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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: 1 hour 30 minutes
Total Marks: 60
Instructions:
- Answer ALL questions.
- Show all working clearly.
- Give non-exact numerical answers correct to 3 significant figures unless otherwise stated.
- The use of an approved graphing calculator is expected, but unsupported answers obtained from a calculator should be supported by appropriate mathematical working.
- Marks are indicated in brackets [ ].
Section A: Probability Concepts and Distributions (15 marks)
1. A bag contains 5 red balls, 3 blue balls, and 2 green balls. Three balls are drawn at random from the bag without replacement.
(a) Find the probability that all three balls are of the same colour. [3]
(b) Find the probability that exactly two of the balls drawn are red. [3]
2. Events A and B are such that P(A) = 0.4, P(B) = 0.5, and P(A ∪ B) = 0.7.
(a) Find P(A ∩ B). [2]
(b) Determine whether A and B are independent. Justify your answer. [2]
3. A fair six-sided die is rolled 8 times. The random variable X represents the number of times a prime number (2, 3, or 5) appears.
(a) State the distribution of X and write down its parameters. [2]
(b) Find P(X = 4). [2]
(c) Find P(3 ≤ X < 6). [3]
4. In a large population, the heights of adult males are normally distributed with mean 172 cm and standard deviation 8 cm.
(a) Find the probability that a randomly selected adult male has a height between 165 cm and 180 cm. [3]
(b) Find the value of h such that 10% of adult males are taller than h cm. [3]
5. A random sample of 25 adult males is taken from the population in Question 4. Find the probability that their mean height exceeds 175 cm. [3]
Section B: Sampling and Hypothesis Testing (15 marks)
6. A company produces light bulbs. The lifetimes of the bulbs are normally distributed with standard deviation 120 hours. A random sample of 40 bulbs is taken and the mean lifetime is found to be 1520 hours.
(a) Calculate the 95% confidence interval for the population mean lifetime. [3]
(b) The manufacturer claims that the mean lifetime of the bulbs is at least 1550 hours. Test this claim at the 5% significance level. State your hypotheses, test statistic, critical region, and conclusion clearly. [6]
7. A researcher wishes to estimate the proportion of students in a college who own a laptop. A random sample of 200 students is surveyed, and 148 of them own a laptop.
(a) Explain what is meant by a random sample in this context. [2]
(b) Calculate an unbiased estimate of the population proportion of students who own a laptop. [1]
(c) Construct an approximate 95% confidence interval for the population proportion. [3]
8. The researcher in Question 7 wants the width of the 95% confidence interval to be at most 0.08. Determine the minimum sample size required. [3]
Section C: Correlation and Regression (15 marks)
9. A biologist is studying the relationship between the temperature x (°C) and the number of eggs y laid by a species of insect. The following data are obtained from 8 observations:
| Temperature, x (°C) | 18 | 20 | 22 | 24 | 26 | 28 | 30 | 32 |
|---|---|---|---|---|---|---|---|---|
| Number of eggs, y | 15 | 22 | 28 | 35 | 42 | 48 | 55 | 62 |
(a) Draw a scatter diagram of the data. Label the axes clearly. [2]
(b) Calculate, correct to 4 decimal places, the value of the product moment correlation coefficient between x and y. [2]
(c) Comment on the relationship between temperature and the number of eggs laid. [1]
10. Using the data from Question 9:
(a) Find the equation of the regression line of y on x. Give your answers correct to 3 significant figures where appropriate. [3]
(b) Use your regression line to estimate the number of eggs laid when the temperature is 25°C. Comment on the reliability of this estimate. [2]
(c) State, with a reason, whether it would be appropriate to use the regression line to estimate the number of eggs laid when the temperature is 10°C. [2]
Section D: Further Probability and Distributions (15 marks)
11. A discrete random variable Y has the following probability distribution:
| y | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| P(Y=y) | 0.2 | 0.3 | 0.4 | 0.1 |
Find E(Y) and Var(Y). [3]
12. The probability that a student passes a driving test on any attempt is 0.6. A student takes the test until they pass.
