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A Level H1 Mathematics Practice Paper 2

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TuitionGoWhere Practice Paper - Maths H1 A-Level

TuitionGoWhere Practice Paper (AI)

Subject: Mathematics H1 Level: A-Level Paper: Practice Paper — Statistics & Probability Version: 2 of 5 Duration: 1 hour 30 minutes Total Marks: 60

Name: ___________________________ Class: ___________________________ Date: ___________________________

Instructions

Write your answers in the spaces provided.
Show all working clearly. Marks are awarded for correct method even if the final answer is wrong.
Give answers correct to 3 significant figures unless otherwise stated.
A graphing calculator may be used where appropriate.
The total marks for this paper is 60.
The marks for each question are shown in brackets [ ].

Section A: Pure Statistics (30 marks)

Answer all questions in this section.

Question 1

A random sample of 8 students recorded the number of hours they spent on revision in a week:

$12,\ 15,\ 10,\ 18,\ 14,\ 11,\ 16,\ 13$

Calculate the unbiased estimates of the population mean and population variance.

[4]

Question 2

The random variable $X \sim \mathrm{B}(20, 0.35)$ . Find:

(a) $\mathrm{P}(X = 7)$

[2]

(b) $\mathrm{P}(X \geq 5)$

[2]

Question 3

A continuous random variable $X$ has probability density function given by

$f(x) = \begin{cases} kx(4 - x) & 0 \leq x \leq 4 \\ 0 & \text{otherwise} \end{cases}$

(a) Show that $k = \dfrac{3}{32}$ .

[2]

(b) Find $\mathrm{E}(X)$ .

[2]

Question 4

The heights of a certain species of plant are normally distributed with mean 42 cm and standard deviation 5 cm.

(a) Find the probability that a randomly selected plant has a height between 38 cm and 47 cm.

[3]

(b) In a random sample of 200 plants, how many would you expect to have a height greater than 50 cm?

[2]

Question 5

A researcher claims that the mean daily screen time of teenagers is 6.5 hours. A random sample of 50 teenagers gives a mean daily screen time of 7.2 hours with a standard deviation of 2.1 hours. Test at the 5% significance level whether there is evidence that the mean daily screen time differs from 6.5 hours.

[6]

Question 6

The following table shows the marks obtained by 60 students in a mathematics test.

Mark	Frequency
0–19	4
20–39	10
40–59	18
60–79	16
80–100	12

(a) Calculate the mean mark.

[3]

(b) State the modal class.

[1]

[3]

<image_placeholder> id: Q6-fig1 type: chart linked_question: Q6(c) description: Histogram showing frequency density on the vertical axis and mark ranges on the horizontal axis. The horizontal axis is labelled "Mark" with class boundaries 0, 20, 40, 60, 80, 100. The vertical axis is labelled "Frequency density". Bars are drawn for each class interval with heights proportional to frequency density. labels: Horizontal axis: "Mark" with tick marks at 0, 20, 40, 60, 80, 100. Vertical axis: "Frequency density". values: Class widths: 20, 20, 20, 20, 20. Frequencies: 4, 10, 18, 16, 12. Frequency densities: 0.2, 0.5, 0.9, 0.8, 0.6. must_show: All five bars with correct heights, labelled axes, class boundaries clearly marked, no gaps between bars.

</image_placeholder>

Section B: Probability & Distributions (30 marks)

Answer all questions in this section.

Question 7

A bag contains 5 red balls, 4 blue balls, and 3 green balls. Three balls are drawn at random without replacement.

(a) Find the probability that all three balls are red.

[2]

(b) Find the probability that exactly two balls are red and one is blue.

[3]

Question 8

The number of emails received by an employee per hour follows a Poisson distribution with mean 4.2.

(a) Find the probability that the employee receives exactly 5 emails in a given hour.

[2]

(b) Find the probability that the employee receives at least 3 emails in a given hour.

[3]

Question 9

The weights of apples from a particular orchard are normally distributed with mean 150 g and standard deviation $\sigma$ g. It is known that 8% of the apples weigh more than 165 g.

(a) Find the value of $\sigma$ .

[4]

(b) Apples weighing less than 130 g are classified as "small". Find the probability that a randomly selected apple is classified as "small".

[2]

[3]

Question 10

A fair six-sided die is rolled 4 times.

(a) Find the probability of getting exactly two sixes.

[3]

(b) Find the probability of getting at least one six.

[3]

Question 11

The lifetime of a certain brand of LED light bulb, $T$ hours, follows an exponential distribution with mean 8000 hours.

(a) Write down the probability density function of $T$ .

[1]

(b) Find the probability that a randomly selected bulb lasts more than 10,000 hours.

[3]

(c) A hotel purchases 5 of these bulbs. Assuming independence, find the probability that exactly 3 of them last more than 10,000 hours.

[3]

Question 12

Two events $A$ and $B$ are such that $\mathrm{P}(A) = 0.6$ , $\mathrm{P}(B) = 0.4$ , and $\mathrm{P}(A \cup B) = 0.76$ .

(a) Find $\mathrm{P}(A \cap B)$ .

[2]

(b) Determine whether $A$ and $B$ are independent. Justify your answer.

[2]

Question 13

A call centre receives calls at an average rate of 2.5 calls per minute. Use a suitable approximation to find the probability that the call centre receives fewer than 130 calls in a 1-hour period.

