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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: 60 Minutes
Total Marks: 60

Instructions:

Answer all 20 questions.
Write your answers in the spaces provided.
You are expected to use an approved graphing calculator. Unsupported answers from a graphing calculator are allowed unless otherwise stated.
Non-exact numerical answers must be given correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question.

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

1. A box contains 5 red balls, 3 blue balls, and 2 green balls. Three balls are drawn at random from the box without replacement. Find the probability that:
(a) all three balls are of different colours,
[2]
 (b) at least two balls are red.
[3]

2. Events $A$ and $B$ are such that $P(A) = 0.4$ , $P(B) = 0.7$ , and $P(A \cup B) = 0.85$ .
(a) Find $P(A \cap B)$ .
[1]
 (b) Determine whether events $A$ and $B$ are independent, giving a reason for your answer.
[2]
 (c) Find $P(A | B')$ .
[2]

3. The discrete random variable $X$ has the probability distribution given by:

$x$	1	2	3	4
$P(X=x)$	$0.1$	$k$	$0.3$	$0.2$

(a) Find the value of $k$ .
[1]
 (b) Calculate $E(X)$ and $Var(X)$ .
[3]
 (c) The random variable $Y$ is defined by $Y = 3X - 2$ . Find $E(Y)$ and $Var(Y)$ .
[2]

4. In a certain factory, 5% of the items produced are defective. A quality control inspector selects a random sample of 20 items.
(a) State the distribution of the number of defective items in the sample, defining any parameters used.
[1]
 (b) Find the probability that exactly 2 items are defective.
[2]
 (c) Find the probability that at least 1 item is defective.
[2]

5. A fair six-sided die is thrown repeatedly until a '6' appears. Let $N$ be the number of throws required.
(a) Write down the probability distribution of $N$ .
[1]
 (b) Find the probability that more than 3 throws are required.
[2]

6. The number of emails received by a manager per hour follows a Poisson distribution with mean 4.
(a) Find the probability that the manager receives exactly 3 emails in a given hour.
[2]
 (b) Find the probability that the manager receives fewer than 2 emails in a given hour.
[2]

7. Two independent discrete random variables $X$ and $Y$ have variances $Var(X) = 4$ and $Var(Y) = 9$ .
Find $Var(2X - 3Y + 5)$ .
[3]

Section B: Normal Distribution and Sampling (Questions 8–14)

8. The heights of adult males in a certain population are normally distributed with mean 175 cm and standard deviation 8 cm. A male is selected at random.
(a) Find the probability that his height is between 170 cm and 185 cm.
[3]
 (b) Find the height $h$ such that 10% of adult males are taller than $h$ .
[3]

9. The masses of bags of rice produced by a machine are normally distributed with mean 5.0 kg and standard deviation $\sigma$ kg. It is known that 5% of the bags have a mass of less than 4.8 kg.
(a) Find the value of $\sigma$ .
[3]
 (b) Find the probability that a randomly selected bag has a mass between 4.9 kg and 5.2 kg.
[3]

10. The lifetime of a certain type of battery is normally distributed with mean 20 hours and standard deviation 4 hours. A pack contains 4 batteries.
(a) Find the probability that the total lifetime of the 4 batteries is less than 70 hours.
[4]
 (b) Find the probability that the mean lifetime of the 4 batteries is greater than 22 hours.
[3]

11. A random sample of size 50 is taken from a population with mean $\mu$ and variance 25.
(a) State the expected value and variance of the sample mean $\bar{X}$ .
[2]
 (b) Explain why the Central Limit Theorem is not strictly necessary to determine the distribution of $\bar{X}$ if the population is known to be normal.
[1]

12. The weights of apples from an orchard are normally distributed with mean 150 g and standard deviation 20 g.
(a) Find the probability that a randomly selected apple weighs more than 160 g.
[2]
 (b) A random sample of 10 apples is selected. Find the probability that the sample mean weight is between 145 g and 155 g.
[3]

13. The time taken by students to complete a test is normally distributed with mean 45 minutes and standard deviation 10 minutes.
(a) Find the probability that a student takes more than 60 minutes.
[2]
 (b) If 100 students take the test, estimate the number of students who will take more than 60 minutes.
[1]

