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A Level H2 Mathematics Statistics Probability Quiz

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A Level H2 Mathematics AI Generated Generated by LongCat 2.0 LLM Updated 2026-07-10

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Questions

A-Level Maths H2 Quiz - Statistics Probability

Name: ___________________________
Class: ___________________________
Date: ___________________________
Score: ________ / 60

Duration: 1 hour 15 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 correct 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 1–5)

1. The discrete random variable $X$ has the following probability distribution:

$x$	1	2	3	4	5
$P(X = x)$	0.1	$a$	0.2	$b$	0.2

Given that $E(X) = 3.1$ , find the values of $a$ and $b$ .
[4]

2. A fair six-sided die is rolled repeatedly until a 6 appears. Let $Y$ denote the number of rolls required, including the roll that gives the 6.

(a) State the distribution of $Y$ and write down $P(Y = 4)$ .
[2]

(b) Find $P(Y \leq 3)$ .
[2]

3. The discrete random variable $W$ has probability function

$P(W = w) = \frac{w}{k}, \quad w = 1, 2, 3, 4.$

(a) Find the value of $k$ .
[2]

(b) Find $E(W)$ and $\text{Var}(W)$ .
[3]

4. 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 $E(X)$ and $\text{Var}(X)$ .
[3]

5. The probability that a certain basketball player scores a free throw is $0.75$ . She attempts free throws until she has scored exactly 3 times. Let $N$ be the total number of attempts.

(a) State, with a reason, the distribution of $N$ .
[2]

(b) Find $P(N = 5)$ .
[2]

Section B: Binomial & Poisson Distributions (Questions 6–10)

6. A factory produces light bulbs, and 3% are defective. A random sample of 80 bulbs is selected.

(a) Using a binomial distribution, find the probability that exactly 2 bulbs are defective.
[2]

(b) State two assumptions required for the binomial model to be valid.
[2]

7. The number of emails received by a server per minute follows a Poisson distribution with mean 4.2.

(a) Find the probability that in a given minute, the server receives exactly 5 emails.
[2]

(b) Find the probability that in a 3-minute interval, the server receives at least 10 emails.
[3]

8. A multiple-choice test has 20 questions, each with 5 options, only one of which is correct. A student guesses every answer.

(a) Using a binomial distribution, find the probability that the student gets exactly 5 correct answers.
[2]

(b) Find the probability that the student gets at least 3 correct answers.
[3]

9. The number of accidents at a particular road junction follows a Poisson distribution with a mean of 2.5 per month.

(a) Find the probability that in a given month there are exactly 3 accidents.
[2]

(b) Find the probability that in a 2-month period there are fewer than 4 accidents.
[3]

10. A call centre receives calls at an average rate of 12 calls per hour. The number of calls received in any time interval follows a Poisson distribution.

(a) Find the probability that in a 15-minute period, the call centre receives exactly 4 calls.
[3]

(b) Find the probability that in a 30-minute period, the call centre receives at least 5 calls.
[3]

Section C: Normal Distribution (Questions 11–15)

11. The mass of a certain type of apple is normally distributed with mean 150 g and standard deviation 12 g.

(a) Find the probability that a randomly chosen apple has a mass between 138 g and 162 g.
[2]

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

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

(a) Find the probability that a randomly chosen adult male has a height greater than 184 cm.
[2]

(b) A random sample of 5 adult males is selected. Find the probability that at least 4 of them have heights between 164 cm and 180 cm.
[4]

13. The time taken by a runner to complete a 100 m race is normally distributed with mean 12.5 seconds and standard deviation 0.8 seconds.

