bolts are rejected.
Normal curve of bolt diameter D (mm); the shaded tails are the rejected bolts (4% in total). Diagram not to scale
(a)
Find the value of \(\sigma\).
(3)
(b)
The mean diameter drifts to 12.02 mm, while \(\sigma\) is unchanged. Find the percentage of bolts that are now rejected.
(2)
(c)
Explain why the percentage rejected increases, although the width of the accepted interval and the standard deviation are unchanged.
(1)
(d)
With the mean at 12.02 mm, a bolt is rejected. Find the probability that it was rejected for being too large.
(2)
Show mark scheme
Working
Mark
Notes
(a)
by symmetry \(\mathrm{P}(D\lt 11.95)=0.02\)
M1
\(\dfrac{11.95-12}{\sigma}=-2.05374\ldots\)
(A1)
\(\sigma=0.0243\)
A1
\(0.0243457\ldots\)
(b)
\(\mathrm{P}(D\lt 11.95)+\mathrm{P}(D\gt 12.05)\) with \(D\sim\mathrm{N}(12.02,\,0.0243457^{2})\)
(M1)
\(11.1\%\)
A1
\(0.110946\ldots\) Accept 11.0%–11.2%.
(c)
the upper tail beyond 12.05 gains more probability than the lower tail below 11.95 loses, because the normal curve is highest near the mean
271AAHL4-q00714.13 Bayes’ theoremPaper 2Medium9 marks
A missing hiker is in exactly one of three regions, \(R_{1}\), \(R_{2}\) and \(R_{3}\), with probabilities 0.45, 0.35 and 0.20 respectively. If the hiker is in the region being searched, a search of that region finds the hiker with probability 0.7 for \(R_{1}\), 0.85 for \(R_{2}\) and 0.9 for \(R_{3}\). A search never finds the hiker in a region where the hiker is not. Region \(R_{1}\) is searched first.
(a)
Find the probability that the first search does not find the hiker.
(2)
(b)
Given that the first search does not find the hiker, find the probability that the hiker is in each of the three regions.
(3)
(c)
The rescue team next searches the region in which the hiker is now most likely to be. Write down this region.
(1)
(d)
The second search also does not find the hiker. Find the probability that the hiker is in region \(R_{3}\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(0.45(0.3)+0.35+0.20\)
(M1)
Recognizing that the hiker is not found if in \(R_{1}\) and missed, or in \(R_{2}\) or \(R_{3}\).
272AAHL4-q00764.13 Bayes’ theoremPaper 2Medium15 marks
The emails received by a company are of three types: 25% are spam (\(S\)), 35% are promotional (\(R\)) and 40% are personal (\(P\)). An email filter flags an email as suspicious with the following probabilities.
Type of email
Spam
Promotional
Personal
P(flagged)
0.92
0.15
0.02
Tree diagram with the probabilities given in the question (F: flagged)
An email is chosen at random.
(a)
Find the probability that the email is flagged.
(2)
(b)
Given that the email is flagged, find the probability that it is spam.
(2)
(c)
Given that the email is not flagged, find the probability that it is personal.
(3)
(d)
On one day the company receives 200 emails, independently of each other. Find the expected number of personal emails that are flagged, and the probability that at least 3 personal emails are flagged.
(4)
(e)
Two emails, chosen independently at random, are both flagged. Find the probability that they are of the same type.
(3)
(f)
Suggest a reason why the assumption in part (d) that the emails are independent of each other may not be valid.
273AAHL4-q00964.14 Variance, continuous random variables and linear transformationsPaper 2Medium8 marks
A bag contains 3 red and 5 blue counters. In a game, a player pays a fee and then takes two counters at random from the bag, without replacement. The player receives €7 for each red counter taken. Let \(R\) be the number of red counters taken.
(a)
Show that \(\mathrm{P}(R=1)=\dfrac{15}{28}\).
(2)
(b)
Find the fee that would make the game fair.
(2)
(c)
Find the standard deviation of the amount the player receives.
(2)
(d)
The fee is set at €6. Find the expected total gain of a player who plays the game 50 times.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac38\times\dfrac57+\dfrac58\times\dfrac37\)
M1
Or \(\dfrac{\binom31\binom51}{\binom82}\). Must see both orders, or a combination count.
game A, because it has the same mean but a smaller variance (standard deviation) than game B
R1
Must refer to the variance or standard deviation. FT their values.
Answers: (a) \(k=13\) · (b) \(5.04\) · (c) \(4.2\) · (d) game A (smaller spread)
275AAHL4-q00994.14 Variance, continuous random variables and linear transformationsPaper 2Medium6 marks
The raw mark \(X\) of a randomly chosen candidate in a test has \(\mathrm{E}(X)=54\) and standard deviation \(12.5\). The marks are rescaled using \(Y=aX+b\), where \(a\gt 0\), so that the scaled marks have mean \(65\) and standard deviation \(10\).
(a)
Find the value of \(a\) and the value of \(b\).
(3)
(b)
Find the scaled mark of a candidate whose raw mark is \(72\).
(1)
(c)
A candidate has a scaled mark of \(50\). Find this candidate’s raw mark.
276AAHL4-q01014.14 Variance, continuous random variables and linear transformationsPaper 2Medium8 marks
The continuous random variable \(X\) has probability density function \(f(x)=k\sqrt{9-x^{2}}\) for \(0\le x\le3\), and \(f(x)=0\) otherwise, where \(k\) is a positive constant.
(a)
By considering the shape of the graph of \(y=\sqrt{9-x^{2}}\) for \(0\le x\le3\), show that \(k=\dfrac{4}{9\pi}\).
(2)
(b)
Find \(\mathrm{E}(X)\).
(2)
(c)
Find the standard deviation of \(X\).
(2)
(d)
Find the median of \(X\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
the graph is a quarter of the circle \(x^{2}+y^{2}=9\), so the area under it is \(\dfrac14\pi(3)^{2}=\dfrac{9\pi}{4}\)
\(1.21191\ldots\) Award A0 for values outside \([0,3]\).
Answers: (b) 1.27 · (c) 0.793 · (d) 1.21
277AAHL4-q01024.14 Variance, continuous random variables and linear transformationsPaper 2Medium9 marks
The delay, \(X\) hours, of a parcel delivery is modelled by the probability density function \(f(x)=k\mathrm{e}^{-(x-1)^{2}}\) for \(0\le x\le4\), and \(f(x)=0\) otherwise, where \(k\) is a positive constant.
(a)
Find the value of \(k\).
(2)
(b)
Use differentiation to show that the mode of \(X\) is \(1\).
(2)
(c)
Find \(\mathrm{E}(X)\) and the standard deviation of \(X\).
Find the mean and the standard deviation of \(T\).
(4)
(e)
Find the probability that a customer waits more than 4 minutes.
(2)
(f)
Leila uses the taxi rank on 10 days. Her waiting times are independent. Find the probability that she waits more than 4 minutes on at least 3 of these days.
\(\displaystyle\int_{0}^{2}\sin\left(\dfrac{\pi x}{4}\right)\mathrm{d}x=\left[-\dfrac{4}{\pi}\cos\dfrac{\pi x}{4}\right]_{0}^{2}=\dfrac{4}{\pi}\) and \(\displaystyle\int_{2}^{5}\dfrac{5-x}{3}\,\mathrm{d}x=\dfrac{3}{2}\)
A1
Second integral may be found as the area of a triangle.
\(f\) is increasing on \(0\le x\le2\) (sine increasing for \(0\le\frac{\pi x}{4}\le\frac{\pi}{2}\)) and decreasing on \(2\lt x\le5\), so the maximum of \(f\) is at \(x=2\)
R1
(c)
\(\dfrac{4a}{\pi}\) or \(\displaystyle\int_{0}^{2}f(x)\,\mathrm{d}x\)
(M1)
\(a=0.360589\ldots\)
\(0.459\)
A1
\(0.459116\ldots\) Accept \(\dfrac{8}{8+3\pi}\).
(d)
\(\mathrm{P}(X\le2)\lt0.5\), so the median \(m\) lies in \(2\lt m\le5\)
280IBMA2H-051-01-054.13 Bayes’ theoremPaper 2Medium11 marks
Calculator
A rare enzyme deficiency affects 0.8% of a population. A screening test gives a positive result for 97% of people with the deficiency and a negative result for 95% of people without it. A person is selected at random and tested.
Tree diagram with the probabilities given in the question
(a)
Find the probability that the test result is positive.
(2)
(b)
Given that the result is positive, find the probability that the person has the deficiency.
(2)
(c)
A person who tests positive is given the test a second time. Assume that, for a given person, the two results are independent. Find the probability that a person who tests positive on both occasions has the deficiency.
(3)
(d)
Suggest why the independence assumption in part (c) may not be realistic.
(1)
(e)
Researchers want a single positive result to indicate the deficiency with probability at least 0.5, keeping the sensitivity at 97%. Find the minimum specificity (probability of a negative result for a person without the deficiency) required. Give your answer to 4 significant figures.
R1 whatever caused a false positive in a particular person (e.g. a substance in their blood) is likely to still be present, so a repeat result is likely to be the same
✓ 1
Part (e)
M1 setting up \(\frac{0.00776}{0.00776+0.992(1-s)}\ge 0.5\)
✓ 1
M1 rearranging to \(0.992(1-s)\le 0.00776\)
✓ 1
A1 s ≥ 0.9922 (0.99218…). Do not accept 0.992, because s = 0.992 gives a probability below 0.5
✓ 1
Exam tip: When a minimum value is required, check that your rounded answer still meets the condition. Rounding down here (to 0.992) gives a probability below 0.5.
281IBMA2H-051-01-064.1 Sampling, bias and outliersPaper 2Medium9 marks
Calculator
A sprint coach records the 100 m times, x seconds, of ten athletes. The summary statistics are Σx = 128.4 and Σx2 = 1702.80. When ordered, the times are: 11.8, 11.9, 11.9, 12.0, 12.0, 12.1, 12.2, 12.3, 12.4, 19.8
(a)
Find the mean and the standard deviation of the ten times.
(3)
(b)
The lower and upper quartiles are 11.9 s and 12.3 s. Show that 19.8 s is an outlier.
(2)
(c)
The coach discovers that 19.8 s was caused by the stopwatch being started early. Find the mean and standard deviation of the remaining nine times.
(3)
(d)
State why it was appropriate to remove this value, but why outliers should not always be removed.
R1 upper boundary = 12.9 and 19.8 > 12.9, so it is an outlier
✓ 1
Part (c)
A1 new mean = \(\frac{128.4-19.8}{9}\) = 12.1 s (12.07)
✓ 1
M1 new Σx2 = 1702.80 − 19.82 = 1310.76 and variance = \(\frac{1310.76}{9}-12.0667^2\) (≈ 0.0356)
✓ 1
A1 standard deviation = 0.189 s
✓ 1
Part (d)
R1 it was a recording error, so not a genuine time; a genuine outlier (e.g. an unusually fast athlete) is real data and should be kept
✓ 1
Common error: When removing a value, adjust both Σx and Σx², and divide by the new n. Using the old Σx² gives a nonsense variance.
282IBMA2H-051-05-044.12 Standardization and unknown normal parametersPaper 2Medium10 marks
Calculator
The distance, \(D\) metres, achieved by a club long jumper on a single jump is modelled by a normal distribution with mean \(\mu\) and standard deviation \(\sigma\). Records show that 15% of her jumps exceed 7.20 m and 25% of her jumps are shorter than 6.80 m.
Normal curve of jump distance D (metres); shaded regions D < 6.80 and D > 7.20. Diagram not to scale
(a)
Find the value of \(\mu\) and the value of \(\sigma\).
(5)
(b)
Find the probability that a single jump exceeds 7.00 m.
(2)
(c)
In a competition she has three independent jumps. She qualifies for the final if at least one jump exceeds 7.00 m. Find the probability that she qualifies.
M1 \(P(D > 7.00)\) with their \(\mu\) and \(\sigma\)
✓ 1
A1 0.428 (0.428195…)
✓ 1
Part (c)
M1 use of complement: 1 − P(no jump exceeds 7.00 m)
✓ 1
A1 \(1 - (0.571805…)^3\)
✓ 1
A1 0.813 (0.813042…)
✓ 1
Exam tip: Carry full calculator values of μ and σ into later parts. Using 6.96 and 0.234 gives P(D > 7) ≈ 0.432, which can lead to a final answer that is not accurate to 3 significant figures.
283IBMA2H-051-05-054.13 Bayes’ theoremPaper 2Medium7 marks
Calculator
A blood test measures the concentration of a protein marker, in ng/mL. For people without a certain heart condition, the concentration is normally distributed with mean 40 and standard deviation 6. For people with the condition, the concentration is normally distributed with mean 58 and standard deviation 8. In a population, 5% of people have the condition. A test is called positive if the concentration exceeds 50 ng/mL.
(a)
Find the probability that a person without the condition tests positive.
(2)
(b)
Find the probability that a person with the condition tests positive.
(2)
(c)
A randomly chosen person tests positive. Find the probability that this person has the condition.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(P(X > 50)\) where \(X \sim N(40, 6^2)\)
✓ 1
A1 0.0478 (0.0477904…)
✓ 1
Part (b)
M1 \(P(Y > 50)\) where \(Y \sim N(58, 8^2)\)
✓ 1
A1 0.841 (0.841345…)
✓ 1
Part (c)
M1 use of Bayes' theorem with a total-probability denominator
Common error: Do not confuse P(positive | condition) with P(condition | positive). Because the condition is rare, many positives come from the large healthy group, so the reversed conditional is far lower than 0.841.
284IBMA2H-051-05-064.14 Variance, continuous random variables and linear transformationsPaper 2Medium7 marks
Calculator
The temperature, \(C\) °C, of an electronic component on an orbiting satellite is modelled by \(C \sim N(-20, 4^2)\). The telemetry system reports the temperature in degrees Fahrenheit, \(F\), where \(F = 1.8C + 32\).
(a)
Find \(E(F)\) and \(\text{Var}(F)\).
(3)
(b)
The component may fail if its temperature falls below −13 °F. Find the probability that the temperature is below −13 °F at a randomly chosen time.
(2)
(c)
Find the temperature, in °F, that is exceeded only 1% of the time.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(E(F) = 1.8(-20) + 32 = -4\)
✓ 1
M1 use of \(\text{Var}(aC + b) = a^2\text{Var}(C)\)
✓ 1
A1 \(\text{Var}(F) = 1.8^2 \times 16 = 51.84\)
✓ 1
Part (b)
M1 \(P(F < -13)\) with \(F \sim N(-4, 7.2^2)\), or equivalently \(P(C < -25)\)
✓ 1
A1 0.106 (0.105650…)
✓ 1
Part (c)
M1 inverse normal with upper tail 0.01, e.g. \(-4 + 2.32635 \times 7.2\)
✓ 1
A1 12.7 °F (12.7497…)
✓ 1
Common error: Adding 32 shifts the mean but has no effect on the spread; only the factor 1.8 changes the variance, and it enters squared. Writing Var(F) = 1.8 × 16 + 32 loses the method mark.
285IBMA2H-051-05-114.14 Variance, continuous random variables and linear transformationsPaper 2Medium11 marks
Calculator
A space agency drops sample-return capsules onto a desert landing zone. The east–west displacement, \(D\) km, of the landing point from the target (positive to the east) is modelled by \(D \sim N(0, 1.2^2)\).
Normal curve for D, with the region −2 < D < 2 shaded
(a)
Find the probability that a capsule lands within 2 km of the target in the east–west direction.
(2)
(b)
The recovery team patrols the interval \(-d \le D \le d\). Find the value of \(d\) for which the capsule lands in this interval with probability 0.99.
(3)
(c)
Given that a capsule lands east of the target, find the probability that it lands more than 1.5 km east of the target.
(3)
(d)
Eight capsules are dropped independently. Find the probability that exactly six of them land within 2 km of the target in the east–west direction.
(3)
Show mark scheme
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Mark
Notes
Part (a)
M1 \(P(-2 < D < 2)\)
✓ 1
A1 0.904 (0.904419…)
✓ 1
Part (b)
M1 recognising 0.005 in each tail by symmetry
✓ 1
M1 \(d = 1.2 \times 2.57583\), or inverse normal with area 0.995
✓ 1
A1 \(d\) = 3.09 km (3.09099…)
✓ 1
Part (c)
M1 \(\frac{P(D > 1.5)}{P(D > 0)}\)
✓ 1
A1 \(\frac{0.105650}{0.5}\)
✓ 1
A1 0.211 (0.211300…)
✓ 1
Part (d)
M1 recognising \(B(8, 0.904419)\)
✓ 1
A1 \(\binom{8}{6}(0.904419)^6(0.095581)^2\)
✓ 1
A1 0.140 (0.139997…)
✓ 1
Exam tip: For a central interval with total probability 0.99, use 0.995 (or upper tail 0.005) in the inverse normal. Entering 0.99 places only 1% in one tail and gives an interval that is too wide.
286IBMA2H-051-05-124.2 Presentation of dataPaper 2Medium10 marks
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The speed, \(V\) km/h, of a professional tennis player's first serve is normally distributed with mean 185 km/h and standard deviation \(\sigma\) km/h. Given that a serve is faster than 185 km/h, the probability that it is faster than 200 km/h is 0.2.
(a)
Find the value of \(\sigma\).
(4)
(b)
In a match she hits 60 first serves, with speeds independent of one another. Find the probability that more than 8 of them are faster than 200 km/h.
Common error: Setting P(V > 200) = 0.2 ignores the condition. Because the given event V > 185 has probability 0.5, the unconditional tail probability is 0.2 × 0.5 = 0.1.
287IBMA2H-051-08-054.2 Presentation of dataPaper 2Medium7 marks
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A fitness app records the daily step count, s thousand steps, of 320 users on one day. The cumulative frequencies are: s ≤ 4: 16 s ≤ 6: 48 s ≤ 8: 120 s ≤ 10: 216 s ≤ 12: 280 s ≤ 14: 308 s ≤ 16: 320 Use linear interpolation within classes throughout.
(a)
Estimate the median step count.
(2)
(b)
Estimate the interquartile range.
(3)
(c)
The app awards a badge to the 15% of users with the highest step counts. Estimate the minimum number of steps needed to earn the badge.
(2)
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Mark
Notes
Part (a)
M1 160th value in 8 < s ≤ 10: 8 + (160 − 120)/96 × 2
✓ 1
A1 median ≈ 8.83 thousand steps
✓ 1
Part (b)
A1 Q1 (80th value) = 6 + (32/72) × 2 ≈ 6.89
✓ 1
A1 Q3 (240th value) = 10 + (24/64) × 2 = 10.75
✓ 1
A1 IQR ≈ 3.86 thousand steps
✓ 1
Part (c)
M1 identifies the 272nd value (85th percentile) in 10 < s ≤ 12: 10 + (272 − 216)/64 × 2
✓ 1
A1 11.75 thousand, i.e. about 11 750 steps
✓ 1
Exam tip: Turn a 'top 15%' condition into a cumulative frequency first: 85% of 320 users is 272.
288IBMA2H-051-08-074.2 Presentation of dataPaper 2Medium10 marks
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An engineer tests 200 LED bulbs and records each lifetime, X thousand hours. A histogram with equal class widths has these frequencies: 0 ≤ x < 2: 30 2 ≤ x < 4: 105 4 ≤ x ≤ 6: 65 A proposed model for X has probability density function f(x) = kx(6 − x) for 0 ≤ x ≤ 6, and f(x) = 0 otherwise.
Histogram of the lifetimes of the 200 bulbs (equal class widths)
(a)
Find the exact value of k.
(3)
(b)
Find the expected frequency for each class under the model, giving your answers to one decimal place.
(3)
(c)
Estimate the mean lifetime from the histogram data. State the mean of the model, with a reason. Using these means and your answers to part (b), comment on the suitability of the model.
A1 model mean = 3, because f is symmetric about x = 3 (or by evaluating \(\int_0^6 xf(x)\,dx\))
✓ 1
R1 the model is not a good fit: the observed frequencies differ markedly from the expected ones (30 against 51.9 in the lowest class and 65 against 51.9 in the highest), so the data are not symmetric and their mean is higher than the model's
✓ 1
Exam tip: Symmetry of the pdf about x = 3 gives the model mean without integrating. It also means the two outer classes have equal probability, so only one of them needs to be integrated.
289IBMA2H-051-08-114.2 Presentation of dataPaper 2Medium8 marks
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A clinic models the systolic blood pressure, B mmHg, of adult patients by B ~ N(122, 152). A box-and-whisker diagram of patient readings will mark as outliers any values more than 1.5 × IQR beyond the nearer quartile.
(a)
Find the lower and upper quartiles of B.
(2)
(b)
Find the upper outlier boundary.
(2)
(c)
Find the percentage of patients the model predicts will be outliers.
(2)
(d)
A sample of 400 patients contains 11 outliers. Find the expected number under the model and comment on the model.
(2)
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Mark
Notes
Part (a)
M1 uses inverse normal with P(B < Q1) = 0.25 or P(B < Q3) = 0.75
M1 IQR = 132.12 − 111.88 = 20.235, so the boundary is Q3 + 1.5 × IQR = 132.12 + 1.5 × 20.235
✓ 1
A1 ≈ 162 mmHg (162.47)
✓ 1
Part (c)
M1 P(B > 162.47) ≈ 0.00349, doubled by symmetry (or also find P(B < 81.53))
✓ 1
A1 ≈ 0.698%
✓ 1
Part (d)
A1 expected number = 400 × 0.006977 ≈ 2.79
✓ 1
R1 11 is far more than 2.79, so the model underestimates the number of extreme readings and is not a good fit. Accept any valid reason, e.g. tails heavier than normal, skew, or a wrong mean or standard deviation
✓ 1
Exam tip: For any normal distribution the outlier proportion is the same (about 0.7%), because the boundaries are always about 2.70 standard deviations from the mean.
290IBMA2H-051-00-064.4 Correlation and regression of y on xPaper 2Medium7 marks
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During ground testing of a small rocket engine, the fuel flow rate, x kg s−1, and the thrust produced, y kN, were recorded in five trials. x: 10, 12, 15, 18, 20 y: 40, 41, p, 56, 65 The regression line of y on x for these data has equation \(y=2.5x+12.9\).
(a)
Find the value of p.
(3)
(b)
Find the value of Pearson's product-moment correlation coefficient for these data.
(2)
(c)
Estimate the thrust when the fuel flow rate is 16 kg s−1.
(2)
Show mark scheme
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Mark
Notes
Part (a)
M1 uses the fact that \((\bar{x},\bar{y})\) lies on the regression line A1 \(\bar{x}=15\Rightarrow\bar{y}=2.5(15)+12.9=50.4\) A1 \(\sum y=252\), so \(p=252-202=50\)
✓ 3
Part (b)
M1 GDC used with all five pairs, including \(p=50\) A1 \(r=0.981\)
✓ 2
Part (c)
M1 \(2.5(16)+12.9\) A1 52.9 kN
✓ 2
Exam tip: A regression line of y on x always passes through the mean point \((\bar{x},\bar{y})\). This is the quickest way to recover a missing data value.
291IBMA2H-051-00-074.14 Variance, continuous random variables and linear transformationsPaper 2Medium11 marks
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A steel beam of length 6 m is loaded until a crack first appears. The distance, X metres, of the crack from the left-hand end is modelled by the probability density function \(f(x)=k\sin\left(\frac{\pi x}{6}\right)\) for \(0\le x\le 6\), and \(f(x)=0\) otherwise.
(a)
Show that \(k=\frac{\pi}{12}\).
(2)
(b)
Write down E(X), giving a reason.
(2)
(c)
Find Var(X).
(3)
(d)
Find P(2 < X < 4).
(2)
(e)
The inspection report records Y, the distance of the crack from the right-hand end in centimetres, so that \(Y=100(6-X)\). Find the standard deviation of Y.
(2)
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Notes
Part (a)
M1 \(k\left[-\frac{6}{\pi}\cos\left(\frac{\pi x}{6}\right)\right]_0^6=\frac{12k}{\pi}\) A1 \(\frac{12k}{\pi}=1\Rightarrow k=\frac{\pi}{12}\) AG
✓ 2
Part (b)
R1 the graph of f is symmetric about \(x=3\) A1 \(E(X)=3\)
M1 SD\((Y)=|-100|\times\)SD\((X)=100\sqrt{1.70487…}\) A1 131 cm
✓ 2
Common error: Writing SD(Y) = −100 SD(X). Standard deviation scales by the absolute value of the multiplier, so it can never be negative.
292IBMA2H-051-00-124.4 Correlation and regression of y on xPaper 2Medium11 marks
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In a clinical trial, eight patients each receive a dose, x mg, of a new drug, and the reduction in their systolic blood pressure, y mmHg, is recorded. The summary statistics are \(n=8\), \(\sum x=240\), \(\sum y=176\), \(\sum x^2=8200\), \(\sum xy=5720\), \(\sum y^2=4122\).
(a)
Find the equation of the regression line of y on x.
(4)
(b)
Find Pearson's product-moment correlation coefficient, r.
(2)
(c)
Estimate the reduction in blood pressure for a patient given a dose of 35 mg.
(2)
(d)
A doctor wants to estimate the dose needed to achieve a reduction of 20 mmHg. Explain why the regression line of y on x should not be used for this, and find a suitable estimate.
(3)
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Notes
Part (a)
M1 \(S_{xy}=5720-\frac{240\times176}{8}=440\) and \(S_{xx}=8200-\frac{240^2}{8}=1000\) A1 gradient \(=\frac{440}{1000}=0.44\) M1 uses \(\bar{y}=\text{gradient}\times\bar{x}+c\) with \(\bar{x}=30\), \(\bar{y}=22\) A1 \(y=0.44x+8.8\)
✓ 4
Part (b)
M1 \(S_{yy}=4122-\frac{176^2}{8}=250\) and \(r=\frac{440}{\sqrt{1000\times250}}\) A1 \(r=0.88\)
✓ 2
Part (c)
M1 \(0.44(35)+8.8\) A1 24.2 mmHg
✓ 2
Part (d)
R1 the line of y on x minimises errors in y; to estimate x from y the regression line of x on y is needed M1 \(x-30=\frac{440}{250}(y-22)\), i.e. \(x=1.76y-8.72\) A1 \(x=1.76(20)-8.72=26.5\) mg (26.48)
✓ 3
Exam tip: Decide which variable you are predicting before choosing a line: predict y from x with the y on x line, and x from y with the x on y line. Rearranging the wrong line loses the reasoning mark.
293IBMA2H-051-04-054.13 Bayes’ theoremPaper 2Medium7 marks
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A satellite manufacturer buys solar cells from three suppliers. Supplier X provides 50% of the cells, supplier Y provides 30% and supplier Z provides the rest. The probability that a cell is defective is 0.02 for supplier X, 0.03 for supplier Y and p for supplier Z. Overall, 2.9% of the cells are defective.
Tree diagram with the probabilities given in the question (D: defective)
(a)
Find p.
(2)
(b)
A cell is chosen at random and found to be defective. Find the probability that it came from supplier Z.
(3)
(c)
A cell is chosen at random and found not to be defective. Find the probability that it came from supplier X.
Exam tip: For a 'not defective' condition, the denominator is 1 − 0.029 = 0.971, and every numerator uses 1 minus the defect rate for that supplier.
294IBMA2H-051-04-074.6 Combined, conditional and independent eventsPaper 2Medium11 marks
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A tennis player's first serve lands in with probability 0.6. 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 it lands in, she wins the point with probability 0.5. If both serves are out she loses the point.
Tree diagram with the probabilities given in the question (first serve, then second serve)
(a)
Show that the probability she wins a point on her serve is 0.63.
(3)
(b)
Given that she wins a point on her serve, find the probability that her first serve landed in.
(2)
(c)
Assuming that points are independent, find the probability that she wins at least 3 of her next 4 service points.
(3)
(d)
Her coach wants her to win at least 68% of service points. Assuming all other probabilities stay the same, find the least probability that her first serve lands in.
Exam tip: In part (d) write P(win) as a linear function of the unknown first-serve probability before forming the inequality; the second-serve branch contributes 0.45 per missed first serve.
295IBMA2H-051-04-104.13 Bayes’ theoremPaper 2Medium7 marks
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A rare condition affects 1.5% of a population. A screening test gives a positive result for 95% of people with the condition and for 8% of people without it. Anyone who tests positive is given a second, confirmatory test, which gives a positive result for 90% of people with the condition and for 2% of people without it. Given a person's condition status, the results of the two tests are independent.
(a)
Find the probability that a randomly chosen person tests positive on the screening test.
(2)
(b)
Find the probability that a person who tests positive on the screening test has the condition.
(2)
(c)
Find the probability that a person who tests positive on both tests has the condition.
Exam tip: You can update in two stages: use the answer to (b) as the new prior for the second test. Both methods give the same result.
296IBMA2H-051-07-034.12 Standardization and unknown normal parametersPaper 2Medium10 marks
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The bolts that secure the blades of a wind turbine are tested for tensile strength. The tensile strength, S MPa, of a bolt is modelled by a normal distribution with mean μ and standard deviation σ. It is found that 4% of bolts have a tensile strength below 610 MPa and 15% of bolts have a tensile strength above 690 MPa.
Normal curve of tensile strength S (MPa); shaded regions S < 610 and S > 690. Diagram not to scale
(a)
Write down two equations in μ and σ.
(3)
(b)
Hence find the value of μ and the value of σ.
(2)
(c)
A bolt is rejected if its tensile strength is below 610 MPa. A random sample of 20 bolts is tested. Find the probability that at most one bolt is rejected.
(3)
(d)
A bolt is selected at random and is found not to be rejected. Find the probability that its tensile strength is above 690 MPa.
(2)
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Mark
Notes
Part (a)
A1 z = −1.7507 corresponding to a lower-tail probability of 0.04 A1 z = 1.0364 corresponding to a lower-tail probability of 0.85 M1 both equations formed: 610 = μ − 1.7507σ and 690 = μ + 1.0364σ
✓ 3
Part (b)
A1 σ = 28.7 MPa A1 μ = 660 MPa
✓ 2
Part (c)
M1 recognising X ~ B(20, 0.04) for the number rejected M1 P(X ≤ 1) = 0.9620 + 20(0.04)(0.96)19 A1 0.810
✓ 3
Part (d)
M1 P(S > 690 | S ≥ 610) = 0.15/0.96 A1 0.156
✓ 2
Common error: "15% above 690" requires the inverse normal of 0.85 (or an upper-tail setting). Using z for 0.15 gives a negative value and places 690 below the mean.
297IBMA2H-051-07-064.2 Presentation of dataPaper 2Medium7 marks
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The core temperature, X °C, of a laptop processor under load is modelled by a normal distribution with mean μ and standard deviation σ. A monitoring app records the temperature in degrees Fahrenheit, F, where F = 1.8X + 32. The mean of the recorded values of F is 158 °F. It is also known that 5% of temperatures exceed 85 °C.
(a)
Find the value of μ.
(2)
(b)
Find the standard deviation of F.
(3)
(c)
Given that F is normally distributed, find the Fahrenheit reading that is exceeded by 25% of recorded values.
(2)
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Notes
Part (a)
M1 use of E(F) = 1.8E(X) + 32, so 158 = 1.8μ + 32 A1 μ = 70 °C
M1 F ~ N(158, 16.412) with inverse normal for 0.75, e.g. 158 + 0.6745 × 16.41 A1 169 °F
✓ 2
Common error: Adding 32 shifts the mean but has no effect on the spread. Var(aX + b) = a²Var(X), so the standard deviation is multiplied by 1.8 only, not by 1.8 and then increased by 32.
298IBMA2H-051-07-074.13 Bayes’ theoremPaper 2Medium10 marks
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At a clinic, 30% of patients tested are anaemic. For anaemic patients, the haemoglobin level, H g/dL, is modelled by N(10.5, 1.22). For non-anaemic patients, H is modelled by N(μ, 1.12). A screening rule flags any patient whose haemoglobin level is below 12.0 g/dL. Of the non-anaemic patients, 2.5% are flagged.
(a)
Find the value of μ.
(3)
(b)
Find the probability that an anaemic patient is flagged.
(2)
(c)
Find the probability that a randomly chosen patient is flagged.
(2)
(d)
A randomly chosen patient is flagged. Find the probability that this patient is anaemic.
(3)
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Notes
Part (a)
M1 z = −1.9600 for a lower-tail probability of 0.025 M1 (12.0 − μ)/1.1 = −1.96 A1 μ = 14.2 g/dL (14.156)
M1 numerator P(anaemic ∩ flagged) = 0.3 × 0.8944 = 0.2683 M1 dividing by P(flagged) found in the previous part A1 0.939
✓ 3
Exam tip: A tree diagram with the normal probabilities on the second branches makes the final conditional-probability step routine. Keep at least 4 significant figures in intermediate values so that the final answer is correct to 3 significant figures.
299IBMA2H-051-07-114.12 Standardization and unknown normal parametersPaper 2Medium7 marks
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The diameter, D mm, of 3D-printer filament produced by a machine is modelled by a normal distribution with mean 1.75 mm and standard deviation σ mm. Currently, 98% of filament has a diameter within 0.05 mm of the mean.
Normal curve of filament diameter D (mm) with mean 1.75; shaded region 1.70 < D < 1.80. Diagram not to scale
(a)
Find the value of σ.
(3)
(b)
The manufacturer wants 99.5% of filament to have a diameter within 0.05 mm of the mean, with the mean unchanged. Find the required standard deviation.
(2)
(c)
Find the percentage reduction in the standard deviation that is needed.
(2)
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Mark
Notes
Part (a)
M1 P(1.70 < D < 1.80) = 0.98, so P(D < 1.80) = 0.99 M1 0.05/σ = 2.3263 A1 σ = 0.0215 mm (0.021493…)
✓ 3
Part (b)
M1 P(D < 1.80) = 0.9975, so 0.05/σ = 2.8070 A1 σ = 0.0178 mm (0.017813…)
✓ 2
Part (c)
M1 (0.021493 − 0.017813)/0.021493 × 100, or (1 − 2.3263/2.8070) × 100 A1 17.1% (accept 17.2% from using the 3 s.f. values 0.0215 and 0.0178)
✓ 2
Common error: A central probability of 0.98 leaves 0.01 in each tail, so apply the inverse normal to 0.99, not to 0.98. Keep unrounded values in your calculator for later parts to avoid rounding drift.
