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A Level H1 Mathematics Statistics Probability Quiz

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A-Level Maths H1 Quiz - Statistics Probability

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

Duration: 1 hour 15 minutes
Total Marks: 60

Instructions:

Answer ALL questions.
Show all working clearly. Marks are awarded for method, not just final answers.
Unless otherwise stated, give non-exact answers to 3 significant figures.
You may use an approved graphing calculator (without CAS) where appropriate.
The number of marks for each question or part is shown in brackets [ ].

Section A: Probability (Questions 1–5)

15 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 both marbles are the same colour. [2]

2. A survey of 200 households found that 120 own a car, 80 own a motorcycle, and 50 own both.

(a) Represent this information on a Venn diagram. [1]

(b) Find the probability that a randomly selected household owns a car but not a motorcycle. [1]

3. A four-digit code is to be formed using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9. No digit may be repeated.

(a) How many different codes can be formed? [1]

(b) How many of these codes are even? [2]

4. Events A and B are such that P(A) = 0.5, P(B) = 0.4, and P(A ∪ B) = 0.7.

(a) Find P(A ∩ B). [1]

(b) Find P(A | B). [1]

5. A company receives components from three suppliers: X supplies 50%, Y supplies 30%, and Z supplies 20%. The proportions of defective components are 2% for X, 3% for Y, and 5% for Z.

(a) Find the probability that a randomly selected component is defective. [2]

(b) Given that a component is defective, find the probability that it came from supplier Z. [2]

Section B: Probability Distributions (Questions 6–12)

22 marks

6. A biased coin is tossed 10 times. The probability of obtaining a head on any toss is 0.35. Find the probability of obtaining:

(a) exactly 4 heads, [1]

(b) at least 6 heads. [2]

7. In a large population, 15% of adults have a particular health condition. A random sample of 20 adults is selected.

(a) State two assumptions required for the number of adults with the condition in the sample to follow a binomial distribution. [2]

(b) Find the probability that fewer than 3 adults in the sample have the condition. [2]

8. The mass of a packet of cereal is normally distributed with mean 500 g and standard deviation 8 g.

(a) Find the probability that a randomly selected packet has a mass less than 492 g. [2]

(b) A packet is classified as underweight if its mass is less than 485 g. In a batch of 1000 packets, estimate the number that are underweight. [2]

9. The time taken by students to complete a puzzle is normally distributed with mean μ minutes and standard deviation σ minutes. It is known that 10% of students take less than 12 minutes, and 5% take more than 25 minutes.

(a) Show that μ − 1.2816σ = 12. [2]

(b) Find the values of μ and σ. [3]

10. The random variable X is normally distributed with mean 60 and variance 25. The random variable Y is normally distributed with mean 45 and variance 16. X and Y are independent.

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

(b) Find P(X + Y > 110). [2]

11. A machine fills bottles with liquid. The volume dispensed is normally distributed. The manufacturer claims the mean volume is 330 ml. A quality control inspector takes a random sample of 40 bottles and finds the mean volume to be 327.5 ml. The population standard deviation is known to be 6 ml.

Test, at the 5% significance level, whether there is evidence that the mean volume is less than 330 ml. State your hypotheses, test statistic, critical value, and conclusion clearly. [5]

12. The random variable X has a binomial distribution B(n, p). Given that E(X) = 8 and Var(X) = 4.8, find the values of n and p. [3]

Section C: Sampling, Correlation, and Regression (Questions 13–20)

23 marks

13. A researcher wishes to estimate the mean weekly expenditure on groceries for households in a city of 50 000 households. She decides to take a sample of 200 households.

(a) Describe how she could obtain a simple random sample. [2]

(b) Explain one advantage of using a larger sample size. [1]

14. A random sample of 8 students recorded the number of hours spent revising (x) and the score obtained in a test (y). The summary statistics are:

Σx = 96, Σy = 520, Σx² = 1280, Σy² = 35 200, Σxy = 6640.

(a) Calculate the product moment correlation coefficient, r. [2]

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

15. For the data in Question 14:

(a) Find the equation of the least squares regression line of y on x, giving the coefficients to 3 significant figures. [3]

(b) Estimate the test score for a student who spent 15 hours revising. Comment on the reliability of this estimate. [2]

16. A random sample of 10 observations of a variable X gave Σx = 245 and Σx² = 6725.

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

17. A population has mean μ and variance σ². A random sample of size n is taken, and the sample mean is X̄.

(a) State the expected value of X̄. [1]

(b) State the variance of X̄. [1]

(c) A sample of size 64 is taken from a population with σ = 12. Find the probability that the sample mean differs from the population mean by more than 2. [3]

18. A company recorded its monthly advertising expenditure (x, in thousands of dollars) and monthly sales revenue (y, in thousands of dollars) for 6 months. The data are shown in the table.