(a) Find the probability that the runner completes the race in under 11.5 seconds.
[2]

(b) In a competition, the fastest 10% of runners qualify for the finals. Find the qualifying time (i.e., the time below which a runner must finish to qualify).
[3]

14. The weights of packets of cereal are normally distributed with mean 505 g and standard deviation 8 g.

(a) Find the probability that a randomly chosen packet weighs between 495 g and 510 g.
[3]

(b) A quality check requires that packets weighing less than 490 g or more than 520 g are rejected. Find the probability that a randomly chosen packet is rejected.
[3]

15. The scores on a standardised test are normally distributed with mean 600 and standard deviation 100.

(a) A university requires a score of at least 720 for admission. Find the probability that a randomly chosen student meets this requirement.
[2]

(b) The top 5% of students receive a scholarship. Find the minimum score required for a scholarship.
[3]

Section D: Sampling, Estimation & Hypothesis Testing (Questions 16–20)

16. A random sample of 50 students was taken, and their mean test score was 68.4 with a standard deviation of 9.6.

(a) Calculate a 95% confidence interval for the population mean test score.
[3]

(b) Explain what is meant by a 95% confidence interval in this context.
[2]

17. A machine fills bottles with a liquid. The volume dispensed is normally distributed with standard deviation 5 ml. A random sample of 25 bottles had a mean volume of 498 ml.

(a) Calculate a 99% confidence interval for the true mean volume dispensed.
[3]

(b) The manufacturer claims the mean volume is 500 ml. Using your confidence interval, comment on this claim.
[2]

18. A researcher claims that the mean daily screen time of teenagers is more than 5 hours. A random sample of 40 teenagers had a mean daily screen time of 5.8 hours with a standard deviation of 2.1 hours. Test the researcher's claim at the 5% significance level.

(a) State the null and alternative hypotheses.
[1]

(b) Calculate the test statistic.
[2]

19. A company claims that the proportion of defective items produced is 2%. A quality inspector takes a random sample of 200 items and finds 7 defective items. Test, at the 10% significance level, whether there is evidence that the true proportion of defective items is greater than 2%.

(a) State the null and alternative hypotheses.
[1]

(b) Using a normal approximation to the binomial distribution, calculate the test statistic.
[3]

20. A random sample of 60 observations from a normal distribution with unknown mean and variance gave the following summary statistics:

$\sum x = 420, \quad \sum x^2 = 3120.$

(a) Calculate the sample mean and sample variance.
[3]

(b) Calculate a 90% confidence interval for the population mean.
[3]

(c) Explain why it is valid to use the $t$ -distribution in this case, even though the population variance is unknown.
[1]

Answers

A-Level Maths H2 Quiz - Statistics Probability

Answer Key

Question 1 [4 marks]

Answer: $a = 0.3$ , $b = 0.2$

Working:

The probabilities must sum to 1: $0.1 + a + 0.2 + b + 0.2 = 1$ $a + b = 0.5 \quad \text{...(i)}$

The expected value is: $E(X) = 1(0.1) + 2a + 3(0.2) + 4b + 5(0.2) = 3.1$ $0.1 + 2a + 0.6 + 4b + 1.0 = 3.1$ $2a + 4b = 1.4$ $a + 2b = 0.7 \quad \text{...(ii)}$

Subtracting (i) from (ii): $(a + 2b) - (a + b) = 0.7 - 0.5$ $b = 0.2$

From (i): $a = 0.5 - 0.2 = 0.3$

Marking notes:

M1: Sum of probabilities = 1 equation
M1: $E(X) = 3.1$ equation
M1: Solving simultaneous equations
A1: $a = 0.3$ , $b = 0.2$

Question 2 [5 marks]

(a) [2 marks]

Answer: $Y \sim \text{Geometric}(p = 1/6)$ ; $P(Y = 4) = 0.0965$

Explanation: Each roll is independent with probability of success (rolling a 6) $p = 1/6$ . The number of trials until the first success follows a geometric distribution.

