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A Level H2 Mathematics Statistics Probability Quiz

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Questions

A-Level Maths H2 Quiz - Statistics Probability

Name: ____________________
Class: ____________________
Date: ____________________
Score: ______ / 60

Duration: 90 minutes
Total Marks: 60

Instructions:

Answer ALL questions.
Show all working clearly. Unsupported answers may not receive full credit.
An approved graphing calculator (without CAS) may be used unless otherwise stated.
Give non-exact answers to 3 significant figures unless otherwise stated.
The number of marks available is shown in brackets [ ] at the end of each question or part-question.

Section A: Discrete Random Variables & Probability Distributions (Questions 1–7)

1. The discrete random variable $X$ has the probability distribution given in the table below.

$x$	0	1	2	3	4
$\mathrm{P}(X = x)$	$a$	$0.1$	$0.3$	$b$	$0.2$

Given that $\mathrm{E}(X) = 2.3$ , find the values of $a$ and $b$ .

[5]

2. A discrete random variable $Y$ has probability function

$\mathrm{P}(Y = y) = \begin{cases} c(y+1) & y = 0, 1, 2, 3, 4 \\ 0 & \text{otherwise} \end{cases}$

(a) Show that $c = \frac{1}{15}$ .
[2]

(b) Find $\mathrm{E}(Y)$ and $\mathrm{Var}(Y)$ .
[4]

3. The random variable $W$ has a binomial distribution with parameters $n = 8$ and $p = 0.3$ .

(a) Find $\mathrm{P}(W = 3)$ .
[2]

(b) Find $\mathrm{E}(W)$ and $\mathrm{Var}(W)$ .
[2]

(c) State, with a reason, whether a normal approximation would be appropriate for a binomial distribution with $n = 80$ and $p = 0.3$ .
[2]

4. The probability that a randomly selected student passes a driving test on any attempt is $0.6$ . Students attempt the test independently.

(a) Find the probability that a student passes the driving test on their third attempt.
[2]

(b) Find the mean and variance of the number of attempts needed for a student to pass.
[3]

5. A fair six-sided die is rolled 5 times. Let $X$ be the number of times a prime number (2, 3, or 5) appears.

(a) State the distribution of $X$ .
[1]

(b) Find $\mathrm{P}(X \geq 4)$ .
[3]

6. The discrete random variable $V$ has the following probability distribution:

$v$	$-1$	0	2	5
$\mathrm{P}(V = v)$	$0.2$	$0.3$	$0.4$	$0.1$

Find:

(a) $\mathrm{E}(V)$
[1]

(b) $\mathrm{Var}(V)$
[2]

7. A bag contains 4 red balls and 6 blue balls. Three balls are drawn at random without replacement. Let $X$ be the number of red balls drawn.

(a) Find the probability distribution of $X$ .
[3]

(b) Find $\mathrm{E}(X)$ and $\mathrm{Var}(X)$ .
[3]

Section B: Continuous Random Variables & Normal Distribution (Questions 8–14)

8. The continuous random variable $X$ has probability density function

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

(a) Show that $k = \frac{3}{32}$ .
[2]

(b) Find $\mathrm{E}(X)$ .
[3]

9. The masses of a certain type of apple are normally distributed with mean $150$ g and standard deviation $20$ g.

(a) Find the probability that a randomly selected apple has mass between $140$ g and $170$ g.
[3]

(b) Find the value of $m$ such that $\mathrm{P}(X < m) = 0.85$ .
[3]

(c) A random sample of 5 apples is selected. Find the probability that exactly 3 of them have mass greater than $160$ g.
[4]

10. The random variable $X \sim \mathrm{N}(\mu, \sigma^2)$ . It is given that $\mathrm{P}(X > 40) = 0.1151$ and $\mathrm{P}(X < 20) = 0.0359$ .

(a) Find $\mu$ and $\sigma$ .
[5]

(b) Find $\mathrm{P}(25 < X < 35)$ .
[2]

11. The time taken, in minutes, for a student to complete a maths problem is normally distributed with mean $12$ and variance $4$ .

(a) Find the probability that a randomly selected student takes more than $15$ minutes.
[3]

(b) In a class of 25 students, find the probability that the mean time taken is less than $12.5$ minutes.
[3]

(c) A second student has completion time $Y \sim \mathrm{N}(10, 9)$ . Assuming independence, find the probability that the total time for both students exceeds $25$ minutes.
[3]

12. The heights of adult males in a country are normally distributed with mean $172$ cm and standard deviation $8$ cm.

(a) A man is selected at random. Find the probability that his height is between $165$ cm and $180$ cm.
[3]

(b) In a random sample of 10 men, find the probability that at least 7 have heights between $165$ cm and $180$ cm.
[4]

(c) A doorway is to be designed so that 99% of adult males can pass through without ducking. Find the minimum height of the doorway.
[3]

13. The random variable $X \sim \mathrm{N}(50, 16)$ . A random sample of 25 observations is taken.

(a) State the distribution of the sample mean $\bar{X}$ .
[2]

(b) Find $\mathrm{P}(\bar{X} > 51.5)$ .
[3]

14. The weights of packages from a production line are normally distributed. A random sample of 16 packages is selected and the sample mean weight is $2.5$ kg. The population standard deviation is known to be $0.4$ kg.