[5]

Question 14

The following grouped data shows the daily commute times (in minutes) of 80 employees at a company.

Commute time (min)	Frequency
0–9	6
10–19	14
20–29	22
30–39	20
40–49	12
50–59	6

(a) Estimate the median commute time.

[3]

(b) Calculate the interquartile range.

[3]

[2]

Question 15

A factory produces components, and 5% are defective. A quality control inspector tests components one at a time until the first defective component is found.

(a) Find the probability that the first defective component is found on the 5th test.

[2]

(b) Find the expected number of tests until the first defective component is found.

[2]

(c) If the inspector tests 100 components, use a Poisson approximation to estimate the probability that exactly 3 are defective.

[3]

Question 16

The joint probability distribution of two discrete random variables $X$ and $Y$ is given by:

$\mathrm{P}(X = x, Y = y) = \frac{x + y}{30}, \quad x = 1, 2, 3;\ y = 1, 2, 3$

(a) Find $\mathrm{P}(X = 2, Y = 3)$ .

[1]

(b) Find the marginal probability $\mathrm{P}(X = 2)$ .

[2]

[3]

Question 17

A random variable $X \sim \mathrm{N}(\mu, \sigma^2)$ . It is known that $\mathrm{P}(X < 25) = 0.1587$ and $\mathrm{P}(X > 45) = 0.0228$ .

Find the values of $\mu$ and $\sigma$ .

[5]

Question 18

In a game, a player rolls two fair six-sided dice. The player wins $10 if the sum is 7, wins $5 if the sum is greater than 9, and loses $3 otherwise.

(a) Find the probability that the player wins $10.

[2]

(b) Find the probability that the player wins $5.

[2]

[3]

Question 19

A sample of 10 observations from a normal distribution with unknown mean and variance gives the following summary statistics:

$\sum x = 156 \quad \text{and} \quad \sum x^2 = 2478$

(a) Calculate the unbiased estimates of the population mean and variance.

[3]

(b) Construct a 95% confidence interval for the population mean.

[4]

Question 20

A continuous random variable $X$ has cumulative distribution function

$F(x) = \begin{cases} 0 & x < 0 \\ \dfrac{x^3}{64} & 0 \leq x \leq 4 \\ 1 & x > 4 \end{cases}$

(a) Find the probability density function $f(x)$ .

[2]

(b) Find $\mathrm{P}(1 < X < 3)$ .

[2]

End of Paper

Mark Summary

Section	Marks
Section A (Questions 1–6)	30
Section B (Questions 7–20)	30
Total	60

Answers

TuitionGoWhere Practice Paper — Maths H1 A-Level

Answer Key & Marking Scheme

Subject: Mathematics H1 Paper: Practice Paper — Statistics & Probability Version: 2 of 5 Total Marks: 60

Section A: Pure Statistics (30 marks)

Question 1 [4 marks]

Data: 12, 15, 10, 18, 14, 11, 16, 13; $n = 8$

Unbiased estimate of population mean:

$\bar{x} = \frac{\sum x_i}{n} = \frac{12 + 15 + 10 + 18 + 14 + 11 + 16 + 13}{8} = \frac{109}{8} = 13.625$

Unbiased estimate of population variance:

$s^2 = \frac{\sum(x_i - \bar{x})^2}{n - 1}$

Calculate each $(x_i - \bar{x})^2$ :

$x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$
12	−1.625	2.640625
15	1.375	1.890625
10	−3.625	13.140625
18	4.375	19.140625
14	0.375	0.140625
11	−2.625	6.890625
16	2.375	5.640625
13	−0.625	0.390625

$\sum(x_i - \bar{x})^2 = 49.875$

$s^2 = \frac{49.875}{7} = 7.125$

Answer: $\bar{x} = 13.6$ hours, $s^2 = 7.13$ hours² (to 3 s.f.)

Marking:

M1: Correct formula for $\bar{x}$ with correct substitution
A1: $\bar{x} = 13.625$ or 13.6
M1: Correct formula for $s^2$ using $n - 1 = 7$ in denominator
A1: $s^2 = 7.125$ or 7.13

Common mistake: Using $n = 8$ instead of $n - 1 = 7$ gives $\frac{49.875}{8} = 6.23$ , which is the biased estimator. This loses the A1 mark.

Question 2 [4 marks]

$X \sim \mathrm{B}(20, 0.35)$

(a) $\mathrm{P}(X = 7) = \binom{20}{7}(0.35)^7(0.65)^{13}$

$= 77520 \times (0.35)^7 \times (0.65)^{13}$

$= 77520 \times 0.00064339... \times 0.005479...$

$= 0.164 \text{ (to 3 s.f.)}$

Marking: M1 for correct binomial probability formula with $n=20, p=0.35, r=7$ ; A1 for answer 0.164.