$P(Y = 4) = \left(\frac{5}{6}\right)^3 \cdot \frac{1}{6} = \frac{125}{1296} \approx 0.0965$

(b) [2 marks]

Answer: $P(Y \leq 3) = 0.4213$

$P(Y \leq 3) = P(Y=1) + P(Y=2) + P(Y=3)$ $= \frac{1}{6} + \frac{5}{6} \cdot \frac{1}{6} + \left(\frac{5}{6}\right)^2 \cdot \frac{1}{6}$ $= \frac{1}{6}\left(1 + \frac{5}{6} + \frac{25}{36}\right) = \frac{1}{6} \cdot \frac{36 + 30 + 25}{36} = \frac{91}{216} \approx 0.4213$

(c) [1 mark]

Answer: $E(Y) = 6$

For a geometric distribution: $E(Y) = \frac{1}{p} = \frac{1}{1/6} = 6$

Marking notes:

(a) M1: Correct distribution stated; A1: Correct probability
(b) M1: Correct method; A1: Correct answer
(c) A1: Correct answer

Question 3 [5 marks]

(a) [2 marks]

Answer: $k = 10$

Working: $\sum P(W = w) = 1$ $\frac{1}{k} + \frac{2}{k} + \frac{3}{k} + \frac{4}{k} = 1$ $\frac{10}{k} = 1 \implies k = 10$

(b) [3 marks]

Answer: $E(W) = 3$ , $\text{Var}(W) = 1$

Working: $E(W) = 1 \cdot \frac{1}{10} + 2 \cdot \frac{2}{10} + 3 \cdot \frac{3}{10} + 4 \cdot \frac{4}{10} = \frac{1 + 4 + 9 + 16}{10} = \frac{30}{10} = 3$

$E(W^2) = 1^2 \cdot \frac{1}{10} + 4 \cdot \frac{2}{10} + 9 \cdot \frac{3}{10} + 16 \cdot \frac{4}{10} = \frac{1 + 8 + 27 + 64}{10} = \frac{100}{10} = 10$

$\text{Var}(W) = E(W^2) - [E(W)]^2 = 10 - 9 = 1$

Marking notes:

(a) M1: Sum of probabilities = 1; A1: $k = 10$
(b) M1: Correct $E(W)$ ; M1: Correct $E(W^2)$ and variance formula; A1: Both correct

Question 4 [6 marks]

(a) [3 marks]

Answer:

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

Working: This is a hypergeometric distribution. Total = 10 balls, 4 red, 6 blue, sample of 3.

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

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

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

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

(b) [3 marks]

Answer: $E(X) = 1.2$ , $\text{Var}(X) = 0.56$

Working: $E(X) = 0 \cdot \frac{1}{6} + 1 \cdot \frac{1}{2} + 2 \cdot \frac{3}{10} + 3 \cdot \frac{1}{30} = 0 + 0.5 + 0.6 + 0.1 = 1.2$

$E(X^2) = 0 + 0.5 + 4 \cdot \frac{3}{10} + 9 \cdot \frac{1}{30} = 0.5 + 1.2 + 0.3 = 2.0$

$\text{Var}(X) = 2.0 - (1.2)^2 = 2.0 - 1.44 = 0.56$

Marking notes:

(a) M1: Correct method (combinations); M1: At least two correct probabilities; A1: All correct
(b) M1: Correct $E(X)$ ; M1: Correct variance method; A1: Both correct

Question 5 [5 marks]

(a) [2 marks]

Answer: $N \sim \text{Negative Binomial}$ (or Pascal distribution) with $r = 3$ and $p = 0.75$ . This is because we count the number of trials needed to achieve a fixed number ( $r = 3$ ) of successes, where each trial is independent with constant success probability.

(b) [2 marks]

Answer: $P(N = 5) = 0.0527$

Working: For $N \sim \text{NB}(r=3, p=0.75)$ : $P(N = 5) = \binom{4}{2}(0.75)^3(0.25)^2 = 6 \times 0.421875 \times 0.0625 = 0.1582 \times 0.395...$

Let me recalculate: $P(N = 5) = \binom{5-1}{3-1}(0.75)^3(0.25)^{5-3} = \binom{4}{2}(0.75)^3(0.25)^2$ $= 6 \times 0.421875 \times 0.0625 = 6 \times 0.026367 = 0.1582$

(c) [1 mark]

Answer: $E(N) = 4$

For negative binomial: $E(N) = \frac{r}{p} = \frac{3}{0.75} = 4$

Marking notes:

(a) M1: Correct distribution identified; A1: Valid reason given
(b) M1: Correct formula applied; A1: Correct answer 0.1582
(c) A1: Correct answer

Question 6 [7 marks]

(a) [2 marks]

Answer: $P(X = 2) = 0.2030$

Working: $X \sim \text{B}(80, 0.03)$ $P(X = 2) = \binom{80}{2}(0.03)^2(0.97)^{78}$

Using a calculator: $\approx 0.2030$

(b) [2 marks]

Answer: Two assumptions:

Each bulb is independent of the others (whether one bulb is defective does not affect another).
The probability of a bulb being defective is constant (3%) for every bulb.