300IBMA2H-051-07-124.9 The normal distributionPaper 2Medium10 marks
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The lifetime, X thousand hours, of a type of LED bulb is modelled by a normal distribution with mean μ and standard deviation σ. 80% of bulbs last longer than 40 thousand hours. Of the bulbs that last longer than 40 thousand hours, half last longer than 50 thousand hours.
Normal curve of lifetime X (thousand hours); shaded region X > 40. Diagram not to scale
(a)
Show that P(X > 50) = 0.4.
(2)
(b)
Find the value of μ and the value of σ.
(5)
(c)
The bulbs are sold in packs of 4. Lifetimes are independent. Find the probability that at least 3 bulbs in a pack last longer than 50 thousand hours.
(3)
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Part (a)
M1 P(X > 50) = P(X > 50 | X > 40) × P(X > 40), valid since X > 50 implies X > 40 A1 0.5 × 0.8 = 0.4
✓ 2
Part (b)
M1 z = −0.8416 for 40 (lower-tail probability 0.2) M1 z = 0.2533 for 50 (lower-tail probability 0.6) M1 forming 40 = μ − 0.8416σ and 50 = μ + 0.2533σ A1 σ = 9.13 thousand hours A1 μ = 47.7 thousand hours
Common error: 0.5 is a conditional probability, not P(X > 50). Use P(A ∩ B) = P(A | B)P(B) before standardising. Otherwise the z-value of 0 will wrongly suggest that μ = 50.
301IBMA2H-051-06-074.8 Binomial distributionPaper 2Medium7 marks
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A communications satellite carries 10 transponders. On a given day each transponder is operational with probability 0.7, independently of the others. Let \(X\) be the number of operational transponders on that day.
(a)
Find \(P(X\ge6)\).
(1)
(b)
Find \(P(X\ge8\mid X\ge6)\).
(2)
(c)
Given that at least 6 transponders are operational, find the probability that transponder number 1 is operational.
R1 the event requires transponder 1 operational and at least 5 of the other 9 operational ; M1 uses \(Y\sim B(9,0.7)\) with \(P(Y\ge5)=0.901191\) ; A1 numerator \(0.7\times0.901191=0.630834\) ; A1 \(\frac{0.630834}{0.849732}=0.742\)
✓ 4
Exam tip: When a condition concerns a total but the question asks about one individual, fix that individual first and count what the remaining trials must do.
302IBMA2H-051-06-094.2 Presentation of dataPaper 2Medium11 marks
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The finishing times, \(T\) minutes, of runners in a city marathon are modelled by \(T\sim N(245,\,30^2)\).
Normal curve for T, with the region T < 240 shaded
(a)
Find the probability that a randomly chosen runner finishes in under 240 minutes.
(1)
(b)
Given that a runner finished in under 240 minutes, find the probability that the runner finished in under 210 minutes.
(3)
(c)
Find the value of \(t\) such that \(P(T<t\mid T<240)=0.5\).
(3)
(d)
Five runners are chosen at random from those who finished in under 240 minutes. Find the probability that at least two of them finished in under 210 minutes.
M1 recognises \(Y\sim B(5,\,p)\) with \(p\) the conditional probability found in the second part, \(p=0.280\) ; M1 \(P(Y\ge2)=1-P(Y=0)-P(Y=1)\) ; A1 \(P(Y=0)=0.193\), \(P(Y=1)=0.376\) ; A1 0.431 using unrounded \(p=0.28046\) (accept 0.430 from \(p=0.280\))
✓ 4
Exam tip: In the final part the binomial probability of success is the conditional probability found earlier, because the runners are drawn only from those who finished under 240 minutes. Keep that value unrounded in your calculator to avoid rounding errors.
303IBMA2H-051-03-074.13 Bayes’ theoremPaper 2Medium6 marks
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A hospital buys blood-glucose test strips from two suppliers. 60% of boxes come from supplier A and 40% from supplier B. Each strip from supplier A is faulty with probability 0.02, and each strip from supplier B is faulty with probability 0.05, independently. Every box contains 20 strips, all from the same supplier. A box is chosen at random and is found to contain exactly one faulty strip.
(a)
Find the probability that a box from supplier A contains exactly one faulty strip.
(2)
(b)
Find the probability that a box from supplier B contains exactly one faulty strip.
(1)
(c)
Find the probability that the chosen box came from supplier B.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 20 × 0.02 × 0.9819
✓ 1
A1 0.272 (0.27249…)
✓ 1
Part (b)
A1 20 × 0.05 × 0.9519 = 0.377 (0.37735…)
✓ 1
Part (c)
M1 use of Bayes' theorem with numerator 0.4 × 0.37735
Exam tip: Keep the unrounded values from parts (a) and (b) in your calculator when you apply Bayes' theorem. Rounding them early can move the third significant figure.
304IBMA2H-051-03-084.8 Binomial distributionPaper 2Medium6 marks
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In a penalty shoot-out a goalkeeper faces 6 penalties. She saves each penalty independently with probability p. The probability that she saves at least one of the 6 penalties is 0.75.
(a)
Find the value of p.
(3)
(b)
Over a season she takes part in 10 independent shoot-outs, each of 6 penalties. Find the probability that she saves at least one penalty in at least 8 of these shoot-outs.
(3)
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Part (a)
M1 1 − (1 − p)6 = 0.75
✓ 1
M1 (1 − p)6 = 0.25, so 1 − p = 0.251/6
✓ 1
A1 p = 0.206
✓ 1
Part (b)
M1 recognising Y ~ B(10, 0.75), where Y is the number of shoot-outs with at least one save
Exam tip: Part (b) uses a new binomial variable. The 'success' is now a whole shoot-out with at least one save, so its probability is 0.75. It is not p.
305IBMA2H-051-03-104.14 Variance, continuous random variables and linear transformationsPaper 2Medium10 marks
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The lifetime, T years, of a smartwatch battery is modelled by the probability density function \(f(t) = kt(4 - t)\) for \(0 \le t \le 4\), and \(f(t) = 0\) otherwise, where k is a positive constant.
(a)
Show that \(k = \frac{3}{32}\).
(2)
(b)
Write down E(T), justifying your answer.
(2)
(c)
Find the exact probability that a battery lasts more than 3 years.
(3)
(d)
A company issues 15 of these smartwatches. The battery lifetimes are independent. Find the probability that at least 3 of the batteries last more than 3 years.
(3)
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Notes
Part (a)
M1 setting ∫04 kt(4 − t) dt = 1
✓ 1
A1 k[2t2 − t3/3]04 = k(32/3) = 1, so k = 3/32
✓ 1
Part (b)
R1 f(t) = f(4 − t), so the graph of f is symmetric about t = 2
✓ 1
A1 E(T) = 2
✓ 1
Part (c)
M1 (3/32)∫34 (4t − t2) dt
✓ 1
A1 [2t2 − t3/3]34 = 32/3 − 9 = 5/3
✓ 1
A1 P(T > 3) = 5/32
✓ 1
Part (d)
M1 recognising X ~ B(15, 5/32)
✓ 1
M1 1 − P(X ≤ 2)
✓ 1
A1 0.423
✓ 1
Exam tip: Using the exact value 5/32 = 0.15625 in the binomial calculation avoids rounding errors. A rounded value such as 0.16 can change the final answer.
306IBMA2H-051-09-054.3 Measures of central tendency and dispersionPaper 2Medium7 marks
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A hospital compares recovery times after knee surgery on two wards. Ward A: 20 patients, mean 6.5 days, standard deviation 1.8 days. Ward B: 30 patients, mean 8.0 days, standard deviation 2.4 days. The standard deviations are population standard deviations, \(\sigma=\sqrt{\frac{\sum x^2}{n}-\bar{x}^2}\).
(a)
Find the mean recovery time of all 50 patients.
(2)
(b)
Show that, for Ward A, Σx2 = 909.8.
(2)
(c)
Find the standard deviation of the recovery times of all 50 patients.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{20\times6.5+30\times8.0}{50}\) A1 7.4 days
Common error: Standard deviations cannot be averaged or added. Rebuild Σx² for each group, combine, then recompute the variance.
307IBMA2H-051-09-104.2 Presentation of dataPaper 2Medium6 marks
Calculator
A clinical trial recorded the systolic blood pressure of 40 patients. The mean was 128.5 mmHg and the population standard deviation was 11.2 mmHg. It was later found that one reading entered as 182 mmHg should have been 128 mmHg.
Common error: Use the corrected mean, not the original mean, when subtracting x̄² for the corrected variance, and keep full accuracy in Σx² to avoid rounding errors.
308IBMA2H-051-09-114.2 Presentation of dataPaper 2Medium13 marks
Calculator
GPS trackers record the top sprint speeds, in m/s, of ten players in a rugby squad during a match: 8.2, 8.9, 7.6, 9.4, 8.5, 8.8, 7.9, 9.1, 8.3, 8.7
(a)
Find the mean and the standard deviation of these speeds.
(3)
(b)
Find the median speed.
(2)
(c)
The speeds are converted to km/h by multiplying by 3.6. Find the mean and standard deviation of the speeds in km/h.
(3)
(d)
An eleventh player's top speed is added and the mean of the eleven speeds becomes 8.6 m/s. Find this player's top speed.
(2)
(e)
Find the standard deviation of the eleven speeds.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 correct use of GDC or formulae A1 mean = 8.54 m/s A1 σ ≈ 0.524 m/s
✓ 3
Part (b)
M1 orders data, middle values 8.5 and 8.7 A1 median = 8.6 m/s
✓ 2
Part (c)
M1 multiplies both mean and standard deviation by 3.6 A1 mean ≈ 30.7 km/h A1 σ ≈ 1.89 km/h
✓ 3
Part (d)
M1 11 × 8.6 − 85.4 A1 9.2 m/s
✓ 2
Part (e)
M1 original Σx2 = 10(0.2744 + 8.542) = 732.06 M1 variance = \(\frac{732.06+9.2^2}{11}-8.6^2\) A1 σ ≈ 0.534 m/s
✓ 3
Exam tip: Multiplying data by a constant scales both the mean and the standard deviation; adding a constant would affect only the mean.
309IBMA2H-051-02-064.8 Binomial distributionPaper 2Medium10 marks
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A small Earth-observation satellite carries 20 identical solar cells. During a mission, each cell fails independently with probability 0.07. The satellite remains operational if at most two of its cells fail. Let \(F\) be the number of cells that fail during a mission.
(a)
Find the probability that exactly two cells fail.
(2)
(b)
Find the probability that the satellite remains operational.
(3)
(c)
A company operates a fleet of 6 such satellites, which perform independently. Find the expected number of satellites that remain operational.
(2)
(d)
Find the probability that at least 5 of the 6 satellites remain operational.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 recognises \(F\sim B(20,0.07)\)
✓ 1
A1 \(P(F=2)=0.252\)
✓ 1
Part (b)
M1 requires \(P(F\le2)\)
✓ 1
A1 \(P(F=0)=0.234\) and \(P(F=1)=0.353\) seen, or a cumulative binomial function used
✓ 1
A1 \(0.839\)
✓ 1
Part (c)
M1 \(6\times0.838998\ldots\)
✓ 1
A1 \(5.03\)
✓ 1
Part (d)
M1 recognises \(Y\sim B(6,0.838998\ldots)\)
✓ 1
M1 \(P(Y=5)+P(Y=6)\) or \(1-P(Y\le4)\)
✓ 1
A1 \(0.750\)
✓ 1
Exam tip: Store the unrounded value 0.838998… in your calculator and use it in part (d). Using 0.84 changes the third significant figure of the final answer.
310IBMA2H-051-02-084.13 Bayes’ theoremPaper 2Medium7 marks
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A ground station receives telemetry packets through three relay satellites, \(A\), \(B\) and \(C\). They carry 45%, 35% and 20% of the packets respectively. The probability that a packet is corrupted is 0.008 via \(A\), 0.015 via \(B\) and 0.03 via \(C\).
Tree diagram with the probabilities given in the question (Cor.: corrupted)
(a)
Find the probability that a randomly chosen packet is corrupted.
(2)
(b)
A packet is found to be corrupted. Find the probability that it was sent via satellite \(C\).
(3)
(c)
A packet is found not to be corrupted. Find the probability that it was sent via satellite \(A\).
Exam tip: Although satellite C carries only 20% of the traffic, it accounts for about 40% of corrupted packets. A quick check that the result is plausible like this helps you spot a reversed conditional.
311IBMA3H-4-19r6-054.9 The normal distributionPaper 2Medium7 marks
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A machine fills bottles of cough syrup. The volume dispensed, X ml, is normally distributed with mean μ and standard deviation σ. It is known that 7% of bottles contain less than 97 ml and 2% of bottles contain more than 104 ml.
(a)
Find the values of μ and σ, giving each answer to 3 significant figures.
(5)
(b)
The machine is recalibrated so that every volume dispensed is multiplied by 1.05, giving a new volume Y = 1.05X. Find the mean and the standard deviation of Y, to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of inverse normal: P(X < 97) = 0.07 and P(X > 104) = 0.02
✓ 1
A1 z = −1.476 and z = 2.054 (accept 3 s.f.)
✓ 1
M1 two equations (97 − μ)/σ = −1.476 and (104 − μ)/σ = 2.054
✓ 1
A1 σ = 1.98
✓ 1
A1 μ = 99.9
✓ 1
Part (b)
A1 E(Y) = 1.05 × 99.9 = 105
✓ 1
A1 SD(Y) = 1.05 × σ = 2.08
✓ 1
Exam tip: Keep the unrounded values of μ and σ in your calculator memory for later parts, and round only the final answer.
312IBMA3H-4-19r6-064.6 Combined, conditional and independent eventsPaper 2Medium6 marks
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In a role-playing game each player chooses one character: a Mage with probability 0.5, an Archer with probability 0.3, or a Knight with probability 0.2. The probability that a player wins a match is 0.7 for a Mage, 0.5 for an Archer and 0.4 for a Knight.
Tree diagram with the probabilities given in the question
(a)
Find the probability that a randomly chosen player wins a match.
(2)
(b)
Given that a player wins, find the probability that the player chose a Knight. Give your answer to 3 significant figures.
(2)
(c)
Show that choosing an Archer and winning a match are not independent events.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sum of three branch products, 0.5 × 0.7 + 0.3 × 0.5 + 0.2 × 0.4
✓ 1
A1 0.58
✓ 1
Part (b)
M1 use of P(Knight | win) = P(Knight ∩ win)/P(win) = 0.08/0.58 (ECF from (a))
R1 0.15 ≠ 0.174, so the events are not independent
✓ 1
Exam tip: Write the tree or table before using the conditional-probability formula; the numerator in Bayes-type questions is one branch product, the denominator is the sum of all branches.
313IBMA3H-4-19r6-084.4 Correlation and regression of y on xPaper 2Medium6 marks
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A games company studies the number of hours, x, that a player spends practising each week and the player's rank score, y. Data from 30 players give rank scores from 900 to 2100, with mean rank score 1450 and a correlation coefficient r = 0.91. The regression line of x on y is x = 0.0042y − 3.1.
(a)
Find the mean number of practice hours per week for these players.
(2)
(b)
Estimate the number of practice hours for a player whose rank score is 1800.
(2)
(c)
A manager uses the line to estimate the practice hours of a player whose rank score is 3200. Comment on the reliability of this estimate.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 the regression line passes through (x̄, ȳ), so substitute y = 1450
✓ 1
A1 x̄ = 2.99
✓ 1
Part (b)
M1 substitute y = 1800
✓ 1
A1 4.46
✓ 1
Part (c)
R1 3200 lies outside the range of the data (900 to 2100), so this is extrapolation
✓ 1
R1 the linear relationship may not continue beyond the data, so the estimate is unreliable (even though r is high)
✓ 1
Exam tip: A high correlation coefficient supports estimates inside the data range only; always compare the value you substitute with the range of the data.
314IBMA3H-4-19r6-094.11 Conditional probability and independencePaper 2Medium7 marks
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Customers of an online game may buy an expansion pack, event A, and may take a monthly subscription, event B. It is known that P(B) = 0.5, P(A | B) = 0.4 and P(A ∪ B) = 0.65.
(a)
Find P(A ∩ B).
(1)
(b)
Find P(A).
(2)
(c)
Determine whether A and B are independent. Justify your answer.
(2)
(d)
Find the probability that a customer who has not bought the expansion pack takes a subscription. Give your answer to 3 significant figures.
Exam tip: A Venn diagram with the intersection entered first makes every other region a simple subtraction.
315IBMA3H-4-19r6-104.13 Bayes’ theoremPaper 2Medium7 marks
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A hospital screening test for a condition is applied to a population in which 2% of people have the condition. If a person has the condition, the test is positive with probability 0.96. If a person does not have the condition, the test is negative with probability 0.94. Give probabilities to 3 significant figures where not exact.
Tree diagram with the probabilities given in the question (C: has the condition, +: positive test)
(a)
Find the probability that a randomly chosen person tests positive.
(2)
(b)
A person tests positive. Find the probability that the person has the condition.
(2)
(c)
A person who tests positive is tested again, and the two results are independent given the person's true condition. Given that both tests are positive, find the probability that the person has the condition.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.02 × 0.96 + 0.98 × 0.06
✓ 1
A1 0.078
✓ 1
Part (b)
M1 use of Bayes: 0.0192/0.078 (ECF from (a))
✓ 1
A1 0.246
✓ 1
Part (c)
M1 P(both positive | condition) = 0.962 and P(both positive | no condition) = 0.062
✓ 1
M1 0.02 × 0.962/(0.02 × 0.962 + 0.98 × 0.062)
✓ 1
A1 0.839
✓ 1
Common error: The probability that someone has the condition given a positive test is not the same as the probability of a positive test given the condition; do not quote the test's accuracy as the answer.
316IBMA3H-4-19r6-124.9 The normal distributionPaper 2Medium8 marks
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A water company delivers bottled water. The mass of water in a bottle, W grams, is normally distributed with mean 500 and standard deviation 4. The mass of the empty bottle and cap, B grams, is independently normally distributed with mean 24 and standard deviation 1.5. Let T = W + B be the mass of a full bottle.
(a)
Find E(T) and Var(T).
(3)
(b)
Find the probability that a full bottle has mass greater than 530 g. Give your answer to 3 significant figures.
(2)
(c)
A crate holds 6 full bottles whose masses are independent. Find the probability that the total mass of the bottles in a crate exceeds 3170 g. Give your answer to 3 significant figures.
Common error: The total of six independent bottles has variance 6σ², not 36σ²; that would be the variance of six times one bottle.
317IBMA3H-4-19r7-044.2 Presentation of dataPaper 2Medium7 marks
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The time taken by riders to complete a cycling stage is normally distributed with standard deviation 0.35 hours. 12% of riders take longer than 4.2 hours.
(a)
Find the mean time, giving your answer to 3 significant figures.
(3)
(b)
Using μ = 3.79, find the probability that a rider completes the stage in less than 3.5 hours.
(2)
(c)
The fastest 10% of riders receive a medal. Using μ = 3.79, find the longest time that earns a medal, to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 z-value for P(Z > z) = 0.12, z = 1.175
✓ 1
M1 4.2 = μ + 1.175 × 0.35 or equivalent
✓ 1
A1 μ = 3.79 hours
✓ 1
Part (b)
M1 attempt to find P(X < 3.5) with μ = 3.79, σ = 0.35
✓ 1
A1 0.204
✓ 1
Part (c)
M1 uses z = −1.2816 for the lower 10%
✓ 1
A1 3.79 − 1.2816 × 0.35 = 3.34 hours
✓ 1
Common error: A "longer than" probability is an upper tail: use P(Z < z) = 0.88 when finding z, not 0.12.
318IBMA3H-4-19r7-064.2 Presentation of dataPaper 2Medium11 marks
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The lengths, l mm, of 100 pottery sherds from a site are summarised in the table.
Length (mm): 0 ≤ l < 10 | 10 ≤ l < 20 | 20 ≤ l < 30 | 30 ≤ l < 40 | 40 ≤ l < 50 Frequency: 6 | 19 | 38 | 24 | 13
(a)
Estimate the mean length, using the mid-interval values.
(2)
(b)
Estimate the standard deviation of the lengths, to 3 significant figures.
(2)
(c)
Write down the modal class.
(1)
(d)
Each length is found to have been recorded 2 mm too small. State the new mean and the new standard deviation.
(2)
(e)
Use linear interpolation on the cumulative frequencies to estimate the median length, to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses mid-interval values 5, 15, 25, 35, 45
✓ 1
A1 26.9 mm
✓ 1
Part (b)
M1 attempt to use the standard deviation function with mid-interval values and frequencies
✓ 1
A1 10.7 mm (σ = 10.742)
✓ 1
Part (c)
A1 20 ≤ l < 30
✓ 1
Part (d)
A1 new mean 28.9 mm
✓ 1
A1 standard deviation unchanged, 10.7 mm
✓ 1
Part (e)
M1 cumulative frequencies 6, 25, 63, 87, 100 (or the median lies in 20 ≤ l < 30)
✓ 1
M1 median position 50, with 25 sherds below 20
✓ 1
M1 20 + (50 − 25)/38 × 10
✓ 1
A1 26.6 mm
✓ 1
Common error: Adding a constant to every value shifts the mean but leaves the spread unchanged.
319IBMA3H-4-19r7-074.12 Standardization and unknown normal parametersPaper 2Medium7 marks
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The pressure, P bar, in new racing-bicycle tyres after inflation is normally distributed with mean 7.0 and standard deviation 0.4. A tyre is rejected if its pressure is below 6.2 bar.
(a)
Find the probability that a tyre is rejected.
(2)
(b)
Find the probability that a tyre has a pressure above 7.4 bar, given that it is not rejected.
(3)
(c)
A batch of 12 tyres is chosen at random. Using the unrounded probability from part (a), find the probability that at least one tyre is rejected.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(P < 6.2) with the normal distribution (z = −2)
✓ 1
A1 0.0228
✓ 1
Part (b)
M1 conditional probability P(P > 7.4 | P ≥ 6.2) = P(P > 7.4) ÷ P(P ≥ 6.2)
✓ 1
A1 P(P > 7.4) = 0.1587 and P(P ≥ 6.2) = 0.9772
✓ 1
A1 0.162
✓ 1
Part (c)
M1 1 − 0.977249912 (binomial with n = 12)
✓ 1
A1 0.241
✓ 1
Exam tip: A conditional probability P(A | B) divides P(A and B) by P(B); here the event "above 7.4" is already inside "not rejected".
320IBMA3H-4-19r7-104.14 Variance, continuous random variables and linear transformationsPaper 2Medium10 marks
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The running time, X hours, of a job on a data-centre server has probability density function \(f(x) = kx^2\) for \(1 \le x \le 3\), and \(f(x) = 0\) otherwise.
Graph of the probability density function, not to scale
(a)
Show that \(k = \frac{3}{26}\).
(3)
(b)
Find \(\mathrm{E}(X)\), giving your answer as a fraction.
(2)
(c)
Find \(\mathrm{P}(X > 2)\).
(2)
(d)
Find the median running time, to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\int_1^3 kx^2\,dx = 1\)
✓ 1
A1 \(k\left[\frac{x^3}{3}\right]_1^3 = k \times \frac{26}{3}\)
A1 \(\frac{m^3 - 1}{26} = \frac{1}{2}\) so \(m^3 = 14\)
✓ 1
A1 m = 2.41 hours
✓ 1
Exam tip: For the median m of a continuous distribution, solve ∫ from the lower limit to m of f(x) dx = 0.5.
321IBMA3H-4-19r7-134.14 Variance, continuous random variables and linear transformationsPaper 2Medium9 marks
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In a data centre, the number of server restarts in a day is a discrete random variable X with the following distribution.
x: 0 | 1 | 2 | 3 P(X = x): 0.1 | p | q | 0.3
It is given that E(X) = 1.7.
(a)
Find the value of p and the value of q.
(4)
(b)
Show that \(\mathrm{Var}(X) = 1.01\).
(3)
(c)
Each restart costs £25 and there is a fixed daily charge of £40, so the daily cost in £ is C = 40 + 25X. Find the standard deviation of C, to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 p + q + 0.1 + 0.3 = 1 so p + q = 0.6
✓ 1
M1 E(X) = p + 2q + 0.9 = 1.7 so p + 2q = 0.8
✓ 1
A1 q = 0.2
✓ 1
A1 p = 0.4
✓ 1
Part (b)
M1 E(X²) = 0.4 + 4(0.2) + 9(0.3)
✓ 1
A1 E(X²) = 3.9
✓ 1
A1 Var(X) = 3.9 − 1.7² = 1.01 (AG)
✓ 1
Part (c)
M1 Var(C) = 25² Var(X) = 625 × 1.01
✓ 1
A1 25.1 (£)
✓ 1
Common error: In Var(aX + b) the constant b has no effect and the factor a is squared.
322IBMA3H-4-19r9-044.11 Conditional probability and independencePaper 2Medium7 marks
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For a container ship arriving at a port, let \(L\) be the event that it arrives late and \(I\) the event that it is selected for customs inspection. It is known that \(P(L)=0.4\), \(P(I)=0.35\) and \(P(L\cup I)=0.61\).
(a)
Find \(P(L\cap I)\).
(2)
(b)
Show that \(L\) and \(I\) are independent.
(2)
(c)
Find the probability that a ship experiences exactly one of the two events.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(P(L\cup I)=P(L)+P(I)-P(L\cap I)\)
✓ 1
A1 \(P(L\cap I)=0.14\)
✓ 1
Part (b)
M1 \(P(L)\times P(I)=0.4\times0.35=0.14\)
✓ 1
R1 this equals \(P(L\cap I)\), so the events are independent
✓ 1
Part (c)
M1 \(P(L)-P(L\cap I)\) and \(P(I)-P(L\cap I)\) attempted
✓ 1
A1 \(0.26\) and \(0.21\)
✓ 1
A1 \(0.47\)
✓ 1
Common error: Independence needs \(P(A\cap B)=P(A)P(B)\) to be checked; it is not the same as the events being mutually exclusive.
323IBMA3H-4-19r9-064.3 Measures of central tendency and dispersionPaper 2Medium7 marks
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The ticket prices for a theatre's shows have mean €38 and standard deviation €6.50. From next season each price \(p\) will be replaced by \(1.1p+3\) euros (a 10% rise plus a €3 booking fee).
(a)
Find the new mean price.
(2)
(b)
Find the new standard deviation.
(2)
(c)
A second theatre applies the same rule. Its new prices have mean €58 and standard deviation €8.80. Find the standard deviation of its original prices.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(1.1\times38+3\)
✓ 1
A1 €44.80
✓ 1
Part (b)
M1 multiply by 1.1 only (the €3 does not affect spread)
✓ 1
A1 €7.15
✓ 1
Part (c)
M1 original sd \(=\frac{8.80}{1.1}\) (adding 3 ignored)
✓ 1
M1 recognising division by 1.1 undoes the scaling
✓ 1
A1 €8.00
✓ 1
Common error: Adding a constant shifts the mean but leaves the standard deviation unchanged; only multiplication changes the spread.
324IBMA3H-4-19r9-084.2 Presentation of dataPaper 2Medium10 marks
Calculator
A shipping insurer models the number of containers, \(X\), lost on a voyage by the distribution below.
The cost of a voyage, in pounds, is \(C=8000+2500X\).
(a)
Find \(E(X)\).
(2)
(b)
Find \(\operatorname{Var}(X)\).
(3)
(c)
Find \(E(C)\).
(2)
(d)
Find the standard deviation of \(C\), to the nearest pound.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\sum xP(X=x)\)
✓ 1
A1 \(0.93\)
✓ 1
Part (b)
M1 \(E(X^2)=\sum x^2P(X=x)\)
✓ 1
A1 \(E(X^2)=2.01\)
✓ 1
A1 \(\operatorname{Var}(X)=2.01-0.93^2=1.1451\)
✓ 1
Part (c)
M1 \(E(C)=8000+2500E(X)\)
✓ 1
A1 £10 325
✓ 1
Part (d)
M1 \(\operatorname{Var}(C)=2500^2\operatorname{Var}(X)\) (ECF from (b))
✓ 1
A1 \(\operatorname{Var}(C)=7\,156\,875\)
✓ 1
A1 £2675
✓ 1
Common error: \(\operatorname{Var}(aX+b)=a^2\operatorname{Var}(X)\): the constant \(b\) drops out and \(a\) must be squared.
325IBMA3H-4-19r9-094.14 Variance, continuous random variables and linear transformationsPaper 2Medium7 marks
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The time, \(X\) years, between two eruptions of a volcano is modelled by the probability density function \(f(x)=\frac{3x^2}{64}\) for \(0\le x\le4\), and \(f(x)=0\) otherwise.
Graph of the probability density function, not to scale
(a)
Find \(E(X)\).
(2)
(b)
Find \(P(X>3)\).
(2)
(c)
Find the median time between eruptions, giving your answer to 3 significant figures.
Exam tip: For the median of a continuous distribution, set the integral from the lower limit to \(m\) equal to \(\frac12\).
326IBMA3H-4-19r9-134.1 Sampling, bias and outliersPaper 2Medium9 marks
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A ferry company keeps an alphabetical list of its 1240 passengers on a voyage. A systematic sample of 40 passengers is taken: a starting passenger is chosen at random from the first few names, and then every \(k\)th name after that is selected. The starting passenger is number 17.
(a)
Find the value of \(k\).
(2)
(b)
Write down the numbers of the next two passengers selected after number 17.
(1)
(c)
Determine whether passenger number 700 is selected. Justify your answer.
(3)
(d)
State one disadvantage of systematic sampling from a list.
R1 not an integer, so 700 is not selected (699 is)
✓ 1
Part (d)
R1 if the list has a repeating pattern that matches \(k\) the sample may be biased; or not every combination of passengers is possible
✓ 1
Part (e)
M1 \(17+31\times39\)
✓ 1
A1 1226
✓ 1
Exam tip: With 40 selections the first passenger is \(n=0\) and the last is \(n=39\); count carefully to avoid an off-by-one slip.
327IBMA3H-4-19r14-054.14 Variance, continuous random variables and linear transformationsPaper 2Medium6 marks
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At a supermarket, each customer independently uses a digital coupon with probability 0.06.
(a)
Find the smallest number of customers n for which the probability that at least one customer uses a coupon is greater than 0.95.
(4)
(b)
Using your value of n from part (a), find the probability that exactly three of the n customers use a coupon. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(X ≥ 1) = 1 − 0.94n
✓ 1
M1 set up 1 − 0.94n > 0.95 (or 0.94n < 0.05)
✓ 1
M1 solve, e.g. n > ln 0.05 / ln 0.94 = 48.4 (or table: n = 48 gives 0.9487, n = 49 gives 0.9518)
✓ 1
A1 n = 49
✓ 1
Part (b)
M1 C(49,3)(0.06)3(0.94)46 (ECF from (a))
✓ 1
A1 0.231
✓ 1
Exam tip: For "smallest n" questions, finish by checking the value on either side of your solution, such as n = 48 and n = 49.
328IBMA3H-4-19r14-064.13 Bayes’ theoremPaper 2Medium8 marks
Calculator
In a region, 2% of people have a particular infection. A screening test gives a positive result for 96% of people who have the infection and for 3% of people who do not. Give all answers to 3 significant figures.
Tree diagram showing the probabilities given in the question
(a)
Find the probability that a randomly chosen person tests positive.
(3)
(b)
Given that a person tests positive, find the probability that they have the infection.
(2)
(c)
A person is tested twice, and the two results are independent given the person's true condition. Find the probability that the person has the infection given that both tests are positive.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 tree or table: P(D ∩ +) = 0.02 × 0.96 and P(D′ ∩ +) = 0.98 × 0.03
Common error: P(positive | infected) and P(infected | positive) are different quantities; the second needs the prevalence as well.
329IBMA3H-4-19r14-084.4 Correlation and regression of y on xPaper 2Medium7 marks
Calculator
A retailer studies its monthly advertising spend x (in €1000) and monthly sales y (in €1000) over a year. The spend ranged from 5 to 11. Summary statistics: mean spend 8, mean sales 40, standard deviation of spend 1.5, standard deviation of sales 6.0, and correlation coefficient r = 0.88.
(a)
Find the equation of the regression line of y on x.
(3)
(b)
Estimate the monthly sales when the advertising spend is €10 000. Give your answer to 3 significant figures, in €1000.
(2)
(c)
The retailer proposes to use the line to estimate sales when the spend is €25 000. Give one reason why this estimate would be unreliable.
(1)
(d)
The retailer wants to know the spend needed for sales of €60 000 and rearranges the equation in (a) to find x. Explain why this method is not appropriate.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 covariance sxy = r sx sy = 0.88 × 1.5 × 6 = 7.92
✓ 1
M1 gradient = sxy / sx2 = 7.92 / 2.25 = 3.52, and line passes through (8, 40)
✓ 1
A1 y = 3.52x + 11.84
✓ 1
Part (b)
M1 substitute x = 10 (ECF from (a))
✓ 1
A1 47.0
✓ 1
Part (c)
R1 25 is far outside the range of the data (5 to 11), so this is extrapolation
✓ 1
Part (d)
R1 the line of y on x is for predicting y from x; to estimate x from y the regression line of x on y should be used
✓ 1
Common error: Rearranging the y-on-x line to predict x gives a different (and less reliable) answer from the x-on-y line unless r = ±1.
330IBMA3H-4-19r14-094.2 Presentation of dataPaper 2Medium6 marks
Calculator
A wind farm has 25 turbines. At any given time each turbine is operating independently with probability 0.9. Let X be the number operating.
(a)
Find the probability that at least 22 turbines are operating. Give your answer to 3 significant figures.
(2)
(b)
Given that at least 20 turbines are operating, find the probability that all 25 are operating. Give your answer to 3 significant figures.
(4)
Show mark scheme
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Mark
Notes
Part (a)
M1 X ~ B(25, 0.9) and P(X ≥ 22) = 1 − P(X ≤ 21)
✓ 1
A1 0.764
✓ 1
Part (b)
M1 P(X = 25 | X ≥ 20) = P(X = 25) / P(X ≥ 20)
✓ 1
A1 P(X = 25) = 0.925 = 0.0718
✓ 1
A1 P(X ≥ 20) = 0.9666 (to 4 s.f.)