14. A continuous random variable $X$ has a normal distribution $N(\mu, \sigma^2)$ . Given that $P(X < 10) = 0.2$ and $P(X < 20) = 0.9$ , find the values of $\mu$ and $\sigma$ .
[4]

Section C: Hypothesis Testing and Correlation (Questions 15–20)

15. A manufacturer claims that the mean weight of their cereal boxes is 500 g. A consumer group suspects the mean weight is less than 500 g. They take a random sample of 36 boxes and find the sample mean to be 498 g. Assume the population standard deviation is known to be 12 g.
(a) State the null and alternative hypotheses.
[2]
 (b) Perform a hypothesis test at the 5% significance level. State your conclusion in context.
[4]

16. In a previous study, the mean score of students in a mathematics test was 70. A teacher introduces a new teaching method and wants to test if the mean score has changed. She takes a random sample of 25 students and finds the sample mean is 72 and the sample standard deviation is 10.
(a) State the null and alternative hypotheses.
[2]
 (b) Perform a hypothesis test at the 5% significance level. (Assume the scores are normally distributed).
[4]

17. The product moment correlation coefficient between variables $x$ and $y$ is found to be $r = -0.85$ based on a sample of size 20.
(a) Interpret the value of $r$ .
[2]
 (b) Explain why a high correlation does not imply causation.
[1]

18. The regression line of $y$ on $x$ is given by $y = 2.5 + 0.8x$ .
(a) Estimate the value of $y$ when $x = 10$ .
[1]
 (b) Explain why it might be inappropriate to use this line to estimate $y$ when $x = 100$ , if the data for $x$ ranged from 1 to 20.
[1]

19. A hypothesis test is conducted with $H_0: \mu = 50$ and $H_1: \mu \neq 50$ . The test statistic is calculated to be $z = 2.1$ .
(a) Find the p-value for this two-tailed test.
[2]
 (b) State whether $H_0$ is rejected at the 5% significance level.
[1]

20. A sample of 100 observations has a mean of 50 and a standard deviation of 5.
(a) Calculate the 95% confidence interval for the population mean.
[3]
 (b) Interpret the meaning of this confidence interval.
[2]

Answers

A-Level Maths H2 Quiz - Statistics Probability (Answer Key)

1.
(a) Total ways to choose 3 balls from 10: $\binom{10}{3} = 120$ .
Ways to choose 1 Red, 1 Blue, 1 Green: $\binom{5}{1}\binom{3}{1}\binom{2}{1} = 5 \times 3 \times 2 = 30$ .
$P(\text{different colours}) = \frac{30}{120} = \frac{1}{4} = 0.25$ .
[2]

(b) Let $R$ be the number of red balls.
$P(R \ge 2) = P(R=2) + P(R=3)$ .
$P(R=2) = \frac{\binom{5}{2}\binom{5}{1}}{\binom{10}{3}} = \frac{10 \times 5}{120} = \frac{50}{120}$ .
$P(R=3) = \frac{\binom{5}{3}\binom{5}{0}}{\binom{10}{3}} = \frac{10 \times 1}{120} = \frac{10}{120}$ .
$P(R \ge 2) = \frac{60}{120} = 0.5$ .
[3]

2.
(a) $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ .
$0.85 = 0.4 + 0.7 - P(A \cap B) \implies P(A \cap B) = 1.1 - 0.85 = 0.25$ .
[1]

(b) If independent, $P(A \cap B) = P(A)P(B)$ .
$P(A)P(B) = 0.4 \times 0.7 = 0.28$ .
Since $0.25 \neq 0.28$ , $A$ and $B$ are not independent.
[2]

(c) $P(A | B') = \frac{P(A \cap B')}{P(B')}$ .
$P(B') = 1 - 0.7 = 0.3$ .
$P(A \cap B') = P(A) - P(A \cap B) = 0.4 - 0.25 = 0.15$ .
$P(A | B') = \frac{0.15}{0.3} = 0.5$ .
[2]

3.
(a) $\sum P(X=x) = 1 \implies 0.1 + k + 0.3 + 0.2 = 1 \implies k = 0.4$ .
[1]