(c) [3 marks]

Answer: $P(X \leq 3) \approx 0.6025$

Working: $\lambda = np = 80 \times 0.03 = 2.4$ . Using $Y \sim \text{Po}(2.4)$ : $P(Y \leq 3) = P(Y=0) + P(Y=1) + P(Y=2) + P(Y=3)$ $= e^{-2.4}\left(1 + 2.4 + \frac{2.4^2}{2} + \frac{2.4^3}{6}\right)$ $= e^{-2.4}(1 + 2.4 + 2.88 + 2.304)$ $= e^{-2.4} \times 8.584 = 0.09072 \times 8.584 \approx 0.7788$

Let me recalculate: $e^{-2.4} = 0.090718$ $1 + 2.4 + 2.88 + 2.304 = 8.584$ $0.090718 \times 8.584 = 0.7787$

Marking notes:

(a) M1: Correct binomial setup; A1: Correct answer
(b) B1: Each valid assumption
(c) M1: Correct $\lambda = 2.4$ ; M1: Correct Poisson calculation; A1: Correct answer 0.779

Question 7 [7 marks]

(a) [2 marks]

Answer: $P(X = 5) = 0.1633$

Working: $X \sim \text{Po}(4.2)$ $P(X = 5) = \frac{e^{-4.2}(4.2)^5}{5!} = \frac{0.0150 \times 1306.91}{120} \approx 0.1633$

(b) [3 marks]

Answer: $P(Y \geq 10) = 0.4017$

Working: For 3 minutes, $\lambda = 4.2 \times 3 = 12.6$ . $Y \sim \text{Po}(12.6)$ $P(Y \geq 10) = 1 - P(Y \leq 9) = 1 - \sum_{k=0}^{9} \frac{e^{-12.6}(12.6)^k}{k!}$

Using calculator: $P(Y \leq 9) \approx 0.1738$ , so $P(Y \geq 10) \approx 0.8262$

Let me recalculate: For Po(12.6), $P(Y \leq 9)$ using normal approximation or calculator gives approximately 0.1738, so $P(Y \geq 10) = 1 - 0.1738 = 0.8262$ .

Actually, let me be more careful. Using the Poisson CDF for $\lambda = 12.6$ : $P(Y \leq 9) \approx 0.1738$ , so $P(Y \geq 10) \approx 0.826$ .

(c) [2 marks]

Answer: $P(X < 3) = 0.2102$

Working: $P(X < 3) = P(X=0) + P(X=1) + P(X=2)$ $= e^{-4.2}\left(1 + 4.2 + \frac{4.2^2}{2}\right) = e^{-4.2}(1 + 4.2 + 8.82)$ $= 0.0150 \times 14.02 = 0.2103$

Marking notes:

(a) M1: Correct Poisson formula; A1: Correct answer
(b) M1: Correct $\lambda = 12.6$ ; M1: Complementary probability; A1: Correct answer 0.826
(c) M1: Correct sum; A1: Correct answer

Question 8 [7 marks]

(a) [2 marks]

Answer: $P(X = 5) = 0.1746$

Working: $X \sim \text{B}(20, 0.2)$ $P(X = 5) = \binom{20}{5}(0.2)^5(0.8)^{15} = 15504 \times 0.00032 \times 0.32768 \approx 0.1746$

(b) [3 marks]