Month	x	y
1	5	42
2	8	50
3	10	58
4	12	65
5	15	72
6	18	80

(a) Draw a scatter diagram of these data. [2]

(b) Describe the relationship between advertising expenditure and sales revenue. [1]

19. For the data in Question 18, the regression line of y on x is y = 26.4 + 3.02x (coefficients given to 3 significant figures).

(a) Interpret the gradient of the regression line in context. [1]

(b) Use the regression line to predict the sales revenue when advertising expenditure is $20 000. Comment on the reliability of this prediction. [2]

20. A random sample of 12 packets of sugar is taken. The masses, in grams, are:

502, 498, 501, 503, 497, 500, 499, 502, 498, 501, 500, 499.

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

(b) The manufacturer claims the mean mass is 500 g. Assuming the masses are normally distributed, test at the 5% significance level whether there is evidence that the mean mass differs from 500 g. State your hypotheses, test statistic, critical value, and conclusion clearly. (Use the sample standard deviation as an estimate of σ.) [5]

END OF QUIZ

Answers

A-Level Maths H1 Quiz - Statistics Probability

Answer Key and Marking Scheme

Total Marks: 60

Section A: Probability (Questions 1–5)

1. Two marbles drawn without replacement from 5R, 3B, 2G (total 10).

(a) Tree diagram [2 marks]

First stage: R (5/10), B (3/10), G (2/10) [1 mark for correct first-stage probabilities]
Second stage: conditional probabilities for each branch (e.g., after R: R 4/9, B 3/9, G 2/9; after B: R 5/9, B 2/9, G 2/9; after G: R 5/9, B 3/9, G 1/9) [1 mark for correct second-stage probabilities]
All branches clearly labelled.

(b) P(same colour) = P(RR) + P(BB) + P(GG) [1 mark for method]

P(RR) = (5/10) × (4/9) = 20/90
P(BB) = (3/10) × (2/9) = 6/90
P(GG) = (2/10) × (1/9) = 2/90
P(same colour) = 28/90 = 14/45 ≈ 0.311 [1 mark for correct answer]

2. Survey: 200 households. C = car (120), M = motorcycle (80), C ∩ M = 50.

(a) Venn diagram [1 mark]

Two overlapping circles labelled C and M.
Intersection: 50.
C only: 120 − 50 = 70.
M only: 80 − 50 = 30.
Outside both: 200 − 70 − 50 − 30 = 50.

(b) P(car but not motorcycle) = 70/200 = 7/20 = 0.35 [1 mark]

(c) Independence check [2 marks]

P(C) = 120/200 = 0.6; P(M) = 80/200 = 0.4; P(C ∩ M) = 50/200 = 0.25 [1 mark for probabilities]
P(C) × P(M) = 0.6 × 0.4 = 0.24 ≠ 0.25, so not independent. [1 mark for correct conclusion with reasoning]

3. Four-digit code from digits 1–9, no repetition.

(a) Total codes = ⁹P₄ = 9 × 8 × 7 × 6 = 3024 [1 mark]

(b) Even codes (last digit even: 2, 4, 6, 8) [2 marks]

Choose last digit: 4 ways [1 mark]
Choose first three digits from remaining 8: ⁸P₃ = 8 × 7 × 6 = 336
Total = 4 × 336 = 1344 [1 mark]

(c) Strictly increasing order [1 mark]

Choose any 4 digits from 9: ⁹C₄ = 126. Only one arrangement is increasing.
Answer: 126 [1 mark]

4. P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.7.

(a) P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.5 + 0.4 − 0.7 = 0.2 [1 mark]

(b) P(A | B) = P(A ∩ B) / P(B) = 0.2 / 0.4 = 0.5 [1 mark]

(c) Mutually exclusive? [1 mark]

No, because P(A ∩ B) = 0.2 ≠ 0. [1 mark for correct conclusion with reason]

5. Suppliers: X (50%, 2% defective), Y (30%, 3% defective), Z (20%, 5% defective).

(a) P(defective) [2 marks]

P(D) = P(X) × P(D|X) + P(Y) × P(D|Y) + P(Z) × P(D|Z) [1 mark for method]
= 0.5 × 0.02 + 0.3 × 0.03 + 0.2 × 0.05
= 0.01 + 0.009 + 0.01 = 0.029 [1 mark]

(b) P(Z | D) [2 marks]

P(Z | D) = P(Z ∩ D) / P(D) = (0.2 × 0.05) / 0.029 [1 mark for method]
= 0.01 / 0.029 ≈ 0.345 (3 s.f.) [1 mark]

Section B: Probability Distributions (Questions 6–12)

6. X ~ B(10, 0.35).

(a) P(X = 4) = ¹⁰C₄ × (0.35)⁴ × (0.65)⁶ = 210 × 0.01500625 × 0.0754189 ≈ 0.238 (3 s.f.) [1 mark]

(b) P(X ≥ 6) = 1 − P(X ≤ 5) [1 mark for method]

Using GC: P(X ≤ 5) ≈ 0.9051
P(X ≥ 6) ≈ 0.0949 (3 s.f.) [1 mark]

7. p = 0.15, n = 20.