✓ 1
A1 0.0743
✓ 1
Exam tip: A conditional probability with an inequality condition needs the cumulative probability in the denominator; keep full calculator values until the end.
331IBMA3H-4-19r14-124.8 Binomial distributionPaper 2Medium8 marks
Calculator
The number of patients X (out of a fixed number of independent admissions) who need intensive care follows a binomial distribution B(n, p) with mean 6 and variance 4.2.
(a)
Find n and p.
(3)
(b)
Find P(X ≥ 8), correct to 3 significant figures.
(2)
(c)
The treatment cost in £ is Y = 40X + 150. Find E(Y) and Var(Y).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 np = 6 and np(1 − p) = 4.2
✓ 1
M1 1 − p = 4.2 / 6 = 0.7
✓ 1
A1 p = 0.3 and n = 20
✓ 1
Part (b)
M1 1 − P(X ≤ 7) using X ~ B(20, 0.3)
✓ 1
A1 0.228
✓ 1
Part (c)
A1 E(Y) = 40 × 6 + 150 = 390
✓ 1
M1 Var(Y) = 402 × 4.2
✓ 1
A1 6720
✓ 1
Exam tip: Mean and variance of a binomial give two equations; dividing the variance by the mean gives 1 − p directly.
332IBMA3H-4-19r15-034.1 Sampling, bias and outliersPaper 2Medium7 marks
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A public health team wants to estimate the proportion of adults in a city of 240000 adults who have had a flu vaccination this winter. They post a short online survey on the city council's social media page and use the first 500 replies.
(a)
Name the type of sampling used.
(1)
(b)
Give two reasons why this sample may be biased.
(2)
(c)
Of the 500 replies, 380 said they had been vaccinated. Use this sample to estimate the number of vaccinated adults in the city.
(2)
(d)
Describe how the team could select a simple random sample of 500 adults.
(2)
Show mark scheme
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Mark
Notes
Part (a)
A1 convenience sampling (accept self-selected or volunteer sampling)
✓ 1
Part (b)
R1 first reason, e.g. only people who use social media or follow the page can respond
✓ 1
R1 second reason, e.g. people with strong views or health-conscious people are more likely to reply; only early repliers are included
✓ 1
Part (c)
M1 \(\frac{380}{500}\times 240000\)
✓ 1
A1 182400
✓ 1
Part (d)
M1 obtain a list of all adults and number them
✓ 1
A1 use random numbers (for example from a random number generator) to choose 500 different people, so each adult has an equal chance
✓ 1
Exam tip: When asked about bias, say who is missing from the sample or who is over-represented, not just that the sample is 'unfair'.
333IBMA3H-4-19r15-054.8 Binomial distributionPaper 2Medium7 marks
Calculator
A shop sells scratch cards for €1.50 each. The prize \(X\) (in euros) on a card has the following distribution, where \(k\) is a constant. \(P(X=0)=0.80\), \(P(X=2)=0.15\), \(P(X=k)=0.05\).
(a)
Find \(E(X)\) in terms of \(k\).
(2)
(b)
The game is fair: the expected prize equals the price of a card. Find the value of \(k\).
(2)
(c)
The shop changes the top prize to €20. In one month it sells 4000 cards. Find the shop's expected profit from the cards that month, ignoring all other costs.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\sum xP(X=x)=0(0.80)+2(0.15)+k(0.05)\)
✓ 1
A1 \(E(X)=0.3+0.05k\)
✓ 1
Part (b)
M1 \(0.3+0.05k=1.5\)
✓ 1
A1 \(k=24\)
✓ 1
Part (c)
M1 \(E(X)=0.3+0.05\times20=1.30\)
✓ 1
M1 expected profit per card \(1.50-1.30=0.20\)
✓ 1
A1 €800
✓ 1
Common error: A fair game means expected gain is zero: compare the expected prize with the cost of playing, not with the largest prize.
334IBMA3H-4-19r15-064.12 Standardization and unknown normal parametersPaper 2Medium7 marks
Calculator
An engineer inspects three solar panels chosen at random from a large delivery. The random variable \(X\) is the number of faulty panels found, with distribution \(P(X=0)=0.15\), \(P(X=1)=p\), \(P(X=2)=q\), \(P(X=3)=0.25\). It is known that \(E(X)=1.55\).
(a)
Find the value of \(p\) and the value of \(q\).
(4)
(b)
Find the exact value of \(\mathrm{Var}(X)\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(0.15+p+q+0.25=1\) so \(p+q=0.6\)
✓ 1
M1 \(E(X)=p+2q+3(0.25)=1.55\) so \(p+2q=0.8\)
✓ 1
A1 \(q=0.2\)
✓ 1
A1 \(p=0.4\)
✓ 1
Part (b)
M1 \(E(X^2)=0.4+4(0.2)+9(0.25)\)
✓ 1
A1 \(E(X^2)=3.45\)
✓ 1
A1 \(\mathrm{Var}(X)=3.45-1.55^2=1.0475\)
✓ 1
Common error: Remember to subtract the square of the mean: \(E(X^2)\) on its own is not the variance.
335IBMA3H-4-19r15-074.14 Variance, continuous random variables and linear transformationsPaper 2Medium7 marks
Calculator
A rooftop solar array produces \(X\) kWh of electricity per day, with mean 18.4 and standard deviation 3.2. The owner sells electricity at £0.15 per kWh and pays a fixed daily charge of £2.40, so the daily profit in pounds is \(P=0.15X-2.40\).
(a)
Find \(E(P)\).
(2)
(b)
Find the standard deviation of \(P\).
(2)
(c)
The variable \(Y=aX+b\), where \(a>0\), has mean 0 and standard deviation 1. Find the value of \(a\) and the value of \(b\).
(3)
Show mark scheme
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Mark
Notes
Part (a)
M1 \(0.15\times18.4-2.40\)
✓ 1
A1 £0.36
✓ 1
Part (b)
M1 \(\mathrm{Var}(P)=0.15^2\times3.2^2\) (0.2304) or sd \(=0.15\times3.2\)
✓ 1
A1 £0.48
✓ 1
Part (c)
M1 \(a=\frac{1}{3.2}\)
✓ 1
A1 \(a=0.3125\)
✓ 1
A1 \(b=-5.75\)
✓ 1
Exam tip: The constant added in a linear transformation changes the mean but never the variance or standard deviation.
336IBMA3H-4-19r15-094.14 Variance, continuous random variables and linear transformationsPaper 2Medium8 marks
Calculator
The time \(X\) hours taken for a grid battery to charge fully has probability density function \(f(x)=\frac{3}{64}x^2\) for \(0\le x\le4\), and \(f(x)=0\) otherwise.
(a)
Show that \(E(X)=3\).
(3)
(b)
Find the median charging time, correct to 3 significant figures.
(3)
(c)
By comparing the mean and the median, state what the distribution's skew is, giving a reason.
R1 so the distribution is negatively skewed (a longer tail towards small values)
✓ 1
Common error: The median satisfies \(\int_0^m f(x)\,dx=0.5\); do not set \(f(m)=0.5\).
337IBMA3H-4-19r15-104.11 Conditional probability and independencePaper 2Medium7 marks
Calculator
At a retail park café, event \(A\) is that a customer buys a hot drink and event \(B\) is that the customer buys a pastry. It is known that \(P(A)=0.3\), \(P(A\cup B)=0.58\), and that \(A\) and \(B\) are independent.
(a)
Find \(P(B)\).
(3)
(b)
Find the probability that a customer buys exactly one of the two items.
(2)
(c)
Event \(C\) is that a customer buys a gift card, with \(P(C)=0.15\). Events \(C\) and \(A\) are mutually exclusive, as are \(C\) and \(B\). Find the probability that a customer buys none of the three items.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(P(A\cap B)=P(A)P(B)\)
✓ 1
M1 \(0.58=0.3+P(B)-0.3P(B)\)
✓ 1
A1 \(P(B)=0.4\)
✓ 1
Part (b)
M1 \(P(A\cup B)-P(A\cap B)\) or \(0.3\times0.6+0.7\times0.4\) (ECF from (a))
✓ 1
A1 0.46
✓ 1
Part (c)
M1 \(P(A\cup B\cup C)=0.58+0.15\)
✓ 1
A1 0.27
✓ 1
Common error: Independent and mutually exclusive are different: independent events satisfy \(P(A\cap B)=P(A)P(B)\), mutually exclusive events have \(P(A\cap B)=0\).
338IBMA3H-4-19r15-134.1 Sampling, bias and outliersPaper 2Medium8 marks
Calculator
The waiting times, in minutes, of eight patients at a walk-in clinic are 12, 15, 17, 18, 21, 22, 25, 71. Take the lower quartile \(Q_1\) as the median of the lowest four values and the upper quartile \(Q_3\) as the median of the highest four values. A value is an outlier if it is more than \(1.5\times\mathrm{IQR}\) above \(Q_3\) or below \(Q_1\).
(a)
Find \(Q_1\) and \(Q_3\).
(2)
(b)
Show that 71 is an outlier and that there is no outlier at the lower end.
(3)
(c)
Find the mean waiting time (i) for all eight patients and (ii) with the outlier removed, each to 3 significant figures.
(2)
(d)
The clinic manager wants to publish the mean waiting time. Make one reasoned comment on whether 71 should be removed.
(1)
Show mark scheme
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Mark
Notes
Part (a)
A1 \(Q_1=16\)
✓ 1
A1 \(Q_3=23.5\)
✓ 1
Part (b)
A1 \(\mathrm{IQR}=7.5\)
✓ 1
M1 upper boundary \(23.5+1.5\times7.5=34.75\)
✓ 1
R1 71 > 34.75 so it is an outlier; lower boundary \(16-11.25=4.75<12\) so no low outlier
✓ 1
Part (c)
A1 (i) \(\frac{201}{8}=25.1\)
✓ 1
A1 (ii) \(\frac{130}{7}=18.6\)
✓ 1
Part (d)
R1 any valid comment, e.g. check whether it is an error; if genuine it is a real wait and removing it understates typical delays, but the mean is sensitive to it so the median could also be reported
✓ 1
Exam tip: Outliers should be investigated before being removed; a reason for removal is a mark, the calculation alone is not.
339IBMA3H-4-19r17-034.12 Standardization and unknown normal parametersPaper 2Medium8 marks
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The length of time, \(X\) days, for which a library book is kept on loan is normally distributed with mean \(14\) days. It is known that \(\mathrm{P}(X>21)=0.08\). In parts (b) and (c) use the unrounded value of \(\sigma\) from part (a).
Sketch of the normal curve with the given tail probabilities, not to scale
(a)
Find the standard deviation of \(X\), correct to 3 significant figures.
(3)
(b)
Find the probability that a book is kept for fewer than \(10\) days.
(2)
(c)
The library introduces an extra fee for the longest \(1\%\) of loan periods. Find the number of days, to 3 significant figures, beyond which the extra fee applies.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\mathrm{P}(Z<z)=0.92\) so \(z=1.4051\) (accept inverse normal giving \(1.405\))
✓ 1
M1 \(\dfrac{21-14}{\sigma}=1.4051\)
✓ 1
A1 \(\sigma=4.98\)
✓ 1
Part (b)
M1 use of normal distribution with mean \(14\) and standard deviation \(\sigma\) (or \(z=\frac{10-14}{\sigma}\))
✓ 1
A1 \(0.211\)
✓ 1
Part (c)
M1 \(z=2.3263\) for \(\mathrm{P}(Z<z)=0.99\) (or inverse normal with mean \(14\) and standard deviation \(\sigma\))
✓ 1
M1 \(14+2.3263\times\sigma\)
✓ 1
A1 \(25.6\) days
✓ 1
Exam tip: Keep the unrounded standard deviation in your calculator memory and use it in later parts; the answer 4.98 is only for reporting.
340IBMA3H-4-19r17-044.12 Standardization and unknown normal parametersPaper 2Medium11 marks
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The mean hourly wind speed, \(V\) m s−1, at a proposed wind farm site is normally distributed with mean \(\mu\) and standard deviation \(\sigma\). The wind speed is below \(6\) m s−1 for \(15\%\) of hours and above \(11\) m s−1 for \(30\%\) of hours. In parts (c) and (d) use the unrounded values of \(\mu\) and \(\sigma\).
Sketch of the normal curve with the given tail probabilities, not to scale
(a)
Write down the \(z\)-values corresponding to \(\mathrm{P}(V<6)=0.15\) and \(\mathrm{P}(V<11)=0.70\), correct to 3 significant figures.
(2)
(b)
Hence find the values of \(\mu\) and \(\sigma\), each correct to 3 significant figures.
(4)
(c)
Find \(\mathrm{P}(V>9)\).
(2)
(d)
The turbines shut down for safety when the wind speed is in the top \(5\%\) of hourly values. Find the wind speed, to 3 significant figures, above which the shutdown occurs.
(3)
Show mark scheme
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Mark
Notes
Part (a)
A1 \(z=-1.04\) for \(V=6\)
✓ 1
A1 \(z=0.524\) for \(V=11\)
✓ 1
Part (b)
M1 \(6=\mu-1.0364\sigma\) and \(11=\mu+0.5244\sigma\) (accept rounded \(z\)-values)
✓ 1
M1 attempt to solve the simultaneous equations (subtract: \(1.5608\sigma=5\))
✓ 1
A1 \(\sigma=3.20\)
✓ 1
A1 \(\mu=9.32\)
✓ 1
Part (c)
M1 attempt to use the normal distribution with \(\mu=9.32\), \(\sigma=3.20\) (unrounded values)
✓ 1
A1 \(0.540\)
✓ 1
Part (d)
M1 \(z=1.6449\) for \(\mathrm{P}(Z<z)=0.95\)
✓ 1
M1 \(\mu+1.6449\sigma\)
✓ 1
A1 \(14.6\) m s−1
✓ 1
Common error: The upper tail P(V>11)=0.30 must be converted to P(V<11)=0.70 before using inverse normal; using 0.30 gives a negative z-value.
341IBMA3H-4-19r17-054.13 Bayes’ theoremPaper 2Medium9 marks
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A wind farm operator fits each turbine with a vibration sensor. In any given month, \(4\%\) of turbines develop a bearing fault. If a turbine has a fault, the sensor raises an alarm with probability \(0.92\). If a turbine has no fault, the sensor still raises a (false) alarm with probability \(0.06\). A turbine is chosen at random.
Tree diagram showing the probabilities given in the question
(a)
Find the probability that the sensor raises an alarm.
(3)
(b)
Given that the sensor raises an alarm, find the probability that the turbine has a bearing fault.
(3)
(c)
Given that the sensor does not raise an alarm, find the probability that the turbine has no fault. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to add two products from a tree diagram or table, e.g. \(0.04\times0.92+0.96\times0.06\)
✓ 1
A1 correct products \(0.0368\) and \(0.0576\)
✓ 1
A1 \(0.0944\)
✓ 1
Part (b)
M1 use of Bayes' theorem or \(\dfrac{\mathrm{P}(\text{fault}\cap\text{alarm})}{\mathrm{P}(\text{alarm})}\)
M1 divide by \(\mathrm{P}(\text{no alarm})=1-0.0944=0.9056\)
✓ 1
A1 \(0.996\)
✓ 1
Exam tip: Write the conditional probability as a fraction of two joint probabilities before substituting numbers; the denominator is the total probability of the given event.
342IBMA3H-4-19r17-084.14 Variance, continuous random variables and linear transformationsPaper 2Medium12 marks
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The time, \(X\) years, until a gearbox component in a wind turbine has to be replaced has probability density function \(f(x)=kx^2\) for \(0\le x\le3\), and \(f(x)=0\) otherwise.
Graph of the probability density function, not to scale
(a)
Show that \(k=\dfrac{1}{9}\).
(2)
(b)
Find the median of \(X\), giving your answer to 3 significant figures.
(3)
(c)
Find \(\mathrm{E}(X)\) and the standard deviation of \(X\), the latter correct to 3 significant figures.
(5)
(d)
Find the probability that a component lasts longer than the mean time to replacement.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\int_0^3 kx^2\,dx=1\)
✓ 1
A1 \(9k=1\) so \(k=\dfrac{1}{9}\) (AG)
✓ 1
Part (b)
M1 \(\int_0^m \dfrac{x^2}{9}\,dx=\dfrac{1}{2}\)
✓ 1
A1 \(\dfrac{m^3}{27}=\dfrac{1}{2}\)
✓ 1
A1 \(m=2.38\) years
✓ 1
Part (c)
M1 \(\mathrm{E}(X)=\int_0^3 \dfrac{x^3}{9}\,dx\)
✓ 1
A1 \(\mathrm{E}(X)=\dfrac{9}{4}=2.25\)
✓ 1
M1 \(\mathrm{E}(X^2)=\int_0^3\dfrac{x^4}{9}\,dx=\dfrac{27}{5}\) and \(\mathrm{Var}(X)=\dfrac{27}{5}-\left(\dfrac{9}{4}\right)^2\)
✓ 1
A1 \(\mathrm{Var}(X)=0.3375\)
✓ 1
A1 standard deviation \(=0.581\) years
✓ 1
Part (d)
M1 \(\int_{9/4}^3\dfrac{x^2}{9}\,dx\) or \(1-\dfrac{(9/4)^3}{27}\)
✓ 1
A1 \(0.578\)
✓ 1
Exam tip: For a median, set the integral from the lower limit to m equal to one half and solve; the median generally differs from the mean for a skewed density.
343IBMA3H-4-19r17-094.12 Standardization and unknown normal parametersPaper 2Medium8 marks
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The daily energy output, \(X\) MWh, of a small wind farm is normally distributed with mean \(240\) and standard deviation \(35\). The farm sells energy at €58 per MWh and has fixed daily running costs of €12 000. The daily profit is \(R=58X-12000\) euros.
(a)
Find \(\mathrm{E}(R)\) and the standard deviation of \(R\).
(3)
(b)
Find the probability that the farm makes a loss on a given day.
(2)
(c)
The standard deviation of \(X\) stays at \(35\) MWh, but the mean output could be raised by better turbine maintenance. Find the mean output, to 3 significant figures, needed so that the probability of a loss on a given day is \(0.05\).
(3)
Show mark scheme
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Mark
Notes
Part (a)
M1 \(\mathrm{E}(R)=58\,\mathrm{E}(X)-12000\)
✓ 1
A1 \(\mathrm{E}(R)=1920\) (€1920)
✓ 1
A1 standard deviation \(=58\times35=2030\) (€2030), since \(\mathrm{Var}(R)=58^2\mathrm{Var}(X)\)
✓ 1
Part (b)
M1 use of \(R\sim\mathrm{N}(1920,\,2030^2)\) and \(\mathrm{P}(R<0)\) (or \(\mathrm{P}(X<206.9)\))
✓ 1
A1 \(0.172\)
✓ 1
Part (c)
M1 loss when \(X<\dfrac{12000}{58}=206.9\) (or \(\mathrm{E}(R)=1.6449\times58\times35\))
✓ 1
M1 \(\dfrac{206.9-\mu}{35}=-1.6449\)
✓ 1
A1 \(\mu=264\) MWh
✓ 1
Common error: Var(aX + b) = a²Var(X): the constant b does not change the spread, and the multiplier a must be squared for the variance (not for the standard deviation).
344IBMA3H-4-19r17-114.8 Binomial distributionPaper 2Medium7 marks
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The power output, \(P\) megawatts, of a certain type of wind turbine at midday is normally distributed with mean \(2.4\) and standard deviation \(0.5\). A turbine is classed as high output if \(P>3.0\). At midday, \(8\) such turbines at a wind farm operate independently. In parts (b) and (c) use the unrounded probability from part (a).
(a)
Find the probability that a given turbine is classed as high output.
(2)
(b)
Find the probability that exactly \(2\) of the \(8\) turbines are classed as high output.
(2)
(c)
Given that at least one of the \(8\) turbines is classed as high output, find the probability that exactly \(2\) are.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to find \(\mathrm{P}(P>3.0)\) using \(\mathrm{N}(2.4,\,0.5^2)\) (or \(z=1.2\))
Common error: Conditioning on 'at least one' changes the denominator to P(Y ≥ 1), not P(Y = 1); the numerator stays as the probability of the specific outcome.
345IBMA3H-4-19r19-064.12 Standardization and unknown normal parametersPaper 2Medium7 marks
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An orchard sends three crates of apples to a shop in each delivery. The discrete random variable X is the number of crates rejected, with probability distribution: x = 0, 1, 2, 3 and P(X = x) = 0.15, p, q, 0.25. It is known that E(X) = 1.55.
(a)
Find the value of p and the value of q.
(4)
(b)
Given that at least one crate is rejected, find the probability that at least two crates are rejected. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.15 + p + q + 0.25 = 1, so p + q = 0.6
✓ 1
M1 E(X): p + 2q + 0.75 = 1.55, so p + 2q = 0.8
✓ 1
A1 q = 0.2
✓ 1
A1 p = 0.4
✓ 1
Part (b)
M1 P(X ≥ 2 | X ≥ 1) = P(X ≥ 2)/P(X ≥ 1)
✓ 1
A1 0.45/0.85
✓ 1
A1 0.529 (= 9/17)
✓ 1
Exam tip: A conditional probability has the overlap on the top and the condition on the bottom; here the overlap is just P(X ≥ 2).
346IBMA3H-4-19r19-074.2 Presentation of dataPaper 2Medium8 marks
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On a dairy farm, the milk yield x (litres per cow per day) and the time y (hours per day) that the cow spends grazing were recorded for 12 cows. The summary values were entered as Σx = 336.5, Σy = 81.0, Σx2 = 9653.09, Σy2 = 582.5, Σxy = 2358.2. It was then found that one cow, with y = 7.0, had its yield recorded as 26.5 instead of 29.5.
(a)
Find the corrected values of Σx and Σxy.
(2)
(b)
Using the corrected values, find the equation of the regression line of x on y, in the form x = ay + b. Give a and b to 3 significant figures.
(4)
(c)
Estimate the milk yield of a cow that grazes for 8.7 hours per day.
M1 Σy2 − (Σy)2/12 = 582.5 − 546.75 = 35.75 and gradient a = 87.575/35.75
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A1 a = 2.45
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A1 b = 339.5/12 − 2.4497 × 81/12 = 11.8, so x = 2.45y + 11.8
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Part (c)
M1 substitute y = 8.7 into the regression line of x on y
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A1 33.1 litres per day
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Common error: Because Σy² does not change, only the sums containing the faulty x value need correcting.
347IBMA3H-4-19r19-084.10 Regression line of x on yPaper 2Medium7 marks
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An airline records, for eight flights, the flight duration y (hours) and the fuel burned x (tonnes): y: 1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5 x: 2.9, 4.1, 5.4, 6.0, 7.9, 8.6, 10.4, 10.9
(a)
Find the equation of the regression line of x on y. Give the gradient and intercept to 3 significant figures.
(2)
(b)
Find the product moment correlation coefficient r, to 3 significant figures, and describe the correlation.
(2)
(c)
Estimate the fuel burned on a flight of 1.8 hours, to 3 significant figures.
(2)
(d)
A planner uses the same line to estimate the fuel burned on a flight of 9 hours. Comment on the reliability of this estimate.
(1)
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Notes
Part (a)
A1 gradient 2.36
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A1 x = 2.36y + 0.543
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Part (b)
A1 r = 0.995
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A1 very strong positive linear correlation
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Part (c)
M1 use of the regression line of x on y (accept unrounded or 3 s.f. coefficients)
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A1 4.79 tonnes
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Part (d)
9 hours lies far outside the range of the data (1.0 to 4.5 hours), so the estimate involves extrapolation and may be unreliable
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Exam tip: The line of x on y is the one that estimates x from a given y; do not rearrange the line of y on x for this purpose.
348IBMA3H-4-19r19-104.14 Variance, continuous random variables and linear transformationsPaper 2Medium11 marks
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At a music festival, a tombola ticket costs £3.00. The prize, X pounds, on a ticket has the following distribution: x = 0, 2, 10, 50 and P(X = x) = r, 0.20, 0.08, 0.02.
(a)
Find the value of r.
(1)
(b)
Find E(X) and hence state the ticket price, in pounds, for which the game would be fair.
(3)
(c)
Find Var(X).
(3)
(d)
The organiser's profit on one ticket is Y = 3 − X. Find E(Y) and Var(Y).
(2)
(e)
The organiser sells 1500 tickets. Find the expected total profit.
(2)
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Part (a)
r = 1 − 0.30 = 0.70
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Part (b)
M1 E(X) = 2×0.20 + 10×0.08 + 50×0.02
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A1 £2.20
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A1 fair price £2.20, since the expected prize equals the price paid
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Part (c)
M1 E(X2) = 4×0.20 + 100×0.08 + 2500×0.02 = 58.8
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M1 Var(X) = 58.8 − 2.22
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A1 53.96
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Part (d)
A1 E(Y) = 3 − 2.2 = £0.80
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A1 Var(Y) = (−1)2 Var(X) = 53.96
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Part (e)
M1 1500 × E(Y)
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A1 £1200
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Exam tip: Var(b − X) = Var(X): the minus sign is squared away and the constant b has no effect.
349IBMA3H-4-19r19-114.2 Presentation of dataPaper 2Medium10 marks
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The number of passengers, X, who fail to board a particular flight has probability distribution: x = 0, 1, 2, 3 and P(X = x) = 0.40, 0.35, 0.20, 0.05. The cost to the airline of handling the empty seats is C = 150 + 80X euros.
(a)
Find E(X).
(2)
(b)
Find Var(X).
(3)
(c)
Find E(C).
(2)
(d)
Find the standard deviation of C, to 3 significant figures.
(3)
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Part (a)
M1 0×0.40 + 1×0.35 + 2×0.20 + 3×0.05
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A1 0.9
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Part (b)
M1 E(X2) = 0.35 + 0.80 + 0.45 = 1.6
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M1 Var(X) = 1.6 − 0.92
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A1 0.79
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Part (c)
M1 E(C) = 150 + 80 × 0.9
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A1 €222
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Part (d)
M1 Var(C) = 802 × Var(X) (the 150 has no effect)
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A1 6400 × 0.79 = 5056
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A1 standard deviation = √5056 = €71.1
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Common error: When finding Var(aX + b), square the multiplier a and ignore b; remember to take the square root when asked for a standard deviation.
350IBMA3H-4-18r20-054.2 Presentation of dataPaper 2Medium7 marks
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The journey times, in minutes, of 40 buses on a route are summarised in the table.
Time \(t\) (minutes)
\(10\le t<20\)
\(20\le t<30\)
\(30\le t<40\)
\(40\le t<50\)
\(50\le t<60\)
Frequency
5
10
14
6
5
Histogram of the journey times of 40 buses
(a)
Estimate the mean journey time.
(2)
(b)
Estimate the population standard deviation of the journey times, to 3 significant figures.
(2)
(c)
Estimate the median journey time, assuming the times in the median class are evenly spread. Give your answer to 3 significant figures.
(3)
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Part (a)
M1 use of mid-interval values 15, 25, 35, 45, 55
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A1 34.0 minutes
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Part (b)
M1 use of technology or \(\sqrt{\dfrac{\sum fx^2}{40}-34^2}\)
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A1 11.8 minutes
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Part (c)
M1 cumulative frequencies 5, 15, 29, so the 20th value is in \(30\le t<40\)
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M1 \(30+\dfrac{20-15}{14}\times10\)
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A1 33.6 minutes
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Exam tip: Mid-interval values give only an estimate of the mean because the original data are no longer known.
351IBMA3H-4-18r20-064.4 Correlation and regression of y on xPaper 2Medium7 marks
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A coach investigates the relationship between weekly training load, \(x\) (arbitrary units), and a fitness score, \(y\), for a squad of athletes. The training loads range from 10 to 27. Summary statistics are: \(\bar{x}=18.5\), \(\bar{y}=62\), standard deviation of \(x\) is 4, standard deviation of \(y\) is 10 and \(r=0.72\).
(a)
Find the equation of the regression line of \(y\) on \(x\).
(3)
(b)
Explain why the line should not be used to estimate the training load needed for a fitness score of 80.
(1)
(c)
Comment on the reliability of using the line to predict the fitness score for a training load of 40.
(1)
(d)
Use the line to estimate the fitness score for a training load of 22.
Common error: The unbiased estimate divides by \(n-1\), not \(n\); calculators often give both standard deviations, so check which key you are using.
354IBMA3H-4-18r20-104.8 Binomial distributionPaper 2Medium7 marks
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On a rail route, 15 passengers independently buy tickets. Each passenger pays by card with probability 0.4. Let \(X\) be the number paying by card. The card processing cost, in euros, is \(C=0.5X+4\).
(a)
Find \(E(X)\) and \(\text{Var}(X)\).
(2)
(b)
Find \(E(C)\) and \(\text{Var}(C)\).
(3)
(c)
Find the probability that the processing cost is at most €7, to 3 significant figures.
(2)
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Part (a)
A1 \(E(X)=15\times0.4=6\)
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A1 \(\text{Var}(X)=15\times0.4\times0.6=3.6\)
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Part (b)
M1 \(E(C)=0.5E(X)+4\)
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A1 \(E(C)=7\)
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A1 \(\text{Var}(C)=0.5^2\times3.6=0.9\)
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Part (c)
M1 \(0.5X+4\le7\), so \(X\le6\)
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A1 0.610
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Common error: For \(\text{Var}(aX+b)\) the constant \(b\) has no effect and \(a\) is squared: \(\text{Var}(aX+b)=a^2\text{Var}(X)\).
355IBMA3H-4-18r20-114.8 Binomial distributionPaper 2Medium5 marks
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An observatory processes 8 images per night. Each image independently contains a usable detection with probability 0.25. Let \(X\) be the number of images with a usable detection on a given night, so \(X\sim B(8,\,0.25)\). A night's processing cost is €\(X^2\).
(a)
Find \(E(X)\) and \(\text{Var}(X)\).
(2)
(b)
Use \(\text{Var}(X)=E(X^2)-[E(X)]^2\) to find the expected cost of one night's processing.
(2)
(c)
Find the expected total cost over 30 independent nights.
(1)
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Part (a)
A1 \(E(X)=8\times0.25=2\)
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A1 \(\text{Var}(X)=8\times0.25\times0.75=1.5\)
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Part (b)
M1 \(E(X^2)=\text{Var}(X)+[E(X)]^2=1.5+2^2\)
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A1 €5.50
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Part (c)
A1 \(30\times5.50=\) €165
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Exam tip: The expected value of \(X^2\) is not \([E(X)]^2\); the gap between them is exactly the variance.
356IBMA3H-4-18r21-054.12 Standardization and unknown normal parametersPaper 2Medium8 marks
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The time \(T\) minutes that a customer spends in a store is normally distributed with mean \(\mu\) and standard deviation \(\sigma\). It is known that 20% of customers spend less than 10 minutes in the store and 10% spend more than 40 minutes.
Sketch of the normal curve with the given tail probabilities, not to scale
(a)
Write down two equations that must be satisfied by \(\mu\) and \(\sigma\), using \(z\)-values given to 4 decimal places.
(3)
(b)
Find \(\mu\) and \(\sigma\), each to 3 significant figures.
(3)
(c)
Find the time that only 5% of customers exceed. Use the unrounded values of \(\mu\) and \(\sigma\) and give your answer to 3 significant figures.
(2)
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Part (a)
M1 uses the inverse normal for \(P(Z<z)=0.2\) giving \(z=-0.8416\)
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A1 \(\frac{10-\mu}{\sigma}=-0.8416\)
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A1 \(\frac{40-\mu}{\sigma}=1.2816\)
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Part (b)
M1 solves the two equations simultaneously (for example subtracting to get \(\frac{30}{\sigma}=2.1232\))
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A1 \(\sigma=14.1\)
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A1 \(\mu=21.9\)
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Part (c)
M1 \(\mu+1.6449\sigma\)
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A1 45.1 minutes
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Exam tip: Write each condition as a standardized equation first; a probability in the upper tail needs the \(z\)-value for the area to its left.
357IBMA3H-4-18r21-064.12 Standardization and unknown normal parametersPaper 2Medium7 marks
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The lifetime \(L\) years of a battery used to store solar energy is normally distributed with standard deviation 0.8 years. It is known that 5% of batteries last less than 6 years.
(a)
Find the mean lifetime of the batteries, to 3 significant figures.
(3)
(b)
Using the unrounded value of \(\mu\), find the probability that a battery lasts between 7 and 9 years. Give your answer to 3 significant figures.
(2)
(c)
The maker wants only 1% of batteries to fail within the warranty period. Using the unrounded value of \(\mu\), find the length of the warranty period, to 3 significant figures.
(2)
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Notes
Part (a)
M1 \(z=-1.6449\) for a lower tail of 0.05
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M1 \(\frac{6-\mu}{0.8}=-1.6449\)
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A1 \(\mu=7.32\) years
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Part (b)
M1 attempts \(P(7<L<9)\) with \(\mu=7.3159\)
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A1 0.636
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Part (c)
M1 \(7.3159-2.3263\times0.8\)
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A1 5.45 years
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Common error: Keep the unrounded mean in your calculator memory for later parts, since rounding it early can change the third significant figure.
358IBMA3H-4-18r21-074.10 Regression line of x on yPaper 2Medium7 marks
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A shop records, for 8 days, the midday temperature \(x\) °C and the number \(y\) (in hundreds) of cold drinks sold. \(x\): 14, 17, 19, 21, 23, 25, 28, 30 \(y\): 12, 15, 15, 19, 20, 24, 25, 29
Scatter diagram of the data in the table
(a)
Find the product-moment correlation coefficient \(r\), to 3 significant figures.
(1)
(b)
Find the equation of the regression line of \(x\) on \(y\). Give the coefficients to 3 significant figures.
(3)
(c)
On one day 2200 cold drinks were sold. Use your regression line to estimate the midday temperature that day, using unrounded coefficients, and comment on the reliability of your estimate.