(a) Find a 95% confidence interval for the population mean weight.
[3]

(b) Find a 99% confidence interval for the population mean weight.
[3]

Section C: Sampling, Estimation & Hypothesis Testing (Questions 15–20)

15. A random sample of 50 students was taken and their scores on a statistics test were recorded. The results are summarised as follows:

$\sum x = 3250, \qquad \sum x^2 = 212{,}400$

(a) Calculate the unbiased estimates of the population mean and variance.
[3]

(b) Explain why the sample mean is an unbiased estimator of the population mean.
[2]

16. A machine fills bags of sugar. The weight of sugar in a bag is normally distributed with standard deviation $5$ g. A random sample of 25 bags is selected and the mean weight is found to be $502$ g.

(a) Test, at the 5% significance level, whether the population mean weight differs from $500$ g. State your hypotheses clearly.
[5]

(b) Find the set of values of the sample mean for which the null hypothesis in part (a) would be rejected at the 5% significance level.
[3]

17. A manufacturer claims that the mean lifetime of its light bulbs is $1200$ hours. A consumer group suspects the mean lifetime is less than claimed. A random sample of 36 light bulbs is tested and the sample mean lifetime is found to be $1175$ hours. Assume the population standard deviation is $100$ hours.

(a) State suitable hypotheses and carry out a test at the 2% significance level.
[5]

(b) State, with a reason, whether it would have made any difference to your conclusion if a two-tailed test had been used at the 5% significance level.
[2]

18. A researcher wishes to estimate the mean height of a certain species of plant. From previous studies, the standard deviation of heights is known to be $15$ cm. The researcher wants a 95% confidence interval with a margin of error of at most $3$ cm.

(a) Calculate the minimum sample size required.
[3]

(b) The researcher takes a random sample of 100 plants and obtains a sample mean of $85$ cm. Construct a 95% confidence interval for the population mean height.
[2]

(c) The researcher claims that the population mean height is $88$ cm. Comment on this claim using your confidence interval from part (b).
[1]

19. A school principal claims that the mean score of students in a mathematics exam is at least 65. A random sample of 40 students is taken and their scores are recorded. The sample mean is $62$ and the sample standard deviation is $12$ .

(a) Carry out a hypothesis test at the 5% significance level to determine whether there is evidence to reject the principal's claim. Assume the population is normally distributed.
[6]

(b) State, with a reason, whether the Central Limit Theorem was needed in your test.
[1]

20. A company produces metal rods whose lengths are normally distributed with mean $\mu$ cm and standard deviation $2$ cm. A random sample of 9 rods is selected and their lengths (in cm) are:

$48.5,\ 51.2,\ 49.8,\ 50.3,\ 52.1,\ 49.0,\ 50.7,\ 48.9,\ 51.5$

(a) Calculate the sample mean and the unbiased estimate of the population variance.
[3]

(b) Construct a 90% confidence interval for $\mu$ .
[4]

(c) The company claims that the mean length is $50$ cm. Using your confidence interval from part (b), comment on this claim.
[1]

END OF QUIZ

Answers

A-Level Maths H2 Quiz - Statistics Probability

Answer Key

Question 1 [5]

Key concept: For any discrete probability distribution, (i) all probabilities sum to 1, and (ii) $\mathrm{E}(X) = \sum x \cdot \mathrm{P}(X=x)$ .

Step 1: Use $\sum \mathrm{P}(X=x) = 1$ : $a + 0.1 + 0.3 + b + 0.2 = 1$ $a + b = 0.4 \quad \text{...(i)}$

Step 2: Use $\mathrm{E}(X) = 2.3$ : $0(a) + 1(0.1) + 2(0.3) + 3(b) + 4(0.2) = 2.3$ $0.1 + 0.6 + 3b + 0.8 = 2.3$ $3b = 0.8$ $b = \frac{4}{15} \approx 0.2667$

Step 3: Substitute into (i): $a = 0.4 - \frac{4}{15} = \frac{6}{15} - \frac{4}{15} = \frac{2}{15} \approx 0.1333$

Answer: $a = \frac{2}{15}$ , $b = \frac{4}{15}$

Marking: 1 mark for probability sum equation, 1 mark for expectation equation, 1 mark for solving $b$ , 1 mark for solving $a$ , 1 mark for both final answers.

Common mistake: Students may forget that probabilities must sum to 1, or may make an arithmetic error in computing $\mathrm{E}(X)$ .

Question 2 [6]

(a) [2]

Key concept: The sum of all probabilities in a probability function must equal 1.

$\sum_{y=0}^{4} \mathrm{P}(Y = y) = 1$ $c(0+1) + c(1+1) + c(2+1) + c(3+1) + c(4+1) = 1$ $c(1 + 2 + 3 + 4 + 5) = 1$ $15c = 1$ $c = \frac{1}{15} \quad \text{(shown)}$

(b) [4]

Key concept: $\mathrm{E}(Y) = \sum y \cdot \mathrm{P}(Y=y)$ and $\mathrm{Var}(Y) = \mathrm{E}(Y^2) - [\mathrm{E}(Y)]^2$ .