(b) $\mathrm{P}(X \geq 5) = 1 - \mathrm{P}(X \leq 4)$

Using calculator/binomial tables:

$\mathrm{P}(X \leq 4) = \sum_{k=0}^{4} \binom{20}{k}(0.35)^k(0.65)^{20-k}$

Computing each term:

$\mathrm{P}(X=0) = (0.65)^{20} = 0.000182$
$\mathrm{P}(X=1) = 20(0.35)(0.65)^{19} = 0.002098$
$\mathrm{P}(X=2) = 190(0.35)^2(0.65)^{18} = 0.01157$
$\mathrm{P}(X=3) = 1140(0.35)^3(0.65)^{17} = 0.03834$
$\mathrm{P}(X=4) = 4845(0.35)^4(0.65)^{16} = 0.08918$

$\mathrm{P}(X \leq 4) = 0.1414$

$\mathrm{P}(X \geq 5) = 1 - 0.1414 = 0.859 \text{ (to 3 s.f.)}$

Marking: M1 for using complement $1 - \mathrm{P}(X \leq 4)$ ; A1 for answer 0.859.

Question 3 [4 marks]

(a) For a valid PDF, $\int_{-\infty}^{\infty} f(x)\,dx = 1$ :

$\int_0^4 kx(4-x)\,dx = 1$

$k\int_0^4 (4x - x^2)\,dx = 1$

$k\left[2x^2 - \frac{x^3}{3}\right]_0^4 = 1$

$k\left[\left(2(16) - \frac{64}{3}\right) - 0\right] = 1$

$k\left[32 - \frac{64}{3}\right] = 1$

$k\left[\frac{96 - 64}{3}\right] = 1$

$k \times \frac{32}{3} = 1$

$k = \frac{3}{32} \quad \text{✓ shown}$

Marking: M1 for setting up the integral equal to 1; M1 for correct integration; A1 for showing $k = \frac{3}{32}$ .

(b) $\mathrm{E}(X) = \int_0^4 x \cdot \frac{3}{32}x(4-x)\,dx = \frac{3}{32}\int_0^4 (4x^2 - x^3)\,dx$

$= \frac{3}{32}\left[\frac{4x^3}{3} - \frac{x^4}{4}\right]_0^4$

$= \frac{3}{32}\left[\frac{4(64)}{3} - \frac{256}{4}\right]$

$= \frac{3}{32}\left[\frac{256}{3} - 64\right]$

$= \frac{3}{32}\left[\frac{256 - 192}{3}\right]$

$= \frac{3}{32} \times \frac{64}{3} = \frac{64}{32} = 2$

Answer: $\mathrm{E}(X) = 2$

Marking: M1 for correct expectation integral setup; M1 for correct integration; A1 for answer 2.

Question 4 [5 marks]

$X \sim \mathrm{N}(42, 5^2)$

(a) $\mathrm{P}(38 < X < 47)$

Standardise: $Z = \dfrac{X - 42}{5}$

$\mathrm{P}(38 < X < 47) = \mathrm{P}\left(\frac{38-42}{5} < Z < \frac{47-42}{5}\right) = \mathrm{P}(-0.8 < Z < 1.0)$

$= \Phi(1.0) - \Phi(-0.8) = \Phi(1.0) - (1 - \Phi(0.8))$

$= 0.8413 - (1 - 0.7881) = 0.8413 - 0.2119 = 0.6294$

Answer: 0.629 (to 3 s.f.)

Marking: M1 for standardising; M1 for using correct probability expression; A1 for answer 0.629.

(b) $\mathrm{P}(X > 50) = \mathrm{P}\left(Z > \frac{50-42}{5}\right) = \mathrm{P}(Z > 1.6) = 1 - \Phi(1.6) = 1 - 0.9452 = 0.0548$

Expected number in 200 plants: $200 \times 0.0548 = 10.96$

Answer: 11 plants (to nearest whole number)

Marking: M1 for finding $\mathrm{P}(X > 50)$ ; M1 for multiplying by 200; A1 for answer 11.

Question 5 [6 marks]

Step 1: State hypotheses

$H_0: \mu = 6.5$ (mean daily screen time is 6.5 hours) $H_1: \mu \neq 6.5$ (mean daily screen time differs from 6.5 hours)

This is a two-tailed test at the 5% significance level.

Step 2: Test statistic

Since $n = 50$ is large, by CLT we use the $z$ -test:

$z = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{7.2 - 6.5}{2.1 / \sqrt{50}} = \frac{0.7}{0.29698} = 2.357$

Step 3: Critical value / p-value

For a two-tailed test at 5%, critical values are $z = \pm 1.96$ .

Since $2.357 > 1.96$ , we reject $H_0$ .

Alternatively, p-value $= 2 \times \mathrm{P}(Z > 2.357) = 2 \times (1 - 0.9908) = 2 \times 0.0092 = 0.0184$

Since $0.0184 < 0.05$ , we reject $H_0$ .

Step 4: Conclusion

There is sufficient evidence at the 5% significance level to conclude that the mean daily screen time of teenagers differs from 6.5 hours.

Marking:

M1: Correct hypotheses stated (two-tailed)
M1: Correct test statistic formula and substitution
A1: $z = 2.36$ (to 3 s.f.)
M1: Comparison with critical value or p-value comparison with 0.05
A1: Correct decision (reject $H_0$ )
B1: Conclusion in context

Question 6 [7 marks]

(a) Calculate the mean:

Midpoints: 9.5, 29.5, 49.5, 69.5, 90

Class	Midpoint $m$	Frequency $f$	$fm$
0–19	9.5	4	38
20–39	29.5	10	295
40–59	49.5	18	891
60–79	69.5	16	1112
80–100	90	12	1080

$\bar{x} = \frac{\sum fm}{\sum f} = \frac{38 + 295 + 891 + 1112 + 1080}{60} = \frac{3416}{60} = 56.93$

Answer: Mean = 56.9 (to 3 s.f.)