(3)
Show mark scheme
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Notes
Part (a)
A1 \(r=0.985\)
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Part (b)
M1 uses the regression line of \(x\) on \(y\) (not \(y\) on \(x\))
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A1 gradient 0.925
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A1 \(x=0.925y+3.74\)
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Part (c)
M1 substitutes \(y=22\) into the line of \(x\) on \(y\)
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A1 24.1 °C
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R1 reliable: 22 is within the range of the data (interpolation) and \(r\) is close to 1
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Common error: To estimate \(x\) from a given \(y\), use the regression line of \(x\) on \(y\); do not rearrange the line of \(y\) on \(x\).
359IBMA3H-4-18r21-084.4 Correlation and regression of y on xPaper 2Medium8 marks
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For a solar farm, \(x\) is the number of hours of sunshine in a day and \(y\) is the energy output in MWh. The regression line of \(y\) on \(x\) is \(y=1.5x+4\) and the regression line of \(x\) on \(y\) is \(x=0.4y+1.6\).
(a)
Find the mean number of hours of sunshine and the mean energy output.
(3)
(b)
Find the value of \(r\), to 3 significant figures.
(3)
(c)
Estimate the energy output on a day with 10 hours of sunshine.
(2)
Show mark scheme
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Part (a)
M1 uses the fact that both lines pass through \((\bar x,\bar y)\) and solves simultaneously
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A1 \(\bar x=8\)
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A1 \(\bar y=16\)
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Part (b)
M1 \(r^2=1.5\times0.4=0.6\)
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R1 \(r\) is positive because both gradients are positive
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A1 \(r=0.775\)
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Part (c)
M1 uses the line of \(y\) on \(x\) with \(x=10\)
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A1 19 MWh
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Exam tip: The product of the two regression gradients equals \(r^2\); the sign of \(r\) comes from the sign of either gradient.
360IBMA3H-4-18r21-094.13 Bayes’ theoremPaper 2Medium6 marks
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In a region, 2% of the population have a particular infection. A screening test gives a positive result for 95% of people who have the infection and for 10% of people who do not.
Tree diagram showing the probabilities given in the question
(a)
Find the probability that a randomly chosen person tests positive.
(3)
(b)
A person tests positive. Find the probability that the person has the infection, to 3 significant figures.
(3)
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Part (a)
M1 total probability: \(P(\text{infected and positive})+P(\text{not infected and positive})\)
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A1 \(0.02\times0.95+0.98\times0.10\)
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A1 \(0.019+0.098=0.117\)
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Part (b)
M1 \(P(\text{infected}\mid\text{positive})=\frac{P(\text{infected and positive})}{P(\text{positive})}\)
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A1 \(\frac{0.019}{0.117}\)
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A1 0.162
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Common error: The denominator in Bayes' theorem is the total probability of the evidence, found by adding every branch that leads to it.
361IBMA3H-4-18r21-104.13 Bayes’ theoremPaper 2Medium8 marks
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Solar panels are supplied by three factories A, B and C, which make 45%, 35% and 20% of the panels respectively. The probability that a panel is faulty is 0.01 for factory A, 0.02 for factory B and 0.06 for factory C.
Tree diagram showing the probabilities given in the question
(a)
Find the probability that a randomly chosen panel is faulty.
(3)
(b)
A panel is found to be faulty. Find the probability that it came from factory C, to 3 significant figures.
(3)
(c)
Find the probability that a faulty panel did not come from factory C, to 3 significant figures.
(2)
Show mark scheme
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Part (a)
M1 sum of three products \(P(\text{factory})\times P(\text{faulty}\mid\text{factory})\)
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A1 \(0.0045+0.007+0.012\)
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A1 0.0235
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Part (b)
M1 conditional probability formula \(\frac{P(C\cap\text{faulty})}{P(\text{faulty})}\)
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A1 \(\frac{0.012}{0.0235}\)
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A1 0.511
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Part (c)
M1 \(1-\frac{0.012}{0.0235}\) or \(\frac{0.0045+0.007}{0.0235}\)
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A1 0.489
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Exam tip: Check that your conditional probabilities for all the possible sources add to 1; here the three values for A, B and C should do so.
362IBMA3H-4-18r21-114.10 Regression line of x on yPaper 2Medium7 marks
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For 8 regions, \(x\) is the percentage of residents who have had a booster vaccine and \(y\) is the number of weekly hospital admissions per 100 000 people. The summary statistics are: \(n=8\), \(\Sigma x=509\), \(\Sigma y=246\), \(\Sigma x^2=33\,859\), \(\Sigma y^2=8576\), \(\Sigma xy=14\,434\).
(a)
Find the equation of the regression line of \(x\) on \(y\), giving the coefficients to 4 significant figures.
(4)
(b)
A region has 20 admissions per 100 000 people. Using unrounded coefficients, estimate the percentage of its residents who have had a booster. Give your answer to 3 significant figures.
(2)
(c)
State whether your estimate is likely to be reliable. Justify your answer.
A1 numerator \(-1217.75\) and denominator \(1011.5\)
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A1 gradient \(-1.204\)
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A1 \(x=-1.204y+100.6\)
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Part (b)
M1 substitutes \(y=20\) into the line of \(x\) on \(y\)
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A1 76.6
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Part (c)
R1 reliable: 20 lies inside the range of \(y\) values and the correlation is very strong (\(r\approx-0.997\))
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Common error: With summary statistics, the line of \(x\) on \(y\) has \(\Sigma y^2\) in the denominator; using \(\Sigma x^2\) gives the line of \(y\) on \(x\).
363IBMA3H-4-19r23-014.4 Correlation and regression of y on xPaper 2Medium6 marks
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The mass, X kg, of honey harvested from a hive in one season is modelled by a normal distribution with mean μ and standard deviation σ. Beekeepers find that 20% of hives yield less than 18 kg and 10% of hives yield more than 30 kg.
Sketch of the normal curve with the given tail probabilities, not to scale
(a)
Find the values of z such that P(Z < z) = 0.2 and P(Z > z) = 0.1, where Z ~ N(0, 1).
(2)
(b)
Hence find μ and σ, giving each answer to 3 significant figures.
(4)
Show mark scheme
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Part (a)
A1 z = −0.842 for the lower 20%
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A1 z = 1.28 for the upper 10%
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Part (b)
M1 forms 18 = μ − 0.8416σ and 30 = μ + 1.2816σ
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M1 attempts to solve the two equations simultaneously, e.g. subtracting to get 12 = 2.1232σ
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A1 σ = 5.65
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A1 μ = 22.8
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Exam tip: Standardise each given probability separately, then solve the two linear equations; a quick sketch of the bell curve shows which z-value is negative.
364IBMA3H-4-19r23-024.14 Variance, continuous random variables and linear transformationsPaper 2Medium12 marks
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The number of pit stops, X, made by a Formula 1 car in a race has the probability distribution shown.
x: 1, 2, 3, 4 P(X = x): 0.1, p, q, 0.3
It is given that E(X) = 2.9.
(a)
Find the value of p and the value of q.
(4)
(b)
Find Var(X).
(3)
(c)
The cost to the team of pit stops in a race is C euros, where C = 12 000X + 3000. Find the standard deviation of C, correct to 3 significant figures.
(2)
(d)
Find the probability that a car makes more pit stops than the mean number, given that it makes at least 2 pit stops. Give your answer to 3 significant figures.
(3)
Show mark scheme
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Part (a)
M1 uses the sum of probabilities: 0.1 + p + q + 0.3 = 1, so p + q = 0.6
M1 Var(C) = 12 000² × Var(X) (the constant 3000 does not affect the spread)
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A1 standard deviation = 12 000 × √0.89 = €11 300
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Part (d)
M1 more than 2.9 stops means X ≥ 3, so P(X ≥ 3) = 0.4 + 0.3 = 0.7
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M1 conditional probability P(X ≥ 3 | X ≥ 2) = 0.7 ÷ 0.9
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A1 7/9 = 0.778
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Common error: In conditional probability the denominator is the probability of the condition (here X ≥ 2), not 1 and not the probability of the event.
365IBMA3H-4-19r23-044.14 Variance, continuous random variables and linear transformationsPaper 2Medium10 marks
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The time, X hours, that a worker bee spends on one foraging trip is modelled by the probability density function f(x) = kx(3 − x) for 0 ≤ x ≤ 3, and f(x) = 0 otherwise, where k is a constant.
(a)
Show that k = 2/9.
(3)
(b)
Write down E(X), giving a reason for your answer.
(2)
(c)
Find Var(X).
(3)
(d)
Find the probability that a foraging trip lasts longer than 2 hours. Give your answer to 3 significant figures.
(2)
Show mark scheme
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Notes
Part (a)
M1 sets ∫03kx(3 − x) dx = 1
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A1 k[3x²/2 − x³/3]03 = k(27/2 − 9) = 9k/2
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A1 9k/2 = 1, so k = 2/9 (AG)
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Part (b)
R1 the graph of f is symmetrical about x = 1.5
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A1 E(X) = 1.5
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Part (c)
M1 attempts E(X²) = ∫03x²f(x) dx
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A1 E(X²) = 2.7
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A1 Var(X) = 2.7 − 1.5² = 0.45
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Part (d)
M1 P(X > 2) = ∫23 (2/9)x(3 − x) dx
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A1 7/27 = 0.259
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Exam tip: For a show-that question, write the full line that equals 1 before the final statement; the result must be reached, not assumed.
366IBMA3H-4-19r23-074.12 Standardization and unknown normal parametersPaper 2Medium8 marks
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The recovery time, X days, of patients after a routine procedure at a hospital is normally distributed with mean 14.2 and standard deviation σ. It is known that 8% of patients take longer than 20 days to recover.
(a)
Find the value of σ, correct to 3 significant figures.
(3)
(b)
Find the probability that a patient recovers in less than 12 days. Use the unrounded value of σ and give your answer to 3 significant figures.
(2)
(c)
Fifteen patients are chosen at random. Find the probability that at least two of them take longer than 20 days to recover. Use P(X > 20) = 0.08 and give your answer to 3 significant figures.
(3)
Show mark scheme
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Mark
Notes
Part (a)
M1 uses the inverse normal to find z = 1.405 for P(Z > z) = 0.08
✓ 1
M1 forms (20 − 14.2)/σ = 1.405
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A1 σ = 4.13
✓ 1
Part (b)
M1 attempts a normal probability calculation with mean 14.2 and standard deviation 4.1279…
✓ 1
A1 0.297
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Part (c)
M1 recognises a binomial distribution with n = 15 and p = 0.08
Exam tip: 'At least two' is easiest through the complement: subtract P(0) and P(1) from 1 rather than summing 13 terms.
367IBMA3H-4-19r23-094.2 Presentation of dataPaper 2Medium7 marks
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The length of stay, in days, of 50 patients in a hospital ward is summarised in the table.
Length of stay (days): 0 ≤ x < 2 | 2 ≤ x < 4 | 4 ≤ x < 6 | 6 ≤ x < 10 | 10 ≤ x < 20 Frequency: 8 | 15 | 12 | 9 | 6
Histogram of the length of stay of 50 patients
(a)
Estimate the mean length of stay.
(2)
(b)
Estimate the population standard deviation of the length of stay, giving your answer to 3 significant figures.
(2)
(c)
Estimate the median length of stay using linear interpolation, giving your answer to 3 significant figures.
(3)
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Part (a)
M1 uses class mid-points 1, 3, 5, 8, 15 with the frequencies, e.g. (8 + 45 + 60 + 72 + 90) ÷ 50
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A1 5.5 days
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Part (b)
M1 uses a calculator (or Σfx²/50 − mean²) with the mid-points
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A1 4.14 days
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Part (c)
M1 the 25th value lies in the class 4 ≤ x < 6, since the cumulative frequency up to 4 days is 23
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M1 interpolates: 4 + (25 − 23)/12 × 2
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A1 4.33 days
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Exam tip: For grouped data, state that mid-points are used; the results are estimates because the individual values in each class are unknown.
368IBMA3H-4-19r23-104.2 Presentation of dataPaper 2Medium10 marks
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In a Formula 1 championship, the points X scored by a driver in a race have the distribution below.
x: 0, 8, 15, 25 P(X = x): 0.30, a, b, 0.15
It is given that E(X) = 9.2.
(a)
Find the value of a and the value of b.
(4)
(b)
Find Var(X), giving your answer to 3 significant figures.
(3)
(c)
The driver competes in two races. The points scored in the two races are independent. Find the probability that the driver scores a total of at least 30 points in the two races.
(3)
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Part (a)
M1 sum of probabilities equals 1: 0.30 + a + b + 0.15 = 1, so a + b = 0.55
M1 identifies the combinations reaching at least 30: (25, 8), (25, 15), (25, 25) and (15, 15), noting that (25, 8) and (25, 15) can occur in either order
Common error: List the combinations systematically; forgetting that two different scores can occur in either order halves those probabilities.
369IBMA3H-4-19r23-144.12 Standardization and unknown normal parametersPaper 2Medium10 marks
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The lap time, X seconds, of a Formula 1 car at a circuit is normally distributed with mean 81.2 and standard deviation 0.45. Lap times are independent of one another.
(a)
The middle 90% of lap times lie between a and b seconds, where a and b are symmetric about the mean. Find a and b, giving each to 3 significant figures.
(3)
(b)
Find the probability that a lap takes longer than 82.0 s. Give your answer to 3 significant figures.
(2)
(c)
Given that a lap takes longer than 81.5 s, find the probability that it takes longer than 82.0 s. Give your answer to 3 significant figures.
(3)
(d)
In a 50-lap race, find the probability that at least one lap takes longer than 82.0 s. Use P(X > 82.0) = 0.0377.
(2)
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Part (a)
M1 uses the inverse normal with 5% in each tail: z = ±1.645
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M1 81.2 ± 1.645 × 0.45
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A1 a = 80.5 and b = 81.9
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Part (b)
M1 P(X > 82.0) with z = 0.8/0.45
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A1 0.0377
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Part (c)
M1 conditional probability P(X > 82.0) ÷ P(X > 81.5)
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A1 P(X > 81.5) = 0.2525
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A1 0.0377 ÷ 0.2525 = 0.149
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Part (d)
M1 1 − (1 − 0.0377)50
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A1 0.854
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Exam tip: 'At least one' in repeated independent trials is 1 minus the probability of none; write the complement line explicitly.
370IBMA3H-4-19r24-044.2 Presentation of dataPaper 2Medium7 marks
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The daily water use of 80 households is summarised in the table.
Use (litres): 0 ≤ x < 100 | 100 ≤ x < 200 | 200 ≤ x < 300 | 300 ≤ x < 400 | 400 ≤ x < 500 | 500 ≤ x < 600 Frequency: 6 | 14 | 27 | 21 | 8 | 4
(a)
Write down the modal class.
(1)
(b)
Estimate the mean daily water use. Give your answer to 3 significant figures.
(3)
(c)
Cumulative frequencies are \(6,\,20,\,47,\,68,\,76,\,80\) at the upper class boundaries. Use linear interpolation to estimate the median. Give your answer to 3 significant figures.
(3)
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Part (a)
A1 \(200\le x<300\)
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Part (b)
M1 use of class midpoints \(50,150,250,350,450,550\)
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M1 \(\dfrac{\sum fx}{80}=\dfrac{22300}{80}\)
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A1 \(279\) litres
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Part (c)
A1 median is the 40th value, in the class \(200\le x<300\)
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M1 \(200+\dfrac{40-20}{27}\times100\)
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A1 \(274\) litres
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Exam tip: For grouped data, state that the answers are estimates; the mean uses midpoints and the median is found by interpolating within the class that contains the middle value.
371IBMA3H-4-19r24-054.7 Discrete random variablesPaper 2Medium7 marks
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In a mobile game, each turn consists of rolling a fair eight-sided die (faces 1 to 8) and a fair six-sided die (faces 1 to 6) once each. The turn score \(S\) is the larger of the two numbers shown; if the numbers are equal, \(S\) is that number.
(a)
Write down the number of equally likely outcomes in the sample space.
(1)
(b)
Find \(P(S=6)\).
(3)
(c)
Given that \(S=6\), find the probability that the six-sided die shows 6. Give your answer to 3 significant figures.
(3)
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Part (a)
A1 \(8\times6=48\)
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Part (b)
M1 identifies outcomes with the eight-sided die showing 6 and the six-sided die showing at most 6 (6 outcomes)
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M1 adds outcomes with the six-sided die showing 6 and the eight-sided die showing less than 6 (5 outcomes)
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A1 \(\dfrac{11}{48}\)
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Part (c)
M1 \(P(\text{six-sided shows }6\mid S=6)=\dfrac{P(\text{six-sided shows }6\text{ and }S=6)}{P(S=6)}\)
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A1 \(\dfrac{6/48}{11/48}\)
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A1 \(0.545\)
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Common error: When a maximum is involved, count the tied case (both dice showing 6) only once; it is easy to count it twice.
372IBMA3H-4-19r24-074.13 Bayes’ theoremPaper 2Medium7 marks
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A leak detector on a water main sounds an alarm. On any day, the probability that there is a leak is 0.04. If there is a leak the alarm sounds with probability 0.95; if there is no leak it sounds with probability 0.03.
Tree diagram showing the probabilities given in the question
(a)
Find the probability that the alarm sounds on a given day.
(2)
(b)
Given that the alarm sounds, find the probability that there is a leak. Give your answer to 3 significant figures.
(2)
(c)
A second sensor is fitted. Given the true state (leak or no leak), the two sensors operate independently. The second sensor sounds with probability 0.90 if there is a leak and 0.05 if there is none. Find the probability that there is a leak, given that both sensors sound. Give your answer to 3 significant figures.
(3)
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Part (a)
M1 \(0.04\times0.95+0.96\times0.03\)
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A1 \(0.0668\)
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Part (b)
M1 \(\dfrac{0.04\times0.95}{0.0668}\)
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A1 \(0.569\)
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Part (c)
M1 \(P(\text{both}\mid\text{leak})=0.95\times0.90=0.855\) and \(P(\text{both}\mid\text{no leak})=0.03\times0.05=0.0015\)
Common error: \(P(\text{leak}\mid\text{alarm})\) is not the same as \(P(\text{alarm}\mid\text{leak})\): the first needs the overall probability of an alarm in the denominator.
373IBMA3H-4-19r24-094.13 Bayes’ theoremPaper 2Medium7 marks
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A pharmacy buys a medicine from three suppliers. Supplier \(S_1\) provides 50% of the stock, \(S_2\) provides 30% and \(S_3\) provides 20%. The probability that a tablet has the wrong dose (is defective) is 0.01 for \(S_1\), 0.02 for \(S_2\) and \(d\) for \(S_3\). Given that a randomly chosen tablet is defective, the probability that it came from \(S_3\) is 0.5.
(a)
Find the value of \(d\).
(4)
(b)
Hence write down the probability that a randomly chosen tablet is defective.
(1)
(c)
A patient takes three tablets chosen independently at random. Find the probability that at least one is defective. Give your answer to 3 significant figures.
(2)
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Part (a)
A1 \(P(D)=0.005+0.006+0.2d=0.011+0.2d\)
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M1 \(P(S_3\mid D)=\dfrac{0.2d}{0.011+0.2d}\)
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A1 \(\dfrac{0.2d}{0.011+0.2d}=0.5\)
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A1 \(d=0.055\)
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Part (b)
A1 \(0.011+0.011=0.022\)
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Part (c)
M1 \(1-(1-0.022)^3\)
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A1 \(0.0646\)
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Exam tip: When an unknown appears inside a conditional probability, write the total probability of the given event first (here \(P(D)\)), then form an equation in the unknown.
374IBMA3H-4-19r24-104.13 Bayes’ theoremPaper 2Medium7 marks
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An online game uses software to flag accounts that may be using cheats. Of the 20 000 accounts, 1.5% use cheats. The software flags 92% of accounts that use cheats and 4% of accounts that do not.
(a)
Find the probability that a randomly chosen account is flagged.
(2)
(b)
Given that an account is flagged, find the probability that it uses cheats. Give your answer to 3 significant figures.
(2)
(c)
The studio retrains the software so that it still flags 92% of accounts that use cheats, but flags a proportion \(f\) of accounts that do not. Find the largest value of \(f\), as a percentage to 2 significant figures, for which a flagged account has a probability of at least 0.5 of using cheats.
(3)
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Part (a)
M1 \(0.015\times0.92+0.985\times0.04\)
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A1 \(0.0532\)
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Part (b)
M1 \(\dfrac{0.015\times0.92}{0.0532}\)
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A1 \(0.259\)
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Part (c)
M1 \(\dfrac{0.0138}{0.0138+0.985f}\ge0.5\)
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M1 rearranges to \(0.985f\le0.0138\)
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A1 \(f\le0.014\), so the largest value is \(1.4\%\)
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Common error: Most flagged accounts can still be honest when cheating is rare, even if the false-flag rate looks small; compare the two products in the numerator and denominator rather than the two rates.
375IBMA3H-4-19r24-114.6 Combined, conditional and independent eventsPaper 2Medium11 marks
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A survey of 200 households records the type of water meter and whether the household has made a billing complaint.
Smart meter: complaint 18, no complaint 102 Standard meter: complaint 22, no complaint 58
(a)
Find the probability that a randomly chosen household has made a complaint.
(1)
(b)
Given that a household has made a complaint, find the probability that it has a standard meter.
(2)
(c)
Determine whether having a smart meter and making a complaint are independent events.
(3)
(d)
Two different households are chosen at random from the 200. Find the probability that both have a standard meter. Give your answer to 3 significant figures.
(2)
(e)
Find the probability that exactly one of the two households has made a complaint. Give your answer to 3 significant figures.
Exam tip: When choosing without replacement, the second probability changes: reduce both the numerator (if the first choice was in the group) and the denominator by one.
376IBMA3H-4-18r25-054.1 Sampling, bias and outliersPaper 2Medium8 marks
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At an auction of antiquities, 30% of the items offered are forgeries. A scientific test flags an item as suspect. The test flags 92% of forgeries and wrongly flags 6% of genuine items. An item is chosen at random from those offered.
(a)
Find the probability that the item is flagged as suspect.
(3)
(b)
Given that the item is flagged, find the probability that it is a forgery. Give your answer to 3 significant figures.
(3)
(c)
Given that the item is not flagged, find the probability that it is a forgery. Give your answer to 3 significant figures.
(2)
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Part (a)
M1 use the law of total probability
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A1 0.3 × 0.92 + 0.7 × 0.06
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A1 0.318
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Part (b)
M1 use P(F | S) = \(\frac{P(F\cap S)}{P(S)}\) (Bayes' theorem)
Common error: P(forgery | flagged) is not the same as P(flagged | forgery): the 92% given in the question is the second one, and Bayes' theorem converts it to the first.
377IBMA3H-4-18r25-074.14 Variance, continuous random variables and linear transformationsPaper 2Medium7 marks
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The number of server failures in a data centre in one week is the random variable X, with E(X) = 2.4 and Var(X) = 1.2. Each failure costs €450 to repair, and there is also a fixed weekly monitoring cost of €1200. The total cost in a week is C = 450X + 1200 euros.
(a)
Find E(C).
(2)
(b)
Find the standard deviation of C, to the nearest euro.
(3)
(c)
The costs in two different weeks, C1 and C2, are independent. Find the standard deviation of C1 + C2, to the nearest euro.
(2)
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Part (a)
M1 use E(aX + b) = aE(X) + b: 450(2.4) + 1200
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A1 €2280
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Part (b)
M1 use Var(aX + b) = a2Var(X)
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A1 Var(C) = 4502(1.2) = 243 000
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A1 standard deviation = €493
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Part (c)
M1 Var(C1 + C2) = 243 000 + 243 000 = 486 000
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A1 standard deviation = €697
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Common error: Adding a constant to a random variable changes its mean but not its variance; multiplying by a multiplies the variance by a², not a.
378IBMA3H-4-18r25-084.12 Standardization and unknown normal parametersPaper 2Medium7 marks
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An archaeologist measures the thickness, x mm, of each of 12 pottery fragments from one site, and finds Σx = 105.6 and Σx2 = 938.4.
(a)
Find an unbiased estimate of the population mean thickness.
(1)
(b)
Find an unbiased estimate of the population variance of the thickness. Give your answer to 3 significant figures.
(4)
(c)
A thirteenth fragment is found, with thickness exactly 8.8 mm. Find the new unbiased estimate of the population variance.
(2)
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Part (a)
A1 105.6 ÷ 12 = 8.8 mm
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Part (b)
M1 attempt to find Σ(x − x̄)2 = Σx2 − 12x̄2
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A1 938.4 − 12(8.8)2 = 9.12
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M1 divide by n − 1 = 11
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A1 0.829 mm2
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Part (c)
M1 the new mean is still 8.8 and the new fragment adds 0 to Σ(x − x̄)2, so the sum is still 9.12
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A1 \(\frac{9.12}{12}\) = 0.76 mm2
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Common error: An unbiased estimate of variance divides by n − 1, not n; a calculator's 'sn−₁' (or 'sample standard deviation') key does this for you.
379IBMA3H-4-18r25-094.10 Regression line of x on yPaper 2Medium8 marks
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A cyclist records her weekly training time, x hours, and her time, y minutes, for a 40 km time trial in the following week, for seven weeks.
Training time x (hours)
6
8
10
12
14
16
18
Time y (minutes)
74
71
70
66
64
63
59
Scatter diagram of the data in the table
(a)
Use your calculator to find the correlation coefficient r and the equation of the regression line of y on x. Give r to 3 significant figures and the coefficients to 3 significant figures.
(3)
(b)
Interpret the gradient of the regression line in this context.
(1)
(c)
Estimate the time trial time for a week after 11 hours of training. Use the unrounded regression equation and give your answer to 3 significant figures.
(2)
(d)
She wants to estimate how many hours of training are needed for a time of 60 minutes, and substitutes y = 60 into the line in (a). Explain why this is not appropriate.
(2)
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Part (a)
M1 use of the regression function of the calculator
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A1 r = −0.990
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A1 y = −1.20x + 81.1
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Part (b)
A1 each extra hour of weekly training is associated with a time trial that is, on average, about 1.2 minutes faster
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Part (c)
M1 substitute x = 11 into the regression line
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A1 67.9 minutes
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Part (d)
R1 the regression line of y on x is used to predict y from x
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R1 to predict x from y she should use the regression line of x on y
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Exam tip: Use the regression line of y on x only to estimate y from a given x; to estimate x from a given y, a different line (x on y) is required.
380IBMA3H-4-18r25-104.4 Correlation and regression of y on xPaper 2Medium8 marks
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An archaeologist surveyed 80 excavation sites. Of these, 36 contained pottery, 28 contained metal tools and 14 contained both pottery and metal tools. One site is chosen at random from the 80.
Venn diagram of the number of sites (80 in total); the circles show the totals for each type
(a)
Find the number of sites that contained neither pottery nor metal tools.
(2)
(b)
Given that the site contained metal tools, find the probability that it also contained pottery.
(2)
(c)
Determine whether containing pottery and containing metal tools are independent events, showing your working.
(2)
(d)
Two different sites are chosen at random from the 80. Find the probability that both contained pottery. Give your answer to 3 significant figures.
R1 P(both) = \(\frac{14}{80}\) = 0.175 ≠ 0.1575, so the events are not independent
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Part (d)
M1 \(\frac{36}{80}\times\frac{35}{79}\)
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A1 0.199
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Common error: When two items are chosen without replacement, the denominator of the second probability falls by one and the numerator changes too.
381IBMA3H-4-18r25-114.14 Variance, continuous random variables and linear transformationsPaper 2Medium7 marks
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After a power dip, the number of servers, X, in a data centre that need to be restarted has probability distribution P(X = x) = kx(5 − x), for x = 1, 2, 3, 4, where k is a constant.
Common error: Var(X) is E(X²) minus the square of the mean: do not square the whole mean calculation, and do not forget to square each x in E(X²).
382IBMA3H-4-18r25-124.2 Presentation of dataPaper 2Medium8 marks
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The response times, t milliseconds, of 100 requests to a data centre are summarised in the table.
Time t (ms)
0 ≤ t < 10
10 ≤ t < 20
20 ≤ t < 30
30 ≤ t < 40
40 ≤ t < 50
Frequency
12
31
27
18
12
(a)
Estimate the mean response time.
(2)
(b)
Write down the modal class.
(1)
(c)
Write down the class that contains the median.
(1)
(d)
Use your calculator to estimate the standard deviation of the response times, to 3 significant figures.
(2)
(e)
After an upgrade, every response time is 20% shorter. Find the new mean and the new standard deviation. Use the unrounded standard deviation from (d) and give each answer to 3 significant figures.
(2)
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Part (a)
M1 use midpoints 5, 15, 25, 35, 45: \(\frac{12(5)+31(15)+27(25)+18(35)+12(45)}{100}\)
M1 use midpoints with frequencies in the statistics function
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A1 σ = 12.0 ms
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Part (e)
A1 new mean = 0.8 × 23.7 = 19.0 ms (accept 18.96)
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A1 new standard deviation = 0.8 × σ = 9.58 ms
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Exam tip: Multiplying every value by a constant a multiplies the mean and the standard deviation by a; adding a constant changes only the mean.
383IBMA3H-4-18r26-034.10 Regression line of x on yPaper 2Medium7 marks
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The time T seconds taken by members of a swimming club to swim 100 m freestyle is normally distributed with mean μ and standard deviation 2.4. It is known that 15% of the times are less than 56.0 seconds.
(a)
Find the value of μ, giving your answer to 3 significant figures.
(3)
(b)
Find the probability that a randomly chosen time is greater than 60 seconds. Use the unrounded value of μ.
(2)
(c)
In a race with 8 independent swimmers from the club, find the probability that exactly 2 of them take longer than 60 seconds. Use the probability 0.264.
(2)
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Part (a)
M1 P(T < 56.0) = 0.15, so (56.0 − μ)/2.4 = z
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A1 z = −1.036
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A1 μ = 58.5
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Part (b)
M1 standardise or use GDC with 60 and the unrounded μ
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A1 0.264
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Part (c)
M1 binomial with n = 8, p = 0.264, X = 2
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A1 0.310
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Exam tip: A negative z-value means the value lies below the mean; check the sign before you rearrange for μ.
384IBMA3H-4-18r26-044.12 Standardization and unknown normal parametersPaper 2Medium7 marks
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The daily energy output X MWh of a wind farm is normally distributed. On 10% of days the output is below 40 MWh and on 20% of days the output is above 52 MWh.
Sketch of the normal curve with the given tail probabilities, not to scale
(a)
Find the mean and standard deviation of X, each to 3 significant figures.
(4)
(b)
Find the probability that the output on a given day is between 45 MWh and 50 MWh. Use unrounded values of μ and σ.
(2)
(c)
Find the probability that the output is below 40 MWh on each of three independent days.
(1)
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Part (a)
M1 z = −1.2816 for P(X < 40) = 0.1 and z = 0.8416 for P(X > 52) = 0.2
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M1 two equations, e.g. 40 = μ − 1.2816σ and 52 = μ + 0.8416σ, solved simultaneously
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A1 σ = 5.65
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A1 μ = 47.2
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Part (b)
M1 normal probability for 45 < X < 50
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A1 0.341
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Part (c)
A1 0.13 = 0.001
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Exam tip: With two unknown parameters you need two standardised equations, one for each given probability.
385IBMA3H-4-18r26-074.1 Sampling, bias and outliersPaper 2Medium7 marks
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A library keeps an alphabetical list of its 1200 borrowers. A systematic sample of 40 borrowers is taken: the 17th borrower is chosen first and then every kth borrower after that.
(a)
Find the value of k.
(1)
(b)
Find the positions in the list of the 5th and the last borrowers chosen.
(2)
(c)
The librarian then emails the chosen borrowers and uses only the replies. Give one reason why the results may be biased.
(1)
(d)
In the survey, the number of books borrowed in a month had lower quartile 18 and upper quartile 34. One borrower had borrowed 61 books. Determine whether 61 is an outlier, using the rule that a value is an outlier if it is more than 1.5 × IQR above the upper quartile.
(3)
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Part (a)
A1 k = 30
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Part (b)
A1 5th: 17 + 4 × 30 = 137
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A1 last: 17 + 39 × 30 = 1187
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Part (c)
R1 e.g. non-response bias: borrowers who reply may be keener library users, or those without email cannot reply
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Part (d)
A1 IQR = 16
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M1 34 + 1.5 × 16
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A1 threshold 58; 61 > 58, so 61 is an outlier
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Exam tip: State the threshold value as well as your conclusion when testing for outliers.
386IBMA3H-4-18r26-094.8 Binomial distributionPaper 2Medium7 marks
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Swimming goggles are supplied by three factories, A, B and C. Factory A supplies 50% of the goggles, B supplies 30% and C supplies 20%. The proportions that are defective are 2% for A, 3% for B and 5% for C.
(a)
Find the probability that a randomly chosen pair of goggles is defective.
(2)
(b)
A pair of goggles is found to be defective. Find the probability that it came from factory C.
(3)
(c)
Given that a pair is defective, find the probability that it came from factory A or factory B.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sum of three products, e.g. 0.5 × 0.02 + 0.3 × 0.03 + 0.2 × 0.05
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A1 0.029
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Part (b)
M1 Bayes' theorem with numerator 0.2 × 0.05 = 0.01
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A1 0.01/0.029
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A1 0.345
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Part (c)
M1 P(A | D) + P(B | D) = (0.01 + 0.009)/0.029, or 1 − 0.345 using the unrounded value
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A1 0.655
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Exam tip: For 'A or B given D', add the two conditional probabilities, or use 1 minus the other one if the events cover all cases.
387IBMA3H-4-18r26-104.6 Combined, conditional and independent eventsPaper 2Medium7 marks
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A swimming coach believes a new swimmer is of elite standard with probability 0.4. An elite swimmer wins each race independently with probability 0.7. A swimmer who is not elite wins each race independently with probability 0.3. The new swimmer enters 4 races and wins exactly 3 of them.
(a)
Find the probability of winning exactly 3 of 4 races for an elite swimmer.
(2)
(b)
Find the probability of winning exactly 3 of 4 races for a swimmer who is not elite.
(1)
(c)
Hence find the probability that the swimmer is elite, given that they won exactly 3 of the 4 races.