(b) $E(X) = 1(0.1) + 2(0.4) + 3(0.3) + 4(0.2) = 0.1 + 0.8 + 0.9 + 0.8 = 2.6$ .
$E(X^2) = 1^2(0.1) + 2^2(0.4) + 3^2(0.3) + 4^2(0.2) = 0.1 + 1.6 + 2.7 + 3.2 = 7.6$ .
$Var(X) = E(X^2) - [E(X)]^2 = 7.6 - 2.6^2 = 7.6 - 6.76 = 0.84$ .
[3]

(c) $E(Y) = E(3X - 2) = 3E(X) - 2 = 3(2.6) - 2 = 7.8 - 2 = 5.8$ .
$Var(Y) = Var(3X - 2) = 3^2 Var(X) = 9(0.84) = 7.56$ .
[2]

4.
(a) Let $D$ be the number of defective items. $D \sim B(20, 0.05)$ .
[1]

(b) $P(D=2) = \binom{20}{2}(0.05)^2(0.95)^{18} \approx 0.1887$ .
[2]

5.
(a) $N$ follows a Geometric distribution. $P(N=n) = (5/6)^{n-1}(1/6)$ for $n=1, 2, \dots$
[1]

(b) $P(N > 3) = P(\text{first 3 are not 6}) = (5/6)^3 = \frac{125}{216} \approx 0.579$ .
[2]

6.
Let $X$ be the number of emails. $X \sim Po(4)$ .
(a) $P(X=3) = \frac{e^{-4} 4^3}{3!} = \frac{64 e^{-4}}{6} \approx 0.1954$ .
[2]

(b) $P(X < 2) = P(X=0) + P(X=1) = e^{-4} + 4e^{-4} = 5e^{-4} \approx 0.0916$ .
[2]

7.
$Var(2X - 3Y + 5) = 2^2 Var(X) + (-3)^2 Var(Y)$ (since $X, Y$ independent and constant 5 has variance 0).
$= 4(4) + 9(9) = 16 + 81 = 97$ .
[3]

8.
Let $H$ be height. $H \sim N(175, 8^2)$ .
(a) $P(170 < H < 185) = P(\frac{170-175}{8} < Z < \frac{185-175}{8}) = P(-0.625 < Z < 1.25)$ .
Using GC: normalcdf(-0.625, 1.25, 0, 1) $\approx 0.8891 - 0.2660 = 0.6231$ .
[3]

(b) $P(H > h) = 0.1 \implies P(H < h) = 0.9$ .
$z_{0.9} \approx 1.2816$ .
$h = 175 + 1.2816(8) \approx 185.25$ cm.
[3]

9.
Let $M$ be mass. $M \sim N(5.0, \sigma^2)$ .
(a) $P(M < 4.8) = 0.05$ .
$Z = \frac{4.8 - 5.0}{\sigma}$ . From tables, $P(Z < -1.645) = 0.05$ .
$\frac{-0.2}{\sigma} = -1.645 \implies \sigma = \frac{0.2}{1.645} \approx 0.1216$ .
[3]

(b) $P(4.9 < M < 5.2)$ .
$Z_1 = \frac{4.9 - 5.0}{0.1216} \approx -0.822$ .
$Z_2 = \frac{5.2 - 5.0}{0.1216} \approx 1.645$ .
$P(-0.822 < Z < 1.645) \approx 0.9500 - 0.2055 = 0.7445$ .
[3]

10.
Let $L_i$ be lifetime of battery $i$ . $L_i \sim N(20, 4^2)$ .
(a) Let $S = \sum_{i=1}^4 L_i$ . $E(S) = 4(20) = 80$ . $Var(S) = 4(4^2) = 64$ . $SD(S) = 8$ .
$S \sim N(80, 64)$ .
$P(S < 70) = P(Z < \frac{70-80}{8}) = P(Z < -1.25) \approx 0.1056$ .
[4]

(b) Let $\bar{L} = \frac{S}{4}$ . $E(\bar{L}) = 20$ . $Var(\bar{L}) = \frac{16}{4} = 4$ . $SD(\bar{L}) = 2$ .
$\bar{L} \sim N(20, 4)$ .
$P(\bar{L} > 22) = P(Z > \frac{22-20}{2}) = P(Z > 1) \approx 0.1587$ .
[3]