Marking: M1 for using midpoints; M1 for correct calculation; A1 for answer 56.9.

(b) The modal class is 40–59 (highest frequency of 18).

Marking: B1 for correct modal class.

(c) Histogram:

Frequency density = frequency ÷ class width. All class widths are 20.

Class	Frequency density
0–19	4/20 = 0.20
20–39	10/20 = 0.50
40–59	18/20 = 0.90
60–79	16/20 = 0.80
80–100	12/20 = 0.60

<image_placeholder> id: Q6-fig1 type: chart linked_question: Q6(c) description: Histogram with 5 bars of equal width representing the mark classes. The tallest bar is at 40-59 with frequency density 0.90. Bars are adjacent with no gaps. labels: Horizontal axis: "Mark" with boundaries at 0, 20, 40, 60, 80, 100. Vertical axis: "Frequency density" from 0 to 1.0. values: Bar heights: 0.20, 0.50, 0.90, 0.80, 0.60. must_show: All five bars with correct heights, labelled axes, class boundaries, no gaps between bars.

</image_placeholder>

Marking: M1 for calculating frequency densities; M1 for drawing bars with correct heights; A1 for fully correct histogram with labels.

Section B: Probability & Distributions (30 marks)

Question 7 [8 marks]

Total balls = 5 red + 4 blue + 3 green = 12 balls. Drawing 3 without replacement.

(a) $\mathrm{P}(\text{all 3 red}) = \frac{\binom{5}{3}}{\binom{12}{3}} = \frac{10}{220} = \frac{1}{22} = 0.0455$

Marking: M1 for using combinations; A1 for $\frac{1}{22}$ or 0.0455.

(b) $\mathrm{P}(\text{2 red, 1 blue}) = \frac{\binom{5}{2} \times \binom{4}{1}}{\binom{12}{3}} = \frac{10 \times 4}{220} = \frac{40}{220} = \frac{2}{11} = 0.182$

Marking: M1 for correct numerator (selecting 2 red from 5 AND 1 blue from 4); A1 for $\frac{2}{11}$ or 0.182.

(c) $\mathrm{P}(\text{all different colours}) = \frac{\binom{5}{1} \times \binom{4}{1} \times \binom{3}{1}}{\binom{12}{3}} = \frac{5 \times 4 \times 3}{220} = \frac{60}{220} = \frac{3}{11} = 0.273$

Marking: M1 for selecting 1 of each colour; A1 for $\frac{3}{11}$ or 0.273.

Question 8 [8 marks]

$X \sim \mathrm{Po}(4.2)$ where $X$ = number of emails per hour.

(a) $\mathrm{P}(X = 5) = \frac{e^{-4.2}(4.2)^5}{5!}$

$(4.2)^5 = 1306.91232$

$5! = 120$

$\mathrm{P}(X = 5) = \frac{e^{-4.2} \times 1306.91232}{120} = \frac{0.014996 \times 1306.91232}{120} = \frac{19.594}{120} = 0.1633$

Answer: 0.163 (to 3 s.f.)

Marking: M1 for correct Poisson formula; A1 for answer 0.163.

(b) $\mathrm{P}(X \geq 3) = 1 - \mathrm{P}(X \leq 2)$

$\mathrm{P}(X = 0) = e^{-4.2} = 0.014996$

$\mathrm{P}(X = 1) = 4.2 \times e^{-4.2} = 0.06298$

$\mathrm{P}(X = 2) = \frac{(4.2)^2}{2} \times e^{-4.2} = \frac{17.64}{2} \times 0.014996 = 8.82 \times 0.014996 = 0.13227$

$\mathrm{P}(X \leq 2) = 0.014996 + 0.06298 + 0.13227 = 0.21025$

$\mathrm{P}(X \geq 3) = 1 - 0.21025 = 0.790$

Answer: 0.790 (to 3 s.f.)

Marking: M1 for using complement; M1 for computing individual probabilities; A1 for answer 0.790.

(c) For 30 minutes, mean = $4.2 \times 0.5 = 2.1$ . Let $Y \sim \mathrm{Po}(2.1)$ .

$\mathrm{P}(Y < 2) = \mathrm{P}(Y = 0) + \mathrm{P}(Y = 1)$

$\mathrm{P}(Y = 0) = e^{-2.1} = 0.12246$

$\mathrm{P}(Y = 1) = 2.1 \times e^{-2.1} = 0.25716$

$\mathrm{P}(Y < 2) = 0.12246 + 0.25716 = 0.380$

Answer: 0.380 (to 3 s.f.)

Marking: M1 for halving the mean; M1 for computing $\mathrm{P}(Y=0) + \mathrm{P}(Y=1)$ ; A1 for answer 0.380.

Question 9 [9 marks]

$X \sim \mathrm{N}(150, \sigma^2)$

(a) $\mathrm{P}(X > 165) = 0.08$

Standardising: $\mathrm{P}\left(Z > \frac{165 - 150}{\sigma}\right) = 0.08$

$\mathrm{P}\left(Z < \frac{15}{\sigma}\right) = 0.92$

From tables, $\Phi(1.405) \approx 0.92$ , so:

$\frac{15}{\sigma} = 1.405$

$\sigma = \frac{15}{1.405} = 10.68$

Answer: $\sigma = 10.7$ g (to 3 s.f.)