(a) Assumptions [2 marks]

Each adult either has the condition or does not (two outcomes). [1 mark]
The probability of having the condition is constant (0.15) for each adult, and the adults are independent (random sample from large population). [1 mark]

(b) X ~ B(20, 0.15). P(X < 3) = P(X ≤ 2) [1 mark for method]

Using GC: P(X ≤ 2) ≈ 0.4049 ≈ 0.405 (3 s.f.) [1 mark]

8. X ~ N(500, 8²).

(a) P(X < 492) [2 marks]

Z = (492 − 500) / 8 = −1 [1 mark for standardisation]
P(Z < −1) = 1 − Φ(1) = 1 − 0.8413 = 0.1587 ≈ 0.159 (3 s.f.) [1 mark]

(b) Underweight: X < 485 [2 marks]

Z = (485 − 500) / 8 = −1.875 [1 mark]
P(X < 485) = P(Z < −1.875) = 1 − Φ(1.875) ≈ 1 − 0.9696 = 0.0304
Number underweight in 1000 = 1000 × 0.0304 ≈ 30.4, so approximately 30 packets. [1 mark]

9. X ~ N(μ, σ²). P(X < 12) = 0.10, P(X > 25) = 0.05.

(a) Show μ − 1.2816σ = 12 [2 marks]

P(X < 12) = 0.10 → P(Z < (12 − μ)/σ) = 0.10
From normal table, P(Z < −1.2816) = 0.10 [1 mark]
So (12 − μ)/σ = −1.2816 → μ − 1.2816σ = 12 [1 mark]

(b) Find μ and σ [3 marks]

P(X > 25) = 0.05 → P(Z > (25 − μ)/σ) = 0.05 → (25 − μ)/σ = 1.6449 [1 mark]
So μ + 1.6449σ = 25 [1 mark]
Solving: subtract equations → 2.9265σ = 13 → σ ≈ 4.44 (3 s.f.)
μ = 12 + 1.2816 × 4.44 ≈ 17.7 (3 s.f.) [1 mark]

10. X ~ N(60, 25), Y ~ N(45, 16), independent.

(a) X + Y ~ N(60 + 45, 25 + 16) = N(105, 41) [1 mark]

(b) P(X + Y > 110) [2 marks]

Z = (110 − 105) / √41 ≈ 5 / 6.4031 ≈ 0.7809 [1 mark]
P(Z > 0.7809) = 1 − Φ(0.7809) ≈ 1 − 0.7826 = 0.2174 ≈ 0.217 (3 s.f.) [1 mark]

11. Hypothesis test: n = 40, x̄ = 327.5, σ = 6, α = 0.05 (one-tail, lower). [5 marks]

H₀: μ = 330 [1 mark]
H₁: μ < 330 [1 mark]
Test statistic: Z = (327.5 − 330) / (6/√40) = −2.5 / 0.9487 ≈ −2.635 [1 mark]
Critical value (one-tail, 5%): −1.645 [1 mark]
Since −2.635 < −1.645, reject H₀.
Conclusion: There is sufficient evidence at the 5% significance level that the mean volume is less than 330 ml. [1 mark for correct conclusion in context]

12. X ~ B(n, p). E(X) = np = 8, Var(X) = np(1 − p) = 4.8. [3 marks]

np(1 − p) = 4.8 → 8(1 − p) = 4.8 → 1 − p = 0.6 → p = 0.4 [2 marks]
np = 8 → n × 0.4 = 8 → n = 20 [1 mark]

Section C: Sampling, Correlation, and Regression (Questions 13–20)

13. (a) Simple random sample [2 marks]

Assign each household a unique number from 1 to 50 000. [1 mark]
Use a random number generator to select 200 distinct numbers. Interview the households corresponding to those numbers. [1 mark]

(b) Advantage of larger sample [1 mark]

A larger sample reduces the standard error of the sample mean (Var(X̄) = σ²/n), giving a more precise estimate of the population mean. [1 mark]

14. n = 8, Σx = 96, Σy = 520, Σx² = 1280, Σy² = 35 200, Σxy = 6640.