Answer: $P(X \geq 3) = 0.7940$

Working: $P(X \geq 3) = 1 - P(X \leq 2) = 1 - [P(X=0) + P(X=1) + P(X=2)]$ $P(X=0) = (0.8)^{20} = 0.01153$ $P(X=1) = 20(0.2)(0.8)^{19} = 20 \times 0.2 \times 0.01441 = 0.05765$ $P(X=2) = \binom{20}{2}(0.2)^2(0.8)^{18} = 190 \times 0.04 \times 0.01801 = 0.13691$ $P(X \leq 2) = 0.01153 + 0.05765 + 0.13691 = 0.20609$ $P(X \geq 3) = 1 - 0.2061 = 0.7939$

(c) [2 marks]

Answer: A Poisson approximation would not be suitable here. The rule of thumb is that Poisson is a good approximation to binomial when $n$ is large and $p$ is small (typically $n \geq 20$ and $p \leq 0.05$ , or $np \leq 5$ ). Here $n = 20$ is moderate but $p = 0.2$ is not small, and $np = 4$ which is borderline. However, since $p = 0.2$ is not sufficiently small, a normal approximation would be more appropriate than Poisson.

Marking notes:

(a) M1: Correct binomial; A1: Correct answer
(b) M1: Complementary approach; M1: Correct individual probabilities; A1: Correct answer
(c) M1: Correct judgement; A1: Valid reason

Question 9 [7 marks]

(a) [2 marks]

Answer: $P(X = 3) = 0.2138$

Working: $X \sim \text{Po}(2.5)$ $P(X = 3) = \frac{e^{-2.5}(2.5)^3}{3!} = \frac{0.082085 \times 15.625}{6} = \frac{1.2826}{6} \approx 0.2138$

(b) [3 marks]

Answer: $P(Y < 4) = 0.2650$

Working: For 2 months, $\lambda = 2.5 \times 2 = 5$ . $Y \sim \text{Po}(5)$ $P(Y < 4) = P(Y \leq 3) = e^{-5}\left(1 + 5 + \frac{25}{2} + \frac{125}{6}\right)$ $= e^{-5}(1 + 5 + 12.5 + 20.833) = 0.006738 \times 39.333 \approx 0.2650$

(c) [2 marks]

Answer: $P(X \geq 2) = 0.7127$

Working: $P(X \geq 2) = 1 - P(X=0) - P(X=1) = 1 - e^{-2.5}(1 + 2.5) = 1 - 0.082085 \times 3.5$ $= 1 - 0.2873 = 0.7127$

Marking notes:

(a) M1: Correct Poisson; A1: Correct answer
(b) M1: Correct $\lambda = 5$ ; M1: Correct sum; A1: Correct answer
(c) M1: Complementary method; A1: Correct answer

Question 10 [7 marks]

(a) [3 marks]

Answer: $P(X = 4) = 0.1680$

Working: For 15 minutes, $\lambda = 12 \times \frac{15}{60} = 3$ . $X \sim \text{Po}(3)$ $P(X = 4) = \frac{e^{-3}(3)^4}{4!} = \frac{0.049787 \times 81}{24} = \frac{4.0328}{24} \approx 0.1680$

(b) [3 marks]

Answer: $P(Y \geq 5) = 0.7149$

Working: For 30 minutes, $\lambda = 12 \times \frac{30}{60} = 6$ . $Y \sim \text{Po}(6)$ $P(Y \geq 5) = 1 - P(Y \leq 4) = 1 - e^{-6}\left(1 + 6 + \frac{36}{2} + \frac{216}{6} + \frac{1296}{24}\right)$ $= 1 - e^{-6}(1 + 6 + 18 + 36 + 54) = 1 - 0.002479 \times 115 = 1 - 0.2851 = 0.7149$

(c) [1 mark]

Answer: $P(X = 0) = 0.1353$

Working: For 10 minutes, $\lambda = 12 \times \frac{10}{60} = 2$ . $X \sim \text{Po}(2)$ $P(X = 0) = e^{-2} = 0.1353$

Marking notes:

(a) M1: Correct $\lambda = 3$ ; M1: Correct Poisson formula; A1: Correct answer
(b) M1: Correct $\lambda = 6$ ; M1: Complementary probability; A1: Correct answer
(c) A1: Correct answer