(a) Find the probability that the student passes on the third attempt. [2]
(b) Find the probability that the student takes at most 4 attempts to pass. [3]
13. The mass of a certain type of apple is normally distributed with mean 150 g and standard deviation 12 g. A random sample of 10 apples is selected.
(a) Find the probability that the total mass of the 10 apples exceeds 1550 g. [3]
(b) Find the probability that exactly 3 of the 10 apples have a mass greater than 160 g. [4]
14. A fair coin is tossed 6 times. Find the probability of obtaining:
(a) exactly 4 heads, [2]
(b) at least 2 heads. [3]
15. Events C and D are independent with P(C) = 0.7 and P(D) = 0.4. Find:
(a) P(C ∩ D), [1]
(b) P(C ∪ D), [2]
(c) P(C | D). [2]
16. A random variable W has a normal distribution with mean 50 and variance 25. Find:
(a) P(W < 45), [2]
(b) the value of w such that P(W > w) = 0.05. [3]
17. A bag contains 4 black marbles and 6 white marbles. Two marbles are drawn at random without replacement. Find the probability that:
(a) both marbles are black, [2]
(b) at least one marble is white. [2]
18. The time taken for a student to complete a puzzle is normally distributed with mean 25 minutes and standard deviation 4 minutes. A random sample of 16 students is selected. Find the probability that the sample mean time is less than 23 minutes. [3]
19. A biased die is such that the probability of rolling a six is 0.25. The die is rolled 10 times. Find the probability of rolling:
(a) exactly 2 sixes, [2]
(b) more than 2 sixes. [3]
20. A continuous random variable T has probability density function given by:
f(t) = kt, for 0 ≤ t ≤ 4, f(t) = 0, otherwise.
(a) Show that k = 1/8. [2]
(b) Find P(1 < T < 3). [3]
END OF QUIZ
Answers
A-Level Maths H2 Quiz - Statistics Probability
ANSWER KEY AND MARKING SCHEME
Section A: Probability Concepts and Distributions (15 marks)
1. (a) P(all same colour) [3 marks]
Total balls = 10. Total ways to choose 3 balls = C(10,3) = 120.
P(all red) = C(5,3)/120 = 10/120 = 1/12
P(all blue) = C(3,3)/120 = 1/120
P(all green) = C(2,3)/120 = 0 (impossible)
P(all same colour) = 1/12 + 1/120 + 0 = 10/120 + 1/120 = 11/120
Answer: 11/120 or 0.0917 (3 s.f.)
Marking: M1 for identifying combinations, M1 for summing probabilities, A1 for correct answer.
(b) P(exactly two red) [3 marks]
Ways to choose 2 red from 5 = C(5,2) = 10
Ways to choose 1 non-red from 5 non-red = C(5,1) = 5
Favourable outcomes = 10 × 5 = 50
P(exactly two red) = 50/120 = 5/12
Answer: 5/12 or 0.417 (3 s.f.)
Marking: M1 for C(5,2), M1 for C(5,1) and multiplication, A1 for correct answer.
2. (a) P(A ∩ B) [2 marks]
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
0.7 = 0.4 + 0.5 - P(A ∩ B)
P(A ∩ B) = 0.9 - 0.7 = 0.2
Answer: 0.2
Marking: M1 for correct formula, A1 for correct answer.
(b) Independence check [2 marks]
For independence: P(A ∩ B) = P(A) × P(B)
P(A) × P(B) = 0.4 × 0.5 = 0.2
Since P(A ∩ B) = 0.2 = P(A) × P(B), A and B are independent.
Answer: Independent, because P(A ∩ B) = P(A) × P(B).
Marking: M1 for comparing P(A ∩ B) with P(A) × P(B), A1 for correct conclusion with justification.
