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A Level H1 Mathematics Statistics Probability Quiz

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Questions

A-Level Maths H1 Quiz - Statistics Probability

Name: _________________________ Class: _________________________ Date: _________________________ Score: ______ / 50

Duration: 1 hour 15 minutes Total Marks: 50

Instructions:

Answer ALL questions.
Show all working clearly.
Where appropriate, give non-exact answers to 3 significant figures.
You may use an approved graphing calculator (GC) unless stated otherwise.
Marks are indicated in brackets [ ].

Section A: Probability (Questions 1–5)

Total: 12 marks

1. A bag contains 5 red marbles, 3 blue marbles, and 2 green marbles. Two marbles are drawn at random without replacement.

(a) Draw a probability tree diagram to represent all possible outcomes. [2]

(b) Find the probability that the two marbles drawn are of different colours. [2]

2. Events $A$ and $B$ are such that $P(A) = 0.35$ , $P(B) = 0.5$ , and $P(A \cup B) = 0.7$ .

(a) Find $P(A \cap B)$ . [1]

(b) Determine whether events $A$ and $B$ are independent. Show your reasoning. [2]

3. A committee of 4 people is to be selected from a group of 7 men and 5 women. Find the number of ways the committee can be formed if:

(a) there are no restrictions, [1]

(b) the committee must contain at least 2 women. [2]

4. A password consists of 4 distinct letters chosen from the 26 letters of the alphabet, followed by 2 distinct digits chosen from 0 to 9. Find the total number of different passwords that can be formed. [2]

Section B: Binomial and Normal Distributions (Questions 5–10)

Total: 15 marks

5. A biased coin is tossed 8 times. The probability of obtaining a head on any toss is 0.3. Find the probability of obtaining:

(a) exactly 3 heads, [1]

(b) more than 5 heads. [2]

6. A factory produces light bulbs. The probability that a bulb is defective is 0.05. A random sample of 20 bulbs is selected.

(a) State two conditions necessary for the number of defective bulbs in the sample to follow a binomial distribution. [2]

(b) Find the probability that at most 2 bulbs are defective. [1]

7. The mass of apples from an orchard is normally distributed with mean 150 g and standard deviation 12 g.

(a) Find the probability that a randomly selected apple has a mass between 140 g and 160 g. [2]

(b) Find the value of $k$ such that $P(X > k) = 0.1$ , where $X$ is the mass of a randomly selected apple. [2]

8. The random variable $X$ is normally distributed with mean $\mu$ and variance $\sigma^2$ . It is given that $P(X < 45) = 0.15$ and $P(X > 62) = 0.08$ . Find $\mu$ and $\sigma$ . [3]

9. The random variable $Y$ has mean 10 and variance 4. The random variable $W$ is defined by $W = 3Y - 5$ . Find $E(W)$ and $\text{Var}(W)$ . [2]

Section C: Sampling, Hypothesis Testing, and Regression (Questions 10–20)

Total: 23 marks

10. A population has mean $\mu$ and variance $\sigma^2$ . A random sample of size $n$ is taken from this population, and the sample mean is denoted by $\bar{X}$ .

(a) State the mean and variance of $\bar{X}$ . [1]

(b) Explain what is meant by the Central Limit Theorem and state when it applies. [2]

11. A researcher wishes to estimate the mean height of students in a school. She takes a random sample of 50 students and records their heights, $x$ cm. The following summary statistics are obtained:

$\sum x = 8250, \quad \sum x^2 = 1\,364\,500$

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

12. A company claims that the mean lifetime of its batteries is at least 120 hours. A consumer group suspects the mean lifetime is less than 120 hours. A random sample of 40 batteries is tested, and the sample mean lifetime is found to be 117.5 hours. The population standard deviation is known to be 8 hours.

(a) State appropriate null and alternative hypotheses for a hypothesis test. [1]

(b) Carry out the test at the 5% significance level. State your conclusion clearly. [3]

13. A machine fills packets of sugar. The mass of sugar in a packet is normally distributed with standard deviation 2.5 g. The manufacturer claims the mean mass is 500 g. A random sample of 10 packets is taken and the masses, $x$ grams, are summarised as follows:

$\sum x = 4970$

Test, at the 1% significance level, whether the mean mass is less than 500 g. State your hypotheses, test statistic, critical value, and conclusion clearly. [4]

14. A teacher records the number of hours, $t$ , that each of 8 students spent revising for a test, and their test scores, $s$ . The data are summarised as follows:

$n = 8, \quad \sum t = 96, \quad \sum s = 520, \quad \sum t^2 = 1280, \quad \sum s^2 = 34\,800, \quad \sum ts = 6560$

(a) Calculate the product moment correlation coefficient, $r$ , between $t$ and $s$ . [2]

(b) Interpret the value of $r$ in the context of the question. [1]

15. For the data in Question 14, find the equation of the least squares regression line of $s$ on $t$ , giving the coefficients to 3 significant figures. [2]

16. Using the regression line from Question 15, estimate the test score of a student who spent 15 hours revising. Comment on the reliability of this estimate. [2]

17. A scatter diagram of revision time ( $t$ hours) against test score ( $s$ ) is drawn. The regression line of $s$ on $t$ is plotted on the diagram. Explain why the regression line must pass through the point $(\bar{t}, \bar{s})$ . [1]

18. A survey is conducted to investigate the relationship between the number of years of education ( $e$ ) and annual income ( $I$ , in thousands of dollars) for a group of 30 employees. The regression line of $I$ on $e$ is found to be $I = 5.2e + 12.8$ .

(a) Interpret the value 5.2 in the context of the question. [1]

(b) Explain why it would not be appropriate to use this regression line to predict the income of someone with 30 years of education. [1]

19. A company wants to estimate the proportion of its customers who are satisfied with its service. It selects a simple random sample of 200 customers and finds that 150 are satisfied.

(a) Explain what is meant by a "simple random sample." [1]

(b) Calculate an unbiased estimate of the proportion of satisfied customers in the population. [1]

20. The random variables $X$ and $Y$ are independent. It is given that $E(X) = 5$ , $\text{Var}(X) = 2$ , $E(Y) = 8$ , and $\text{Var}(Y) = 3$ . Find:

(a) $E(2X - 3Y)$ , [1]

(b) $\text{Var}(2X - 3Y)$ . [2]

END OF QUIZ

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Answers

A-Level Maths H1 Quiz - Statistics Probability — Answer Key

Total Marks: 50

Section A: Probability (Questions 1–5)

1. (a) Tree diagram showing:

First draw: R (5/10), B (3/10), G (2/10)
Second draw branches with updated probabilities (without replacement)
All branches correctly labelled [2 marks]
- Award 1 mark for correct first-stage probabilities and structure
- Award 1 mark for correct second-stage conditional probabilities

(b) $P(\text{different colours}) = 1 - P(\text{same colour})$ $P(\text{same}) = P(RR) + P(BB) + P(GG)$ $= \frac{5}{10} \times \frac{4}{9} + \frac{3}{10} \times \frac{2}{9} + \frac{2}{10} \times \frac{1}{9}$ $= \frac{20}{90} + \frac{6}{90} + \frac{2}{90} = \frac{28}{90} = \frac{14}{45}$ $P(\text{different}) = 1 - \frac{14}{45} = \frac{31}{45} \approx 0.689$ [2 marks]

Award 1 mark for correct method (complement or direct calculation)
Award 1 mark for correct answer

2. (a) $P(A \cap B) = P(A) + P(B) - P(A \cup B)$ $= 0.35 + 0.5 - 0.7 = 0.15$ [1 mark]

(b) For independence: $P(A) \times P(B) = 0.35 \times 0.5 = 0.175$ Since $P(A \cap B) = 0.15 \neq 0.175$ , events $A$ and $B$ are not independent. [2 marks]

Award 1 mark for calculating $P(A) \times P(B)$
Award 1 mark for correct conclusion with comparison

3. (a) Total people = 12, choose 4: $\binom{12}{4} = 495$ [1 mark]

(b) At least 2 women means 2, 3, or 4 women: $\binom{5}{2}\binom{7}{2} + \binom{5}{3}\binom{7}{1} + \binom{5}{4}\binom{7}{0}$ $= 10 \times 21 + 10 \times 7 + 5 \times 1$ $= 210 + 70 + 5 = 285$ [2 marks]

Award 1 mark for correct cases
Award 1 mark for correct total

4. Letters: $^{26}P_4 = 26 \times 25 \times 24 \times 23 = 358\,800$ Digits: $^{10}P_2 = 10 \times 9 = 90$ Total passwords: $358\,800 \times 90 = 32\,292\,000$ [2 marks]

Award 1 mark for correct permutation for letters or digits
Award 1 mark for correct multiplication and final answer

Section B: Binomial and Normal Distributions (Questions 5–9)