Finding $\mathrm{E}(Y)$ : $\mathrm{E}(Y) = \sum_{y=0}^{4} y \cdot \frac{y+1}{15} = \frac{1}{15}\sum_{y=0}^{4} y(y+1)$ $= \frac{1}{15}[0(1) + 1(2) + 2(3) + 3(4) + 4(5)]$ $= \frac{1}{15}[0 + 2 + 6 + 12 + 20] = \frac{40}{15} = \frac{8}{3}$

Finding $\mathrm{E}(Y^2)$ : $\mathrm{E}(Y^2) = \sum_{y=0}^{4} y^2 \cdot \frac{y+1}{15} = \frac{1}{15}\sum_{y=0}^{4} y^2(y+1)$ $= \frac{1}{15}[0 + 1(2) + 4(3) + 9(4) + 16(5)]$ $= \frac{1}{15}[0 + 2 + 12 + 36 + 80] = \frac{130}{15} = \frac{26}{3}$

Finding $\mathrm{Var}(Y)$ : $\mathrm{Var}(Y) = \frac{26}{3} - \left(\frac{8}{3}\right)^2 = \frac{26}{3} - \frac{64}{9} = \frac{78 - 64}{9} = \frac{14}{9}$

Answer: $\mathrm{E}(Y) = \frac{8}{3}$ , $\mathrm{Var}(Y) = \frac{14}{9}$

Marking (a): 1 mark for summing probabilities, 1 mark for correct $c$ .
Marking (b): 1 mark for $\mathrm{E}(Y)$ formula/substitution, 1 mark for $\mathrm{E}(Y)$ value, 1 mark for $\mathrm{E}(Y^2)$ and $\mathrm{Var}(Y)$ formula, 1 mark for correct $\mathrm{Var}(Y)$ .

Question 3 [6]

(a) [2]

$W \sim \mathrm{B}(8, 0.3)$

$\mathrm{P}(W = 3) = \binom{8}{3}(0.3)^3(0.7)^5 = 56 \times 0.027 \times 0.16807 = 0.2541 \quad (4 \text{ s.f.})$

(b) [2]

For $W \sim \mathrm{B}(n, p)$ : $\mathrm{E}(W) = np = 8 \times 0.3 = 2.4$

$\mathrm{Var}(W) = np(1-p) = 8 \times 0.3 \times 0.7 = 1.68$

(c) [2]

For $n = 80$ , $p = 0.3$ : $np = 24$ and $n(1-p) = 56$ . Both are greater than 5, so a normal approximation would be appropriate.

Answer: (a) 0.2541 (b) $\mathrm{E}(W) = 2.4$ , $\mathrm{Var}(W) = 1.68$ (c) Yes, because $np = 24 > 5$ and $n(1-p) = 56 > 5$ .

Marking (a): 1 mark for correct binomial formula, 1 mark for correct value.
Marking (b): 1 mark each for $\mathrm{E}(W)$ and $\mathrm{Var}(W)$ .
Marking (c): 1 mark for stating "yes/appropriate", 1 mark for checking $np > 5$ and $n(1-p) > 5$ .

Question 4 [5]

(a) [2]

This is a geometric distribution. The student passes on the 3rd attempt means: fail, fail, pass.

$\mathrm{P}(\text{pass on 3rd}) = (0.4)^2 \times 0.6 = 0.16 \times 0.6 = 0.096$

(b) [3]

For a geometric distribution with success probability $p = 0.6$ :

$\mathrm{E}(Y) = \frac{1}{p} = \frac{1}{0.6} = \frac{5}{3} \approx 1.667$

$\mathrm{Var}(Y) = \frac{1-p}{p^2} = \frac{0.4}{0.36} = \frac{10}{9} \approx 1.111$

Answer: (a) 0.096 (b) Mean $= \frac{5}{3}$ , Variance $= \frac{10}{9}$

Marking (a): 1 mark for identifying geometric distribution, 1 mark for correct answer.
Marking (b): 1 mark for mean formula, 1 mark for variance formula, 1 mark for both correct values.

Question 5 [6]

(a) [1]

Each roll: probability of prime (2, 3, or 5) $= \frac{3}{6} = 0.5$ . With 5 independent rolls:

$X \sim \mathrm{B}(5, 0.5)$

(b) [3]

$\mathrm{P}(X \geq 4) = \mathrm{P}(X = 4) + \mathrm{P}(X = 5)$ $= \binom{5}{4}(0.5)^4(0.5)^1 + \binom{5}{5}(0.5)^5$ $= 5 \times \frac{1}{32} + 1 \times \frac{1}{32} = \frac{5}{32} + \frac{1}{32} = \frac{6}{32} = \frac{3}{16} = 0.1875$

(c) [2]

$\mathrm{E}(X) = np = 5 \times 0.5 = 2.5$ $\mathrm{Var}(X) = np(1-p) = 5 \times 0.5 \times 0.5 = 1.25$

Answer: (a) $X \sim \mathrm{B}(5, 0.5)$ (b) $\frac{3}{16}$ or 0.1875 (c) $\mathrm{E}(X) = 2.5$ , $\mathrm{Var}(X) = 1.25$

Marking (a): 1 mark for correct distribution.
Marking (b): 1 mark for $\mathrm{P}(X=4)$ , 1 mark for $\mathrm{P}(X=5)$ , 1 mark for correct sum.
Marking (c): 1 mark each.