3. (a) Distribution of X [2 marks]
X ~ B(8, 0.5)
Reason: 8 independent trials, each with probability of prime = 3/6 = 0.5.
Answer: X ~ B(8, 0.5)
Marking: B1 for binomial, B1 for correct parameters.
(b) P(X = 4) [2 marks]
P(X = 4) = C(8,4) × (0.5)^4 × (0.5)^4 = 70 × (0.5)^8 = 70/256 = 35/128 ≈ 0.273
Answer: 0.273 (3 s.f.)
Marking: M1 for correct binomial formula, A1 for correct answer.
(c) P(3 ≤ X < 6) [3 marks]
P(3 ≤ X < 6) = P(X = 3) + P(X = 4) + P(X = 5)
P(X = 3) = C(8,3) × (0.5)^8 = 56/256 = 7/32
P(X = 4) = 70/256 = 35/128
P(X = 5) = C(8,5) × (0.5)^8 = 56/256 = 7/32
Total = (56 + 70 + 56)/256 = 182/256 = 91/128 ≈ 0.711
Answer: 0.711 (3 s.f.)
Marking: M1 for identifying correct values, M1 for summing, A1 for correct answer.
4. (a) P(165 < H < 180) [3 marks]
H ~ N(172, 8²)
Z₁ = (165 - 172)/8 = -0.875
Z₂ = (180 - 172)/8 = 1.0
P(-0.875 < Z < 1.0) = Φ(1.0) - Φ(-0.875) = Φ(1.0) - [1 - Φ(0.875)]
= 0.8413 - (1 - 0.8092) = 0.8413 - 0.1908 = 0.6505
Answer: 0.651 (3 s.f.)
Marking: M1 for standardising, M1 for correct use of normal tables/GC, A1 for correct answer.
(b) Value of h [3 marks]
P(H > h) = 0.10 → P(H < h) = 0.90
Z₀.₉₀ = 1.2816 (or from GC)
h = 172 + 1.2816 × 8 = 172 + 10.2528 = 182.2528
Answer: h = 182 cm (3 s.f.)
Marking: M1 for identifying inverse normal, M1 for correct calculation, A1 for correct answer.
5. P(X̄ > 175) [3 marks]
X̄ ~ N(172, 8²/25) = N(172, 1.6²)
Z = (175 - 172)/1.6 = 3/1.6 = 1.875
P(Z > 1.875) = 1 - Φ(1.875) = 1 - 0.9696 = 0.0304
Answer: 0.0304 (3 s.f.)
Marking: M1 for distribution of sample mean, M1 for standardising, A1 for correct answer.
Section B: Sampling and Hypothesis Testing (15 marks)
6. (a) 95% confidence interval [3 marks]
X̄ = 1520, σ = 120, n = 40
95% CI: X̄ ± z₀.₀₂₅ × σ/√n
= 1520 ± 1.96 × 120/√40
= 1520 ± 1.96 × 18.9737
= 1520 ± 37.19
= (1482.81, 1557.19)
Answer: (1480, 1560) to 3 s.f. or (1482.8, 1557.2)
Marking: M1 for correct formula, M1 for substitution, A1 for correct interval.
(b) Hypothesis test [6 marks]
H₀: μ = 1550 (or μ ≥ 1550)
H₁: μ < 1550 (one-tailed test)
Significance level: α = 0.05
Test statistic: Z = (X̄ - μ₀)/(σ/√n) = (1520 - 1550)/(120/√40) = -30/18.9737 = -1.581
Critical value: z_critical = -1.645 (for one-tailed test at 5%)
Since -1.581 > -1.645, the test statistic does not fall in the critical region.
Conclusion: Do not reject H₀. There is insufficient evidence at the 5% significance level to reject the manufacturer's claim that the mean lifetime is at least 1550 hours.
Answer: Do not reject H₀; insufficient evidence to dispute the claim.