5. $X \sim B(8, 0.3)$

(a) $P(X = 3) = \binom{8}{3}(0.3)^3(0.7)^5 = 56 \times 0.027 \times 0.16807 = 0.254$ (3 s.f.) [1 mark]

(b) $P(X > 5) = P(X = 6) + P(X = 7) + P(X = 8)$ $= \binom{8}{6}(0.3)^6(0.7)^2 + \binom{8}{7}(0.3)^7(0.7)^1 + \binom{8}{8}(0.3)^8(0.7)^0$ $= 28(0.000729)(0.49) + 8(0.0002187)(0.7) + 1(0.00006561)$ $= 0.0100 + 0.00122 + 0.0000656 = 0.0113$ (3 s.f.) [2 marks]

Award 1 mark for correct method (sum or $1 - P(X \leq 5)$ )
Award 1 mark for correct answer

6. (a) Conditions:

Each bulb is either defective or not (two possible outcomes).
The probability of a bulb being defective is constant (0.05) for each bulb.
The bulbs are selected independently (random sample). (Any two of the above) [2 marks]

(b) $Y \sim B(20, 0.05)$ $P(Y \leq 2) = 0.924$ (3 s.f.) [using GC] [1 mark]

7. $X \sim N(150, 12^2)$

(a) $P(140 < X < 160) = P\left(\frac{140-150}{12} < Z < \frac{160-150}{12}\right)$ $= P(-0.8333 < Z < 0.8333)$ $= 2 \times P(0 < Z < 0.8333) = 2 \times 0.2976 = 0.595$ (3 s.f.) [2 marks]

Award 1 mark for correct standardisation
Award 1 mark for correct probability

(b) $P(X > k) = 0.1 \implies P\left(Z > \frac{k-150}{12}\right) = 0.1$ $\frac{k-150}{12} = 1.28155$ (inverse normal) $k = 150 + 12(1.28155) = 165.4$ (3 s.f.) [2 marks]

Award 1 mark for correct z-value
Award 1 mark for correct k

8. $P(X < 45) = 0.15 \implies \frac{45 - \mu}{\sigma} = -1.03643$ $P(X > 62) = 0.08 \implies P(X < 62) = 0.92 \implies \frac{62 - \mu}{\sigma} = 1.40507$

Solving simultaneously: $45 - \mu = -1.03643\sigma$ ... (1) $62 - \mu = 1.40507\sigma$ ... (2)

Subtract (1) from (2): $17 = 2.4415\sigma \implies \sigma = 6.96$ (3 s.f.) Substitute into (1): $45 - \mu = -1.03643(6.963) \implies \mu = 45 + 7.217 = 52.2$ (3 s.f.) [3 marks]

Award 1 mark for each correct z-value
Award 1 mark for solving correctly

9. $E(W) = E(3Y - 5) = 3E(Y) - 5 = 3(10) - 5 = 25$ $\text{Var}(W) = \text{Var}(3Y - 5) = 3^2\text{Var}(Y) = 9 \times 4 = 36$ [2 marks]

Award 1 mark for correct mean
Award 1 mark for correct variance

Section C: Sampling, Hypothesis Testing, and Regression (Questions 10–20)

10. (a) $E(\bar{X}) = \mu$ , $\text{Var}(\bar{X}) = \frac{\sigma^2}{n}$ [1 mark]

(b) The Central Limit Theorem states that for a random sample of size $n$ from any population with mean $\mu$ and variance $\sigma^2$ , the distribution of the sample mean $\bar{X}$ is approximately normal with mean $\mu$ and variance $\frac{\sigma^2}{n}$ , provided $n$ is sufficiently large (typically $n > 30$ ). [2 marks]

Award 1 mark for stating approximate normality
Award 1 mark for stating condition $n > 30$ (or large sample)

11. $n = 50$ $\bar{x} = \frac{\sum x}{n} = \frac{8250}{50} = 165$ Unbiased estimate of $\mu$ : $\bar{x} = 165$ cm

$s^2 = \frac{1}{n-1}\left[\sum x^2 - \frac{(\sum x)^2}{n}\right]$ $= \frac{1}{49}\left[1\,364\,500 - \frac{8250^2}{50}\right]$ $= \frac{1}{49}[1\,364\,500 - 1\,361\,250]$ $= \frac{3250}{49} = 66.3$ cm² (3 s.f.) [3 marks]

Award 1 mark for correct mean
Award 1 mark for correct formula
Award 1 mark for correct variance