Question 6 [5]

(a) [1]

$\mathrm{E}(V) = (-1)(0.2) + (0)(0.3) + (2)(0.4) + (5)(0.1) = -0.2 + 0 + 0.8 + 0.5 = 1.1$

(b) [2]

$\mathrm{E}(V^2) = (-1)^2(0.2) + (0)^2(0.3) + (2)^2(0.4) + (5)^2(0.1) = 0.2 + 0 + 1.6 + 2.5 = 4.3$

$\mathrm{Var}(V) = \mathrm{E}(V^2) - [\mathrm{E}(V)]^2 = 4.3 - (1.1)^2 = 4.3 - 1.21 = 3.09$

(c) [2]

Using the linear transformation rules $\mathrm{E}(aV + b) = a\mathrm{E}(V) + b$ and $\mathrm{Var}(aV + b) = a^2\mathrm{Var}(V)$ :

$\mathrm{E}(3V - 4) = 3(1.1) - 4 = 3.3 - 4 = -0.7$

$\mathrm{Var}(3V - 4) = 3^2 \times 3.09 = 9 \times 3.09 = 27.81$

Answer: (a) 1.1 (b) 3.09 (c) $\mathrm{E}(3V - 4) = -0.7$ , $\mathrm{Var}(3V - 4) = 27.81$

Marking (a): 1 mark.
Marking (b): 1 mark for $\mathrm{E}(V^2)$ , 1 mark for $\mathrm{Var}(V)$ .
Marking (c): 1 mark each.

Question 7 [6]

(a) [3]

Total balls = 10. Drawing 3 without replacement. $X$ = number of red balls drawn. This is a hypergeometric distribution.

$\mathrm{P}(X = 0) = \frac{\binom{4}{0}\binom{6}{3}}{\binom{10}{3}} = \frac{1 \times 20}{120} = \frac{1}{6}$

$\mathrm{P}(X = 1) = \frac{\binom{4}{1}\binom{6}{2}}{\binom{10}{3}} = \frac{4 \times 15}{120} = \frac{60}{120} = \frac{1}{2}$

$\mathrm{P}(X = 2) = \frac{\binom{4}{2}\binom{6}{1}}{\binom{10}{3}} = \frac{6 \times 6}{120} = \frac{36}{120} = \frac{3}{10}$

$\mathrm{P}(X = 3) = \frac{\binom{4}{3}\binom{6}{0}}{\binom{10}{3}} = \frac{4 \times 1}{120} = \frac{1}{30}$

Check: $\frac{1}{6} + \frac{1}{2} + \frac{3}{10} + \frac{1}{30} = \frac{5+15+9+1}{30} = \frac{30}{30} = 1$ ✓

$x$	0	1	2	3
$\mathrm{P}(X=x)$	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{3}{10}$	$\frac{1}{30}$

(b) [3]

$\mathrm{E}(X) = 0\left(\frac{1}{6}\right) + 1\left(\frac{1}{2}\right) + 2\left(\frac{3}{10}\right) + 3\left(\frac{1}{30}\right) = 0 + \frac{1}{2} + \frac{3}{5} + \frac{1}{10} = \frac{5+6+1}{10} = \frac{12}{10} = 1.2$

(Alternatively, for hypergeometric: $\mathrm{E}(X) = \frac{nK}{N} = \frac{3 \times 4}{10} = 1.2$ )

$\mathrm{E}(X^2) = 0 + 1\left(\frac{1}{2}\right) + 4\left(\frac{3}{10}\right) + 9\left(\frac{1}{30}\right) = \frac{1}{2} + \frac{6}{5} + \frac{3}{10} = \frac{5+12+3}{10} = \frac{20}{10} = 2$

$\mathrm{Var}(X) = 2 - (1.2)^2 = 2 - 1.44 = 0.56$

Answer: (a) Distribution as table above. (b) $\mathrm{E}(X) = 1.2$ , $\mathrm{Var}(X) = 0.56$

Marking (a): 1 mark each for $\mathrm{P}(X=0)$ , $\mathrm{P}(X=1)$ , $\mathrm{P}(X=2)$ (or equivalent correct probabilities).
Marking (b): 1 mark for $\mathrm{E}(X)$ , 1 mark for $\mathrm{E}(X^2)$ , 1 mark for $\mathrm{Var}(X)$ .

Question 8 [8]

(a) [2]

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

$\int_0^4 kx(4-x)\,dx = 1$ $k\int_0^4 (4x - x^2)\,dx = 1$ $k\left[2x^2 - \frac{x^3}{3}\right]_0^4 = 1$ $k\left[\left(2(16) - \frac{64}{3}\right) - 0\right] = 1$ $k\left(32 - \frac{64}{3}\right) = 1$ $k\left(\frac{96-64}{3}\right) = 1$ $k \cdot \frac{32}{3} = 1$ $k = \frac{3}{32} \quad \text{(shown)}$

(b) [3]

$\mathrm{E}(X) = \int_0^4 x \cdot \frac{3}{32}x(4-x)\,dx = \frac{3}{32}\int_0^4 x^2(4-x)\,dx$ $= \frac{3}{32}\int_0^4 (4x^2 - x^3)\,dx = \frac{3}{32}\left[\frac{4x^3}{3} - \frac{x^4}{4}\right]_0^4$ $= \frac{3}{32}\left[\frac{4(64)}{3} - \frac{256}{4}\right] = \frac{3}{32}\left[\frac{256}{3} - 64\right]$ $= \frac{3}{32}\left[\frac{256 - 192}{3}\right] = \frac{3}{32} \times \frac{64}{3} = \frac{64}{32} = 2$

(c) [3]

The median $m$ satisfies $\int_0^m f(x)\,dx = 0.5$ :

$\frac{3}{32}\int_0^m (4x - x^2)\,dx = 0.5$ $\frac{3}{32}\left[2x^2 - \frac{x^3}{3}\right]_0^m = 0.5$ $\frac{3}{32}\left(2m^2 - \frac{m^3}{3}\right) = 0.5$ $2m^2 - \frac{m^3}{3} = \frac{16}{3}$ $6m^2 - m^3 = 16$ $m^3 - 6m^2 + 16 = 0$

Testing $m = 2$ : $8 - 24 + 16 = 0$ ✓

So $(m-2)$ is a factor. Factoring: $(m-2)(m^2 - 4m - 8) = 0$

$m = 2$ or $m = 2 \pm 2\sqrt{3}$ . Only $m = 2$ lies in $[0, 4]$ (note: $2 + 2\sqrt{3} \approx 5.46 > 4$ and $2 - 2\sqrt{3} < 0$ ).