Marking: B1 for correct hypotheses, B1 for significance level, M1 for test statistic, M1 for critical value, A1 for correct comparison, A1 for conclusion in context.
7. (a) Random sample explanation [2 marks]
A random sample means that every student in the college has an equal probability of being selected for the survey, and the selections are independent of each other. This ensures the sample is representative of the population.
Answer: Each student has equal chance of selection; selections are independent.
Marking: B1 for equal probability, B1 for independence/representativeness.
(b) Unbiased estimate of proportion [1 mark]
p̂ = 148/200 = 0.74
Answer: 0.74
Marking: B1 for correct answer.
(c) 95% confidence interval for proportion [3 marks]
p̂ = 0.74, n = 200
95% CI: p̂ ± z₀.₀₂₅ × √[p̂(1 - p̂)/n]
= 0.74 ± 1.96 × √(0.74 × 0.26 / 200)
= 0.74 ± 1.96 × √(0.1924/200)
= 0.74 ± 1.96 × √0.000962
= 0.74 ± 1.96 × 0.03102
= 0.74 ± 0.0608
= (0.6792, 0.8008)
Answer: (0.679, 0.801) to 3 s.f.
Marking: M1 for correct formula, M1 for substitution, A1 for correct interval.
8. Minimum sample size [3 marks]
Width = 2 × z₀.₀₂₅ × √[p̂(1 - p̂)/n] ≤ 0.08
2 × 1.96 × √(0.74 × 0.26 / n) ≤ 0.08
3.92 × √(0.1924/n) ≤ 0.08
√(0.1924/n) ≤ 0.020408
0.1924/n ≤ 0.0004165
n ≥ 0.1924/0.0004165 = 462.0
Therefore, minimum sample size = 463.
Answer: 463
Marking: M1 for setting up inequality, M1 for solving, A1 for correct answer (rounded up).
Section C: Correlation and Regression (15 marks)
9. (a) Scatter diagram [2 marks]
Axes labelled: x-axis "Temperature (°C)", y-axis "Number of eggs"
Points plotted correctly: (18,15), (20,22), (22,28), (24,35), (26,42), (28,48), (30,55), (32,62)
Appropriate scales used.
Answer: Scatter diagram showing strong positive linear correlation.
Marking: B1 for correct axes labels and scales, B1 for correctly plotted points.
(b) Product moment correlation coefficient [2 marks]
Using calculator: r = 0.9987 (to 4 d.p.)
Answer: r = 0.9987
Marking: M1 for correct method, A1 for correct value to 4 d.p.
(c) Comment on relationship [1 mark]
There is a very strong positive linear correlation between temperature and the number of eggs laid. As temperature increases, the number of eggs laid tends to increase.
Answer: Very strong positive linear correlation.
Marking: B1 for correct interpretation with reference to strength and direction.
10. (a) Regression line of y on x [3 marks]
Using calculator:
x̄ = 25, ȳ = 38.375
S_xx = Σ(x - x̄)² = 168
S_xy = Σ(x - x̄)(y - ȳ) = 798
b = S_xy / S_xx = 798/168 = 4.75
a = ȳ - b x̄ = 38.375 - 4.75 × 25 = 38.375 - 118.75 = -80.375
Regression line: y = -80.4 + 4.75x (to 3 s.f.)
Answer: y = -80.4 + 4.75x
Marking: M1 for calculating S_xx and S_xy, M1 for finding a and b, A1 for correct equation.
(b) Estimate at 25°C and reliability [2 marks]
When x = 25: y = -80.375 + 4.75 × 25 = -80.375 + 118.75 = 38.375 ≈ 38.4
This estimate is reliable because x = 25 is within the range of observed data (interpolation), and the correlation coefficient is very high (r = 0.9987), indicating a strong linear relationship.
Answer: 38.4 eggs; reliable as it is interpolation with strong correlation.
Marking: B1 for correct estimate, B1 for comment on reliability with reasoning.