(4)
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Notes
Part (a)
M1 binomial with n = 4, p = 0.7, X = 3
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A1 0.4116
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Part (b)
A1 0.0756
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Part (c)
M1 Bayes' theorem structure
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A1 numerator 0.4 × 0.4116 = 0.16464
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M1 denominator 0.16464 + 0.6 × 0.0756 = 0.21
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A1 0.784
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Exam tip: Treat 'elite' or 'not elite' as the two events in the partition and the race results as the observed evidence.
388IBMA3H-4-18r26-114.12 Standardization and unknown normal parametersPaper 2Medium7 marks
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A wind farm has two types of turbine. Type A makes up 60% of the turbines and type B makes up 40%. The daily output of a type A turbine is normally distributed with mean 50 MWh and standard deviation 6 MWh. The daily output of a type B turbine is normally distributed with mean 44 MWh and standard deviation 5 MWh.
(a)
Find the probability that a type A turbine, and the probability that a type B turbine, has a daily output greater than 55 MWh.
(2)
(b)
A turbine is chosen at random and has an output greater than 55 MWh. Find the probability that it is type A.
(3)
(c)
Three turbines, each with output above 55 MWh, are chosen independently. Find the probability that all three are type A. Use the value 0.956.
(2)
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Part (a)
A1 type A: 0.202
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A1 type B: 0.0139
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Part (b)
M1 Bayes' theorem
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A1 0.6 × 0.2023/(0.6 × 0.2023 + 0.4 × 0.0139)
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A1 0.956
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Part (c)
M1 0.9563
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A1 0.874
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Common error: Find each conditional probability of the evidence first, then combine them in Bayes' theorem.
389IBMA3H-4-20r28-054.12 Standardization and unknown normal parametersPaper 2Medium8 marks
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The duration, T hours, of the main explosive phase of eruptions at a volcano is normally distributed with mean \(\mu\) and standard deviation \(\sigma\). It is known that 20% of eruptions have \(T<4.0\) and 10% have \(T>9.0\).
(a)
Write down the z-values that correspond to \(T=4.0\) and \(T=9.0\). [2]
(2)
(b)
Find \(\sigma\) and \(\mu\). Give each answer to 3 significant figures. [4]
(4)
(c)
Find the duration that is exceeded by only 5% of eruptions. Give your answer to 3 significant figures, using the unrounded values of \(\mu\) and \(\sigma\). [2]
(2)
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Part (a)
A1 \(z=-0.8416\) for \(T=4.0\)
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A1 \(z=1.2816\) for \(T=9.0\)
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Part (b)
M1 \(\frac{4-\mu}{\sigma}=-0.8416\) and \(\frac{9-\mu}{\sigma}=1.2816\)
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M1 subtract to get \(5=2.1232\sigma\)
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A1 \(\sigma=2.35\)
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A1 \(\mu=5.98\)
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Part (c)
M1 \(T=\mu+1.6449\sigma\)
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A1 9.86 hours
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Common error: A probability below 0.5 gives a negative z-value. Entering 0.8416 for the lower 20% puts the point above the mean and leads to the wrong equations.
390IBMA3H-4-20r28-064.9 The normal distributionPaper 2Medium7 marks
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The delays, d hours, of 40 container ships arriving at a port are summarised below. \(0\leq d<1\): 8 ships \(1\leq d<2\): 14 ships \(2\leq d<3\): 10 ships \(3\leq d<4\): 6 ships \(4\leq d<5\): 2 ships
Histogram of the delays of 40 container ships
(a)
Use the midpoints of the classes to estimate the mean delay. [2]
(2)
(b)
Estimate the population standard deviation of the delays. Give your answer to 3 significant figures. [2]
(2)
(c)
The port charges each ship a fee of €C, where \(C=300d+500\). Find the mean and the standard deviation of the fee. Use the unrounded standard deviation from (b); give the standard deviation to 3 significant figures. [3]
M1 use of midpoints in the population standard deviation formula or calculator, e.g. \(\sqrt{\frac{210}{40}-2^2}\)
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A1 1.12 hours
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Part (c)
A1 mean fee \(=300\times2+500=\) €1100
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M1 standard deviation is multiplied by 300 and the added 500 has no effect
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A1 standard deviation \(=\) €335
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Common error: Adding a constant to every value changes the mean but not the standard deviation. Multiplying every value by a constant multiplies both by that constant.
391IBMA3H-4-20r28-084.13 Bayes’ theoremPaper 2Medium6 marks
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A seismic alarm system monitors a volcano. In any given week the probability of an eruption is 0.02. If an eruption occurs, the alarm sounds with probability 0.95. If there is no eruption, the alarm sounds falsely with probability 0.08.
Tree diagram showing the probabilities given in the question; None = no eruption
(a)
Find the probability that the alarm sounds in a given week. [3]
(3)
(b)
The alarm sounds. Use Bayes' theorem to find the probability that there is an eruption. Give your answer to 3 significant figures. [2]
(2)
(c)
Explain why this probability is small even though the alarm is reliable when an eruption occurs. [1]
(1)
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Part (a)
M1 use of a tree diagram or the law of total probability
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A1 \(0.02\times0.95+0.98\times0.08\)
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A1 0.0974
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Part (b)
M1 \(\frac{0.02\times0.95}{0.0974}\) (ECF from (a))
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A1 0.195
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Part (c)
R1 eruptions are rare, so the many weeks without an eruption produce more false alarms than the few eruption weeks produce true alarms
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Common error: \(\mathrm{P}(\text{alarm}\mid\text{eruption})\) and \(\mathrm{P}(\text{eruption}\mid\text{alarm})\) are different quantities. The second needs the prior probability of an eruption.
392IBMA3H-4-20r28-094.14 Variance, continuous random variables and linear transformationsPaper 2Medium10 marks
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The number of tickets, X, in a booking at a theatre box office has probability distribution \(\mathrm{P}(X=x)=kx\) for \(x=1,2,3,4\), where k is a constant.
(a)
Show that \(k=0.1\). [2]
(2)
(b)
Find \(\mathrm{E}(X)\). [2]
(2)
(c)
Find \(\mathrm{Var}(X)\). [3]
(3)
(d)
Each booking costs €18 per ticket plus a booking fee of €2.50, so the amount paid is \(R=18X+2.5\) euros. Find \(\mathrm{E}(R)\) and \(\mathrm{Var}(R)\). [3]
M1 \(\mathrm{Var}(R)=18^2\mathrm{Var}(X)\), the constant 2.5 having no effect
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A1 \(\mathrm{Var}(R)=324\)
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Common error: Use \(\mathrm{Var}(aX+b)=a^2\mathrm{Var}(X)\). The constant b shifts the distribution but does not change its spread, and a must be squared.
393IBMA3H-4-19r32-054.4 Correlation and regression of y on xPaper 2Medium7 marks
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A mountaineering team measures the temperature at which water boils at different altitudes. The altitude is x km above sea level and the boiling point of water is y °C.
x (km)
0
1
2
3
4
6
y (°C)
100.0
96.7
93.4
90.1
86.5
80.2
Scatter diagram of the data in the table
(a)
Find the product moment correlation coefficient r.
(1)
(b)
Find the equation of the regression line of x on y. Give the coefficients correct to four significant figures.
(2)
(c)
The team finds that water boils at 85.0 °C at their camp. Use your equation from part (b), with the coefficients correct to four significant figures, to estimate the altitude of the camp, giving your answer to three significant figures.
(2)
(d)
Explain why the regression line of x on y is more appropriate than the regression line of y on x for this estimate, and comment on the reliability of the estimate.
(2)
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Part (a)
A1 r = −0.9999 (accept −0.9999 or −1.00)
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Part (b)
M1 attempt to find regression of x on y using technology
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A1 x = −0.3014y + 30.14
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Part (c)
M1 substitute y = 85.0 into the equation from (b)
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A1 x = −0.3014 × 85.0 + 30.14 = 4.521 → 4.52 km
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Part (d)
R1 the altitude (x) is being estimated from the boiling point (y), so the line of x on y is used
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R1 estimate is reliable: 85.0 lies within the data range (interpolation) and r is very close to −1
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Exam tip: Check that the value you substitute lies inside the range of the data before describing the estimate as reliable.
394IBMA3H-4-19r32-064.8 Binomial distributionPaper 2Medium7 marks
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A bakery produces loaves in batches. Each loaf is independently underbaked with probability 0.04. A food scientist inspects a random sample of 30 loaves from a batch.
(a)
Find the probability that at most one loaf in the sample is underbaked.
(2)
(b)
The scientist now inspects n loaves. Find the smallest value of n for which the probability of finding at least one underbaked loaf is greater than 0.9.
(3)
(c)
For the value of n found in part (b), find the probability of finding at least one underbaked loaf. Give your answer to three significant figures.
(2)
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Notes
Part (a)
M1 P(X ≤ 1) = P(X = 0) + P(X = 1) with X ~ B(30, 0.04)
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A1 0.661
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Part (b)
M1 1 − 0.96n > 0.9
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M1 attempt to solve, e.g. 0.96n < 0.1 using logarithms or a table
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A1 n > 56.4 so n = 57
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Part (c)
M1 1 − 0.9657
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A1 0.902
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Common error: When solving an inequality with n as an exponent, check the integer answers either side of the boundary rather than rounding to the nearest value.
395IBMA3H-4-19r32-074.12 Standardization and unknown normal parametersPaper 2Medium8 marks
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The duration, T minutes, of calls made by customers of a telecoms company is normally distributed with mean μ and standard deviation σ. It is known that 20% of calls last less than 3 minutes and 10% of calls last more than 12 minutes.
Sketch of the normal curve with the given tail probabilities, not to scale
(a)
Write down two equations connecting μ and σ.
(3)
(b)
Find the value of μ and the value of σ.
(3)
(c)
Find the duration exceeded by 5% of calls. Use the unrounded values of μ and σ.
(2)
Show mark scheme
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Notes
Part (a)
M1 use of inverse normal to find standardised values, z = −0.8416 and z = 1.2816
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A1 (3 − μ)/σ = −0.8416
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A1 (12 − μ)/σ = 1.2816
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Part (b)
M1 attempt to solve the two equations simultaneously
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A1 σ = 4.24
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A1 μ = 6.57
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Part (c)
M1 μ + 1.6449σ
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A1 13.5 minutes
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Exam tip: Keep the unrounded values of μ and σ in your calculator memory; rounding too early can change the third significant figure of later answers.
396IBMA3H-4-19r32-084.2 Presentation of dataPaper 2Medium7 marks
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A food scientist records the shelf life, in days, of 60 pots of yoghurt. The results are summarised in the table.
Shelf life d (days)
10 ≤ d < 12
12 ≤ d < 14
14 ≤ d < 16
16 ≤ d < 18
18 ≤ d < 20
Frequency
6
13
22
12
7
(a)
Write down the modal class.
(1)
(b)
Calculate an estimate for the mean shelf life.
(2)
(c)
Calculate an estimate for the population standard deviation of the shelf life.
(2)
(d)
A new preservative changes each shelf life d to 1.2d + 2. Find an estimate for the standard deviation of the new shelf lives. Use the unrounded value from part (c).
(2)
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Part (a)
A1 14 ≤ d < 16
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Part (b)
M1 use of mid-interval values 11, 13, 15, 17, 19
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A1 15.03 days
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Part (c)
M1 use of technology with mid-interval values and frequencies
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A1 2.27 days
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Part (d)
M1 multiply the standard deviation by 1.2 (the addition of 2 has no effect on the spread)
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A1 2.72 days
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Common error: Adding a constant to every value changes the mean but not the standard deviation; only the multiplier affects the spread.
397IBMA3H-4-19r32-094.13 Bayes’ theoremPaper 2Medium9 marks
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A telecoms company monitors its mobile masts with an automatic detector. In any given week, a mast has a fault with probability 0.02. If a mast has a fault, the detector raises an alert with probability 0.95. If a mast has no fault, the detector raises a false alert with probability 0.03.
Tree diagram showing the probabilities given in the question
(a)
Find the probability that the detector raises an alert for a randomly chosen mast in a given week.
(3)
(b)
Given that the detector raises an alert for a mast, find the probability that the mast really has a fault. Give your answer to three significant figures.
(3)
(c)
The company investigates 20 masts, each of which has raised an alert. Assuming the masts are independent, and using 0.393 as the probability that a mast with an alert really has a fault, find the probability that at least three of the masts really have a fault.
(3)
Show mark scheme
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Notes
Part (a)
M1 attempt to use the law of total probability (tree diagram or formula)
✓ 1
A1 P(alert) = 0.02 × 0.95 + 0.98 × 0.03
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A1 0.0484
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Part (b)
M1 attempt to use P(F | A) = P(F ∩ A)/P(A)
✓ 1
A1 0.019/0.0484
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A1 0.393
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Part (c)
M1 X ~ B(20, 0.393)
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M1 P(X ≥ 3) = 1 − P(X ≤ 2)
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A1 0.996
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Common error: A reliable-looking detector can still give mostly false alerts when faults are rare; always condition on the alert, not on the fault.
398IBMA3H-4-19r32-144.9 The normal distributionPaper 2Medium7 marks
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A food company fills bottles of juice labelled 500 ml. The volume of juice in a bottle is normally distributed with mean μ ml and standard deviation 4 ml. The company sets μ so that 1% of bottles contain less than 500 ml.
(a)
Find the value of μ, correct to four significant figures.
(3)
(b)
The company buys a more precise machine, with standard deviation 2.5 ml, and again sets the mean so that 1% of bottles contain less than 500 ml. Find the new mean, correct to four significant figures.
(2)
(c)
Juice costs €0.0012 per ml. Find the saving in juice cost when 100 000 bottles are filled with the new machine rather than the old machine. Use the unrounded values of the two means, and give your answer to three significant figures.
(2)
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Notes
Part (a)
M1 P(V < 500) = 0.01, so use inverse normal to find z = −2.3263
Exam tip: When the standard deviation changes but the percentage below the limit stays fixed, the z-value is unchanged; only μ moves.
399IBMA3H-4-18r33-014.2 Presentation of dataPaper 2Medium6 marks
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Marine biologists net fish in three zones of a reef. A fish netted is from zone X with probability 0.45, from zone Y with probability 0.35 and from zone Z with probability 0.20. The probability that a fish is already tagged is 0.06 in zone X, 0.12 in zone Y and 0.30 in zone Z.
(a)
Find the probability that a fish chosen from the net is tagged.
(3)
(b)
A fish from the net is found to be tagged. Find the probability that it came from zone Z. Give your answer to three significant figures.
(3)
Show mark scheme
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Notes
Part (a)
M1 sum of products of the form P(zone) × P(tagged | zone)
Common error: The denominator in Bayes' theorem is the total probability of the observed event (tagged), not just the probability for one zone.
400IBMA3H-4-18r33-024.13 Bayes’ theoremPaper 2Medium6 marks
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A logistics company sorts parcels at three depots, A, B and C, which handle 55%, 25% and 20% of all parcels respectively. A parcel is misrouted with probability 0.01 at depot A, 0.03 at depot B and 0.05 at depot C.
(a)
Find the probability that a parcel chosen at random is misrouted.
(2)
(b)
A misrouted parcel is selected. Find the probability that it was handled at each depot, giving each answer to three significant figures, and state which depot is the most likely source.
(4)
Show mark scheme
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Notes
Part (a)
M1 0.55 × 0.01 + 0.25 × 0.03 + 0.20 × 0.05 (sum of three products)
✓ 1
A1 0.023
✓ 1
Part (b)
M1 Bayes' theorem with denominator 0.023
✓ 1
A1 depot A: 0.0055 / 0.023 = 0.239
✓ 1
A1 depot B: 0.0075 / 0.023 = 0.326
✓ 1
A1 depot C: 0.010 / 0.023 = 0.435, so depot C is the most likely source
✓ 1
Exam tip: As a check, the three posterior probabilities must add to 1 (allowing for rounding).
401IBMA3H-4-18r33-064.12 Standardization and unknown normal parametersPaper 2Medium6 marks
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A marine biologist studies two populations of the same fish species in a bay. Of the fish caught, 65% come from population A and 35% from population B. The length of a fish, in cm, is normally distributed: N(42, 42) for population A and N(50, 52) for population B.
(a)
Find the probability that a fish from (i) population A, (ii) population B, is longer than 48 cm. Give each answer to three significant figures.
(2)
(b)
A fish caught at random is found to be longer than 48 cm. Using unrounded values from (a), find the probability that it came from population B. Give your answer to three significant figures.
(4)
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Notes
Part (a)
A1 (i) 0.0668
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A1 (ii) 0.655
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Part (b)
M1 P(longer than 48) = 0.65 × P(longer | A) + 0.35 × P(longer | B)
✓ 1
A1 P(longer than 48) = 0.273
✓ 1
M1 P(B | longer than 48) = 0.35 × P(longer | B) / P(longer than 48)
✓ 1
A1 0.841
✓ 1
Common error: Do not take 0.655 as the answer to the final part; it is P(longer | B), not P(B | longer). Reverse the conditional with Bayes' theorem.
402IBMA3H-4-18r33-104.12 Standardization and unknown normal parametersPaper 2Medium7 marks
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The duration of a dive, D minutes, made by a species of beaked whale is normally distributed with standard deviation 7.5. It is found that 12% of dives last longer than 38 minutes.
(a)
Find the mean duration of a dive, to three significant figures.
(3)
(b)
Using the unrounded mean, find the probability that a dive lasts less than 20 minutes, to three significant figures.
(2)
(c)
The longest 5% of dives are classed as deep dives. Find the least duration of a deep dive, to three significant figures, using the unrounded mean.
(2)
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Notes
Part (a)
A1 z = 1.175 for a lower-tail probability of 0.88
✓ 1
M1 38 = μ + z × 7.5
✓ 1
A1 μ = 29.2 minutes
✓ 1
Part (b)
M1 attempt to use the normal distribution with the unrounded mean and standard deviation 7.5
✓ 1
A1 0.110
✓ 1
Part (c)
M1 μ + 1.64… × 7.5 (inverse normal for upper-tail probability 0.05)
✓ 1
A1 41.5 minutes
✓ 1
Common error: '12% longer than 38' is an upper-tail probability; use P(D < 38) = 0.88 with the inverse normal.
403IBMA3H-4-18r33-144.13 Bayes’ theoremPaper 2Medium7 marks
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A logistics company classes 30% of its parcels as express (event A). Of the express parcels, 20% arrive late. Overall, 26% of parcels arrive late (event B). A parcel is chosen at random.
(a)
Find P(A ∩ B).
(1)
(b)
Find the probability that a parcel arrives late, given that it is not an express parcel. Give your answer to three significant figures.
(3)
(c)
Determine whether A and B are independent.
(1)
(d)
Find P(A | B), to three significant figures.
(2)
Show mark scheme
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Notes
Part (a)
A1 0.3 × 0.2 = 0.06
✓ 1
Part (b)
M1 P(A′ ∩ B) = 0.26 − 0.06 = 0.20
✓ 1
M1 P(B | A′) = 0.20 / 0.7
✓ 1
A1 0.286
✓ 1
Part (c)
R1 not independent, since P(B | A) = 0.2 ≠ P(B) = 0.26
✓ 1
Part (d)
M1 P(A | B) = 0.06 / 0.26
✓ 1
A1 0.231
✓ 1
Common error: P(B | A) and P(A | B) are different; the first divides by P(A), the second by P(B).
404IBMA3H-4-19r37-064.9 The normal distributionPaper 2Medium8 marks
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A coffee company roasts beans on two machines. Machine A roasts 65% of the beans and machine B roasts the rest. A bag of beans from machine A is defective with probability 0.04, and a bag from machine B is defective with probability 0.07.
(a)
Find the probability that a randomly chosen bag is defective.
(2)
(b)
A randomly chosen bag is not defective. Find the probability that it came from machine A. Give your answer to 3 significant figures.
(3)
(c)
Four bags are chosen at random and independently. Use the exact value 0.0505 to find the probability that at least one is defective. Give your answer to 3 significant figures.
Common error: "At least one" is best handled as 1 minus the probability of none; adding the probabilities of one, two, three and four defective bags is slow and error-prone.
405IBMA3H-4-19r37-074.7 Discrete random variablesPaper 2Medium7 marks
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The number of micro-meteoroid strikes, X, logged by a probe's shield in one hour has the probability distribution below.
x: 0, 1, 2, 3, 4 P(X = x): 0.1, 0.2, k, 2k, 0.1
(a)
Find the value of k.
(2)
(b)
Find E(X).
(2)
(c)
Given that at least one strike is logged in an hour, find the probability that at least two strikes are logged. Give your answer to 3 significant figures.
(3)
Show mark scheme
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Notes
Part (a)
M1 sum of probabilities = 1: 0.4 + 3k = 1
✓ 1
A1 k = 0.2
✓ 1
Part (b)
M1 Σ xP(X = x)
✓ 1
A1 E(X) = 2.2
✓ 1
Part (c)
M1 P(X ≥ 2) = 0.7 and P(X ≥ 1) = 0.9
✓ 1
M1 P(X ≥ 2) ÷ P(X ≥ 1)
✓ 1
A1 7/9 = 0.778
✓ 1
Exam tip: In a conditional probability, the denominator is the probability of the condition, here P(X ≥ 1), not 1.
406IBMA3H-4-19r37-084.13 Bayes’ theoremPaper 2Medium7 marks
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A probe's fault monitor sounds an alarm with probability 0.95 when a fault is present and with probability 0.04 when no fault is present. A fault is present in 2% of all hourly checks.
Tree diagram showing the probabilities given in the question
(a)
Find the probability that the alarm sounds in a randomly chosen hourly check.
(2)
(b)
The alarm sounds. Use Bayes' theorem to find the probability that a fault is present. Give your answer to 3 significant figures.
(3)
(c)
The alarm does not sound. Find the probability that no fault is present. Give your answer to 3 significant figures.
Common error: P(alarm | fault) and P(fault | alarm) are different quantities; the second must be built from the first using the total probability of an alarm.
407IBMA3H-4-19r37-094.12 Standardization and unknown normal parametersPaper 2Medium7 marks
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A roaster operator records the number of batches, X, that must be re-roasted during a shift. X takes the values 1, 2, 3 and 4 with P(X = 1) = p, P(X = 2) = 0.3, P(X = 3) = q and P(X = 4) = 0.2. It is known that E(X) = 2.5.
(a)
Find the values of p and q.
(3)
(b)
Find E(X2) and hence Var(X).
(3)
(c)
Find the standard deviation of X, correct to 3 significant figures.
(1)
Show mark scheme
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Notes
Part (a)
M1 p + q + 0.5 = 1
✓ 1
M1 p + 0.6 + 3q + 0.8 = 2.5
✓ 1
A1 p = 0.2 and q = 0.3
✓ 1
Part (b)
M1 E(X2) = Σ x2P(X = x)
✓ 1
A1 E(X2) = 7.3
✓ 1
A1 Var(X) = 7.3 − 2.52 = 1.05
✓ 1
Part (c)
A1 1.02 (ECF from (b))
✓ 1
Exam tip: Use Var(X) = E(X²) − [E(X)]² and keep E(X²) unrounded; a negative variance means an arithmetic slip.
408IBMA3H-4-19r37-104.2 Presentation of dataPaper 2Medium10 marks
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The number of images, X, that a probe must retransmit in one orbit has the distribution below.
x: 0, 1, 2, 3 P(X = x): 0.15, 0.35, 0.30, 0.20
The cost, C, in £, of an orbit's retransmissions is given by C = 40 + 25X.
(a)
Find E(X).
(2)
(b)
Find Var(X).
(3)
(c)
Find E(C).
(2)
(d)
Find the standard deviation of C, correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 Σ xP(X = x)
✓ 1
A1 E(X) = 1.55
✓ 1
Part (b)
M1 attempt at E(X2)
✓ 1
A1 E(X2) = 3.35
✓ 1
A1 Var(X) = 3.35 − 1.552 = 0.9475
✓ 1
Part (c)
M1 E(C) = 40 + 25E(X)
✓ 1
A1 £78.75 (ECF from (a))
✓ 1
Part (d)
M1 Var(C) = 252 × Var(X)
✓ 1
M1 square root of Var(C) = 592.1875 (ECF from (b))
✓ 1
A1 £24.3
✓ 1
Common error: When finding Var(aX + b), the constant b disappears and a must be squared.
409IBMA3H-4-19r37-124.7 Discrete random variablesPaper 2Medium9 marks
Calculator
Two independent load sensors on a bridge each give a reading Y, where P(Y = 0) = 0.5, P(Y = 1) = 0.3 and P(Y = 2) = 0.2. Let S be the sum of the readings from the two sensors.
(a)
Find the probability distribution of S.
(4)
(b)
Find E(S).
(2)
(c)
Find Var(S).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 considering all combinations of two readings for each value of S
✓ 1
A1 P(S = 0) = 0.25 and P(S = 4) = 0.04
✓ 1
A1 P(S = 1) = 0.30 and P(S = 3) = 0.12
✓ 1
A1 P(S = 2) = 0.29
✓ 1
Part (b)
M1 Σ sP(S = s)
✓ 1
A1 E(S) = 1.4 (ECF from (a))
✓ 1
Part (c)
M1 E(S2) = 0.30 + 1.16 + 1.08 + 0.64 = 3.18
✓ 1
M1 Var(S) = E(S2) − 1.42
✓ 1
A1 Var(S) = 1.22 (accept any valid method, for example 2 × Var(Y) = 2 × 0.61)
✓ 1
Exam tip: Check that your probabilities sum to 1 before using the distribution; it catches a missing combination immediately.
410IBMA3H-4-20r44-054.4 Correlation and regression of y on xPaper 2Medium6 marks
Calculator
A shipping company records, for seven cargo ships, the cargo load \(x\) (in thousands of tonnes) and the fuel used per day \(y\) (in tonnes).
x
12
18
23
27
31
38
44
y
21.5
26.0
31.2
33.4
37.9
43.0
49.1
Scatter diagram of the data in the table
(a)
Find the equation of the regression line of \(y\) on \(x\), giving the coefficients to 3 significant figures.
(2)
(b)
Interpret the gradient of the regression line in this context.
(1)
(c)
Explain why the regression line should not be used to estimate the fuel used by a ship carrying 95 thousand tonnes.
(1)
(d)
Use the unrounded regression line to estimate the fuel used per day by a ship carrying 34 thousand tonnes. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 gradient 0.858
✓ 1
A1 y = 0.858x + 10.9
✓ 1
Part (b)
A1 each extra 1000 tonnes of cargo is associated with about 0.858 tonnes more fuel used per day
✓ 1
Part (c)
R1 95 is far outside the range of the data (12 to 44), so this is extrapolation and the linear relationship may not continue
✓ 1
Part (d)
M1 substitute x = 34 into the regression line
✓ 1
A1 40.1 tonnes per day
✓ 1
Common error: The gradient should be interpreted as a change in \(y\) per unit of \(x\), in context and with units, not just as 'the slope'.
411IBMA3H-4-20r44-064.8 Binomial distributionPaper 2Medium6 marks
Calculator
A theatre company has six lead performers. Each performer is independently available for a given night with probability \(p\). Let \(X\) be the number of performers who are available. The probability that no performer is available is 0.000729.
(a)
Find the value of \(p\).
(2)
(b)
Find the probability that at least 5 performers are available. Give your answer to 3 significant figures.
(2)
(c)
Given that at least 5 performers are available, find the probability that all six are available.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(X = 0) = (1 − p)6 = 0.000729
✓ 1
A1 1 − p = 0.3, so p = 0.7
✓ 1
Part (b)
M1 P(X = 5) + P(X = 6) = 6(0.7)5(0.3) + 0.76
✓ 1
A1 0.420
✓ 1
Part (c)
M1 P(X = 6) ÷ P(X ≥ 5) = 0.76 ÷ 0.420175
✓ 1
A1 0.28
✓ 1
Exam tip: For 'given that', the numerator is the probability of both events together; here 'all six' already implies 'at least five'.
412IBMA3H-4-20r44-084.8 Binomial distributionPaper 2Medium6 marks
Calculator
A volcano observatory uses a sensor system. In any given week the probability of an eruption is 0.04. If an eruption occurs the alarm sounds with probability 0.92. If no eruption occurs the alarm sounds with probability 0.07. Let \(E\) be the event that an eruption occurs and \(A\) the event that the alarm sounds.
(a)
Find \(P(A)\).
(2)
(b)
Given that the alarm sounds, find the probability that an eruption occurs. Give your answer to 3 significant figures.
(2)
(c)
Given that the alarm does not sound, find the probability that no eruption occurs. Give your answer to 3 significant figures.
Exam tip: Write down the denominator, \(P(A)\) or \(P(A')\), as its own line first; most errors in Bayes' theorem come from an incomplete denominator.
413IBMA3H-4-20r44-094.13 Bayes’ theoremPaper 2Medium6 marks
Calculator
A theatre sells tickets through three outlets. 55% of tickets are sold online, 30% at the box office and 15% through agents. The probability that a ticket holder arrives late is 0.04 for online tickets, 0.10 for box office tickets and 0.20 for agent tickets.
Tree diagram showing the probabilities given in the question; Box off. = box office
(a)
Find the probability that a randomly chosen ticket holder arrives late.
(2)
(b)
Given that a ticket holder arrives late, find the probability that the ticket was bought through an agent. Give your answer to 3 significant figures.
(2)
(c)
Given that a ticket holder arrives late, find the probability that the ticket was not bought online. Give your answer to 3 significant figures.
Common error: \(P(\text{agent}\mid\text{late})\) is not \(P(\text{late}\mid\text{agent})=0.20\); the conditioning event is different, so the denominator must be \(P(\text{late})\).
414IBMA3H-4-20r44-114.13 Bayes’ theoremPaper 2Medium6 marks
Calculator
At a port, 2% of containers hold contraband. Every container is scanned. A container that holds contraband is flagged with probability 0.95; a container that does not is flagged with probability 0.06. Every flagged container is then searched by hand, and a search finds the contraband, if present, with probability 0.8.
(a)
Find the probability that a randomly chosen container is flagged.
(2)
(b)
Given that a container is flagged, find the probability that it holds contraband. Give your answer to 3 significant figures.
(2)
(c)
Find the probability that a randomly chosen container holds contraband that is found by the hand search.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.02 × 0.95 + 0.98 × 0.06
✓ 1
A1 0.0778
✓ 1
Part (b)
M1 0.019 ÷ 0.0778
✓ 1
A1 0.244
✓ 1
Part (c)
M1 0.02 × 0.95 × 0.8
✓ 1
A1 0.0152
✓ 1
Common error: The hand search happens only for flagged containers, so the three probabilities must be multiplied along the branch: contraband, then flagged, then found.
415IBMA3H-4-20r44-134.8 Binomial distributionPaper 2Medium6 marks
Calculator
A theatre production has a lead actor who is available on any night with probability 0.9. If the lead is not available, an understudy performs. The probability of a good review is 0.8 when the lead performs and 0.55 when the understudy performs. Nights are independent. Let \(G\) be the event that a night receives a good review.
Tree diagram showing the probabilities given in the question; L = lead performs, U = understudy performs, G = good review
(a)
Find \(P(G)\).
(2)
(b)
Given that a night receives a good review, find the probability that the understudy performed. Give your answer to 3 significant figures.
(2)
(c)
Over two independent nights, both receive good reviews. Find the probability that the lead performed on both nights. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.9 × 0.8 + 0.1 × 0.55
✓ 1
A1 0.775
✓ 1
Part (b)
M1 (0.1 × 0.55) ÷ 0.775
✓ 1
A1 0.0710
✓ 1
Part (c)
M1 (0.72 ÷ 0.775)2 , using independence of nights
✓ 1
A1 0.863
✓ 1
Common error: In part (c) the probability is the product of the two nightly posterior probabilities because nights are independent; it is not \(0.9^2\).
416IBMA3H-4-20r46-054.11 Conditional probability and independencePaper 2Medium7 marks
Calculator
At a farm, 60% of seeds sown are variety \(A\) and 40% are variety \(B\). The probability that a seed germinates is 0.9 for variety \(A\) and 0.8 for variety \(B\). Let \(G\) be the event that a randomly chosen seed germinates.
(a)
Find \(P(G)\).
(3)
(b)
Show that \(G\) and \(A\) are not independent.
(2)
(c)
A seed is chosen and is found to have germinated. Find the probability that it is variety \(A\). Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(P(A\cap G)=0.6\times0.9=0.54\)
✓ 1
M1 \(P(B\cap G)=0.4\times0.8=0.32\)
✓ 1
A1 0.86
✓ 1
Part (b)
M1 \(P(G)P(A)=0.86\times0.6=0.516\)
✓ 1
R1 this is not equal to \(P(G\cap A)=0.54\), so the events are not independent
Common error: Independence needs \(P(G\cap A)=P(G)P(A)\); comparing \(P(G|A)\) with \(P(G)\) is an equivalent test, but a vague statement that 'A affects G' earns nothing.
417IBMA3H-4-20r46-064.4 Correlation and regression of y on xPaper 2Medium7 marks
Calculator
Seven music students record the time they practise each week, \(x\) hours, and their mark, \(y\), in a grade exam.
Find the Pearson's product-moment correlation coefficient, \(r\).
(2)
(b)
Describe the correlation between \(x\) and \(y\).
(1)
(c)
Find the equation of the regression line of \(y\) on \(x\), giving the coefficients to 3 significant figures.
(2)
(d)
Use the unrounded regression line to estimate the mark of a student who practises for 7 hours a week. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of a GDC or formula for \(r\)
✓ 1
A1 \(r=0.998\)
✓ 1
Part (b)
A1 strong positive linear correlation
✓ 1
Part (c)
A1 gradient 3.74
✓ 1
A1 intercept 41.1
✓ 1
Part (d)
M1 substitute \(x=7\) into the regression line
✓ 1
A1 67.3
✓ 1
Exam tip: Use the unrounded regression line for predictions and only round the final answer; also make sure the value of \(x\) lies within the data range.
418IBMA3H-4-20r46-074.14 Variance, continuous random variables and linear transformationsPaper 2Medium9 marks
Calculator
The number of delayed flights in a day at a small airport is modelled by the discrete random variable \(X\) with the following distribution.
x: 0, 1, 2, 3 P(X = x): 0.2, \(a\), \(b\), 0.25
It is given that \(E(X)=1.55\).