Answer: (a) $k = \frac{3}{32}$ (shown) (b) $\mathrm{E}(X) = 2$ (c) Median $= 2$

Marking (a): 1 mark for setting up integral = 1, 1 mark for correct $k$ .
Marking (b): 1 mark for $\mathrm{E}(X)$ integral setup, 1 mark for correct integration, 1 mark for answer.
Marking (c): 1 mark for setting up equation, 1 mark for solving cubic, 1 mark for correct median.

Question 9 [10]

Let $X \sim \mathrm{N}(150, 20^2)$ .

(a) [3]

$Z = \frac{X - 150}{20}$ $\mathrm{P}(140 < X < 170) = \mathrm{P}\left(\frac{140-150}{20} < Z < \frac{170-150}{20}\right) = \mathrm{P}(-0.5 < Z < 1.0)$ $= \Phi(1.0) - \Phi(-0.5) = \Phi(1.0) - [1 - \Phi(0.5)]$ $= 0.8413 - 1 + 0.6915 = 0.5328$

(b) [3]

$\mathrm{P}(X < m) = 0.85$ , so $\Phi\left(\frac{m-150}{20}\right) = 0.85$

From tables: $\Phi(1.036) \approx 0.85$

$\frac{m - 150}{20} = 1.036$ $m = 150 + 20.72 = 170.72 \approx 171 \text{ g (3 s.f.)}$

(c) [4]

First find $\mathrm{P}(X > 160)$ for one apple:

$\mathrm{P}(X > 160) = \mathrm{P}\left(Z > \frac{160-150}{20}\right) = \mathrm{P}(Z > 0.5) = 1 - 0.6915 = 0.3085$

Let $Y$ = number of apples (out of 5) with mass $> 160$ g. Then $Y \sim \mathrm{B}(5, 0.3085)$ .

$\mathrm{P}(Y = 3) = \binom{5}{3}(0.3085)^3(1-0.3085)^2 = 10 \times 0.02936 \times 0.4782 = 0.1404$

Answer: (a) 0.5328 (b) 171 g (c) 0.140

Marking (a): 1 mark for standardising, 1 mark for using $\Phi$ , 1 mark for correct answer.
Marking (b): 1 mark for setting up equation, 1 mark for inverse $\Phi$ , 1 mark for answer.
Marking (c): 1 mark for $\mathrm{P}(X > 160)$ , 1 mark for identifying binomial, 1 mark for formula, 1 mark for answer.

Question 10 [7]

(a) [5]

$\mathrm{P}(X > 40) = 0.1151$ , so $\mathrm{P}(X \leq 40) = 0.8849$

$\Phi\left(\frac{40-\mu}{\sigma}\right) = 0.8849$ . From tables: $\Phi(1.20) = 0.8849$

$\frac{40 - \mu}{\sigma} = 1.20 \quad \text{...(i)}$

$\mathrm{P}(X < 20) = 0.0359$

$\Phi\left(\frac{20-\mu}{\sigma}\right) = 0.0359$ , so $\frac{20-\mu}{\sigma} = -1.80$ (since $\Phi(-1.80) = 1 - 0.9641 = 0.0359$ )

$\frac{20 - \mu}{\sigma} = -1.80 \quad \text{...(ii)}$

From (i): $40 - \mu = 1.20\sigma$
From (ii): $20 - \mu = -1.80\sigma$

Subtracting: $20 = 3.00\sigma$ , so $\sigma = \frac{20}{3} = 6.667$

From (i): $\mu = 40 - 1.20 \times 6.667 = 40 - 8.00 = 32.0$

(b) [2]

$\mathrm{P}(25 < X < 35) = \mathrm{P}\left(\frac{25-32}{6.667} < Z < \frac{35-32}{6.667}\right) = \mathrm{P}(-1.05 < Z < 0.45)$ $= \Phi(0.45) - \Phi(-1.05) = 0.6736 - (1 - 0.8531) = 0.6736 - 0.1469 = 0.5267$

Answer: (a) $\mu = 32.0$ , $\sigma = 6.67$ (b) 0.5267

Marking (a): 1 mark for each equation from given probabilities (2 marks), 1 mark for solving the simultaneous equations, 1 mark for $\sigma$ , 1 mark for $\mu$ .
Marking (b): 1 mark for standardising, 1 mark for answer.

Question 11 [9]

Let $X \sim \mathrm{N}(12, 4)$ , so $\sigma = 2$ .

(a) [3]

$\mathrm{P}(X > 15) = \mathrm{P}\left(Z > \frac{15-12}{2}\right) = \mathrm{P}(Z > 1.5) = 1 - 0.9332 = 0.0668$

(b) [3]

By the Central Limit Theorem, $\bar{X} \sim \mathrm{N}\left(12, \frac{4}{25}\right) = \mathrm{N}(12, 0.16)$ , so $\sigma_{\bar{X}} = \sqrt{0.16} = 0.4$ .