Marking: M1 for standardising; M1 for using $\Phi^{-1}(0.92) \approx 1.405$ ; A1 for $\sigma = 10.7$ .

(b) $\mathrm{P}(X < 130) = \mathrm{P}\left(Z < \frac{130 - 150}{10.68}\right) = \mathrm{P}(Z < -1.873) = 1 - \Phi(1.873) = 1 - 0.9695 = 0.0305$

Answer: 0.0305 (to 3 s.f.)

Marking: M1 for standardising with found $\sigma$ ; A1 for answer 0.0305.

(c) Let $W$ = number of "small" apples in sample of 10. $W \sim \mathrm{B}(10, 0.0305)$ .

$\mathrm{P}(W \geq 2) = 1 - \mathrm{P}(W = 0) - \mathrm{P}(W = 1)$

$\mathrm{P}(W = 0) = (0.9695)^{10} = 0.7374$

$\mathrm{P}(W = 1) = 10 \times 0.0305 \times (0.9695)^9 = 10 \times 0.0305 \times 0.7606 = 0.2320$

$\mathrm{P}(W \geq 2) = 1 - 0.7374 - 0.2320 = 0.0306$

Answer: 0.0306 (to 3 s.f.)

Marking: M1 for identifying binomial with $n=10, p=0.0305$ ; M1 for using complement; A1 for answer 0.0306.

Question 10 [6 marks]

Let $X$ = number of sixes in 4 rolls. $X \sim \mathrm{B}(4, \frac{1}{6})$ .

(a) $\mathrm{P}(X = 2) = \binom{4}{2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^2 = 6 \times \frac{1}{36} \times \frac{25}{36} = \frac{150}{1296} = \frac{25}{216} = 0.1157$

Answer: 0.116 (to 3 s.f.)

Marking: M1 for correct binomial formula; A1 for answer 0.116.

(b) $\mathrm{P}(X \geq 1) = 1 - \mathrm{P}(X = 0) = 1 - \left(\frac{5}{6}\right)^4 = 1 - \frac{625}{1296} = \frac{671}{1296} = 0.5177$

Answer: 0.518 (to 3 s.f.)

Marking: M1 for using complement; A1 for answer 0.518.

Question 11 [7 marks]

$T \sim \mathrm{Exp}(\lambda)$ with mean $\mathrm{E}(T) = \frac{1}{\lambda} = 8000$ , so $\lambda = \frac{1}{8000} = 0.000125$ .

(a) $f(t) = \lambda e^{-\lambda t} = 0.000125\, e^{-0.000125t}$ for $t \geq 0$

Marking: B1 for correct PDF with correct $\lambda$ .

(b) $\mathrm{P}(T > 10000) = e^{-0.000125 \times 10000} = e^{-1.25} = 0.2865$

Answer: 0.287 (to 3 s.f.)

Marking: M1 for using survival function of exponential; A1 for answer 0.287.

(c) Let $Y$ = number of bulbs (out of 5) lasting more than 10,000 hours.

$p = 0.2865$ , $Y \sim \mathrm{B}(5, 0.2865)$

$\mathrm{P}(Y = 3) = \binom{5}{3}(0.2865)^3(1 - 0.2865)^2 = 10 \times 0.02352 \times 0.5091 = 0.1197$

Answer: 0.120 (to 3 s.f.)

Marking: M1 for identifying binomial; M1 for correct substitution; A1 for answer 0.120.

Question 12 [6 marks]

(a) $\mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B)$

$0.76 = 0.6 + 0.4 - \mathrm{P}(A \cap B)$

$\mathrm{P}(A \cap B) = 1.0 - 0.76 = 0.24$

Answer: $\mathrm{P}(A \cap B) = 0.24$

Marking: M1 for correct addition rule; A1 for answer 0.24.

(b) If independent: $\mathrm{P}(A \cap B) = \mathrm{P}(A) \times \mathrm{P}(B) = 0.6 \times 0.4 = 0.24$

Since $\mathrm{P}(A \cap B) = 0.24 = \mathrm{P}(A) \times \mathrm{P}(B)$ , yes, A and B are independent.

Marking: M1 for computing $\mathrm{P}(A) \times \mathrm{P}(B)$ ; A1 for correct conclusion with justification.

(c) $\mathrm{P}(A' \mid B) = \frac{\mathrm{P}(A' \cap B)}{\mathrm{P}(B)}$

Since $A$ and $B$ are independent, $\mathrm{P}(A' \mid B) = \mathrm{P}(A') = 1 - 0.6 = 0.4$

Alternatively: $\mathrm{P}(A' \cap B) = \mathrm{P}(B) - \mathrm{P}(A \cap B) = 0.4 - 0.24 = 0.16$

$\mathrm{P}(A' \mid B) = \frac{0.16}{0.4} = 0.4$

Answer: $\mathrm{P}(A' \mid B) = 0.4$

Marking: M1 for correct conditional probability formula or independence argument; A1 for answer 0.4.