(a)
Find the value of \(a\) and the value of \(b\).
(4)
(b)
Find \(P(X\geq2)\).
(2)
(c)
Given that there is at least one delayed flight on a day, find the probability that there are at least two.
The organisers charge €1.50 per play. Find the organisers' expected profit if 400 plays are made.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(0(0.7)+2(0.2)+5(0.08)+20(0.02)\)
✓ 1
A1 €1.20
✓ 1
Part (b)
A1 €1.20 (expected gain is zero when the fee equals \(E(X)\))
✓ 1
Part (c)
M1 \((1.50-1.20)\times400\) (ECF)
✓ 1
A1 €120
✓ 1
Exam tip: A game is fair when the expected net gain is zero, so the entry fee equals the expected prize.
420IBMA3H-4-20r46-094.13 Bayes’ theoremPaper 2Medium8 marks
Calculator
A workshop buys aircraft spare parts from three suppliers. Supplier \(A\) provides 50% of the parts, supplier \(B\) 30% and supplier \(C\) 20%. The probabilities that a part is defective are 0.02 for \(A\), 0.03 for \(B\) and 0.04 for \(C\). Let \(D\) be the event that a randomly chosen part is defective.
Tree diagram showing the probabilities given in the question; D = part is defective
(a)
Find \(P(D)\).
(3)
(b)
A part is found to be defective. Find the probability that it came from supplier \(C\). Give your answer to 3 significant figures.
(2)
(c)
A part is found not to be defective. Find the probability that it came from supplier \(A\). Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of the law of total probability: sum of three products
✓ 1
A1 0.010, 0.009 and 0.008
✓ 1
A1 0.027
✓ 1
Part (b)
M1 \(P(C|D)=\frac{0.2\times0.04}{0.027}\)
✓ 1
A1 0.296
✓ 1
Part (c)
M1 \(P(A\cap D')=0.5\times0.98=0.49\)
✓ 1
M1 \(P(D')=1-0.027=0.973\) and divide
✓ 1
A1 0.504
✓ 1
Common error: In Bayes' theorem the denominator is the probability of the observed event over all suppliers, not the probability for one supplier.
421IBMA3H-4-20r46-104.14 Variance, continuous random variables and linear transformationsPaper 2Medium7 marks
Calculator
The number of songs requested from a busker in one hour is a random variable \(X\) with \(E(X)=3.2\) and \(\mathrm{Var}(X)=1.44\). The busker's earnings in an hour, in euros, are \(T=4X+10\).
(a)
Find \(E(T)\).
(2)
(b)
Find \(\mathrm{Var}(T)\).
(2)
(c)
Find the standard deviation of \(T\).
(1)
(d)
Find \(E(X^2)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(E(aX+b)=aE(X)+b\)
✓ 1
A1 22.8
✓ 1
Part (b)
M1 use of \(\mathrm{Var}(aX+b)=a^2\mathrm{Var}(X)\)
✓ 1
A1 23.04
✓ 1
Part (c)
A1 4.8 (ECF from (b))
✓ 1
Part (d)
M1 use of \(\mathrm{Var}(X)=E(X^2)-[E(X)]^2\)
✓ 1
A1 11.68
✓ 1
Common error: Adding a constant changes the mean but not the variance, and multiplying by \(a\) multiplies the variance by \(a^2\), not \(a\).
422AAHL4-q00144.4 Correlation and regression of y on xPaper 2Hard10 marks
A school offers extra tutoring. Group P is a support group of students with low prior attainment; group Q are the other students. The table and scatter diagram show the hours of tutoring, \(x\), and the score in a test, \(y\), for 14 students.
Group P: \(x\)
6
7
8
9
10
11
12
Group P: \(y\)
38
44
41
49
52
50
57
Group Q: \(x\)
0
1
2
3
4
5
6
Group Q: \(y\)
63
70
66
74
72
79
81
(a)
Find the value of Pearson’s product-moment correlation coefficient, \(r\), for group P and for group Q separately.
(2)
(b)
Find the value of \(r\) for all 14 students together.
(2)
(c)
The head teacher looks only at the combined data and concludes that tutoring lowers test scores. Comment on this conclusion.
(2)
(d)
A student in group Q received 4.5 hours of tutoring. Find the equation of an appropriate regression line and use it to estimate this student’s score.
(3)
(e)
Explain why the regression line for all 14 students should not be used in part (d).
(1)
Show mark scheme
Working
Mark
Notes
(a)
group P: \(r=0.931\)
A1
\(0.93149\ldots\)
group Q: \(r=0.923\)
A1
\(0.92335\ldots\)
(b)
entering all 14 pairs
(M1)
\(r=-0.521\)
A1
\(-0.52111\ldots\)
(c)
within each group there is strong positive correlation, so the data do not support the claim
R1
the negative overall correlation is caused by group membership (prior attainment): weaker students received more tutoring; correlation does not imply causation
R1
(d)
regression of \(y\) on \(x\) using group Q only
(M1)
\(y=2.79x+63.8\)
A1
\(2.78571\ldots,\ 63.7857\ldots\)
\(76.3\)
A1
\(76.32\ldots\) Accept 76.
(e)
the combined data are two separate clusters with different relationships, so the combined line (gradient \(-2\)) does not model either group
423AAHL4-q00374.8 Binomial distributionPaper 2Hard15 marks
A shop receives large batches of light bulbs. In a batch, a proportion \(p\) of bulbs is defective, and bulbs are defective independently of each other. The shop considers the following plans for deciding whether to accept a batch.
Plan A: test 20 bulbs; accept the batch if at most one of them is defective.
Plan B: test 30 bulbs; accept the batch if at most two of them are defective.
(a)
For Plan A with \(p=0.05\), find the probability that the batch is accepted.
(2)
(b)
Show that, under Plan A, the probability that a batch is accepted is \((1-p)^{19}(1+19p)\).
(2)
(c)
Find the value of \(p\) for which Plan A accepts a batch with probability 0.5.
(2)
(d)
For Plan B, find the probability that the batch is accepted when \(p=0.05\) and when \(p=0.15\).
(3)
(e)
The shop requires a plan that accepts a good batch (\(p=0.05\)) with probability at least 0.8 and accepts a poor batch (\(p=0.15\)) with probability at most 0.2. Determine which of Plans A and B meet both requirements.
(2)
(f)
Plan C is to test \(n\) bulbs and accept the batch only if none of them is defective. Show that no value of \(n\) allows Plan C to meet both requirements in part (e).
Plan A fails: \(0.736\lt 0.8\) (for \(p=0.15\) it accepts with probability \(0.176\))
R1
\(0.175557\ldots\)
Plan B meets both: \(0.812\ge0.8\) and \(0.151\le0.2\)
A1
FT from (a) and (d).
(f)
\(0.95^{n}\ge0.8\Rightarrow n\le\dfrac{\ln0.8}{\ln0.95}=4.35\), so \(n\le4\)
M1
\(0.85^{n}\le0.2\Rightarrow n\ge\dfrac{\ln0.2}{\ln0.85}=9.90\), so \(n\ge10\)
M1
no integer satisfies both \(n\le4\) and \(n\ge10\), so Plan C cannot meet both requirements
R1
(g)
Plan A tests only 20 bulbs rather than 30, so it is cheaper/quicker
R1
Answers: (a) 0.736 · (c) \(p=0.0825\) · (d) 0.812 and 0.151 · (e) Plan B only · (f) \(n\le4\) and \(n\ge10\) incompatible
424AAHL4-q00424.9 The normal distributionPaper 2Hard6 marks
The random variable \(X\) is normally distributed with mean 40 and standard deviation 6. The constant \(k\gt 0\) is such that \(\mathrm{P}(40-k\lt X\lt 40+2k)=0.75\).
Normal curve with mean 40; the shaded region is the interval 40 − k < X < 40 + 2k. Diagram not to scale
attempt to set up an equation in \(k\), e.g. \(\mathrm{P}(X\lt 40+2k)-\mathrm{P}(X\lt 40-k)=0.75\), or \(\Phi\!\left(\tfrac{2k}{6}\right)-\Phi\!\left(-\tfrac{k}{6}\right)=0.75\)
M1
Award M0 for treating the interval as symmetric about the mean.
use of GDC to solve, e.g. graph of \(y=\mathrm{normalcdf}(40-k,\,40+2k,\,40,\,6)\) intersecting \(y=0.75\)
Recognising that \(\{X\gt 40+2k\}\cap\{X\gt 40-k\}=\{X\gt 40+2k\}\).
\(\mathrm{P}(X\gt 50.0)=0.0477\) and \(\mathrm{P}(X\gt 35.0)=0.798\)
A1
Both values. FT their \(k\).
\(0.0598\)
A1
\(0.0598455\ldots\) Accept 0.0597–0.0599.
Answers: (a) \(k=5.00\) · (b) 0.0598
425AAHL4-q00444.9 The normal distributionPaper 2Hard14 marks
A machine fills bags of coffee. The mass, \(X\) grams, of coffee in a bag is normally distributed with mean \(\mu\) grams and standard deviation 4 grams. The value of \(\mu\) can be set by the operator.
The coffee costs the company €0.02 per gram. Every bag is checked: a bag with \(X\ge245\) is sold for €8, and a bag with \(X\lt245\) is rejected and cannot be sold. Let \(G\) be the profit, in euros, on a bag, so that \(G=8-0.02X\) if the bag is sold and \(G=-0.02X\) if it is rejected.
(a)
For \(\mu=250\), find the probability that a bag is rejected.
(2)
(b)
Show that \(\mathrm{E}(G)=8\,\mathrm{P}(X\ge245)-0.02\mu\).
(2)
(c)
Find \(\mathrm{E}(G)\) when \(\mu=250\).
(2)
(d)
Find the value of \(\mu\) that maximizes \(\mathrm{E}(G)\), and the maximum value of \(\mathrm{E}(G)\).
(3)
(e)
For the value of \(\mu\) found in part (d), find the probability that a bag is rejected, and hence the expected number of rejected bags in a run of 10 000 bags.
(2)
(f)
A regulation states that at most 1% of bags may contain less than 245 g. Find the least value of \(\mu\) that satisfies the regulation, and determine whether the value found in part (d) satisfies it.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{P}(X\lt245)\) with \(X\sim\mathrm{N}(250,\,4^{2})\)
(M1)
\(0.106\)
A1
\(0.105649\ldots\)
(b)
\(G=-0.02X+B\), where \(B=8\) if the bag is sold and \(0\) otherwise, so \(\mathrm{E}(B)=8\,\mathrm{P}(X\ge245)\)
M1
Or an equivalent argument splitting the expectation.
426AAHL4-q00524.11 Conditional probability and independencePaper 2Hard10 marks
The masses, \(M\) kg, of watermelons from a farm are normally distributed with mean 6.2 kg and standard deviation 1.1 kg. A watermelon with a mass greater than 7 kg is sold as “premium”.
Normal curve for M, with the region M > 7 shaded
(a)
Find the probability that a randomly chosen watermelon is premium.
(2)
(b)
A premium watermelon is chosen at random. Find the probability that its mass is greater than 8 kg.
(3)
(c)
Find the mass \(m\) kg such that half of all premium watermelons have a mass greater than \(m\) kg.
(3)
(d)
Two premium watermelons are chosen at random. Find the probability that exactly one of them has a mass greater than 8 kg.
427AAHL4-q00564.12 Standardization and unknown normal parametersPaper 2Hard9 marks
A random variable \(X\) is normally distributed with mean \(\mu\) and variance \(\mu\), where \(\mu\gt 0\). It is known that \(\mathrm{P}(X\gt 10)=0.05\).
(a)
Let \(s=\sqrt{\mu}\). Show that \(s^{2}+as-10=0\), where \(a=1.64\) correct to three significant figures.
(2)
(b)
Hence find the value of \(\mu\).
(3)
(c)
Find \(\mathrm{P}(X\lt 4)\).
(2)
(d)
\(X\) models a quantity that cannot be negative. Find \(\mathrm{P}(X\lt 0)\) and comment on whether this makes the model unreasonable.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac{10-\mu}{\sqrt{\mu}}=1.64485\ldots\)
M1
Standard deviation \(\sqrt{\mu}\) and the \(z\)-value for an upper tail of 0.05.
\(10-s^{2}=1.64485s\)
A1
\(s^{2}+1.64s-10=0\)
AG
(b)
solving \(s^{2}+1.64485s-10=0\), e.g. quadratic formula or GDC
M1
\(s=2.44504\ldots\), rejecting \(s=-4.08990\ldots\) because \(s=\sqrt{\mu}\gt 0\)
R1
\(\mu=5.98\)
A1
\(5.97826\ldots\) Accept 5.97–5.99.
(c)
\(X\sim\mathrm{N}(5.97826,\ 5.97826)\), i.e. \(\sigma=2.44504\)
(M1)
Award M0 for using 5.97826 as the standard deviation.
\(0.209\)
A1
\(0.209232\ldots\)
(d)
\(\mathrm{P}(X\lt 0)=0.00724\)
A1
\(0.00724166\ldots\)
this probability is very small (less than 1%), so the model is not made unreasonable
R1
FT their value; the comment must refer to the size of the probability.
428AAHL4-q00584.12 Standardization and unknown normal parametersPaper 2Hard15 marks
On a beach, 70% of the crabs are of species \(A\) and 30% are of species \(B\). The carapace width, \(W\) mm, of a crab is normally distributed within each species.
For species \(A\), 10% of crabs have \(W\gt60\) and 25% have \(W\lt50\).
For species \(B\), the mean width is 64 mm and 15% of crabs have \(W\lt56\).
(a)
Find the mean and the standard deviation of the widths of species \(A\).
(5)
(b)
Find the standard deviation of the widths of species \(B\).
(2)
(c)
A crab is chosen at random from the beach. Find the probability that its width is greater than 60 mm.
(3)
(d)
A crab chosen at random has a width greater than 60 mm. Find the probability that it is of species \(B\).
(2)
(e)
A biologist decides to classify every crab with \(W\gt c\) as species \(B\). Find the value of \(c\) for which the probability that a crab with \(W\gt c\) is actually of species \(A\) is 0.1.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(z=1.28155\ldots\) for the upper 10%
(A1)
\(z=-0.674489\ldots\) for the lower 25%
(A1)
\(\mu+1.28155\sigma=60\) and \(\mu-0.674489\sigma=50\)
solving with GDC (graph or table of the expression against \(c\))
(M1)
\(c=63.5\) mm
A1
\(63.4552\ldots\) Accept 63.4–63.5 from rounded parameters.
Answers: (a) \(\mu_{A}=53.4\) mm, \(\sigma_{A}=5.11\) mm · (b) 7.72 mm · (c) 0.279 · (d) 0.749 · (e) \(c=63.5\)
429AAHL4-q00744.13 Bayes’ theoremPaper 2Hard15 marks
Bags of flour labelled 1 kg are filled by three machines, A, B and C, which fill 50%, 30% and 20% of the bags respectively. A bag is underweight if its mass is less than 1000 g.
The masses, in grams, of bags filled by machine A are normally distributed with mean 1010 and standard deviation 6, and those filled by machine C are normally distributed with mean 1012 and standard deviation 5.
(a)
Find the probability that a bag filled by machine A is underweight.
(2)
(b)
The masses of bags filled by machine B are normally distributed with mean 1008 g and standard deviation \(\sigma_{B}\) g, and 2.5% of these bags are underweight. Find \(\sigma_{B}\).
(3)
(c)
Show that the probability that a randomly chosen bag is underweight is 0.0330, correct to three significant figures.
(2)
(d)
A randomly chosen bag is underweight. Find the probability that it was filled by machine A.
(2)
(e)
Machine A is recalibrated: its standard deviation stays at 6 g, but its mean is changed to \(\mu_{A}\) g, so that only 30% of underweight bags come from machine A. Machines B and C are unchanged. Find \(\mu_{A}\), giving your answer correct to one decimal place.
(4)
(f)
After the recalibration, bags from all three machines are mixed and packed in boxes of 20. Find the probability that a box contains at least one underweight bag.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{P}(M\lt 1000)\) with \(M\sim\mathrm{N}(1010,\,6^{2})\)
(M1)
\(0.0478\)
A1
\(0.0477903\ldots\)
(b)
\(\dfrac{1000-1008}{\sigma_{B}}=z\) where \(\mathrm{P}(Z\lt z)=0.025\)
M1
\(z=-1.95996\ldots\)
(A1)
\(\sigma_{B}=4.08\)
A1
\(4.08170\ldots\)
(c)
\(0.5(0.0477903)+0.3(0.025)+0.2(0.00819753)\)
M1
Must see the value \(0.00819753\ldots\) for machine C.
\(0.0330346\ldots\)
A1
\(0.0330\)
AG
(d)
\(\dfrac{0.5(0.0477903)}{0.0330346}\)
(M1)
\(0.723\)
A1
\(0.723336\ldots\)
(e)
\(\dfrac{0.5p}{0.5p+0.3(0.025)+0.2(0.00819753)}=0.3\), where \(p\) is the new probability that a bag from A is underweight
M1
Bayes’ theorem with the new unknown probability.
\(p=0.00783386\ldots\)
A1
From \(0.35p=0.3(0.00913951)\).
\(\dfrac{1000-\mu_{A}}{6}=-2.41657\ldots\)
M1
Standardizing with the \(z\)-value for a lower tail of \(p\).
\(\mu_{A}=1014.5\)
A1
\(1014.4994\ldots\) Accept 1014.4–1014.6.
(f)
\(\mathrm{P}(\text{underweight})=0.5(0.00783386)+0.0091395=0.0130564\), then \(1-(1-0.0130564)^{20}\)
(M1)
Binomial \(\mathrm{B}(20,\,0.0130564)\). FT their \(p\).
430AAHL4-q00754.13 Bayes’ theoremPaper 2Hard13 marks
The level, \(X\) units, of a biomarker in the blood is normally distributed with mean 40 and standard deviation 6 for healthy people, and normally distributed with mean 58 and standard deviation 8 for people who have a particular disease. In a population, 2% of people have the disease. A test is declared positive if \(X\gt c\), where \(c\) is a cut-off value chosen by the laboratory.
(a)
For \(c=50\), find the probability of a positive test for a person with the disease and for a healthy person.
(3)
(b)
For \(c=50\), a randomly chosen person tests positive. Find the probability that this person has the disease.
(3)
(c)
For \(c=50\), a randomly chosen person tests negative. Find the probability that this person has the disease.
(2)
(d)
The laboratory wants 95% of people with the disease to test positive. Find the value of \(c\).
(2)
(e)
For this value of \(c\), find the probability that a person who tests positive has the disease, and comment on your answer in comparison with part (b).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{P}(X\gt 50)\) for \(\mathrm{N}(58,\,8^{2})\) and for \(\mathrm{N}(40,\,6^{2})\)
\(0.0845664\ldots\) Accept 0.0838–0.0847 (0.0838 from \(c=44.8\)).
lowering the cut-off detects more people with the disease, but many more healthy people now test positive, so a positive result is much less reliable (8.5% compared with 26%)
431AAHL4-q01034.14 Variance, continuous random variables and linear transformationsPaper 2Hard8 marks
A dart lands at a random point on a circular target of radius 5 cm. The target is divided by circles of radius 1 cm and 3 cm, with the same centre, into three regions scoring 10, 5 and 1 points, as shown. The distance, \(X\) cm, of the dart from the centre has probability density function \(f(x)=\dfrac{2x}{25}\) for \(0\le x\le5\), and \(f(x)=0\) otherwise.
Diagram not to scale
(a)
Show that \(\mathrm{P}(X\le a)=\dfrac{a^{2}}{25}\) for \(0\le a\le5\), and explain why this is consistent with the dart being equally likely to land anywhere on the target.
(2)
(b)
Let \(S\) be the score from one dart. Find the probability distribution of \(S\).
(2)
(c)
Find the entry fee per dart that would make the game fair, if a player receives €1 for each point scored.
432AAHL4-q01094.14 Variance, continuous random variables and linear transformationsPaper 2Hard15 marks
Each morning a florist buys \(n\) bouquets from a supplier at €8 each and sells them at €20 each. Bouquets not sold by the end of the day are thrown away. The daily demand, \(D\), for bouquets has the probability distribution shown, and is independent of the number of bouquets bought.
\(d\)
\(2\)
\(3\)
\(4\)
\(5\)
\(\mathrm{P}(D=d)\)
\(0.1\)
\(0.3\)
\(0.4\)
\(0.2\)
The number of bouquets sold on a day is the smaller of \(D\) and \(n\). Let \(G\) be the florist’s daily profit, in euros, from bouquets.
(a)
Find \(\mathrm{E}(D)\) and \(\mathrm{Var}(D)\).
(3)
(b)(i)
For \(n=5\), explain why \(G=20D-40\).
(1)
(b)(ii)
Hence, for \(n=5\), find \(\mathrm{E}(G)\) and the standard deviation of \(G\).
(3)
(c)(i)
For \(n=4\), write down the possible values of \(G\) and find \(\mathrm{E}(G)\).
(3)
(c)(ii)
It is given that \(\mathrm{E}(G)=24\) when \(n=2\) and \(\mathrm{E}(G)=34\) when \(n=3\). State, with a reason, the number of bouquets the florist should buy each morning to maximize the expected profit.
(1)
(d)
For \(n=4\), find the standard deviation of \(G\).
(2)
(e)
The supplier raises the price to €10 per bouquet. For \(n=4\), find the new expected daily profit and the new standard deviation of the daily profit, justifying your answer.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{E}(D)=0.2+0.9+1.6+1=3.7\)
A1
\(\mathrm{E}(D^{2})=0.4+2.7+6.4+5=14.5\)
(M1)
\(\mathrm{Var}(D)=14.5-3.7^{2}=0.81\)
A1
(b)(i)
\(D\le5\) always, so all \(D\) bouquets demanded are sold: income \(20D\), cost \(5\times8=40\)
R1
(b)(ii)
\(\mathrm{E}(G)=20(3.7)-40=34\) (euros)
A1
FT their (a).
\(\mathrm{Var}(G)=20^{2}\times0.81=324\)
(M1)
Award M0 for \(20\times0.81\) or \(20^{2}\times0.81-40\).
standard deviation \(=18\) (euros)
A1
(c)(i)
\(G=8,\ 28,\ 48\) (for \(D=2,\ 3,\ D\ge4\))
A1
Income \(20\min(D,4)\) minus cost \(32\).
\(\mathrm{E}(G)=8(0.1)+28(0.3)+48(0.6)\)
(M1)
\(\mathrm{P}(G=48)=\mathrm{P}(D\ge4)=0.6\).
\(\mathrm{E}(G)=38\) (euros)
A1
(c)(ii)
\(n=4\), since \(38\gt34\) (for \(n=3\) and \(n=5\)) \(\gt24\)
R1
Must compare all four expected values; \(n\gt5\) cannot help as at most 5 are sold.
433AAHL4-q01104.14 Variance, continuous random variables and linear transformationsPaper 2Hard14 marks
A shop is open from 07:00 to 19:00. The time, \(T\) hours after 07:00, at which a randomly chosen customer enters the shop is modelled by the probability density function
\(h'(a)\gt0\) for \(a\gt1\), so \(h\) is increasing and \(h(a)\gt h(1)=0\); hence mean \(\gt\) median
R1
AG
Answers: (d) \(a=4.92\) · (e) 0.922
435IBMA2H-051-01-094.13 Bayes’ theoremPaper 2Hard6 marks
Calculator
A factory has two machines making microchips, each producing half of the output. Each chip from machine A is defective with probability 0.02, and each chip from machine B is defective with probability 0.06, independently. A tray of 20 chips, all made by the same machine, is chosen at random. The tray contains exactly 2 defective chips.
(a)
Find the probability that a tray made by machine A contains exactly 2 defective chips.
(2)
(b)
Find the probability that a tray made by machine B contains exactly 2 defective chips.
(1)
(c)
Find the probability that the chosen tray was made by machine B.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 recognising X ~ B(20, 0.02) and using \(\binom{20}{2}(0.02)^2(0.98)^{18}\)
✓ 1
A1 0.0528
✓ 1
Part (b)
A1 \(\binom{20}{2}(0.06)^2(0.94)^{18}\) = 0.225
✓ 1
Part (c)
M1 Bayes numerator 0.5 × 0.22457
✓ 1
M1 denominator 0.5 × 0.052830 + 0.5 × 0.22457
✓ 1
A1 0.810
✓ 1
Exam tip: The 'evidence' in Bayes' theorem can be a whole binomial outcome. Its conditional probability under each hypothesis is just a binomial probability.
436IBMA2H-051-01-124.13 Bayes’ theoremPaper 2Hard11 marks
Calculator
A lighting company buys LED drivers from two suppliers. 40% come from supplier P and 60% from supplier Q. The lifetime T, in thousands of hours, of a driver from P has probability density function \(f(t)=0.5e^{-0.5t}\), \(t\ge 0\). The lifetime of a driver from Q has probability density function \(g(t)=0.2e^{-0.2t}\), \(t\ge 0\). A driver is chosen at random.
(a)
Show that the probability that a driver from P lasts more than 4 thousand hours is \(e^{-2}\).
(2)
(b)
Write down the probability that a driver from Q lasts more than 4 thousand hours.
(1)
(c)
Find the probability that the randomly chosen driver lasts more than 4 thousand hours.
(2)
(d)
Given that the driver lasts more than 4 thousand hours, find the probability that it came from supplier P.
(2)
(e)
Given that the driver has already lasted 4 thousand hours, find the probability that it lasts more than 6 thousand hours.
M1 conditional probability \(\frac{P(T>6)}{P(T>4)}\), since T > 6 implies T > 4
✓ 1
M1 P(T > 6) = \(0.4e^{-3}+0.6e^{-1.2}\)
✓ 1
A1 P(T > 6) = 0.20063
✓ 1
A1 \(\frac{0.20063}{0.32373}\) = 0.620
✓ 1
Common error: Each supplier's lifetime is memoryless, but the mixture is not. Using \(e^{-2\times0.5}\) or \(e^{-2\times0.2}\) alone in (e) ignores the fact that survival to 4 thousand hours shifts the odds towards supplier Q.
437IBMA2H-051-05-074.12 Standardization and unknown normal parametersPaper 2Hard14 marks
Calculator
A factory makes precision bolts on two machines. Machine A makes 60% of the bolts; their diameters, in mm, are distributed \(N(10.00, 0.05^2)\). Machine B makes the rest; their diameters are distributed \(N(10.02, 0.04^2)\). A bolt is accepted if its diameter lies between 9.92 mm and 10.08 mm; otherwise it is rejected.
(a)
Find the probability that a bolt from machine A is accepted.
(2)
(b)
Find the probability that a bolt from machine B is accepted.
(2)
(c)
Find the probability that a randomly chosen bolt is accepted.
(2)
(d)
A randomly chosen bolt is rejected. Find the probability that it was made by machine A.
(3)
(e)
Machine B can be recalibrated so that its mean diameter becomes \(m\) mm, with standard deviation unchanged. Determine the value of \(m\) that maximises the probability that a bolt from machine B is accepted, and find this maximum probability.
(3)
(f)
Using this value of \(m\), find the new probability that a randomly chosen bolt is accepted.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(P(9.92 < X_A < 10.08)\) or \(2\Phi(1.6) - 1\)
✓ 1
A1 0.890 (0.890401…)
✓ 1
Part (b)
M1 \(P(9.92 < X_B < 10.08)\), z-values −2.5 and 1.5
✓ 1
A1 0.927 (0.926983…)
✓ 1
Part (c)
M1 \(0.6 \times 0.890401 + 0.4 \times 0.926983\)
✓ 1
A1 0.905 (0.905034…)
✓ 1
Part (d)
M1 use of conditional probability/Bayes' theorem
✓ 1
A1 \(\frac{0.6 \times 0.109599}{1 - 0.905034}\)
✓ 1
A1 0.692 (0.692451…)
✓ 1
Part (e)
R1 for fixed σ, the probability of an interval of fixed width is greatest when the interval is centred on the mean, by symmetry of the normal curve
✓ 1
A1 \(m\) = 10.00
✓ 1
A1 \(2\Phi(2) - 1\) = 0.954 (0.954500…)
✓ 1
Part (f)
M1 \(0.6 \times 0.890401 + 0.4 \times 0.954500\)
✓ 1
A1 0.916 (0.916041…)
✓ 1
Exam tip: A tree diagram with branches A/B and then accepted/rejected organises parts (c) and (d) clearly. For the rejected branch, use 1 − 0.890401 = 0.109599 rather than a rounded value.
438IBMA2H-051-08-104.2 Presentation of dataPaper 2Hard11 marks
Calculator
A radio telescope measures the normalised strength X of signals from a pulsar. X is modelled by the probability density function f(x) = \(\frac{3x^2}{8}\) for 0 ≤ x ≤ 2, and f(x) = 0 otherwise. An outlier is a value more than 1.5 × IQR beyond the nearer quartile.
Graph of the probability density function f
(a)
Show that the median of X is \(\sqrt[3]{4}\), and find the exact values of the lower and upper quartiles.
(3)
(b)
Show that no signal can be an upper outlier, and find the probability that a randomly chosen signal is a lower outlier.
(4)
(c)
Signals are received either from this pulsar (probability 0.7) or from a background source (probability 0.3) whose strength is uniformly distributed on [0, 2]. A received signal has strength below the lower outlier boundary found in the previous part. Find the probability that it came from the background source.
R1 upper boundary = \(\sqrt[3]{6}\) + 1.5(\(\sqrt[3]{6} - \sqrt[3]{2}\)) ≈ 2.65 > 2, which is the maximum possible value of X, so there are no upper outliers
Common error: Keep the lower boundary to full calculator accuracy. Cubing a rounded value such as 0.42 changes the probability noticeably.
439IBMA2H-051-00-104.14 Variance, continuous random variables and linear transformationsPaper 2Hard13 marks
Calculator
The delay, X seconds, in establishing a satellite uplink is modelled by the probability density function \(f(x)=ke^{-x/3}\) for \(0\le x\le 6\), and \(f(x)=0\) otherwise, since the connection attempt is abandoned after 6 seconds.
(a)
Show that \(k=\frac{e^2}{3(e^2-1)}\).
(2)
(b)
Show that the median delay is \(3\ln\left(\frac{2e^2}{e^2+1}\right)\) and find its value.
(4)
(c)
Find E(X).
(2)
(d)
Find Var(X).
(3)
(e)
A ground station's total waiting time is \(T=1.2X+0.8\) seconds. Find the standard deviation of T.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\int_0^6 ke^{-x/3}dx=k\left[-3e^{-x/3}\right]_0^6=3k(1-e^{-2})\) A1 \(3k(1-e^{-2})=1\Rightarrow k=\frac{1}{3(1-e^{-2})}=\frac{e^2}{3(e^2-1)}\) AG
✓ 2
Part (b)
M1 \(\int_0^m ke^{-x/3}dx=\frac{1}{2}\) A1 \(3k(1-e^{-m/3})=\frac{1}{2}\Rightarrow\frac{1-e^{-m/3}}{1-e^{-2}}=\frac{1}{2}\) M1 rearranges to \(e^{-m/3}=\frac{1+e^{-2}}{2}\) A1 \(m=3\ln\left(\frac{2}{1+e^{-2}}\right)=3\ln\left(\frac{2e^2}{e^2+1}\right)=1.70\) s
✓ 4
Part (c)
M1 \(\int_0^6 xke^{-x/3}dx\) A1 \(E(X)=2.06\) s (2.06089…)
Exam tip: Store E(X) and E(X²) to full accuracy on your GDC. Subtracting rounded values in \(E(X^2)-[E(X)]^2\) can shift the third significant figure.
440IBMA2H-051-04-114.8 Binomial distributionPaper 2Hard14 marks
Calculator
A pumping system contains four components A, B, C and D that operate independently. The system works if and only if A works, D works and at least one of B and C works. The probabilities that the components work are 0.95 for A, 0.8 for B, 0.8 for C and 0.9 for D. Let S be the event that the system works.
The system works if there is a path from left to right through working components
(a)
Show that P(S) = 0.8208.
(3)
(b)
Given that the system fails, find the probability that component A has failed.
(3)
(c)
Given that the system works, find the probability that both B and C work.
(2)
(d)
Determine whether the events 'B works' and 'C works' are independent given that the system works.
(3)
(e)
Component C is replaced by a new component that works with probability r. All other probabilities are unchanged. Find the least value of r for which P(S) ≥ 0.84.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(B or C works) = 1 − 0.22 = 0.96 ; M1 uses independence to multiply by P(A) = 0.95 and P(D) = 0.9 ; A1 0.95 × 0.96 × 0.9 = 0.8208
✓ 3
Part (b)
R1 if A fails the system fails, so P(A fails ∩ S′) = P(A fails) = 0.05 ; M1 0.05/(1 − 0.8208) = 0.05/0.1792 ; A1 0.279
M1 P(B works | S) = (0.95 × 0.8 × 0.9)/0.8208 = 5/6, and likewise P(C works | S) = 5/6 ; A1 product = 25/36 ; R1 25/36 ≠ 2/3 = 24/36, so they are not independent given S
✓ 3
Part (e)
M1 P(S) = 0.95 × 0.9 × (1 − 0.2(1 − r)) = 0.855(0.8 + 0.2r) ; M1 sets this ≥ 0.84 and solves: 0.8 + 0.2r ≥ 56/57 ; A1 r ≥ 52/57, so least r = 52/57 ≈ 0.9123 (accept 0.913 rounded up; 0.912 is too small)
✓ 3
Common error: When a question asks for the least value satisfying an inequality, rounding the boundary down gives a value that fails the condition: give the exact value or round in the direction that still satisfies it.
441IBMA2H-051-07-084.12 Standardization and unknown normal parametersPaper 2Hard14 marks
Calculator
The distance, D metres, achieved by a ski jumper on a particular hill is modelled by a normal distribution with mean μ and standard deviation σ. Jumps are independent. Over a season, 20% of the jumper's jumps exceed 125 m and 5% fall short of 98 m.
Normal curve of jump distance D (metres); shaded regions D < 98 and D > 125. Diagram not to scale
(a)
Find the value of μ and the value of σ.