12. (a) $H_0: \mu = 120$ (or $\mu \geq 120$ ) $H_1: \mu < 120$ (one-tail test) [1 mark]

(b) Test statistic: $Z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}} = \frac{117.5 - 120}{8/\sqrt{40}} = \frac{-2.5}{1.2649} = -1.976$ Critical value at 5% level (one-tail): $z_{0.05} = -1.645$ Since $-1.976 < -1.645$ , we reject $H_0$ . There is sufficient evidence at the 5% level to conclude that the mean lifetime is less than 120 hours. [3 marks]

Award 1 mark for correct test statistic
Award 1 mark for correct critical value
Award 1 mark for correct conclusion in context

13. $H_0: \mu = 500$ $H_1: \mu < 500$ (one-tail test) $\bar{x} = \frac{4970}{10} = 497$ Test statistic: $Z = \frac{497 - 500}{2.5/\sqrt{10}} = \frac{-3}{0.7906} = -3.795$ Critical value at 1% level (one-tail): $z_{0.01} = -2.326$ Since $-3.795 < -2.326$ , we reject $H_0$ . There is sufficient evidence at the 1% level to conclude that the mean mass is less than 500 g. [4 marks]

Award 1 mark for correct hypotheses
Award 1 mark for correct test statistic
Award 1 mark for correct critical value
Award 1 mark for correct conclusion in context

14. (a) $r = \frac{n\sum ts - (\sum t)(\sum s)}{\sqrt{[n\sum t^2 - (\sum t)^2][n\sum s^2 - (\sum s)^2]}}$ $= \frac{8(6560) - (96)(520)}{\sqrt{[8(1280) - 96^2][8(34\,800) - 520^2]}}$ $= \frac{52\,480 - 49\,920}{\sqrt{[10\,240 - 9216][278\,400 - 270\,400]}}$ $= \frac{2560}{\sqrt{1024 \times 8000}} = \frac{2560}{\sqrt{8\,192\,000}} = \frac{2560}{2862.2} = 0.894$ (3 s.f.) [2 marks]

Award 1 mark for correct substitution
Award 1 mark for correct answer

(b) The value $r = 0.894$ is close to 1, indicating a strong positive linear correlation between revision time and test score. Students who spent more time revising tended to score higher. [1 mark]

15. $b = \frac{n\sum ts - (\sum t)(\sum s)}{n\sum t^2 - (\sum t)^2} = \frac{2560}{1024} = 2.5$ $a = \bar{s} - b\bar{t} = \frac{520}{8} - 2.5\left(\frac{96}{8}\right) = 65 - 2.5(12) = 65 - 30 = 35$ Equation: $s = 2.50t + 35.0$ (3 s.f.) [2 marks]

Award 1 mark for correct gradient
Award 1 mark for correct intercept

16. When $t = 15$ : $s = 2.50(15) + 35.0 = 37.5 + 35.0 = 72.5$ This is an interpolation since $t = 15$ lies within the range of the data (assuming $t$ ranges from approximately 8 to 16 based on $\bar{t}=12$ and $n=8$ ). The estimate is reasonably reliable as it is within the observed data range. [2 marks]

Award 1 mark for correct estimate
Award 1 mark for comment on interpolation/reliability

17. The regression line of $s$ on $t$ always passes through the point $(\bar{t}, \bar{s})$ because the equation is derived from $s - \bar{s} = b(t - \bar{t})$ . When $t = \bar{t}$ , $s = \bar{s}$ . [1 mark]

18. (a) The gradient 5.2 means that for each additional year of education, annual income is estimated to increase by $5,200 (since income is in thousands of dollars). [1 mark]

(b) 30 years of education is likely beyond the range of the data used to construct the regression line (extrapolation). The linear relationship may not hold for such extreme values, making the prediction unreliable. [1 mark]

19. (a) A simple random sample is one where every member of the population has an equal chance of being selected, and selections are made independently. [1 mark]

(b) Unbiased estimate of proportion: $\hat{p} = \frac{150}{200} = 0.75$ [1 mark]

20. (a) $E(2X - 3Y) = 2E(X) - 3E(Y) = 2(5) - 3(8) = 10 - 24 = -14$ [1 mark]

(b) Since $X$ and $Y$ are independent: $\text{Var}(2X - 3Y) = 2^2\text{Var}(X) + (-3)^2\text{Var}(Y)$ $= 4(2) + 9(3) = 8 + 27 = 35$ [2 marks]

Award 1 mark for correct formula
Award 1 mark for correct answer

END OF ANSWER KEY