$\mathrm{P}(\bar{X} < 12.5) = \mathrm{P}\left(Z < \frac{12.5 - 12}{0.4}\right) = \mathrm{P}(Z < 1.25) = 0.8944$

(c) [3]

Let $T = X + Y$ be the total time. Since $X$ and $Y$ are independent normal variables:

$\mathrm{E}(T) = 12 + 10 = 22$ $\mathrm{Var}(T) = 4 + 9 = 13$ $T \sim \mathrm{N}(22, 13)$

$\mathrm{P}(T > 25) = \mathrm{P}\left(Z > \frac{25-22}{\sqrt{13}}\right) = \mathrm{P}\left(Z > \frac{3}{3.606}\right) = \mathrm{P}(Z > 0.832)$ $= 1 - 0.7972 = 0.2028$

Answer: (a) 0.0668 (b) 0.8944 (c) 0.203

Marking (a): 1 mark for standardising, 1 mark for using $\Phi$ , 1 mark for answer.
Marking (b): 1 mark for distribution of $\bar{X}$ , 1 mark for standardising, 1 mark for answer.
Marking (c): 1 mark for mean of $T$ , 1 mark for variance of $T$ , 1 mark for final answer.

Question 12 [10]

Let $X \sim \mathrm{N}(172, 8^2)$ .

(a) [3]

$\mathrm{P}(165 < X < 180) = \mathrm{P}\left(\frac{165-172}{8} < Z < \frac{180-172}{8}\right) = \mathrm{P}(-0.875 < Z < 1.0)$ $= \Phi(1.0) - \Phi(-0.875) = 0.8413 - (1 - 0.8092) = 0.8413 - 0.1908 = 0.6505$

(b) [4]

Let $p = \mathrm{P}(165 < X < 180) = 0.6505$ . Let $Y$ = number of men (out of 10) with height in range. $Y \sim \mathrm{B}(10, 0.6505)$ .

$\mathrm{P}(Y \geq 7) = \mathrm{P}(Y = 7) + \mathrm{P}(Y = 8) + \mathrm{P}(Y = 9) + \mathrm{P}(Y = 10)$

$\mathrm{P}(Y = 7) = \binom{10}{7}(0.6505)^7(0.3495)^3 = 120 \times 0.04907 \times 0.04269 = 0.2513$

$\mathrm{P}(Y = 8) = \binom{10}{8}(0.6505)^8(0.3495)^2 = 45 \times 0.03192 \times 0.1222 = 0.1755$

$\mathrm{P}(Y = 9) = \binom{10}{9}(0.6505)^9(0.3495)^1 = 10 \times 0.02076 \times 0.3495 = 0.07256$

$\mathrm{P}(Y = 10) = (0.6505)^{10} = 0.01351$

$\mathrm{P}(Y \geq 7) = 0.2513 + 0.1755 + 0.07256 + 0.01351 = 0.5129$

(c) [3]

Find $h$ such that $\mathrm{P}(X < h) = 0.99$ :

$\Phi\left(\frac{h-172}{8}\right) = 0.99$ . From tables: $\Phi(2.326) = 0.99$

$\frac{h - 172}{8} = 2.326$ $h = 172 + 18.61 = 190.6 \text{ cm}$

Answer: (a) 0.6505 (b) 0.513 (c) 191 cm (3 s.f.)

Marking (a): 1 mark for standardising, 1 mark for $\Phi$ values, 1 mark for answer.
Marking (b): 1 mark for identifying binomial, 1 mark for computing at least 2 probabilities correctly, 1 mark for summing, 1 mark for final answer.
Marking (c): 1 mark for setting up equation, 1 mark for inverse $\Phi$ , 1 mark for answer.

Question 13 [8]

$X \sim \mathrm{N}(50, 16)$ , so $\sigma = 4$ . Sample size $n = 25$ .

(a) [2]

$\bar{X} \sim \mathrm{N}\left(\mu, \frac{\sigma^2}{n}\right) = \mathrm{N}\left(50, \frac{16}{25}\right) = \mathrm{N}(50, 0.64)$

(b) [3]

$\mathrm{P}(\bar{X} > 51.5) = \mathrm{P}\left(Z > \frac{51.5 - 50}{\sqrt{0.64}}\right) = \mathrm{P}\left(Z > \frac{1.5}{0.8}\right) = \mathrm{P}(Z > 1.875)$ $= 1 - \Phi(1.875) = 1 - 0.9696 = 0.0304$

(c) [3]

$\mathrm{P}(\bar{X} < k) = 0.95$ , so $\Phi\left(\frac{k-50}{0.8}\right) = 0.95$

From tables: $\Phi(1.645) = 0.95$

$\frac{k - 50}{0.8} = 1.645$ $k = 50 + 1.316 = 51.32$

Answer: (a) $\bar{X} \sim \mathrm{N}(50, 0.64)$ (b) 0.0304 (c) 51.3

Marking (a): 1 mark for correct mean, 1 mark for correct variance.
Marking (b): 1 mark for standardising, 1 mark for $\Phi$ value, 1 mark for answer.
Marking (c): 1 mark for setting up equation, 1 mark for inverse $\Phi$ , 1 mark for answer.

Question 14 [9]

$\bar{x} = 2.5$ , $\sigma = 0.4$ , $n = 16$ . Since $\sigma$ is known, use the $z$ -interval.