11.
(a) $E(\bar{X}) = \mu$ . $Var(\bar{X}) = \frac{25}{50} = 0.5$ .
[2]

(b) If the population is normal, the sample mean $\bar{X}$ is exactly normally distributed for any sample size $n$ . The CLT is an approximation for large $n$ when the population distribution is unknown or non-normal.
[1]

12.
Let $W$ be weight. $W \sim N(150, 20^2)$ .
(a) $P(W > 160) = P(Z > \frac{160-150}{20}) = P(Z > 0.5) \approx 0.3085$ .
[2]

(b) $\bar{W} \sim N(150, \frac{20^2}{10}) = N(150, 40)$ . $SD(\bar{W}) = \sqrt{40} \approx 6.325$ .
$P(145 < \bar{W} < 155) = P(\frac{145-150}{6.325} < Z < \frac{155-150}{6.325}) = P(-0.79 < Z < 0.79)$ .
$\approx 0.7852 - 0.2148 = 0.5704$ .
[3]

13.
Let $T$ be time. $T \sim N(45, 10^2)$ .
(a) $P(T > 60) = P(Z > \frac{60-45}{10}) = P(Z > 1.5) \approx 0.0668$ .
[2]

(b) Expected number $= 100 \times 0.0668 \approx 7$ students.
[1]

14.
$P(X < 10) = 0.2 \implies \frac{10-\mu}{\sigma} = -0.8416$ (1).
$P(X < 20) = 0.9 \implies \frac{20-\mu}{\sigma} = 1.2816$ (2).
From (1): $10 - \mu = -0.8416 \sigma \implies \mu = 10 + 0.8416 \sigma$ .
Sub into (2): $\frac{20 - (10 + 0.8416 \sigma)}{\sigma} = 1.2816$ .
$\frac{10 - 0.8416 \sigma}{\sigma} = 1.2816 \implies 10 = 2.1232 \sigma \implies \sigma \approx 4.71$ .
$\mu = 10 + 0.8416(4.71) \approx 13.96$ .
[4]

15.
(a) $H_0: \mu = 500$ . $H_1: \mu < 500$ .
[2]

(b) Test statistic $Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} = \frac{498 - 500}{12/\sqrt{36}} = \frac{-2}{2} = -1$ .
Critical value for 1-tail 5%: $-1.645$ .
Since $-1 > -1.645$ , we do not reject $H_0$ .
Conclusion: There is insufficient evidence at the 5% level to suggest the mean weight is less than 500 g.
[4]

16.
(a) $H_0: \mu = 70$ . $H_1: \mu \neq 70$ .
[2]

(b) Since $\sigma$ is unknown and $n < 30$ (though $n=25$ is borderline, t-test is appropriate), use t-test.
$t = \frac{\bar{x} - \mu}{s/\sqrt{n}} = \frac{72 - 70}{10/\sqrt{25}} = \frac{2}{2} = 1$ .
Degrees of freedom $= 24$ .
Critical value for 2-tail 5%: $t_{0.025, 24} \approx 2.064$ .
Since $1 < 2.064$ , we do not reject $H_0$ .
Conclusion: There is insufficient evidence to suggest the mean score has changed.
[4]

17.
(a) Strong negative linear correlation between $x$ and $y$ .
[2]

(b) Correlation measures association, not cause. A third variable could influence both, or the relationship could be coincidental.
[1]

18.
(a) $y = 2.5 + 0.8(10) = 10.5$ .
[1]

(b) $x=100$ is far outside the range of the data (extrapolation). The linear relationship may not hold outside the observed range.
[1]

19.
(a) p-value $= 2 \times P(Z > 2.1) = 2 \times (1 - 0.9821) = 2 \times 0.0179 = 0.0358$ .
[2]

(b) Since $0.0358 < 0.05$ , reject $H_0$ .
[1]

20.
(a) 95% CI: $\bar{x} \pm z_{0.025} \frac{s}{\sqrt{n}}$ .
$50 \pm 1.96 \frac{5}{\sqrt{100}} = 50 \pm 1.96(0.5) = 50 \pm 0.98$ .
Interval: $(49.02, 50.98)$ .
[3]

(b) We are 95% confident that the true population mean lies within this interval.
[2]

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