(a) Correlation coefficient r [2 marks]

r = [nΣxy − (Σx)(Σy)] / √[(nΣx² − (Σx)²)(nΣy² − (Σy)²)] [1 mark for formula]
= [8(6640) − 96(520)] / √[(8×1280 − 96²)(8×35200 − 520²)]
= [53120 − 49920] / √[(10240 − 9216)(281600 − 270400)]
= 3200 / √[1024 × 11200] = 3200 / √11468800 ≈ 3200 / 3386.6 ≈ 0.945 (3 s.f.) [1 mark]

(b) Interpretation [1 mark]

r ≈ 0.945 indicates a strong positive linear correlation between hours spent revising and test score. [1 mark]

15. (a) Regression line y on x [3 marks]

b = [nΣxy − (Σx)(Σy)] / [nΣx² − (Σx)²] = 3200 / 1024 ≈ 3.125 [1 mark]
a = ȳ − b x̄ = (520/8) − 3.125(96/8) = 65 − 3.125(12) = 65 − 37.5 = 27.5 [1 mark]
Equation: y = 27.5 + 3.13x (3 s.f.) [1 mark]

(b) Estimate for x = 15 [2 marks]

y = 27.5 + 3.125(15) = 27.5 + 46.875 = 74.375 ≈ 74.4 [1 mark]
Comment: x = 15 is within the range of the data (x values range from approximately 8 to 16 based on x̄ = 12 and typical spread), so this is interpolation. The estimate is reasonably reliable given the strong correlation (r ≈ 0.945). [1 mark]

16. n = 10, Σx = 245, Σx² = 6725. [3 marks]

Sample mean x̄ = 245/10 = 24.5 [1 mark]
Unbiased variance s² = [Σx² − (Σx)²/n] / (n − 1) [1 mark for formula]
= [6725 − (245²/10)] / 9 = [6725 − 6002.5] / 9 = 722.5 / 9 ≈ 80.3 (3 s.f.) [1 mark]

17. (a) E(X̄) = μ [1 mark]

(b) Var(X̄) = σ²/n [1 mark]

X̄ ~ N(μ, σ²/n) = N(μ, 144/64) = N(μ, 2.25) [1 mark]
Standard error = √2.25 = 1.5
P(|X̄ − μ| > 2) = P(X̄ − μ > 2) + P(X̄ − μ < −2) = 2 × P(Z > 2/1.5) = 2 × P(Z > 1.3333) [1 mark]
= 2 × (1 − Φ(1.3333)) ≈ 2 × (1 − 0.9088) = 2 × 0.0912 = 0.1824 ≈ 0.182 (3 s.f.) [1 mark]

18. (a) Scatter diagram [2 marks]

Axes labelled: x (Advertising expenditure, $000) and y (Sales revenue,$ 000). [1 mark]
Six points plotted accurately: (5,42), (8,50), (10,58), (12,65), (15,72), (18,80). [1 mark]
Appropriate scale on both axes.

(b) Description [1 mark]

There is a strong positive linear relationship between advertising expenditure and sales revenue. As advertising expenditure increases, sales revenue tends to increase. [1 mark]

19. Regression line: y = 26.4 + 3.02x.

(a) Interpretation of gradient [1 mark]

For every additional $1000 spent on advertising, sales revenue increases by approximately$ 3020, on average. [1 mark]

(b) Prediction for x = 20 [2 marks]

y = 26.4 + 3.02(20) = 26.4 + 60.4 = 86.8 ($86 800) [1 mark]
Comment: x = 20 is outside the range of the data (5 to 18), so this is extrapolation. The prediction may be unreliable as the linear relationship may not hold beyond the observed range. [1 mark]

20. Data: 502, 498, 501, 503, 497, 500, 499, 502, 498, 501, 500, 499. n = 12.

(a) Sample mean and unbiased variance [3 marks]

Σx = 502 + 498 + 501 + 503 + 497 + 500 + 499 + 502 + 498 + 501 + 500 + 499 = 6000
x̄ = 6000/12 = 500 [1 mark]
Σx² = 502² + 498² + ... + 499² = 3 000 018 (or calculate deviations)
Unbiased variance s² = [Σx² − (Σx)²/n] / (n − 1) = [3 000 018 − 6000²/12] / 11 [1 mark for method]
= [3 000 018 − 3 000 000] / 11 = 18/11 ≈ 1.636 (4 s.f.) [1 mark]

(b) Hypothesis test (two-tail, α = 0.05) [5 marks]

H₀: μ = 500 [1 mark]
H₁: μ ≠ 500 [1 mark]
Sample standard deviation s = √1.63636 ≈ 1.279
Test statistic: t = (500 − 500) / (1.279/√12) = 0 / 0.3693 = 0 [1 mark]
Critical value: For two-tail test at 5% with df = 11, t_crit ≈ ±2.201 [1 mark]
Since |0| < 2.201, do not reject H₀.
Conclusion: There is insufficient evidence at the 5% significance level that the mean mass differs from 500 g. [1 mark for correct conclusion in context]

END OF ANSWER KEY