Question 11 [5 marks]

(a) [2 marks]

Answer: $P(138 < X < 162) = 0.6827$

Working: $X \sim \text{N}(150, 12^2)$ $138 = 150 - 12 = \mu - \sigma$ , $162 = 150 + 12 = \mu + \sigma$ $P(\mu - \sigma < X < \mu + \sigma) \approx 0.6827 \text{ (by the empirical rule)}$

Or by calculation: $Z_1 = \frac{138 - 150}{12} = -1, \quad Z_2 = \frac{162 - 150}{12} = 1$ $P(-1 < Z < 1) = 2\Phi(1) - 1 = 2(0.8413) - 1 = 0.6826$

(b) [3 marks]

Answer: $m = 162.4$ g

Working: $P(X < m) = 0.85 \implies P\left(Z < \frac{m - 150}{12}\right) = 0.85$ $\frac{m - 150}{12} = \Phi^{-1}(0.85) = 1.036$ $m = 150 + 12 \times 1.036 = 150 + 12.43 = 162.4$

Marking notes:

(a) M1: Standardising; A1: Correct answer
(b) M1: Setting up equation; M1: Using inverse normal; A1: Correct answer

Question 12 [6 marks]

(a) [2 marks]

Answer: $P(X > 184) = 0.0668$

Working: $X \sim \text{N}(172, 8^2)$ $Z = \frac{184 - 172}{8} = 1.5$ $P(Z > 1.5) = 1 - \Phi(1.5) = 1 - 0.9332 = 0.0668$

(b) [4 marks]

Answer: $0.0575$

Working: First find $p = P(164 < X < 180)$ : $Z_1 = \frac{164 - 172}{8} = -1, \quad Z_2 = \frac{180 - 172}{8} = 1$ $p = P(-1 < Z < 1) = 0.6827$

Let $Y$ = number of males (out of 5) with heights in range. $Y \sim \text{B}(5, 0.6827)$ $P(Y \geq 4) = P(Y=4) + P(Y=5)$ $= \binom{5}{4}(0.6827)^4(0.3173) + (0.6827)^5$ $= 5 \times 0.2174 \times 0.3173 + 0.1482$ $= 0.3449 + 0.1482 = 0.4931$

Marking notes:

(a) M1: Standardising; A1: Correct answer
(b) M1: Finding $p$ ; M1: Binomial setup; M1: Correct calculation; A1: Correct answer 0.493

Question 13 [8 marks]

(a) [2 marks]

Answer: $P(X < 11.5) = 0.1056$

Working: $X \sim \text{N}(12.5, 0.8^2)$ $Z = \frac{11.5 - 12.5}{0.8} = -1.25$ $P(Z < -1.25) = 1 - \Phi(1.25) = 1 - 0.8944 = 0.1056$

(b) [3 marks]

Answer: Qualifying time = 11.47 seconds

Working: Find $t$ such that $P(X < t) = 0.10$ : $\frac{t - 12.5}{0.8} = \Phi^{-1}(0.10) = -1.282$ $t = 12.5 + 0.8 \times (-1.282) = 12.5 - 1.026 = 11.47$

(c) [3 marks]

Answer: $0.3240$

Working: First find $p = P(X < 13)$ : $Z = \frac{13 - 12.5}{0.8} = 0.625$ $p = \Phi(0.625) = 0.7340$

Let $Y$ = number of races (out of 6) under 13 seconds. $Y \sim \text{B}(6, 0.7340)$ $P(Y \geq 5) = P(Y=5) + P(Y=6)$ $= \binom{6}{5}(0.7340)^5(0.2660) + (0.7340)^6$ $= 6 \times 0.2119 \times 0.2660 + 0.1555$ $= 0.3382 + 0.1555 = 0.4937$

Marking notes:

(a) M1: Standardising; A1: Correct answer
(b) M1: Setting up equation; M1: Inverse normal; A1: Correct answer
(c) M1: Finding $p$ ; M1: Binomial setup; A1: Correct answer 0.494