Question 13 [5 marks]

Let $X$ = number of calls in 1 hour. $X \sim \mathrm{Po}(2.5 \times 60) = \mathrm{Po}(150)$ .

Since $\lambda = 150$ is large, use normal approximation:

$X \approx \mathrm{N}(150, 150)$

With continuity correction:

$\mathrm{P}(X < 130) = \mathrm{P}(X \leq 129) \approx \mathrm{P}\left(Z < \frac{129.5 - 150}{\sqrt{150}}\right)$

$= \mathrm{P}\left(Z < \frac{-20.5}{12.247}\right) = \mathrm{P}(Z < -1.674)$

$= 1 - \Phi(1.674) = 1 - 0.9530 = 0.0470$

Answer: 0.0470 (to 3 s.f.)

Marking:

M1: Correct Poisson mean $\lambda = 150$
M1: Normal approximation $X \approx \mathrm{N}(150, 150)$
M1: Continuity correction (using 129.5)
A1: Correct $z$ -value
A1: Final answer 0.0470

Question 14 [8 marks]

Class	Frequency	Cumulative frequency
0–9	6	6
10–19	14	20
20–29	22	42
30–39	20	62
40–49	12	74
50–59	6	80

$n = 80$

(a) Median position = $\frac{n}{2} = \frac{80}{2} = 40$ th value.

The 40th value lies in the class 20–29 (cumulative frequency reaches 42).

Using linear interpolation:

$\text{Median} = 20 + \frac{40 - 20}{22} \times 10 = 20 + \frac{20}{22} \times 10 = 20 + 9.09 = 29.09$

Answer: Median ≈ 29.1 minutes (to 3 s.f.)

Marking: M1 for identifying median class; M1 for interpolation formula; A1 for answer 29.1.

(b) Lower quartile $Q_1$ : position = $\frac{n}{4} = \frac{80}{4} = 20$ th value.

The 20th value lies at the upper boundary of class 10–19 (cumulative frequency = 20).

$Q_1 = 19.5 \text{ (or by interpolation: } 10 + \frac{20 - 6}{14} \times 10 = 10 + 10 = 20\text{)}$

Using interpolation: $Q_1 = 10 + \frac{20 - 6}{14} \times 10 = 10 + 10 = 20.0$

Upper quartile $Q_3$ : position = $\frac{3n}{4} = \frac{3 \times 80}{4} = 60$ th value.

The 60th value lies in class 30–39 (cumulative frequency reaches 62).

$Q_3 = 30 + \frac{60 - 42}{20} \times 10 = 30 + \frac{18}{20} \times 10 = 30 + 9 = 39.0$

$\mathrm{IQR} = Q_3 - Q_1 = 39.0 - 20.0 = 19.0$

Answer: IQR = 19.0 minutes

Marking: M1 for identifying $Q_1$ and $Q_3$ classes; M1 for interpolation; A1 for $Q_1$ ; A1 for $Q_3$ ; A1 for IQR = 19.0.

(c) The distribution is approximately symmetric (or very slightly right-skewed). The median (29.1) is roughly in the middle of the range, and the frequencies rise to a central peak at 20–29 then decrease in a similar pattern. $Q_3 - \text{median} = 39.0 - 29.1 = 9.9$ and $\text{median} - Q_1 = 29.1 - 20.0 = 9.1$ , which are approximately equal, suggesting approximate symmetry.

Marking: B1 for stating shape; B1 for justification using quartiles or frequency pattern.

Question 15 [7 marks]

$p = 0.05$ (probability of defective). Let $X$ = number of tests until first defective. $X \sim \mathrm{Geometric}(p = 0.05)$ .

(a) $\mathrm{P}(X = 5) = (1 - p)^{4} \times p = (0.95)^4 \times 0.05 = 0.8145 \times 0.05 = 0.0407$

Answer: 0.0407 (to 3 s.f.)

Marking: M1 for geometric distribution formula; A1 for answer 0.0407.

(b) $\mathrm{E}(X) = \frac{1}{p} = \frac{1}{0.05} = 20$

Answer: Expected number of tests = 20

Marking: B1 for correct formula and answer.

(c) Let $Y$ = number of defectives in 100 components. $Y \sim \mathrm{B}(100, 0.05)$ .

Using Poisson approximation with $\lambda = np = 100 \times 0.05 = 5$ :

$\mathrm{P}(Y = 3) \approx \frac{e^{-5}(5^3)}{3!} = \frac{0.006738 \times 125}{6} = \frac{0.84225}{6} = 0.1404$

Answer: 0.140 (to 3 s.f.)

Marking: M1 for identifying Poisson approximation with $\lambda = 5$ ; M1 for correct Poisson formula; A1 for answer 0.140.

Question 16 [6 marks]

$\mathrm{P}(X = x, Y = y) = \frac{x + y}{30}$ , for $x = 1, 2, 3$ and $y = 1, 2, 3$ .

(a) $\mathrm{P}(X = 2, Y = 3) = \frac{2 + 3}{30} = \frac{5}{30} = \frac{1}{6}$

Answer: $\frac{1}{6}$ or 0.167

Marking: B1 for correct substitution.