(c) Appropriateness of estimate at 10°C [2 marks]
It would NOT be appropriate to use the regression line to estimate the number of eggs at 10°C because this temperature is outside the range of the observed data (extrapolation). The relationship may not hold outside the observed range.
Answer: Not appropriate; extrapolation beyond the data range.
Marking: B1 for stating not appropriate, B1 for correct reasoning (extrapolation).
Section D: Further Probability and Distributions (15 marks)
11. E(Y) and Var(Y) [3 marks]
E(Y) = Σ y P(Y=y) = 1(0.2) + 2(0.3) + 3(0.4) + 4(0.1) = 0.2 + 0.6 + 1.2 + 0.4 = 2.4
E(Y²) = 1²(0.2) + 2²(0.3) + 3²(0.4) + 4²(0.1) = 0.2 + 1.2 + 3.6 + 1.6 = 6.6
Var(Y) = E(Y²) - [E(Y)]² = 6.6 - (2.4)² = 6.6 - 5.76 = 0.84
Answer: E(Y) = 2.4, Var(Y) = 0.84
Marking: M1 for E(Y), M1 for E(Y²), A1 for both correct.
12. (a) Pass on third attempt [2 marks]
Geometric distribution: P(X = 3) = (0.4)² × 0.6 = 0.096
Answer: 0.096
Marking: M1 for correct geometric probability, A1 for correct answer.
(b) At most 4 attempts [3 marks]
P(X ≤ 4) = 1 - P(fail first 4) = 1 - (0.4)⁴ = 1 - 0.0256 = 0.9744
Alternatively: P(X=1) + P(X=2) + P(X=3) + P(X=4) = 0.6 + 0.4×0.6 + (0.4)²×0.6 + (0.4)³×0.6 = 0.6 + 0.24 + 0.096 + 0.0384 = 0.9744
Answer: 0.974 (3 s.f.)
Marking: M1 for correct method, M1 for calculation, A1 for correct answer.
13. (a) Total mass exceeds 1550 g [3 marks]
Let M ~ N(150, 12²) for one apple.
Total mass T = M₁ + ... + M₁₀ ~ N(10×150, 10×12²) = N(1500, 1440)
σ_T = √1440 = 37.9473
P(T > 1550) = P(Z > (1550 - 1500)/37.9473) = P(Z > 1.3176) = 1 - Φ(1.3176) = 1 - 0.9062 = 0.0938
Answer: 0.0938 (3 s.f.)
Marking: M1 for distribution of total, M1 for standardising, A1 for correct answer.
(b) Exactly 3 apples > 160 g [4 marks]
P(one apple > 160) = P(Z > (160-150)/12) = P(Z > 0.8333) = 1 - Φ(0.8333) = 1 - 0.7977 = 0.2023
Let Y ~ B(10, 0.2023)
P(Y = 3) = C(10,3) × (0.2023)³ × (0.7977)⁷ = 120 × 0.00828 × 0.2097 = 0.208
Answer: 0.208 (3 s.f.)
Marking: M1 for probability of >160g, M1 for binomial distribution, M1 for binomial formula, A1 for correct answer.
14. (a) Exactly 4 heads [2 marks]
X ~ B(6, 0.5)
P(X = 4) = C(6,4) × (0.5)⁴ × (0.5)² = 15 × (0.5)⁶ = 15/64 = 0.234375
Answer: 0.234 (3 s.f.)
Marking: M1 for binomial formula, A1 for correct answer.
(b) At least 2 heads [3 marks]
P(X ≥ 2) = 1 - P(X = 0) - P(X = 1)
P(X = 0) = C(6,0) × (0.5)⁶ = 1/64
P(X = 1) = C(6,1) × (0.5)⁶ = 6/64
P(X ≥ 2) = 1 - 7/64 = 57/64 = 0.890625
Answer: 0.891 (3 s.f.)