(a) [3]

95% CI: $\bar{x} \pm z_{0.025} \cdot \frac{\sigma}{\sqrt{n}} = 2.5 \pm 1.96 \times \frac{0.4}{\sqrt{16}} = 2.5 \pm 1.96 \times 0.1 = 2.5 \pm 0.196$

95% CI: $(2.304, 2.696)$

(b) [3]

99% CI: $\bar{x} \pm z_{0.005} \cdot \frac{\sigma}{\sqrt{n}} = 2.5 \pm 2.576 \times 0.1 = 2.5 \pm 0.2576$

99% CI: $(2.242, 2.758)$

(c) [3]

Width $= 2 \times z_{0.025} \times \frac{\sigma}{\sqrt{n}} \leq 0.3$

$2 \times 1.96 \times \frac{0.4}{\sqrt{n}} \leq 0.3$ $\frac{1.568}{\sqrt{n}} \leq 0.3$ $\sqrt{n} \geq \frac{1.568}{0.3} = 5.227$ $n \geq 27.32$

So minimum $n = 28$ .

Answer: (a) $(2.30, 2.70)$ (b) $(2.24, 2.76)$ (c) 28

Marking (a): 1 mark for formula, 1 mark for correct $z$ -value and substitution, 1 mark for interval.
Marking (b): 1 mark for formula, 1 mark for correct $z$ -value and substitution, 1 mark for interval.
Marking (c): 1 mark for setting up inequality, 1 mark for solving, 1 mark for rounding up.

Question 15 [5]

(a) [3]

Unbiased estimate of population mean: $\bar{x} = \frac{\sum x}{n} = \frac{3250}{50} = 65$

Unbiased estimate of population variance: $s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right) = \frac{1}{49}\left(212400 - \frac{3250^2}{50}\right)$ $= \frac{1}{49}\left(212400 - \frac{10{,}562{,}500}{50}\right) = \frac{1}{49}(212400 - 211250) = \frac{1150}{49} = 23.47$

(b) [2]

The sample mean $\bar{X}$ is an unbiased estimator of the population mean $\mu$ because $\mathrm{E}(\bar{X}) = \mu$ . This follows from:

$\mathrm{E}(\bar{X}) = \mathrm{E}\left(\frac{1}{n}\sum_{i=1}^n X_i\right) = \frac{1}{n}\sum_{i=1}^n \mathrm{E}(X_i) = \frac{1}{n} \cdot n\mu = \mu$

In other words, the expected value of the sample mean equals the population mean, so on average the sample mean neither overestimates nor underestimates $\mu$ .

Answer: (a) $\bar{x} = 65$ , $s^2 = 23.5$ (b) $\mathrm{E}(\bar{X}) = \mu$ , so it is unbiased.

Marking (a): 1 mark for $\bar{x}$ , 1 mark for formula for $s^2$ , 1 mark for correct value.
Marking (b): 1 mark for stating $\mathrm{E}(\bar{X}) = \mu$ , 1 mark for explanation.

Question 16 [8]

(a) [5]

$H_0: \mu = 500$ (population mean weight is 500 g)
$H_1: \mu \neq 500$ (population mean weight differs from 500 g)

Significance level: $\alpha = 0.05$ (two-tailed)

Test statistic (since $\sigma = 5$ is known, use $z$ -test): $z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} = \frac{502 - 500}{5 / \sqrt{25}} = \frac{2}{1} = 2.00$

Critical value: $z_{0.025} = 1.96$

Since $|z| = 2.00 > 1.96$ , we reject $H_0$ .

There is sufficient evidence at the 5% significance level to conclude that the population mean weight differs from 500 g.

(b) [3]

Reject $H_0$ when $\left|\frac{\bar{x} - 500}{1}\right| > 1.96$

$|\bar{x} - 500| > 1.96$ $\bar{x} - 500 > 1.96 \quad \text{or} \quad \bar{x} - 500 < -1.96$ $\bar{x} > 501.96 \quad \text{or} \quad \bar{x} < 498.04$

Answer: (a) Reject $H_0$ ; there is evidence that the mean differs from 500 g. (b) $\bar{x} < 498.04$ or $\bar{x} > 501.96$

Marking (a): 1 mark for correct hypotheses, 1 mark for test statistic formula, 1 mark for correct $z$ -value, 1 mark for comparison with critical value, 1 mark for conclusion.
Marking (b): 1 mark for setting up inequality, 1 mark for solving, 1 mark for both critical values.

Question 17 [7]

(a) [5]

$H_0: \mu = 1200$
$H_1: \mu < 1200$ (one-tailed, since consumer group suspects less than claimed)

Significance level: $\alpha = 0.02$

Test statistic: $z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} = \frac{1175 - 1200}{100 / \sqrt{36}} = \frac{-25}{100/6} = \frac{-25}{16.667} = -1.50$

Critical value (one-tailed, lower): $z_{0.02} = -2.054$ (since $\Phi(-2.054) = 0.02$ )

Since $z = -1.50 > -2.054$ , we do not reject $H_0$ .

There is insufficient evidence at the 2% significance level to support the claim that the mean lifetime is less than 1200 hours.

(b) [2]

For a two-tailed test at 5%: $H_1: \mu \neq 1200$ , critical values $= \pm 1.96$ .

Since $|z| = 1.50 < 1.96$ , we would still not reject $H_0$ . The conclusion would be the same.

Answer: (a) Do not reject $H_0$ ; insufficient evidence. (b) Same conclusion; $|z| = 1.50 < 1.96$ .