(b) $\mathrm{P}(X = 2) = \sum_{y=1}^{3} \mathrm{P}(X = 2, Y = y)$

$= \frac{2+1}{30} + \frac{2+2}{30} + \frac{2+3}{30} = \frac{3}{30} + \frac{4}{30} + \frac{5}{30} = \frac{12}{30} = \frac{2}{5}$

Answer: $\frac{2}{5}$ or 0.4

Marking: M1 for summing over all $y$ values; A1 for answer $\frac{2}{5}$ .

(c) First find the full marginal distribution of $X$ :

$\mathrm{P}(X = 1) = \frac{1+1}{30} + \frac{1+2}{30} + \frac{1+3}{30} = \frac{2+3+4}{30} = \frac{9}{30} = \frac{3}{10}$

$\mathrm{P}(X = 2) = \frac{12}{30} = \frac{2}{5}$ (from part b)

$\mathrm{P}(X = 3) = \frac{3+1}{30} + \frac{3+2}{30} + \frac{3+3}{30} = \frac{4+5+6}{30} = \frac{15}{30} = \frac{1}{2}$

Check: $\frac{3}{10} + \frac{4}{10} + \frac{5}{10} = \frac{12}{10}$ ... Let me recheck.

$\mathrm{P}(X=1) = \frac{9}{30}$ , $\mathrm{P}(X=2) = \frac{12}{30}$ , $\mathrm{P}(X=3) = \frac{15}{30}$

Sum: $\frac{9+12+15}{30} = \frac{36}{30} = \frac{6}{5} \neq 1$

Wait — let me verify the total probability over all 9 cells:

$\sum_{x=1}^{3}\sum_{y=1}^{3} \frac{x+y}{30} = \frac{1}{30}\sum_{x=1}^{3}\sum_{y=1}^{3}(x+y)$

For each $x$ : $\sum_{y=1}^3(x+y) = 3x + (1+2+3) = 3x + 6$

Total: $\sum_{x=1}^3(3x+6) = (3+6)+(6+6)+(9+6) = 9+12+15 = 36$

So total probability = $\frac{36}{30} = \frac{6}{5} > 1$ . This is not a valid joint probability distribution as stated.

Correction for the question: The distribution should be $\mathrm{P}(X = x, Y = y) = \frac{x + y}{36}$ for the probabilities to sum to 1.

With the corrected denominator of 36:

(a) $\mathrm{P}(X = 2, Y = 3) = \frac{2 + 3}{36} = \frac{5}{36}$

(b) $\mathrm{P}(X = 2) = \frac{3}{36} + \frac{4}{36} + \frac{5}{36} = \frac{12}{36} = \frac{1}{3}$

(c) $\mathrm{P}(X = 1) = \frac{2+3+4}{36} = \frac{9}{36} = \frac{1}{4}$

$\mathrm{P}(X = 2) = \frac{12}{36} = \frac{1}{3}$

$\mathrm{P}(X = 3) = \frac{4+5+6}{36} = \frac{15}{36} = \frac{5}{12}$

Check: $\frac{9+12+15}{36} = \frac{36}{36} = 1$ ✓

$\mathrm{E}(X) = 1 \times \frac{9}{36} + 2 \times \frac{12}{36} + 3 \times \frac{15}{36} = \frac{9 + 24 + 45}{36} = \frac{78}{36} = \frac{13}{6} = 2.167$

Answer: $\mathrm{E}(X) = \frac{13}{6}$ or 2.17 (to 3 s.f.)

Marking:

M1: Correcting the denominator to 36 (or noting the distribution must sum to 1)
M1: Finding marginal probabilities by summing over $y$
A1: Correct marginal probabilities
M1: Using $\mathrm{E}(X) = \sum x \cdot \mathrm{P}(X=x)$
A1: $\mathrm{E}(X) = \frac{13}{6}$

Note to student: Always verify that a joint probability distribution sums to 1 over all possible values. If it doesn't, there may be an error in the question or the normalising constant.

Question 17 [5 marks]

$X \sim \mathrm{N}(\mu, \sigma^2)$

$\mathrm{P}(X < 25) = 0.1587$

From standard normal tables, $\Phi(-1.00) = 0.1587$ , so:

$\frac{25 - \mu}{\sigma} = -1.00 \quad \Rightarrow \quad 25 - \mu = -\sigma \quad \Rightarrow \quad \mu - \sigma = 25 \quad \text{...(i)}$

$\mathrm{P}(X > 45) = 0.0228$

$\mathrm{P}(X < 45) = 1 - 0.0228 = 0.9772$

From tables, $\Phi(2.00) = 0.9772$ , so:

$\frac{45 - \mu}{\sigma} = 2.00 \quad \Rightarrow \quad 45 - \mu = 2\sigma \quad \text{...(ii)}$

From (i): $\mu = 25 + \sigma$

Substitute into (ii): $45 - (25 + \sigma) = 2\sigma$

$20 - \sigma = 2\sigma$

$20 = 3\sigma$

$\sigma = \frac{20}{3} = 6.667$

$\mu = 25 + \frac{20}{3} = \frac{75 + 20}{3} = \frac{95}{3} = 31.67$

Answer: $\mu = 31.7$ , $\sigma = 6.67$ (to 3 s.f.)

Marking:

M1: Converting to $z$ -scores using standard normal table values
A1: Correct $z$ -values (−1.00 and 2.00)
M1: Setting up simultaneous equations
M1: Solving the equations
A1: $\mu = 31.7$ , $\sigma = 6.67$

Question 18 [7 marks]

Sample space for sum of two dice: 36 outcomes.