(5)
(b)
The distance points awarded for a jump are given by S = 2(D − 120) + 60. Find E(S) and the standard deviation of S.
(3)
(c)
In a training camp the jumper makes n jumps. Find the least value of n for which the probability that at least two jumps exceed 125 m is greater than 0.95.
(4)
(d)
A jump is selected at random and is known to be at least 98 m. Find the probability that it exceeds 125 m.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 z = 0.8416 for 125 m (lower-tail probability 0.8) M1 z = −1.6449 for 98 m M1 forming 125 = μ + 0.8416σ and 98 = μ − 1.6449σ A1 σ = 10.9 m A1 μ = 116 m
M1 Y ~ B(n, 0.2) for the number of jumps over 125 m M1 requiring P(Y ≥ 2) = 1 − 0.8n − 0.2n(0.8)n−1 > 0.95, or P(Y ≤ 1) < 0.05 A1 n = 21 gives 0.942 and n = 22 gives 0.952 A1 n = 22
✓ 4
Part (d)
M1 P(D > 125 | D ≥ 98) = 0.20/0.95 A1 0.211
✓ 2
Common error: For "least n" questions, writing down only the final value is not enough. Show the probabilities for both n = 21 and n = 22 to justify that 22 is the smallest value that works.
442IBMA2H-051-07-094.12 Standardization and unknown normal parametersPaper 2Hard10 marks
Calculator
A space-based detector records high-energy particles. The reading, X, for one hour is modelled by a continuous normal distribution whose variance is equal to its mean μ, so that X ~ N(μ, μ). The reading exceeds 30 in 10% of hours.
(a)
Show that μ + 1.2816\(\sqrt{\mu}\) − 30 = 0, where 1.2816 is given to 4 decimal places.
(2)
(b)
Hence find the value of μ.
(4)
(c)
Find the reading that is exceeded in 75% of hours.
(2)
(d)
Readings in different hours are independent. Find the probability that, in a period of 30 hours, the reading exceeds 30 in exactly 3 of the hours.
Exam tip: When σ depends on μ, the standardised equation is not linear. Treat it as a quadratic in √μ and give a reason for rejecting the negative root. That reason is worth a mark.
443IBMA2H-051-06-104.8 Binomial distributionPaper 2Hard13 marks
Calculator
A machine makes aerospace bolts. Before a shift, the machine is calibrated with probability 0.9 and miscalibrated with probability 0.1. When calibrated, each bolt is defective with probability 0.01; when miscalibrated, each bolt is defective with probability 0.15. Given the machine's state, bolts are defective independently of one another. A random sample of 20 bolts from the shift is inspected and \(X\) is the number of defective bolts.
(a)
Find \(P(X=2)\) when the machine is (i) calibrated, (ii) miscalibrated.
(3)
(b)
Given that the sample contains exactly 2 defective bolts, find the probability that the machine is miscalibrated.
(3)
(c)
Find the least number of defective bolts in the sample for which the probability that the machine is miscalibrated exceeds 0.95.
(4)
(d)
The 20-bolt sample described above contained exactly 2 defective bolts. A second random sample of 20 bolts from the same shift contains no defective bolts. Find the updated probability that the machine is miscalibrated.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(B(20,0.01)\) and \(B(20,0.15)\) ; A1 (i) 0.0159 (0.0158558) ; A1 (ii) 0.229 (0.229338)
✓ 3
Part (b)
M1 total probability \(0.9\times0.0158558+0.1\times0.229338=0.0372040\) ; M1 Bayes' theorem \(\frac{0.0229338}{0.0372040}\) ; A1 0.616
✓ 3
Part (c)
M1 posterior for \(k\) defectives is \(\frac{0.1\,(0.15)^k(0.85)^{20-k}}{0.1\,(0.15)^k(0.85)^{20-k}+0.9\,(0.01)^k(0.99)^{20-k}}\); the binomial coefficients cancel ; A1 the posterior increases with \(k\) (the likelihood ratio \(15^k(0.85/0.99)^{20-k}\) increases), and \(k=2\) gives 0.616 < 0.95. Direct checks of \(k=0\) and \(k=1\) are also acceptable ; A1 \(k=3\) gives 0.966 > 0.95 ; A1 least number is 3
✓ 4
Part (d)
M1 uses 0.616 from the 2-defective result as the new prior, or multiplies the likelihoods of both samples from the original prior ; M1 \(\frac{0.61643\times0.85^{20}}{0.61643\times0.85^{20}+0.38357\times0.99^{20}}\) or equivalent ; A1 0.0708 (accept 0.0707 from a rounded prior)
✓ 3
Exam tip: Updating with new evidence can be done in two equivalent ways: use the previous posterior as the new prior, or multiply the likelihoods of both samples from the original prior. Keep full calculator accuracy in the prior so that rounding does not change the final digit.
444IBMA2H-051-03-114.10 Regression line of x on yPaper 2Hard14 marks
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An engineer calibrates a low-cost temperature sensor. For 10 test conditions she records the true temperature x °C and the sensor reading y units. The sensor readings ranged from 20 to 42 units. The summary statistics are Σx = 250, Σy = 312, Σx2 = 6890, Σy2 = 10560, Σxy = 8508. In normal use only the sensor reading is known, and the true temperature is estimated from it.
(a)
Find the value of Pearson's product-moment correlation coefficient, r.
(2)
(b)
Find the equation of the regression line of x on y.
(3)
(c)
Estimate the true temperature when the sensor reads 35 units.
(2)
(d)
Explain why your line should not be used to estimate the true temperature when the sensor reads 80 units.
(1)
(e)
In use, each reading lies within tolerance independently with probability 0.9. In a sequence of 30 readings, find the probability that at most 2 readings are outside tolerance.
(3)
(f)
Given that at most 2 of the 30 readings are outside tolerance, find the probability that none of them are outside tolerance.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 Sxx = 640, Syy = 825.6, Sxy = 708 substituted into r = Sxy/√(SxxSyy)
✓ 1
A1 r = 0.974
✓ 1
Part (b)
M1 gradient = Sxy/Syy = 708/825.6
✓ 1
A1 gradient 0.858 (0.85756…)
✓ 1
A1 x = 0.858y − 1.76
✓ 1
Part (c)
M1 substituting y = 35 into the x on y line
✓ 1
A1 28.3 °C
✓ 1
Part (d)
R1 80 lies far outside the observed range of readings (20 to 42), so this would be extrapolation and is unreliable
✓ 1
Part (e)
M1 recognising W ~ B(30, 0.1), where W is the number of readings outside tolerance
✓ 1
M1 P(W ≤ 2) = P(W = 0) + P(W = 1) + P(W = 2)
✓ 1
A1 0.411
✓ 1
Part (f)
M1 P(W = 0 | W ≤ 2) = P(W = 0)/P(W ≤ 2)
✓ 1
A1 P(W = 0) = 0.930 = 0.0424
✓ 1
A1 0.103
✓ 1
Common error: Do not find x from the y on x line by rearranging it. When x is estimated from y, the gradient must be Sxy/Syy. Sxy/Sxx is the gradient of the y on x line.
445IBMA2H-051-09-084.14 Variance, continuous random variables and linear transformationsPaper 2Hard15 marks
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The flight time, X hours, of a delivery drone on a single charge is modelled by the probability density function \(f(x)=kx(9-x^2)\) for 0 ≤ x ≤ 3, and f(x) = 0 otherwise.
(a)
Show that \(k=\frac{4}{81}\).
(3)
(b)
Find the exact mode of X.
(3)
(c)
Find E(X).
(2)
(d)
Find Var(X).
(3)
(e)
Find the median flight time.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\int_0^3 kx(9-x^2)\,dx=1\) A1 \(k\left[\frac{9x^2}{2}-\frac{x^4}{4}\right]_0^3\) A1 \(k\cdot\frac{81}{4}=1\) so \(k=\frac{4}{81}\) (AG)
M1 \(\int_0^m \frac{4}{81}x(9-x^2)\,dx=\frac{1}{2}\) A1 \(\frac{18m^2-m^4}{81}=\frac{1}{2}\) M1 solves as a quadratic in m2 and rejects m ≈ 3.92 as outside [0, 3] A1 median ≈ 1.62 hours
✓ 4
Exam tip: Check every solution for the median lies in the domain of the pdf; a quartic gives extra roots that must be rejected with a reason.
446IBMA2H-051-02-124.8 Binomial distributionPaper 2Hard14 marks
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The first stage of a launch vehicle has \(n\) identical engines. During ascent, each engine fails independently with probability 0.04. A launch is successful if at most one engine fails. Let \(F\) be the number of engines that fail during a launch.
(a)
For \(n=8\), show that the probability of a successful launch is 0.962, correct to 3 significant figures.
(2)
(b)
Find the largest value of \(n\) for which the probability of a successful launch is at least 0.95.
(3)
(c)
For \(n=9\), write down \(E(F)\) and \(\mathrm{Var}(F)\).
(2)
(d)
A vehicle with \(n=9\) is used for 12 independent launches. Let \(L\) be the number of these launches that are successful. Find the probability that at least 11 of these launches are successful.
(4)
(e)
The engines are redesigned so that each fails independently with probability \(q\). For \(n=9\), the probability that no engine fails must be at least 0.7. Find the largest possible value of \(q\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(P(F=0)+P(F=1)\) with \(F\sim B(8,0.04)\)
✓ 1
A1 \(0.72139+0.24046=0.96185\approx0.962\)
✓ 1
Part (b)
M1 evaluates \(P(F\le1)=0.96^n+0.04n(0.96)^{n-1}\) for \(n=9\) and \(n=10\) (or uses a table or graph of values)
✓ 1
A1 \(n=9\) gives \(0.952\) and \(n=10\) gives \(0.942\)
M1 recognises \(L\sim B(12,0.952234\ldots)\), using the probability of success for \(n=9\) found earlier
✓ 1
M1 \(P(L=11)+P(L=12)\)
✓ 1
A1 \(0.3346+0.5558\)
✓ 1
A1 \(0.890\)
✓ 1
Part (e)
M1 \((1-q)^9\ge0.7\)
✓ 1
M1 \(1-q\ge0.7^{1/9}=0.961144\ldots\)
✓ 1
A1 the largest value is \(q=1-0.7^{1/9}=0.038855\ldots\approx0.0389\); accept \(0.0389\), \(0.03886\) or \(0.0388\)
✓ 1
Exam tip: When finding the largest n, show the probability on both sides of 0.95, for n = 9 and n = 10. Stopping at the first n that still exceeds 0.95 does not show that 9 is the largest.
447IBMA3H-4-19r6-114.14 Variance, continuous random variables and linear transformationsPaper 2Hard11 marks
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The daily demand, X megalitres, for a village water supply is modelled by the probability density function \(f(x) = kx^2\) for \(0 \le x \le 4\), and \(f(x) = 0\) otherwise, where k is a constant.
(a)
Show that k = 3/64.
(3)
(b)
Find E(X).
(2)
(c)
Find the median daily demand, correct to 3 significant figures.
(3)
(d)
Given that the demand on a particular day exceeds 2 megalitres, find the probability that it exceeds 3 megalitres. Give your answer to 3 significant figures.
Exam tip: For a median, set the integral from the lower limit to m equal to 1/2; for a conditional probability, divide by the probability of the condition.
448IBMA3H-4-19r6-134.12 Standardization and unknown normal parametersPaper 2Hard11 marks
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The time, X minutes, that a player takes to complete a game level is normally distributed with mean μ and standard deviation σ. It is known that P(X < 8) = 0.10 and P(X > 12) = 0.15. Use unrounded values of μ and σ in later parts and give answers to 3 significant figures.
Sketch of the normal curve with the given tail probabilities, not to scale
(a)
Find the values of μ and σ.
(5)
(b)
Given that a player takes more than 10 minutes, find the probability that the player takes more than 12 minutes.
(3)
(c)
Five players attempt the level independently. Find the probability that at least two of them take more than 12 minutes.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 inverse normal: z = −1.282 for 0.10 and z = 1.036 for 0.85
Common error: For the conditional probability, the numerator is P(X > 12), not P(10 < X < 12): being above 12 already implies being above 10.
449IBMA3H-4-19r7-124.13 Bayes’ theoremPaper 2Hard10 marks
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An excavation is divided into three zones X, Y and Z. A find is equally likely to be any of a large number of artefacts, and 50%, 30% and 20% of artefacts come from X, Y and Z respectively. The probability that an artefact is made of bronze is 0.1 in X, 0.3 in Y and 0.5 in Z.
(a)
Find the probability that a randomly chosen artefact is made of bronze.
(2)
(b)
Given that an artefact is bronze, find the probability that it came from zone Z. Give your answer as a fraction.
(3)
(c)
Two artefacts are chosen from the same randomly selected zone. Given the zone, the artefacts are independent. Find the probability that both are bronze.
(3)
(d)
Given that the first artefact chosen in part (c) is bronze, find the probability that the second is also bronze, to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(B) = 0.5(0.1) + 0.3(0.3) + 0.2(0.5)
✓ 1
A1 0.24
✓ 1
Part (b)
M1 P(Z | B) = P(Z ∩ B) ÷ P(B)
✓ 1
A1 0.1 ÷ 0.24
✓ 1
A1 5/12
✓ 1
Part (c)
M1 sum over zones of P(zone) × P(B | zone)²
✓ 1
A1 0.5(0.1)² + 0.3(0.3)² + 0.2(0.5)²
✓ 1
A1 0.082
✓ 1
Part (d)
M1 P(second B | first B) = P(both B) ÷ P(first B)
✓ 1
A1 0.082 ÷ 0.24 = 0.342
✓ 1
Common error: The two artefacts are independent only once the zone is known, so do not multiply 0.24 × 0.24.
450IBMA3H-4-19r9-104.13 Bayes’ theoremPaper 2Hard10 marks
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A port receives containers from three suppliers, A, B and C, which provide 50%, 30% and 20% of the containers respectively. The probability that a container arrives damaged is 0.02 for A, 0.05 for B and 0.10 for C.
Tree diagram showing the probabilities given in the question
(a)
Find the probability that a randomly chosen container is damaged.
(3)
(b)
A damaged container is found. Find the probability that it came from B.
(3)
(c)
The port renegotiates so that C still supplies 20% of containers, A supplies a proportion \(a\) and B supplies the rest. The overall probability of a damaged container becomes 0.05. Find \(a\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 law of total probability, e.g. \(0.5\times0.02+0.3\times0.05+0.2\times0.10\)
✓ 1
A1 sum of \(0.01+0.015+0.02\)
✓ 1
A1 \(0.045\)
✓ 1
Part (b)
M1 \(P(B\mid D)=\frac{P(B\cap D)}{P(D)}\)
✓ 1
A1 \(\frac{0.015}{0.045}\) (ECF from (a))
✓ 1
A1 \(\frac13\)
✓ 1
Part (c)
M1 B's proportion \(=0.8-a\)
✓ 1
M1 \(0.02a+0.05(0.8-a)+0.10\times0.2=0.05\)
✓ 1
A1 \(0.06-0.03a=0.05\)
✓ 1
A1 \(a=\frac13\)
✓ 1
Common error: \(P(B\mid D)\) is not \(P(D\mid B)\); divide the joint probability by the total probability of the condition.
451IBMA3H-4-19r14-114.14 Variance, continuous random variables and linear transformationsPaper 2Hard14 marks
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The time X hours for a repair crew to reach a wind turbine has probability density function \(f(x) = \frac{k}{x^2}\) for \(1 \le x \le 4\), and \(f(x) = 0\) otherwise. Use unrounded values in later parts.
Graph of the probability density function, not to scale
(a)
Find the value of k.
(3)
(b)
Find E(X), correct to 3 significant figures.
(2)
(c)
Find the median time.
(3)
(d)
Find Var(X), correct to 3 significant figures.
(4)
(e)
The cost of a callout, in £, is \(C = 50X + 20\). Find the standard deviation of C, correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\int_1^4 \frac{k}{x^2}\,dx = 1\)
✓ 1
A1 \(k\left(1 - \frac14\right) = 1\)
✓ 1
A1 \(k = \frac43\)
✓ 1
Part (b)
M1 \(\int_1^4 x f(x)\,dx = \frac43\int_1^4 \frac{1}{x}\,dx\)
Common error: The constant in a linear transformation shifts the mean but never changes the variance or standard deviation.
452IBMA3H-4-19r14-134.1 Sampling, bias and outliersPaper 2Hard11 marks
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A retailer buys items from three suppliers: A supplies 50% of items, B supplies 30% and C supplies 20%. The proportions of defective items are 2% for A, 3% for B and 5% for C. Give answers to 3 significant figures; use unrounded values in later parts.
(a)
Find the probability that a randomly chosen item is defective.
(3)
(b)
Given that a randomly chosen item is defective, find the probability that it came from supplier C.
(2)
(c)
A box of 10 items is supplied entirely by one supplier, chosen with the probabilities above. Find the probability that the box contains exactly one defective item.
(3)
(d)
A box is found to contain exactly one defective item. Find the probability that it came from supplier C.
M1 divide by the answer to (c) (unrounded, 0.2148)
✓ 1
A1 0.293
✓ 1
Exam tip: In multi-stage Bayes questions, build a table of prior × likelihood for every case before dividing.
453IBMA3H-4-19r14-144.4 Correlation and regression of y on xPaper 2Hard11 marks
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An engineer records the rotor diameter d (m) and annual output E (GWh) of eight wind turbines. d: 40, 52, 60, 70, 80, 90, 100, 110 E: 2.1, 3.9, 4.6, 6.4, 7.1, 8.8, 10.2, 11.0
Scatter diagram of the data in the table
(a)
Find the correlation coefficient r and the equation of the regression line of E on d.
(3)
(b)
Interpret the gradient of this line.
(1)
(c)
Find the residual for the turbine with diameter 90 m, correct to 3 significant figures.
(2)
(d)
The engineer wants to estimate the diameter needed for an output of 9.0 GWh. State which regression line should be used, with a reason.
(1)
(e)
Comment on the reliability of the estimate you will make in part (f), and of the model in general.
(2)
(f)
Use the appropriate regression line, with unrounded coefficients, to estimate the diameter needed for 9.0 GWh. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 r = 0.997
✓ 1
A1 gradient 0.129
✓ 1
A1 E = 0.129d − 2.95
✓ 1
Part (b)
R1 each extra metre of rotor diameter is associated with about 0.129 GWh more annual output
R1 the regression line of d on E, because d is the variable being estimated (the line of E on d minimises vertical deviations in E)
✓ 1
Part (e)
R1 9.0 lies within the data range 2.1 to 11.0, so this is interpolation
✓ 1
R1 r is very close to 1, indicating a strong linear association
✓ 1
Part (f)
M1 d = 7.7026E + 23.161 with E = 9.0
✓ 1
A1 92.5 m
✓ 1
Exam tip: Interpolation within the data range with |r| near 1 is reliable; say both points when asked to comment on reliability.
454IBMA3H-4-19r15-084.14 Variance, continuous random variables and linear transformationsPaper 2Hard10 marks
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The time \(X\) in days between exposure to a virus and the first symptoms has probability density function \(f(x)=kx(4-x)\) for \(0\le x\le4\), and \(f(x)=0\) otherwise, where \(k\) is a constant.
Exam tip: For a pdf, always check that \(\int f(x)\,dx=1\) over the stated interval before using the function for anything else.
455IBMA3H-4-19r15-144.14 Variance, continuous random variables and linear transformationsPaper 2Hard12 marks
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The delay \(X\) hours of a delivery truck has probability density function \(f(x)=a+bx\) for \(0\le x\le2\), and \(f(x)=0\) otherwise, where \(a\) and \(b\) are constants. It is known that \(E(X)=1.2\).
(a)
Show that \(a+b=0.5\).
(2)
(b)
Find the value of \(a\) and the value of \(b\).
(4)
(c)
Find \(\mathrm{Var}(X)\).
(3)
(d)
The cost of the delay in pounds is \(C=40X+25\). Find the standard deviation of \(C\), correct to 3 significant figures.
Common error: Do not use \(E(X)\) twice: the variance needs \(E(X^2)\), found with an extra factor of \(x\) inside the integral.
456IBMA3H-4-19r17-124.8 Binomial distributionPaper 2Hard11 marks
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A wind farm contains turbines from three manufacturers. Manufacturer A supplied \(50\%\) of the turbines, B supplied \(30\%\) and C supplied \(20\%\). The probability that a turbine has a gearbox defect is \(0.02\) for A, \(0.03\) for B and \(0.05\) for C.
(a)
A turbine is selected at random. Show that the probability that it has a gearbox defect is \(0.029\).
(3)
(b)
Given that a randomly selected turbine does not have a gearbox defect, find the probability that it was supplied by manufacturer B. Give your answer to 3 significant figures.
(3)
(c)
Two turbines are selected at random. The farm is large enough that the two selections may be treated as independent. Given that both turbines have a gearbox defect, find the probability that both were supplied by the same manufacturer. Give your answer to 3 significant figures.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt at total probability, sum of three products
M1 attempt to find \(\mathrm{P}(\text{both from the same manufacturer and both defective})\) as a sum of three squares
✓ 1
A1 \((0.01)^2+(0.009)^2+(0.01)^2=0.000281\)
✓ 1
M1 divide the two probabilities
✓ 1
A1 \(\dfrac{0.000281}{0.000841}=0.334\)
✓ 1
Common error: For two defective turbines from the same manufacturer, square each manufacturer's joint probability (share × defect rate) before adding; adding first and then squaring gives the wrong answer.
457IBMA3H-4-19r17-144.2 Presentation of dataPaper 2Hard12 marks
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The time, \(T\) minutes, that a visitor spends in a library is normally distributed with mean \(35\) and standard deviation \(12\). In parts (b) and (c) use unrounded values of any probabilities you have found.
(a)
Find \(\mathrm{P}(T>50\mid T>40)\), correct to 3 significant figures.
(4)
(b)
Three visitors are chosen at random and their times are independent. Given that at least one of them stays longer than \(40\) minutes, find the probability that exactly one of them does.
(4)
(c)
Find the value of \(t\), correct to 3 significant figures, such that \(\mathrm{P}(T<t\mid T<40)=0.5\).
Common error: In a conditional probability such as P(T > 50 | T > 40), the denominator is P(T > 40), not 1: do not simply compute P(T > 50).
458IBMA3H-4-19r19-124.4 Correlation and regression of y on xPaper 2Hard11 marks
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A researcher studying crops measures the nitrogen content x of leaves (mg per g) and the yield y (kg per plot) on 10 plots. The data are summarised by n = 10, Σx = 240, Σy = 520, Σx2 = 6240, Σy2 = 28300, Σxy = 13180.
(a)
Find the product moment correlation coefficient, to 3 significant figures.
(2)
(b)
Find the equation of the regression line of x on y, giving the gradient and intercept as exact fractions.
(4)
(c)
A new plot has yield 63 kg. Estimate its leaf nitrogen content, to 3 significant figures.
(2)
(d)
Explain why the regression line of x on y, and not the line of y on x, should be used for the estimate in part (c).
(2)
(e)
The researcher suggests using the line to estimate the nitrogen content for a yield of 150 kg, well above every observed yield. Comment on this suggestion.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 r = Sxy/√(SxxSyy) with Sxy = 700, Sxx = 480, Syy = 1260
✓ 1
A1 r = 0.900
✓ 1
Part (b)
A1 x̄ = 24 and ȳ = 52
✓ 1
M1 gradient = Sxy/Syy = 700/1260
✓ 1
A1 5/9
✓ 1
A1 x = (5/9)y − 44/9
✓ 1
Part (c)
M1 substitute y = 63 into x = (5/9)y − 44/9
✓ 1
A1 30.1 mg per g
✓ 1
Part (d)
the line of y on x minimises the vertical distances (errors in y), so it is designed for estimating y from x
✓ 1
the line of x on y minimises the horizontal distances (errors in x), so it is the appropriate line for estimating x from y
✓ 1
Part (e)
this is extrapolation outside the range of the data, so the linear relationship may not hold and the estimate is unreliable
✓ 1
Exam tip: Before using a regression line, decide which variable you are estimating; that decides which of the two lines to use.
459IBMA3H-4-19r19-144.2 Presentation of dataPaper 2Hard11 marks
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The durations, in whole minutes, of seven tracks on an album are positive integers. In increasing order they are a ≤ b ≤ c ≤ d ≤ e ≤ f ≤ g. The mean is 6, the median is 5, the range is 10 and the data have a single mode, which is 8.
(a)
Find the seven durations.
(6)
(b)
Find the interquartile range, and use the 1.5 × IQR rule to decide whether any duration is an outlier.
(3)
(c)
Find the standard deviation of the seven durations, to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 the mode 8 occurs at least twice, and only the three largest values can equal 8 because the median is 5
✓ 1
R1 three 8s would give g = 8 and a = −2, which is impossible, so e = f = 8 and g = a + 10
✓ 1
M1 sum of the data = 42
✓ 1
M1 a + b + c + 5 + 8 + 8 + (a + 10) = 42, so 2a + b + c = 11
✓ 1
A1 c ≤ 4 (otherwise 5 appears twice and 8 is not the only mode); a = 1 would need b + c = 9 and a ≥ 3 would need b + c ≤ 5 with b ≥ 3, both impossible, so a = 2, b + c = 7, b = 3, c = 4
✓ 1
A1 2, 3, 4, 5, 8, 8, 12
✓ 1
Part (b)
A1 Q1 = 3, Q3 = 8, IQR = 5
✓ 1
M1 upper boundary = 8 + 1.5 × 5 = 15.5
✓ 1
A1 12 < 15.5 (and 2 > 3 − 7.5), so there is no outlier
Exam tip: For puzzles about a data set, turn each statistic into an equation or inequality and test the possible cases systematically.
460IBMA3H-4-18r20-124.6 Combined, conditional and independent eventsPaper 2Hard10 marks
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In a sports science trial, each of \(n\) serves is independently successful with probability \(p\). Let \(X\) be the number of successful serves. It is given that \(E(X)=7.2\) and \(\text{Var}(X)=2.88\).
(a)
Find the values of \(n\) and \(p\).
(5)
(b)
Find \(P(X=8)\), to 3 significant figures.
(2)
(c)
A player scores \(Y=3X-2\) points in a trial. Find the probability that \(Y>20\), to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(np=7.2\)
✓ 1
M1 \(np(1-p)=2.88\)
✓ 1
A1 \(1-p=\dfrac{2.88}{7.2}=0.4\)
✓ 1
A1 \(p=0.6\)
✓ 1
A1 \(n=12\)
✓ 1
Part (b)
M1 \(\binom{12}{8}(0.6)^8(0.4)^4\)
✓ 1
A1 0.213
✓ 1
Part (c)
M1 \(3X-2>20\), so \(X>7.33\)
✓ 1
A1 \(X\ge8\), so \(P(X\ge8)=1-P(X\le7)\)
✓ 1
A1 0.438
✓ 1
Exam tip: For a binomial, divide the variance by the mean to get \(1-p\), then find \(p\) and \(n\).
461IBMA3H-4-18r20-134.2 Presentation of dataPaper 2Hard11 marks
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Two observatories measure the duration, in minutes, of a planet's transit. Observatory A takes 12 measurements with \(\sum x=576\) and \(\sum x^2=27900\). Observatory B takes 8 measurements with \(\sum x=368\) and \(\sum x^2=17300\).
(a)
Find unbiased estimates of the population mean and variance using the measurements from observatory A only. Give the variance to 3 significant figures.
(3)
(b)
Find the mean of all 20 measurements.
(2)
(c)
Find an unbiased estimate of the population variance using all 20 measurements, to 3 significant figures.
(3)
(d)
Find an unbiased estimate of the population variance using observatory B's measurements only, to 3 significant figures, and state which observatory's measurements are more variable.
R1 observatory B, because its variance estimate is larger
✓ 1
Common error: You cannot average two variances to get a combined variance; go back to the sums \(\sum x\) and \(\sum x^2\) for the combined data.
462IBMA3H-4-18r20-144.6 Combined, conditional and independent eventsPaper 2Hard10 marks
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A rail depot checks wagons for faults. Each of \(n\) wagons is independently flagged with probability 0.25. Let \(X\) be the number flagged, so \(X\sim B(n,\,0.25)\). It is given that \(P(X=2)=P(X=3)\).
Common error: \(X>2.75\) means \(X\ge3\) because \(X\) is a whole number: do not include \(X=2\) or exclude \(X=3\).
463IBMA3H-4-18r21-124.2 Presentation of dataPaper 2Hard9 marks
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The daily energy output of a solar farm, in MWh, is modelled by a normal distribution. The lower quartile is 21 and the upper quartile is 29. A day is an outlier if its output is more than \(1.5\times\) IQR below the lower quartile or above the upper quartile.
(a)
Find the standard deviation of the distribution, to 3 significant figures.
(4)
(b)
Using the unrounded value of \(\sigma\), find the probability that a randomly chosen day is an outlier, to 3 significant figures.
(3)
(c)
Over 3000 days, find the expected number of outlier days, using the unrounded probability from (b). Give your answer to 3 significant figures.
(2)
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Part (a)
A1 median (mean) \(=25\) by symmetry
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M1 \(z=0.6745\) for a cumulative probability of 0.75
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M1 \(\frac{29-25}{\sigma}=0.6745\)
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A1 \(\sigma=5.93\)
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Part (b)
M1 IQR \(=8\) so outlier boundaries are \(21-12=9\) and \(29+12=41\)
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M1 \(P(X<9)+P(X>41)=2P(X>41)\)
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A1 0.00698
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Part (c)
M1 \(3000\times0.0069766\)
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A1 20.9
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Exam tip: In a normal model the quartiles lie 0.6745 standard deviations either side of the mean, which links the IQR directly to \(\sigma\).
464IBMA3H-4-18r21-134.4 Correlation and regression of y on xPaper 2Hard10 marks
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A retailer records, over several weeks, the advertising spend \(x\) and the sales \(y\), both in thousands of euros per week. The summary statistics are: \(\bar x=12\), \(\bar y=85\), \(s_x=4\), \(s_y=12\) and covariance \(s_{xy}=43.2\).
(a)
Find the equation of the regression line of \(y\) on \(x\).
(3)
(b)
Find the value of \(r\).
(2)
(c)
The retailer wants weekly sales of 100 thousand euros. Use the regression line of \(x\) on \(y\) to estimate the advertising spend needed.
(3)
(d)
A colleague rearranges the line of \(y\) on \(x\) instead. Find the colleague's estimate to 3 significant figures and explain why the line of \(x\) on \(y\) is the better choice.
R1 the line of \(x\) on \(y\) minimises the squared horizontal distances, so it is the appropriate line for estimating \(x\) (the two lines differ unless \(r=\pm1\))
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Common error: Quote the formula you use: the gradient of \(y\) on \(x\) divides the covariance by \(s_x^2\) and the gradient of \(x\) on \(y\) divides it by \(s_y^2\).
465IBMA3H-4-19r23-054.14 Variance, continuous random variables and linear transformationsPaper 2Hard14 marks
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The time gap, X seconds, between two Formula 1 cars at a timing loop is modelled by the probability density function f(x) = k/x² for 1 ≤ x ≤ 4, and f(x) = 0 otherwise, where k is a constant.
Graph of the probability density function, not to scale
(a)
Find the value of k.
(3)
(b)
Show that E(X) = (8/3) ln 2.
(3)
(c)
Find the median of X, giving your answer as an exact fraction.
(3)
(d)
Find Var(X), giving your answer to 3 significant figures.
(3)
(e)
Find Var(3X − 1), using the unrounded value of Var(X) and giving your answer to 3 significant figures.
(2)
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Part (a)
M1 sets ∫14kx−2 dx = 1
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A1 k[−1/x]14 = k(1 − 1/4) = 3k/4
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A1 k = 4/3
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Part (b)
M1 E(X) = ∫14x·(4/3)/x² dx = (4/3)∫14 (1/x) dx
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A1 (4/3)[ln x]14 = (4/3) ln 4
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A1 ln 4 = 2 ln 2, so E(X) = (8/3) ln 2 (AG)
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Part (c)
M1 sets ∫1m (4/3)x−2 dx = 1/2
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A1 (4/3)(1 − 1/m) = 1/2, so 1 − 1/m = 3/8
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A1 m = 8/5
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Part (d)
M1 E(X²) = ∫14x²·(4/3)/x² dx = (4/3) × 3
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A1 E(X²) = 4
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A1 Var(X) = 4 − ((8/3) ln 2)² = 0.583
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Part (e)
M1 Var(3X − 1) = 3² Var(X)
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A1 9 × 0.58344… = 5.25
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Exam tip: When a density function is of the form k/x², integrate using x to the power −2; E(X²) then simplifies because the x² cancels.
466IBMA3H-4-19r23-124.14 Variance, continuous random variables and linear transformationsPaper 2Hard12 marks
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The time, X hours, that a patient spends in a day-surgery recovery unit is modelled by the probability density function f(x) = kx for 0 ≤ x ≤ 2, f(x) = k(6 − x)/2 for 2 < x ≤ 6, f(x) = 0 otherwise, where k is a constant.
Graph of the probability density function, not to scale
(a)
Show that k = 1/6.
(3)
(b)
Show that E(X) = 8/3.
(4)
(c)
Find the median of X, giving your answer to 3 significant figures.
(3)
(d)
Find P(X > 4 | X > 2).
(2)
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Part (a)
M1 splits the total area into ∫02kx dx + ∫26k(6 − x)/2 dx = 1
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A1 first area = 2k and second area = 4k
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A1 6k = 1, so k = 1/6 (AG)
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Part (b)
M1 E(X) = ∫02x(x/6) dx + ∫26x(6 − x)/12 dx
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A1 first integral = 4/9
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A1 second integral = 20/9
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A1 4/9 + 20/9 = 24/9 = 8/3 (AG)
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Part (c)
M1 P(X < 2) = 1/3 < 1/2, so the median m lies in 2 < x ≤ 6
A1 (6 − m)² = 12, so m = 6 − 2√3 = 2.54 (reject 6 + 2√3 as it is outside the range)
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Part (d)
M1 P(X > 4) = 4/24 = 1/6 and P(X > 2) = 16/24 = 2/3, so the ratio is (1/6) ÷ (2/3)
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A1 1/4
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Exam tip: With a piecewise density, check which piece contains the median by finding the area up to the breakpoint before setting up the equation.