Marking (a): 1 mark for hypotheses, 1 mark for test statistic, 1 mark for critical value, 1 mark for comparison, 1 mark for conclusion.
Marking (b): 1 mark for stating same conclusion, 1 mark for reason.

Question 18 [6]

(a) [3]

Margin of error: $E = z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}$

For 95% CI: $z_{0.025} = 1.96$ , $\sigma = 15$ , $E = 3$

$n \geq \left(\frac{z_{\alpha/2} \cdot \sigma}{E}\right)^2 = \left(\frac{1.96 \times 15}{3}\right)^2 = \left(\frac{29.4}{3}\right)^2 = (9.8)^2 = 96.04$

Minimum $n = 97$ .

(b) [2]

95% CI: $\bar{x} \pm 1.96 \times \frac{15}{\sqrt{100}} = 85 \pm 1.96 \times 1.5 = 85 \pm 2.94$

95% CI: $(82.06, 87.94)$

(c) [1]

The claimed value of 88 cm lies outside the 95% confidence interval $(82.06, 87.94)$ . This suggests that the researcher's claim of $\mu = 88$ cm is not supported by the sample data at the 95% confidence level.

Answer: (a) 97 (b) $(82.06, 87.94)$ (c) 88 is outside the CI, so the claim is not supported.

Marking (a): 1 mark for formula, 1 mark for substitution, 1 mark for rounding up.
Marking (b): 1 mark for formula, 1 mark for interval.
Marking (c): 1 mark for correct comment.

Question 19 [7]

(a) [6]

$H_0: \mu = 65$ (or $\mu \geq 65$ )
$H_1: \mu < 65$ (principal's claim is "at least 65", so we test if it is less)

Significance level: $\alpha = 0.05$ (one-tailed)

Since $\sigma$ is unknown and $n = 40$ is large, use the $t$ -distribution (or $z$ -approximation). With $s = 12$ :

$t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{62 - 65}{12 / \sqrt{40}} = \frac{-3}{12/6.325} = \frac{-3}{1.897} = -1.581$

Degrees of freedom $= 39$ . Critical value: $t_{0.05, 39} \approx -1.685$ (one-tailed, lower)

Since $t = -1.581 > -1.685$ , we do not reject $H_0$ .

There is insufficient evidence at the 5% significance level to reject the principal's claim that the mean score is at least 65.

(Note: Using $z$ -approximation, critical value $= -1.645$ , and $z = -1.581 > -1.645$ , same conclusion.)

(b) [1]

The Central Limit Theorem was not strictly needed because the population is assumed to be normally distributed. However, since $n = 40$ is large ( $>30$ ), the CLT would justify using the $z$ - or $t$ -test even if the population were not exactly normal.

Answer: (a) Do not reject $H_0$ ; insufficient evidence to reject the principal's claim. (b) Not needed since population is assumed normal, but CLT provides additional justification.

Marking (a): 1 mark for hypotheses, 1 mark for test statistic formula, 1 mark for correct value, 1 mark for critical value, 1 mark for comparison, 1 mark for conclusion.
Marking (b): 1 mark for correct statement with reason.

Question 20 [8]

(a) [3]

Sample data: $48.5, 51.2, 49.8, 50.3, 52.1, 49.0, 50.7, 48.9, 51.5$

$\sum x = 48.5 + 51.2 + 49.8 + 50.3 + 52.1 + 49.0 + 50.7 + 48.9 + 51.5 = 452.0$

$\bar{x} = \frac{452.0}{9} = 50.22$

$\sum x^2 = 48.5^2 + 51.2^2 + 49.8^2 + 50.3^2 + 52.1^2 + 49.0^2 + 50.7^2 + 48.9^2 + 51.5^2$ $= 2352.25 + 2621.44 + 2480.04 + 2530.09 + 2714.41 + 2401.00 + 2570.49 + 2391.21 + 2652.25 = 22713.18$

Unbiased estimate of population variance: $s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right) = \frac{1}{8}\left(22713.18 - \frac{452^2}{9}\right)$ $= \frac{1}{8}\left(22713.18 - \frac{204304}{9}\right) = \frac{1}{8}(22713.18 - 22700.44) = \frac{12.74}{8} = 1.593$

(b) [4]

For a 90% CI with unknown $\sigma$ , use $t$ -distribution with $n - 1 = 8$ degrees of freedom.

$t_{0.05, 8} = 1.860$

90% CI: $\bar{x} \pm t_{0.05,8} \times \frac{s}{\sqrt{n}} = 50.22 \pm 1.860 \times \frac{\sqrt{1.593}}{\sqrt{9}}$ $= 50.22 \pm 1.860 \times \frac{1.262}{3} = 50.22 \pm 1.860 \times 0.4207 = 50.22 \pm 0.7825$

90% CI: $(49.44, 51.00)$

(c) [1]

The claimed value of 50 cm lies within the 90% confidence interval $(49.44, 51.00)$ . This is consistent with the company's claim that the mean length is 50 cm.

Answer: (a) $\bar{x} = 50.2$ , $s^2 = 1.59$ (b) $(49.4, 51.0)$ (c) 50 is within the CI, so the claim is consistent with the data.

Marking (a): 1 mark for $\bar{x}$ , 1 mark for $\sum x^2$ or formula, 1 mark for $s^2$ .
Marking (b): 1 mark for correct $t$ -value, 1 mark for standard error, 1 mark for margin of error, 1 mark for interval.
Marking (c): 1 mark for correct comment.

END OF ANSWER KEY

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