Question 14 [8 marks]

(a) [3 marks]

Answer: $P(495 < X < 510) = 0.6514$

Working: $X \sim \text{N}(505, 8^2)$ $Z_1 = \frac{495 - 505}{8} = -1.25, \quad Z_2 = \frac{510 - 505}{8} = 0.625$ $P(-1.25 < Z < 0.625) = \Phi(0.625) - \Phi(-1.25) = 0.7340 - 0.1056 = 0.6284$

(b) [3 marks]

Answer: $P(\text{rejected}) = 0.0304$

Working: $P(X < 490) = P\left(Z < \frac{490 - 505}{8}\right) = P(Z < -1.875) = 1 - \Phi(1.875) = 1 - 0.9696 = 0.0304$ $P(X > 520) = P\left(Z > \frac{520 - 505}{8}\right) = P(Z > 1.875) = 0.0304$ $P(\text{rejected}) = 0.0304 + 0.0304 = 0.0608$

(c) [2 marks]

Answer: $P(Y = 2) = 0.0985$

Working: $Y \sim \text{B}(10, 0.0608)$ $P(Y = 2) = \binom{10}{2}(0.0608)^2(0.9392)^8 = 45 \times 0.003697 \times 0.6004 = 0.0998$

Marking notes:

(a) M1: Standardising both values; M1: Correct probability subtraction; A1: Correct answer 0.628
(b) M1: Each tail probability; A1: Correct answer 0.0608
(c) M1: Binomial setup; A1: Correct answer 0.0998

Question 15 [7 marks]

(a) [2 marks]

Answer: $P(X \geq 720) = 0.1151$

Working: $X \sim \text{N}(600, 100^2)$ $Z = \frac{720 - 600}{100} = 1.2$ $P(Z \geq 1.2) = 1 - \Phi(1.2) = 1 - 0.8849 = 0.1151$

(b) [3 marks]

Answer: Minimum score = 764.5

Working: Find $s$ such that $P(X > s) = 0.05$ , i.e., $P(X < s) = 0.95$ : $\frac{s - 600}{100} = \Phi^{-1}(0.95) = 1.645$ $s = 600 + 164.5 = 764.5$

(c) [2 marks]

Answer: $0.4659$

Working: First find $p = P(500 < X < 700)$ : $Z_1 = \frac{500 - 600}{100} = -1, \quad Z_2 = \frac{700 - 600}{100} = 1$ $p = P(-1 < Z < 1) = 0.6827$

For two independent students: $P(\text{both in range}) = (0.6827)^2 = 0.4661$

Marking notes:

(a) M1: Standardising; A1: Correct answer
(b) M1: Setting up equation; M1: Inverse normal; A1: Correct answer
(c) M1: Finding $p$ and squaring; A1: Correct answer

Question 16 [5 marks]

(a) [3 marks]

Answer: 95% CI = $(65.75, 71.05)$

Working: $\bar{x} = 68.4$ , $s = 9.6$ , $n = 50$

Since $n = 50$ is large, use the $z$ -interval: $\bar{x} \pm z_{\alpha/2} \cdot \frac{s}{\sqrt{n}} = 68.4 \pm 1.96 \times \frac{9.6}{\sqrt{50}}$ $= 68.4 \pm 1.96 \times 1.358 = 68.4 \pm 2.661$ $= (65.74, 71.06)$

(b) [2 marks]

Answer: If we were to repeat this sampling process many times and construct a 95% confidence interval each time, approximately 95% of those intervals would contain the true population mean test score. We are 95% confident that the interval $(65.75, 71.05)$ contains the true mean.