Marking: M1 for complement method, M1 for correct probabilities, A1 for correct answer.
15. (a) P(C ∩ D) [1 mark]
Since independent: P(C ∩ D) = P(C) × P(D) = 0.7 × 0.4 = 0.28
Answer: 0.28
Marking: B1 for correct answer.
(b) P(C ∪ D) [2 marks]
P(C ∪ D) = P(C) + P(D) - P(C ∩ D) = 0.7 + 0.4 - 0.28 = 0.82
Answer: 0.82
Marking: M1 for correct formula, A1 for correct answer.
(c) P(C | D) [2 marks]
Since independent: P(C | D) = P(C) = 0.7
Answer: 0.7
Marking: M1 for recognising independence, A1 for correct answer.
16. (a) P(W < 45) [2 marks]
W ~ N(50, 5²)
Z = (45 - 50)/5 = -1.0
P(Z < -1.0) = 1 - Φ(1.0) = 1 - 0.8413 = 0.1587
Answer: 0.159 (3 s.f.)
Marking: M1 for standardising, A1 for correct answer.
(b) Value of w [3 marks]
P(W > w) = 0.05 → P(W < w) = 0.95
Z₀.₉₅ = 1.6449
w = 50 + 1.6449 × 5 = 50 + 8.2245 = 58.2245
Answer: w = 58.2 (3 s.f.)
Marking: M1 for inverse normal, M1 for correct calculation, A1 for correct answer.
17. (a) Both marbles black [2 marks]
P(BB) = (4/10) × (3/9) = 12/90 = 2/15 ≈ 0.133
Answer: 2/15 or 0.133 (3 s.f.)
Marking: M1 for correct multiplication, A1 for correct answer.
(b) At least one white [2 marks]
P(at least one white) = 1 - P(both black) = 1 - 2/15 = 13/15 ≈ 0.867
Answer: 13/15 or 0.867 (3 s.f.)
Marking: M1 for complement method, A1 for correct answer.
18. Sample mean less than 23 minutes [3 marks]
X̄ ~ N(25, 4²/16) = N(25, 1²)
Z = (23 - 25)/1 = -2.0
P(Z < -2.0) = 1 - Φ(2.0) = 1 - 0.9772 = 0.0228
Answer: 0.0228
Marking: M1 for distribution of sample mean, M1 for standardising, A1 for correct answer.
19. (a) Exactly 2 sixes [2 marks]
X ~ B(10, 0.25)
P(X = 2) = C(10,2) × (0.25)² × (0.75)⁸ = 45 × 0.0625 × 0.1001 = 0.2816
Answer: 0.282 (3 s.f.)
Marking: M1 for binomial formula, A1 for correct answer.
(b) More than 2 sixes [3 marks]
P(X > 2) = 1 - P(X ≤ 2) = 1 - [P(X=0) + P(X=1) + P(X=2)]
P(X=0) = (0.75)¹⁰ = 0.0563
P(X=1) = 10 × 0.25 × (0.75)⁹ = 0.1877
P(X=2) = 0.2816
P(X > 2) = 1 - (0.0563 + 0.1877 + 0.2816) = 1 - 0.5256 = 0.4744
Answer: 0.474 (3 s.f.)
Marking: M1 for complement method, M1 for correct probabilities, A1 for correct answer.
20. (a) Show k = 1/8 [2 marks]
∫₀⁴ kt dt = 1
k [t²/2]₀⁴ = k × (16/2) = 8k = 1
k = 1/8
Answer: k = 1/8
Marking: M1 for setting integral to 1, A1 for correct value.
(b) P(1 < T < 3) [3 marks]
P(1 < T < 3) = ∫₁³ (1/8)t dt = (1/8) [t²/2]₁³ = (1/16) (9 - 1) = 8/16 = 1/2 = 0.5
Answer: 0.5
Marking: M1 for correct integral, M1 for evaluation, A1 for correct answer.
END OF ANSWER KEY