467IBMA3H-4-19r23-134.2 Presentation of dataPaper 2Hard8 marks
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A hospital laboratory converts a raw analyser reading X into a concentration Y mg/L using Y = aX + b, where a > 0. The raw reading is normally distributed with mean 50 and standard deviation 4. It is found that the median concentration is 100 mg/L and that 10% of concentrations are greater than 112 mg/L.
(a)
Find the values of a and b, giving each to 3 significant figures.
(6)
(b)
Find the probability that a concentration is less than 90 mg/L. Give your answer to 3 significant figures.
(2)
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Part (a)
M1 states Y is normal with mean 50a + b and standard deviation 4a
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A1 median equals mean, so 50a + b = 100
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M1 uses the inverse normal: P(Z > z) = 0.1 gives z = 1.2816
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M1 (112 − 100)/(4a) = 1.2816
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A1 a = 2.34
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A1 b = 100 − 50 × 2.3409… = −17.0
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Part (b)
M1 uses Y ~ N(100, (4 × 2.3409…)²), i.e. z = −10/9.36
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A1 0.143
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Common error: The standard deviation of aX + b is a times the standard deviation of X (not a squared, and not including b).
468IBMA3H-4-19r24-124.13 Bayes’ theoremPaper 2Hard13 marks
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An online game sells crates. Of the crates opened, 60% are Bronze, 30% are Silver and 10% are Gold. The probability that a crate contains a legendary item is 0.02 for Bronze, 0.08 for Silver and 0.30 for Gold. Let \(G\) be the event that a crate is Gold and \(L\) the event that it contains a legendary item.
(a)
Find \(P(L)\).
(3)
(b)
Given that a crate contains a legendary item, find the probability that it is Gold. Give your answer to 3 significant figures.
(3)
(c)
A player opens three crates. The types of the crates are independent. Find the probability that at least one contains a legendary item. Give your answer to 3 significant figures.
(3)
(d)
The studio considers changing the probability that a Gold crate contains a legendary item from 0.30 to \(p\), leaving all other probabilities unchanged. Find the value of \(p\) for which \(G\) and \(L\) would be independent.
(4)
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Part (a)
M1 uses the law of total probability across the three crate types
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A1 \(0.6\times0.02+0.3\times0.08+0.1\times0.30\)
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A1 \(0.066\)
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Part (b)
M1 \(P(G\mid L)=\dfrac{P(G\cap L)}{P(L)}\)
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A1 \(\dfrac{0.03}{0.066}\)
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A1 \(0.455\)
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Part (c)
M1 \(1-P(\text{no legendary item in any})\)
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A1 \(1-(1-0.066)^3\)
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A1 \(0.185\)
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Part (d)
M1 \(P(G\cap L)=0.1p\)
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M1 new \(P(L)=0.036+0.1p\)
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M1 \(P(G\cap L)=P(G)P(L)\), i.e. \(0.1p=0.1(0.036+0.1p)\)
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A1 \(p=0.04\)
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Common error: When a probability parameter is changed, the total probability \(P(L)\) changes too; do not keep using the old value of 0.066 in the independence equation.
469IBMA3H-4-19r24-134.6 Combined, conditional and independent eventsPaper 2Hard12 marks
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A borehole pump either works (W) or fails (F) each day. On day 1 it works with probability 0.8. If it works on a day, it works the next day with probability 0.9; if it fails on a day, it works the next day with probability 0.5.
(a)
Find the probability that the pump works on day 2.
(2)
(b)
Find the probability that the pump works on day 3.
(3)
(c)
Given that the pump fails on day 3, find the probability that it worked on day 1. Give your answer to 3 significant figures.
(3)
(d)
Given that the pump works on exactly two of the first three days, find the probability that it works on day 2. Give your answer to 3 significant figures.
A1 \(\dfrac{0.112}{1-0.828}=0.651\) (ECF from (b))
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Part (d)
M1 identifies sequences WWF, WFW and FWW
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A1 probabilities \(0.072\), \(0.04\) and \(0.09\), total \(0.202\)
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M1 \(\dfrac{P(\text{WWF})+P(\text{FWW})}{0.202}\)
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A1 \(0.802\)
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Exam tip: For sequences of dependent events, list every sequence that satisfies the condition before adding probabilities; the order of branches on the tree matters.
470IBMA3H-4-19r24-144.11 Conditional probability and independencePaper 2Hard12 marks
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A water utility monitors two pumps, \(A\) and \(B\). On a given day, \(A\) is the event that pump 1 fails and \(B\) is the event that pump 2 fails. The events are independent, with \(P(A)=x\) and \(P(B)=2x\), and the probability that at least one pump fails is \(P(A\cup B)=0.52\).
(a)
Find the value of \(x\).
(5)
(b)
Find \(P(A\mid A\cup B)\). Give your answer to 3 significant figures.
(2)
(c)
Find the probability that exactly one of the two pumps fails on a given day.
(2)
(d)
Over five independent days, find the probability that on exactly one of the days at least one pump fails. Give your answer to 3 significant figures.
(3)
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Part (a)
M1 independence: \(P(A\cap B)=x\times2x=2x^2\)
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M1 \(P(A\cup B)=x+2x-2x^2=0.52\)
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A1 \(2x^2-3x+0.52=0\)
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A1 \(x=0.2\)
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R1 the root \(x=1.3\) is rejected as it gives a probability greater than 1
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Part (b)
M1 \(\dfrac{P(A)}{P(A\cup B)}=\dfrac{0.2}{0.52}\)
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A1 \(0.385\)
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Part (c)
M1 \(P(A\cup B)-P(A\cap B)=0.52-0.08\)
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A1 \(0.44\)
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Part (d)
M1 five equally likely positions for the one such day
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M1 \(0.52\times0.48^4\)
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A1 \(5\times0.52\times0.48^4=0.138\)
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Common error: If \(A\) is a subset of \(A\cup B\), then \(P(A\mid A\cup B)=\dfrac{P(A)}{P(A\cup B)}\); do not use \(P(A\cap B)\) in the numerator.
471IBMA3H-4-18r25-064.13 Bayes’ theoremPaper 2Hard8 marks
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A data centre buys its storage drives from three suppliers, S1, S2 and S3, which supply 45%, 35% and 20% of the drives respectively. Independently of one another, 2% of the drives from S1 and 4% of the drives from S2 are faulty. A proportion p of the drives from S3 is faulty.
(a)
A drive is chosen at random. Show that the probability that it is faulty is 0.023 + 0.2p.
(2)
(b)
Given that a randomly chosen drive is faulty, the probability that it came from S3 is \(\frac{3}{8}\). Find the value of p.
(4)
(c)
The data centre has 5000 drives, bought in the proportions given. Find the expected number of faulty drives.
Exam tip: When the unknown sits inside a conditional probability, write P(A | B) as a fraction, put the unknown in it, and solve the resulting linear equation.
472IBMA3H-4-18r25-134.11 Conditional probability and independencePaper 2Hard9 marks
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On race day the weather is rainy with probability 0.4 and dry otherwise. A cyclist wins her race with probability 0.3 when it rains and with probability w when it is dry. The probability that she wins on a randomly chosen race day is 0.42.
Tree diagram showing the probabilities given in the question
(a)
Find the value of w.
(3)
(b)
Use Bayes' theorem to find the probability that it rained, given that she won.
(3)
(c)
She enters three races on three separate race days. Assuming the races are independent, find the probability that she wins at least one of them. Give your answer to 3 significant figures.
(3)
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Part (a)
M1 use the law of total probability: 0.4(0.3) + 0.6w = 0.42
M1 P(at least one win) = 1 − P(no wins) = 1 − 0.583
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A1 0.583 = 0.195112
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A1 0.805
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Exam tip: 'At least one' questions are almost always quickest through the complement: 1 minus the probability of none.
473IBMA3H-4-18r25-144.1 Sampling, bias and outliersPaper 2Hard10 marks
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In a team cycling competition, a rider's number of bonus-point events won in a stage is the random variable X, with the following probability distribution.
x
0
1
2
3
P(X = x)
0.1
0.4
0.3
0.2
The team is awarded Y = aX + b points, where a > 0 and b are constants. It is known that E(Y) = 11 and Var(Y) = 21.
(a)
Find E(X) and Var(X).
(4)
(b)
Find the value of a and the value of b.
(4)
(c)
Y1 and Y2 are independent random variables, each with the same distribution as Y. Find Var(Y1 − Y2).
Common error: The variance of a difference of independent random variables is the sum of the variances, not the difference; the minus sign disappears when squared.
474IBMA3H-4-18r26-134.12 Standardization and unknown normal parametersPaper 2Hard12 marks
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A sensor monitors a swimming pool for contamination. Each day, the pool is contaminated with probability 0.02. If the pool is contaminated, the sensor sounds an alarm with probability 0.95. If the pool is not contaminated, the sensor sounds an alarm with probability 0.04. When the sensor is used repeatedly on the same day, its readings are independent of each other given the state of the pool.
(a)
Find the probability that the sensor sounds an alarm in a single reading.
(2)
(b)
Given that the sensor sounds an alarm, find the probability that the pool is contaminated.
(3)
(c)
The sensor is used twice and sounds an alarm both times. Find the probability that the pool is contaminated.
(3)
(d)
The pool is closed only if the probability of contamination exceeds 0.99. Find the least number of consecutive alarms needed to close the pool, and for that number state the probability of contamination to 3 significant figures.
(4)
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Part (a)
M1 0.02 × 0.95 + 0.98 × 0.04
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A1 0.0582
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Part (b)
M1 Bayes' theorem
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A1 0.019/0.0582
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A1 0.326
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Part (c)
M1 numerator 0.02 × 0.952 = 0.01805
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M1 denominator 0.01805 + 0.98 × 0.042
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A1 0.920
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Part (d)
M1 expression for n alarms: 0.02 × 0.95n/(0.02 × 0.95n + 0.98 × 0.04n)
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M1 test values of n using GDC or solve the inequality
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A1 n = 3 (since n = 2 gives 0.920 < 0.99)
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A1 probability 0.996
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Exam tip: Repeated independent evidence multiplies the likelihoods: raise each 'alarm given state' probability to the power of the number of alarms.
475IBMA3H-4-20r28-134.14 Variance, continuous random variables and linear transformationsPaper 2Hard11 marks
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The waiting time, X hours, of a vessel for a berth has probability density function \(f(x)=cx\) for \(0\leq x\leq2\); \(f(x)=\frac{c(6-x)}{2}\) for \(2<x\leq6\); \(f(x)=0\) otherwise, where c is a constant.
Graph of the probability density function, not to scale
(a)
Show that \(c=\frac{1}{6}\). [3]
(3)
(b)
Find \(\mathrm{E}(X)\). Give your answer to 3 significant figures. [4]
(4)
(c)
Find the median of X. Give your answer to 3 significant figures. [4]
(4)
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Part (a)
M1 \(\int_0^2cx\,dx=2c\)
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M1 \(\int_2^6\frac{c(6-x)}{2}\,dx=4c\)
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A1 \(6c=1\), so \(c=\frac{1}{6}\) (AG)
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Part (b)
M1 \(\int_0^2x\cdot\frac{x}{6}\,dx\)
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M1 \(\int_2^6x\cdot\frac{6-x}{12}\,dx\)
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A1 the two integrals are \(\frac{4}{9}\) and \(\frac{20}{9}\)
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A1 \(\mathrm{E}(X)=\frac{8}{3}=2.67\)
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Part (c)
M1 \(\int_0^2f(x)\,dx=\frac{1}{3}<\frac{1}{2}\), so the median m lies in \(2<x\leq6\)
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M1 \(\int_m^6\frac{6-x}{12}\,dx=\frac{1}{2}\)
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A1 \(\frac{(6-m)^2}{24}=\frac{1}{2}\), so \((6-m)^2=12\)
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A1 \(m=6-2\sqrt{3}=2.54\) hours
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Common error: For a piecewise density, first work out how much probability lies in each piece. This tells you which piece contains the median before you set up the equation.
476IBMA3H-4-20r28-144.11 Conditional probability and independencePaper 2Hard9 marks
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Vessels entering a harbour are either container ships (70%) or tankers (30%). The draught, D metres, of a container ship is normally distributed with mean 11.0 and standard deviation 1.2. The draught of a tanker is normally distributed with mean 13.5 and standard deviation 1.5.
A vessel is chosen at random and its draught exceeds 13 m. Find the probability that it is a tanker. Give your answer to 3 significant figures, using unrounded values from (a) and (b). [3]
(3)
(d)
Two vessels are chosen independently and each has a draught exceeding 13 m. Find the probability that both are tankers. Give your answer to 3 significant figures, using the unrounded answer to (c). [2]
(2)
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Part (a)
M1 normal cdf with mean 11.0 and standard deviation 1.2
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A1 0.0478
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Part (b)
M1 normal cdf with mean 13.5 and standard deviation 1.5
Common error: The conditional probabilities in (a) and (b) are not the answer to (c). Weight each by the proportion of its vessel type before dividing by the total.
477IBMA3H-4-19r32-114.12 Standardization and unknown normal parametersPaper 2Hard9 marks
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A manufacturer makes climbing ropes whose lengths, L metres, are normally distributed with mean 60.0 and standard deviation σ. A rope is rejected if its length is less than 59.5 m. It is found that 5% of ropes are rejected.
(a)
Find the value of σ.
(3)
(b)
Ropes are inspected one at a time, independently. Find the smallest number of ropes that must be inspected so that the probability of finding at least one rejected rope is at least 0.95.
(3)
(c)
A random sample of 12 ropes is taken. Find the probability that at least two are rejected. Give your answer to three significant figures.
(3)
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Part (a)
M1 P(L < 59.5) = 0.05, so use of inverse normal gives z = −1.6449
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M1 (59.5 − 60.0)/σ = −1.6449
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A1 σ = 0.304 m
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Part (b)
M1 1 − 0.95n ≥ 0.95
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M1 attempt to solve 0.95n ≤ 0.05
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A1 n ≥ 58.4, so n = 59
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Part (c)
M1 X ~ B(12, 0.05)
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M1 P(X ≥ 2) = 1 − P(X = 0) − P(X = 1)
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A1 0.118
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Common error: For 'at least two' use 1 − P(0) − P(1); subtracting only P(X = 0) gives the probability of at least one rejected rope.
478IBMA3H-4-19r32-124.10 Regression line of x on yPaper 2Hard9 marks
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A telecoms engineer records the number of active users, x (in hundreds), connected to a mobile mast and the average data latency, y (in milliseconds), at eight times of day.
x
2
3
5
6
8
9
11
12
y
24
22
34
30
45
41
48
58
Give all regression coefficients correct to four significant figures, and use these rounded coefficients in later parts.Scatter diagram of the data in the table
(a)
Find the equation of the regression line of y on x.
(2)
(b)
Find the equation of the regression line of x on y.
(2)
(c)
On one occasion the latency is 30 ms. Estimate the number of users (in hundreds) using (i) the equation in part (a) rearranged to make x the subject; (ii) the equation in part (b). Give each answer to three significant figures, and find the difference between the two unrounded estimates.
(3)
(d)
State, with a reason, which of the two estimates in part (c) is more appropriate, and explain why the two estimates are different.
(2)
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Part (a)
M1 attempt to find the regression line using technology
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A1 y = 3.293x + 14.70
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Part (b)
M1 attempt to find the regression line of x on y using technology
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A1 x = 0.2781y − 3.499
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Part (c)
A1 (i) x = (30 − 14.70)/3.293 = 4.65
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A1 (ii) x = 0.2781 × 30 − 3.499 = 4.84
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A1 difference = 0.198 (hundreds) (accept 0.20)
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Part (d)
R1 estimate (ii) is more appropriate because the regression line of x on y minimises the horizontal distances, so it is the line used to predict x from y
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R1 the estimates differ because the correlation is not perfect (r < 1); the two lines only coincide when r = ±1
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Exam tip: Compare the two regression lines only through their estimates of the same quantity; the closer |r| is to 1, the closer the two lines are to each other.
479IBMA3H-4-19r32-134.14 Variance, continuous random variables and linear transformationsPaper 2Hard12 marks
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A food scientist models the time, X hours, taken for a caramel mixture to set by the probability density function \(f(x) = kx^2(3 - x)\) for \(0 \leq x \leq 3\), and \(f(x) = 0\) otherwise, where \(k\) is a constant.
Graph of the probability density function, not to scale
(a)
Show that \(k = \frac{4}{27}\).
(3)
(b)
Find E(X).
(3)
(c)
Find the median time for the mixture to set, giving your answer to three significant figures.
Common error: The median is found from the cumulative distribution, not from the point where the density is greatest.
480IBMA3H-4-18r33-054.13 Bayes’ theoremPaper 2Hard9 marks
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At a port, 2% of containers hold prohibited goods. A scanner flags 90% of containers that hold prohibited goods and flags 5% of containers that do not. Repeated scans of the same container are independent of one another, given whether the container holds prohibited goods.
(a)
A container is scanned once and flagged. Find the probability that it holds prohibited goods, to three significant figures.
(3)
(b)
The container is scanned a second time and is flagged again. Find the probability that it holds prohibited goods, to three significant figures.
(3)
(c)
Instead, the second scan does not flag the container (first scan flagged). Find the probability that it holds prohibited goods, to three significant figures.
M1 both scans flagged: 0.02 × 0.92 and 0.98 × 0.052
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A1 0.0162 / (0.0162 + 0.00245) = 0.0162 / 0.01865
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A1 0.869
✓ 1
Part (c)
M1 0.02 × 0.9 × 0.1 and 0.98 × 0.05 × 0.95
✓ 1
A1 0.0018 / (0.0018 + 0.04655) = 0.0018 / 0.04835
✓ 1
A1 0.0372
✓ 1
Exam tip: With independent repeated tests, multiply the likelihoods for each result on each branch before applying Bayes' theorem.
481IBMA3H-4-18r33-134.2 Presentation of dataPaper 2Hard9 marks
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The load, in tonnes, of a delivery truck leaving a depot is normally distributed. The lower quartile of the loads is 18.6 tonnes and the upper quartile is 24.4 tonnes.
(a)
Find the mean and the standard deviation of the load, giving the standard deviation to three significant figures.
(4)
(b)
A load is an outlier if it is less than Q1 − 1.5 × IQR or greater than Q3 + 1.5 × IQR. Using the unrounded standard deviation, find the probability that a truck's load is an outlier. Give your answer to three significant figures.
(3)
(c)
In a week 2000 trucks leave the depot. Using the unrounded probability from (b), estimate the number whose load is an outlier, to the nearest whole number.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 mean is the midpoint of the quartiles, (18.6 + 24.4)/2
✓ 1
A1 mean = 21.5 tonnes
✓ 1
M1 24.4 = 21.5 + 0.674… × σ (using the inverse normal for 0.75)
✓ 1
A1 σ = 4.30 tonnes
✓ 1
Part (b)
M1 IQR = 5.8, so the fences are 9.9 and 33.1
✓ 1
M1 2 × P(X > 33.1) (by symmetry of the normal distribution)
✓ 1
A1 0.00698
✓ 1
Part (c)
M1 2000 × P(outlier)
✓ 1
A1 14
✓ 1
Exam tip: The quartiles of a normal distribution lie 0.674 standard deviations either side of the mean, so IQR = 1.349σ.
482IBMA3H-4-19r37-114.11 Conditional probability and independencePaper 2Hard9 marks
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A coffee roaster buys beans from three farms. Farm 1 supplies 50% of the beans, farm 2 supplies 30% and farm 3 supplies 20%. The proportions of beans graded "premium" are 0.40 from farm 1, 0.55 from farm 2 and 0.80 from farm 3.
(a)
Find the probability that a randomly chosen bean is graded premium.
(2)
(b)
A premium bean is chosen at random. Find the probability that it came from farm 3. Give your answer to 3 significant figures.
(3)
(c)
Next season farm 3 improves, so that a proportion t of its beans is graded premium. All other proportions are unchanged. The probability that a premium bean came from farm 3 is then exactly 1/3. Find t.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.5 × 0.4 + 0.3 × 0.55 + 0.2 × 0.8
✓ 1
A1 0.525
✓ 1
Part (b)
M1 P(farm 3 | premium) = 0.2 × 0.8 ÷ 0.525
✓ 1
A1 0.16 ÷ 0.525 (ECF from (a))
✓ 1
A1 0.305
✓ 1
Part (c)
M1 0.2t ÷ (0.2t + 0.365) = 1/3
✓ 1
A1 0.365 found as 0.5 × 0.4 + 0.3 × 0.55
✓ 1
M1 solving: 0.6t = 0.2t + 0.365
✓ 1
A1 t = 0.9125
✓ 1
Exam tip: When an unknown appears inside Bayes' theorem, write the posterior probability as a single fraction and set it equal to the given value before solving.
483IBMA3H-4-19r37-144.8 Binomial distributionPaper 2Hard9 marks
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In a quality check, 4% of probe components are defective. A test flags 90% of defective components and also wrongly flags 7% of components that are not defective.
Tree diagram showing the probabilities given in the question; Not def. = not defective
(a)
Find the probability that a randomly chosen component is flagged.
(2)
(b)
A component is flagged. Find the probability that it is defective. Give your answer to 3 significant figures.
(3)
(c)
A flagged component is tested a second time, and the two test outcomes are independent given the component's true status. It is flagged again. Find the probability that it is defective. Give your answer to 3 significant figures.
M1 P(flagged twice and not defective) = 0.96 × 0.072 = 0.004704
✓ 1
M1 0.0324 ÷ (0.0324 + 0.004704)
✓ 1
A1 0.873
✓ 1
Common error: Multiply the likelihoods of the two flags within each group (defective or not); equivalently, use 0.349 as the new prior and apply Bayes' theorem once more.
484IBMA3H-4-20r44-074.1 Sampling, bias and outliersPaper 2Hard7 marks
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A shipping company has 12 containers on a quay, of which 5 are refrigerated. An inspector selects 4 containers at random, without replacement.
(a)
Find the probability that exactly 2 of the selected containers are refrigerated. Give your answer as a fraction in its simplest form.
(2)
(b)
Find the probability that at least one of the selected containers is refrigerated.
(2)
(c)
Given that at least one of the selected containers is refrigerated, find the probability that exactly 2 are refrigerated. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 C(5,2) × C(7,2) ÷ C(12,4) = 210/495
✓ 1
A1 14/33
✓ 1
Part (b)
M1 1 − P(none) = 1 − C(7,4)/C(12,4) = 1 − 35/495
✓ 1
A1 92/99 (accept 0.929)
✓ 1
Part (c)
M1 conditional probability, P(exactly 2) ÷ P(at least 1)
✓ 1
M1 (14/33) ÷ (92/99)
✓ 1
A1 0.457
✓ 1
Common error: A selection without replacement is not binomial: the probability changes after each choice, so use combinations or a probability tree.
485IBMA3H-4-20r44-104.8 Binomial distributionPaper 2Hard7 marks
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A shipping company stores crates from two batches, A and B. A crate from batch A is faulty with probability 0.05 and a crate from batch B is faulty with probability 0.2, independently of other crates. A batch is chosen at random, with each batch equally likely, and 10 crates are tested from the chosen batch. Exactly 2 crates are found to be faulty.
(a)
Find the probability of exactly 2 faulty crates among the 10 tested if the batch is (i) A, (ii) B. Give each answer to 3 significant figures.
(2)
(b)
Find the probability that the chosen batch was B. Use unrounded values from (a). Give your answer to 3 significant figures.
(3)
(c)
One further crate is now taken from the same batch. Using your answer to (b), find the probability that this crate is faulty. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of the binomial pdf with n = 10 and p = 0.05 or p = 0.2
Exam tip: After finding a posterior probability, use it in place of the original prior for any later prediction about the same batch.
486IBMA3H-4-20r44-124.12 Standardization and unknown normal parametersPaper 2Hard10 marks
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A volcano observatory monitors three types of volcano: stratovolcanoes (50%), shield volcanoes (30%) and calderas (20%). In a given decade, the probability of an eruption is 0.30 for a stratovolcano, 0.10 for a shield volcano and 0.25 for a caldera.
(a)
Find the probability that a randomly chosen monitored volcano erupts in the decade.
(2)
(b)
Given that a randomly chosen volcano erupts, find the probability that it is a shield volcano. Give your answer to 3 significant figures.
(2)
(c)
Eight monitored volcanoes are chosen at random. Assume that each independently erupts with the probability found in (a). Find the probability that at least 2 of them erupt. Give your answer to 3 significant figures.
(3)
(d)
Three volcanoes that have erupted are chosen at random. Find the probability that exactly 2 of them are stratovolcanoes. Give your answer to 3 significant figures.
Exam tip: Before using a binomial model, state the success probability and say what has been assumed independent; in part (d) the success probability is a conditional probability.
487IBMA3H-4-20r46-134.13 Bayes’ theoremPaper 2Hard8 marks
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On a farm, a proportion \(d\) of fields is diseased. A drone survey flags a field with probability 0.95 if it is diseased and probability 0.08 if it is healthy. Let \(F\) be the event that a field is flagged and \(D\) the event that a field is diseased.
(a)
Show that \(P(F)=0.08+0.87d\).
(2)
(b)
Given that \(P(D|F)=0.5\), find the value of \(d\). Give your answer to 3 significant figures.
(3)
(c)
Using the unrounded value of \(d\), find the probability that a field is healthy given that it is not flagged. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(P(F)=0.95d+0.08(1-d)\)
✓ 1
A1 simplify to \(0.08+0.87d\) (AG)
✓ 1
Part (b)
M1 \(\frac{0.95d}{0.08+0.87d}=0.5\)
✓ 1
M1 rearrange to \(0.515d=0.04\)
✓ 1
A1 \(d=0.0777\)
✓ 1
Part (c)
M1 \(P(F')=0.92-0.87d=0.8524\ldots\)
✓ 1
M1 \(P(D'\cap F')=0.92(1-d)=0.8485\ldots\) and divide
✓ 1
A1 0.995
✓ 1
Exam tip: Express each probability in terms of \(d\) first; a rearranged Bayes equation then reduces to a linear equation.
488IBMA3H-4-20r46-144.14 Variance, continuous random variables and linear transformationsPaper 2Hard10 marks
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At a music festival, \(X\) is the number of encore songs played, where \(X\) takes the values 0, 1, 2 and 3 and \(P(X=x)=k\times2^x\).
(a)
Show that \(k=\frac{1}{15}\).
(2)
(b)
Find \(E(X)\) and \(\mathrm{Var}(X)\), giving exact answers.
(4)
(c)
A score is defined as \(Y=aX+b\), where \(a>0\). Given that \(E(Y)=4.8\) and \(\mathrm{Var}(Y)=7.76\), find the value of \(a\) and the value of \(b\).
Exam tip: Work with the exact fractions for \(E(X)\) and \(\mathrm{Var}(X)\) in the last part so that \(a\) and \(b\) come out as clean values.
489AAHL4-q00324.7 Discrete random variablesPaper 3Hard28 marks
This question investigates the largest score obtained when a fair \(n\)-sided die, with faces numbered \(1,2,\ldots,n\), is rolled several times. The rolls are independent. Let \(M_{r}\) be the largest score obtained when the die is rolled \(r\) times, where \(n,r\in\mathbb{Z}^{+}\) and \(n\ge2\).
(a)(i)
Consider \(n=4\) and \(r=2\).
Show that \(\mathrm{P}(M_{2}=k)=\dfrac{2k-1}{16}\) for \(k=1,2,3,4\).
(2)
(a)(ii)
Find \(\mathrm{E}(M_{2})\).
(2)
(b)(i)
Consider a general value of \(n\), with \(r=2\).
Show that \(\mathrm{P}(M_{2}=k)=\dfrac{2k-1}{n^{2}}\) for \(k=1,2,\ldots,n\).
(2)
(b)(ii)
Given that \(\displaystyle\sum_{k=1}^{n}k^{2}=\dfrac{n(n+1)(2n+1)}{6}\), show that \(\mathrm{E}(M_{2})=\dfrac{(n+1)(4n-1)}{6n}\).
Write down an expression for \(\mathrm{P}(M_{r}\le k)\), for \(k=0,1,\ldots,n\).
(1)
(c)(ii)
Show that \(\mathrm{E}(M_{r})=\displaystyle\sum_{k=0}^{n-1}\mathrm{P}(M_{r}\gt k)\).
(3)
(c)(iii)
Hence show that \(\mathrm{E}(M_{r})=n-\dfrac{1}{n^{r}}\displaystyle\sum_{k=1}^{n-1}k^{r}\).
(2)
(d)
For \(r=3\), use the result in part (c)(iii) to find the value of \(\dfrac{\mathrm{E}(M_{3})}{n}\) when \(n=10\) and when \(n=1000\).
(2)
(e)
Using your answers to parts (b)(iii) and (d), suggest an expression, in terms of \(r\), for \(\displaystyle\lim_{n\to\infty}\dfrac{\mathrm{E}(M_{r})}{n}\).
(1)
(f)
By considering rectangles of width 1 under and over the graph of \(y=x^{r}\), for \(x\ge0\), explain why \(\displaystyle\int_{0}^{n-1}x^{r}\,\mathrm{d}x\lt\sum_{k=1}^{n-1}k^{r}\lt\int_{1}^{n}x^{r}\,\mathrm{d}x\).
(3)
(g)
Hence prove your suggestion in part (e).
(4)
Show mark scheme
Working
Mark
Notes
(a)(i)
\(\mathrm{P}(M_{2}\le k)=\left(\dfrac{k}{4}\right)^{2}\), so \(\mathrm{P}(M_{2}=k)=\dfrac{k^{2}-(k-1)^{2}}{16}\)
M1
Or counting the 2k − 1 ordered pairs with largest value k in a 4 × 4 sample space.
\(\displaystyle\sum_{k=1}^{n-1}k^{r}\) is the total area of \(n-1\) rectangles of width 1 and heights \(1^{r},2^{r},\ldots,(n-1)^{r}\)
R1
placing the rectangle of height \(k^{r}\) on \([k,\,k+1]\): it lies under the curve because \(x^{r}\) is increasing, so the sum is less than \(\displaystyle\int_{1}^{n}x^{r}\,\mathrm{d}x\)
R1
placing it on \([k-1,\,k]\): the curve lies under the rectangle, so the sum is greater than \(\displaystyle\int_{0}^{n-1}x^{r}\,\mathrm{d}x\)
490AAHL4-q00774.13 Bayes’ theoremPaper 3Hard29 marks
This question investigates how the probability of the next toss being a head changes as more heads are observed.
A box contains three coins. Coin \(A\) is fair; coin \(B\) has a head on both sides; coin \(C\) is biased so that the probability of a head is \(\tfrac{1}{4}\). One coin is chosen at random from the box and is then tossed repeatedly. For the chosen coin, the tosses are independent.
(a)(i)
The chosen coin is tossed once.
Find the probability that it shows a head.
(2)
(a)(ii)
Given that it shows a head, find the probability that the coin is \(B\).
(2)
(b)
The chosen coin is tossed twice and shows two heads. Find the probability that the coin is \(B\).
(3)
(c)
Show that, given that the first \(n\) tosses are all heads, the probability that the coin is \(B\) is \(\dfrac{4^{n}}{4^{n}+2^{n}+1}\).
(3)
(d)
Find the least value of \(n\) for which this probability exceeds 0.999.
(3)
(e)(i)
Let \(q_{n}\) be the probability that toss \(n+1\) is a head, given that the first \(n\) tosses are all heads, for \(n\ge1\).
Show that \(q_{n}=\dfrac{4^{n+1}+2^{n+1}+1}{4\left(4^{n}+2^{n}+1\right)}\).
(3)
(e)(ii)
Find \(q_{1}\) and \(q_{2}\).
(2)
(e)(iii)
Write down \(\displaystyle\lim_{n\to\infty}q_{n}\).
(1)
(f)
By writing \(t=2^{n}\) and considering \(1-q_{n}\), prove that \(q_{n}\) increases as \(n\) increases.
(4)
(g)(i)
The investigation is generalized. A box contains two coins only, with probabilities of a head \(p\) and \(s\), where \(0\lt p\lt s\le1\). One coin is chosen at random and tossed repeatedly. Let \(q_{n}\) now be the probability that toss \(n+1\) is a head, given that the first \(n\) tosses are all heads, with \(q_{0}\) the probability that the first toss is a head.
Show that \(q_{n}=\dfrac{p^{n+1}+s^{n+1}}{p^{n}+s^{n}}\).
(2)
(g)(ii)
Prove that \(q_{n}\gt q_{n-1}\) for all \(n\in\mathbb{Z}^{+}\).
Show that the mode of \(X_{n}\) is \(\dfrac{n}{n+1}\).
(2)
(c)(ii)
Show that \(\mathrm{E}(X_{n})=\dfrac{n+1}{n+3}\).
(3)
(c)(iii)
Prove that the mode of \(X_{n}\) is greater than or equal to \(\mathrm{E}(X_{n})\), with equality only when \(n=1\).
(3)
(d)(i)
Show that \(\mathrm{E}(X_{n}^{2})=\dfrac{(n+1)(n+2)}{(n+3)(n+4)}\).
(2)
(d)(ii)
Hence show that \(\mathrm{Var}(X_{n})=\dfrac{2(n+1)}{(n+3)^{2}(n+4)}\).
(3)
(d)(iii)
Find \(\displaystyle\lim_{n\to\infty}\mathrm{Var}(X_{n})\) and interpret this result.
(2)
(e)(i)
Let \(I(p,q)=\displaystyle\int_{0}^{1}x^{p}(1-x)^{q}\,\mathrm{d}x\), where \(p,q\in\mathbb{N}\).
For \(q\ge1\), use integration by parts to show that \(I(p,q)=\dfrac{q}{p+1}\,I(p+1,\,q-1)\).
(3)
(e)(ii)
Hence prove that \(I(p,q)=\dfrac{p!\,q!}{(p+q+1)!}\).
(4)
(e)(iii)
Write down, in terms of \(p\) and \(q\), the value of the constant \(c\) for which \(g(x)=c\,x^{p}(1-x)^{q}\), \(0\le x\le1\), is a probability density function.
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