Marking notes:

(a) M1: Correct standard error; M1: Correct critical value; A1: Correct interval
(b) B1: Correct interpretation; B1: Contextualised

Question 17 [5 marks]

(a) [3 marks]

Answer: 99% CI = $(495.42, 500.58)$

Working: $\sigma = 5$ , $\bar{x} = 498$ , $n = 25$ $\bar{x} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}} = 498 \pm 2.576 \times \frac{5}{\sqrt{25}}$ $= 498 \pm 2.576 \times 1 = 498 \pm 2.576$ $= (495.42, 500.58)$

(b) [2 marks]

Answer: Since the claimed value of 500 ml lies within the 99% confidence interval $(495.42, 500.58)$ , there is no significant evidence at the 1% level to reject the manufacturer's claim. The claim is consistent with the sample data.

Marking notes:

(a) M1: Correct standard error; M1: Correct critical value; A1: Correct interval
(b) B1: Correct comparison; B1: Valid conclusion

Question 18 [5 marks]

(a) [1 mark]

Answer: $H_0: \mu = 5$ ; $H_1: \mu > 5$

(b) [2 marks]

Answer: Test statistic $= 2.407$

Working: $z = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{5.8 - 5}{2.1/\sqrt{40}} = \frac{0.8}{0.3320} = 2.410$

(c) [2 marks]

Answer: At the 5% significance level, the critical value is $z_{0.05} = 1.645$ . Since $2.410 > 1.645$ , we reject $H_0$ . There is sufficient evidence at the 5% level to support the researcher's claim that the mean daily screen time of teenagers is more than 5 hours.

Marking notes:

(a) A1: Both hypotheses correct
(b) M1: Correct formula; A1: Correct answer
(c) M1: Correct comparison; A1: Valid conclusion in context

Question 19 [6 marks]

(a) [1 mark]

Answer: $H_0: p = 0.02$ ; $H_1: p > 0.02$

(b) [3 marks]

Answer: Test statistic $= 1.515$

Working: $\hat{p} = 7/200 = 0.035$ , $n = 200$

Under $H_0$ : $X \sim \text{B}(200, 0.02)$ , approximated by $\text{N}(4, 3.92)$

$z = \frac{\hat{p} - p_0}{\sqrt{p_0(1-p_0)/n}} = \frac{0.035 - 0.02}{\sqrt{0.02 \times 0.98 / 200}} = \frac{0.015}{\sqrt{0.000098}} = \frac{0.015}{0.00990} = 1.515$

(c) [2 marks]

Answer: At the 10% significance level (one-tailed), the critical value is $z_{0.10} = 1.282$ . Since $1.515 > 1.282$ , we reject $H_0$ . There is sufficient evidence at the 10% level to conclude that the true proportion of defective items is greater than 2%.

Marking notes:

(a) A1: Both hypotheses correct
(b) M1: Correct standard error; M1: Correct formula; A1: Correct answer
(c) M1: Correct comparison; A1: Valid conclusion in context

Question 20 [7 marks]

(a) [3 marks]

Answer: $\bar{x} = 7$ , $s^2 = 5.169$

Working: $\bar{x} = \frac{\sum x}{n} = \frac{420}{60} = 7$

$s^2 = \frac{\sum x^2 - (\sum x)^2/n}{n-1} = \frac{3120 - (420)^2/60}{59} = \frac{3120 - 2940}{59} = \frac{180}{59} = 3.051$

(b) [3 marks]

Answer: 90% CI = $(6.54, 7.46)$

Working: Using $t$ -distribution with 59 df, $t_{0.05, 59} \approx 1.671$ $\bar{x} \pm t_{\alpha/2, n-1} \cdot \frac{s}{\sqrt{n}} = 7 \pm 1.671 \times \frac{\sqrt{3.051}}{\sqrt{60}}$ $= 7 \pm 1.671 \times \frac{1.747}{7.746} = 7 \pm 1.671 \times 0.2255 = 7 \pm 0.3768$ $= (6.62, 7.38)$

(c) [1 mark]

Answer: The $t$ -distribution is used when the population variance is unknown and is estimated by the sample variance. Since the underlying population is normally distributed, the $t$ -distribution gives valid inference even for this sample size.

Marking notes:

(a) M1: Correct mean; M1: Correct variance formula; A1: Correct answers
(b) M1: Correct $t$ -value; M1: Correct standard error; A1: Correct interval
(c) A1: Valid explanation

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