Sum	Outcomes	Count
2	(1,1)	1
3	(1,2),(2,1)	2
4	(1,3),(2,2),(3,1)	3
5	(1,4),(2,3),(3,2),(4,1)	4
6	(1,5),(2,4),(3,3),(4,2),(5,1)	5
7	(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)	6
8	(2,6),(3,5),(4,4),(5,3),(6,2)	5
9	(3,6),(4,5),(5,4),(6,3)	4
10	(4,6),(5,5),(6,4)	3
11	(5,6),(6,5)	2
12	(6,6)	1

(a) $\mathrm{P}(\text{sum} = 7) = \frac{6}{36} = \frac{1}{6}$

Answer: $\frac{1}{6}$

Marking: B1 for correct probability.

(b) $\mathrm{P}(\text{sum} > 9) = \mathrm{P}(\text{sum} = 10, 11, \text{or } 12) = \frac{3 + 2 + 1}{36} = \frac{6}{36} = \frac{1}{6}$

Answer: $\frac{1}{6}$

Marking: B1 for correct probability.

(c) Let $W$ = winnings.

Outcome	Winnings	Probability
Sum = 7	$10	$\frac{6}{36}$
Sum > 9	$5	$\frac{6}{36}$
Otherwise	−$3	$\frac{24}{36}$

$\mathrm{P}(\text{otherwise}) = 1 - \frac{6}{36} - \frac{6}{36} = \frac{24}{36} = \frac{2}{3}$

$\mathrm{E}(W) = 10 \times \frac{6}{36} + 5 \times \frac{6}{36} + (-3) \times \frac{24}{36}$

$= \frac{60}{36} + \frac{30}{36} - \frac{72}{36} = \frac{18}{36} = 0.50$

Answer: Expected winnings = $0.50 per game

Marking: M1 for identifying all three outcomes and probabilities; M1 for correct expectation formula; A1 for answer $0.50.

Question 19 [7 marks]

$n = 10$ , $\sum x = 156$ , $\sum x^2 = 2478$

(a) Unbiased estimate of mean:

$\bar{x} = \frac{\sum x}{n} = \frac{156}{10} = 15.6$

Unbiased estimate of variance:

$s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right) = \frac{1}{9}\left(2478 - \frac{156^2}{10}\right)$

$= \frac{1}{9}\left(2478 - \frac{24336}{10}\right) = \frac{1}{9}(2478 - 2433.6) = \frac{1}{9}(44.4) = 4.933$

Answer: $\bar{x} = 15.6$ , $s^2 = 4.93$ (to 3 s.f.)

Marking: M1 for correct mean; M1 for correct variance formula (using $n-1$ ); A1 for $\bar{x} = 15.6$ ; A1 for $s^2 = 4.93$ .

(b) 95% confidence interval for $\mu$ :

Since $\sigma$ is unknown and $n = 10$ is small, use $t$ -distribution with $n - 1 = 9$ degrees of freedom.

$t_{0.025, 9} = 2.262$

$\text{CI} = \bar{x} \pm t_{0.025,9} \times \frac{s}{\sqrt{n}} = 15.6 \pm 2.262 \times \frac{\sqrt{4.933}}{\sqrt{10}}$

$= 15.6 \pm 2.262 \times \frac{2.221}{3.162} = 15.6 \pm 2.262 \times 0.7024$

$= 15.6 \pm 1.589$

$= (14.01, 17.19)$

Answer: 95% CI = $(14.0, 17.2)$ (to 3 s.f.)

Marking: M1 for using $t$ -distribution with 9 d.f.; M1 for correct critical value 2.262; M1 for correct standard error; A1 for correct interval.

Question 20 [6 marks]

$F(x) = \begin{cases} 0 & x < 0 \\ \dfrac{x^3}{64} & 0 \leq x \leq 4 \\ 1 & x > 4 \end{cases}$

(a) $f(x) = F'(x)$

For $0 \leq x \leq 4$ : $f(x) = \frac{d}{dx}\left(\frac{x^3}{64}\right) = \frac{3x^2}{64}$

$f(x) = \begin{cases} \dfrac{3x^2}{64} & 0 \leq x \leq 4 \\ 0 & \text{otherwise} \end{cases}$

Marking: M1 for differentiating $F(x)$ ; A1 for correct PDF.

(b) $\mathrm{P}(1 < X < 3) = F(3) - F(1) = \frac{27}{64} - \frac{1}{64} = \frac{26}{64} = \frac{13}{32} = 0.40625$

Answer: 0.406 (to 3 s.f.)

Marking: M1 for using $F(3) - F(1)$ ; A1 for answer 0.406.

(c) Median $m$ satisfies $F(m) = 0.5$ :

$\frac{m^3}{64} = 0.5$

$m^3 = 32$

$m = \sqrt[3]{32} = 2\sqrt[3]{4} = 3.1748$

Answer: Median = 3.17 (to 3 s.f.)

Marking: M1 for setting $F(m) = 0.5$ ; M1 for solving $m^3 = 32$ ; A1 for answer 3.17.

Mark Summary

Section	Marks
Section A (Questions 1–6)	30
Section B (Questions 7–20)	30
Total	60