A Level H1 Mathematics Practice Paper 3

A-Level Maths H1 Quiz - Statistics Probability

ANSWER KEY AND MARKING SCHEME

Total Marks: 50

Section A: Probability and Counting (Questions 1–5)

Question 1

(a) Number of ways = ¹²C₄ = 495 [1 mark]

(b) Choose 2 women from 5: ⁵C₂ = 10
Choose 2 men from 7: ⁷C₂ = 21
Total ways = 10 × 21 = 210 [2 marks]

(c) At least 3 men means 3 men + 1 woman OR 4 men + 0 women.
3 men, 1 woman: ⁷C₃ × ⁵C₁ = 35 × 5 = 175
4 men, 0 women: ⁷C₄ × ⁵C₀ = 35 × 1 = 35
Total ways = 175 + 35 = 210 [2 marks]

Question 2

Total balls = 6 + 4 + 5 = 15. Two drawn without replacement.

(a) P(both red) = (6/15) × (5/14) = 30/210 = 1/7 ≈ 0.143 [2 marks]

(b) P(different colours) = 1 − P(same colour)
P(both red) = 1/7
P(both blue) = (4/15) × (3/14) = 12/210 = 2/35
P(both green) = (5/15) × (4/14) = 20/210 = 2/21
P(same colour) = 1/7 + 2/35 + 2/21 = 15/105 + 6/105 + 10/105 = 31/105
P(different colours) = 1 − 31/105 = 74/105 ≈ 0.705 [2 marks]

Question 3

(a) P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.45 + 0.6 − 0.78 = 0.27 [1 mark]

(b) For independence: P(A) × P(B) = 0.45 × 0.6 = 0.27
Since P(A ∩ B) = 0.27 = P(A) × P(B), A and B are independent. [2 marks]

Question 4

(a) P(X) = P(sum = 7) = 6/36 = 1/6 [1 mark]
(Favourable outcomes: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1))

(b) P(Y) = P(first roll even) = 3/6 = 1/2 [1 mark]

(c) X ∩ Y: sum = 7 AND first roll even.
Favourable: (2,5), (4,3), (6,1) → 3 outcomes.
P(X ∩ Y) = 3/36 = 1/12 [2 marks]

(d) For independence: P(X) × P(Y) = (1/6) × (1/2) = 1/12
Since P(X ∩ Y) = 1/12 = P(X) × P(Y), X and Y are independent. [2 marks]

Question 5

(a) Letters: 26³, Digits: 10²
Total = 26³ × 10² = 17 576 × 100 = 1 757 600 [1 mark]

(b) Exactly one vowel among 3 letters.
Vowels: 5, Consonants: 21.
Choose position for vowel: 3 ways.
Vowel choice: 5, Consonant choices: 21 × 21.
Letters: 3 × 5 × 21 × 21 = 6615
Digits: 10² = 100
Total = 6615 × 100 = 661 500 [3 marks]

Section B: Binomial and Normal Distributions (Questions 6–10)

Question 6

(a) Yes, binomial is appropriate because:

• Fixed number of trials (n = 20)
• Each bulb is either defective or not (two outcomes)
• Probability of defective is constant (p = 0.08)
• Bulbs are independent (random sample) [2 marks]

(b) X ~ B(20, 0.08)
P(X = 2) = ²⁰C₂ × (0.08)² × (0.92)¹⁸ = 190 × 0.0064 × 0.92¹⁸ ≈ 0.271 [2 marks]

(c) P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= 0.92²⁰ + 20 × 0.08 × 0.92¹⁹ + 190 × 0.0064 × 0.92¹⁸ + 1140 × 0.000512 × 0.92¹⁷
≈ 0.1887 + 0.3282 + 0.2711 + 0.1414 = 0.929 (using GC) [2 marks]

Question 7

X ~ N(150, 20²)

(a) P(140 < X < 165) = P(−0.5 < Z < 0.75)
= Φ(0.75) − Φ(−0.5) = Φ(0.75) − [1 − Φ(0.5)]
= 0.7734 − (1 − 0.6915) = 0.7734 − 0.3085 = 0.4649 ≈ 0.465 [2 marks]

(b) Heaviest 10%: P(X > x₀) = 0.10 → P(Z > z₀) = 0.10 → z₀ = 1.2816
x₀ = μ + z₀σ = 150 + 1.2816 × 20 = 150 + 25.632 = 175.632
Minimum mass ≈ 176 g (3 s.f.) [2 marks]

Question 8

X ~ N(50, 4²)

(a) P(X < 55) = P(Z < 1.25) = 0.8944 ≈ 0.894 [1 mark]

(b) P(46 < X < 54) = P(−1 < Z < 1) = Φ(1) − Φ(−1) = 0.8413 − 0.1587 = 0.6826 ≈ 0.683 [2 marks]

(c) P(X > k) = 0.05 → P(Z > (k − 50)/4) = 0.05 → (k − 50)/4 = 1.6449
k = 50 + 4 × 1.6449 = 50 + 6.5796 = 56.5796 ≈ 56.6 [2 marks]

Question 9

X ~ N(μ, 8²). P(X > 45) = 0.15 → P(Z > (45 − μ)/8) = 0.15
z-value for upper 15%: z = 1.0364
(45 − μ)/8 = 1.0364 → 45 − μ = 8.2912 → μ = 45 − 8.2912 = 36.7088
μ ≈ 36.7 minutes (3 s.f.) [3 marks]

Question 10

X ~ N(100, 225), n = 36.

(a) X̄ ~ N(μ, σ²/n) = N(100, 225/36) = N(100, 6.25)
i.e., X̄ ~ N(100, 2.5²) [2 marks]

(b) P(X̄ > 105) = P(Z > (105 − 100)/2.5) = P(Z > 2) = 1 − 0.9772 = 0.0228 ≈ 0.0228 [2 marks]

Section C: Sampling, Estimation, and Hypothesis Testing (Questions 11–15)

Question 11

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

(b) Unbiased estimate of μ: x̄ = 168.2 cm
Unbiased estimate of σ²: s² = (n/(n−1)) × sample variance = (50/49) × 9.5² = (50/49) × 90.25 = 92.0918...
s² ≈ 92.1 cm² (3 s.f.) [2 marks]

(c) 95% CI: x̄ ± z₀.₀₂₅ × s/√n
z₀.₀₂₅ = 1.96
s = √92.0918 = 9.5964...
Margin of error = 1.96 × 9.5964/√50 = 1.96 × 1.3573 = 2.6603...
CI: 168.2 ± 2.66 → (165.5, 170.9) cm (1 d.p.) [3 marks]

Question 12

(a) H₀: μ = 120 (or μ ≥ 120)
H₁: μ < 120 (one-tail test) [2 marks]

(b) Test statistic: z = (x̄ − μ₀)/(σ/√n) = (117.5 − 120)/(12/√40) = −2.5/1.8974 = −1.3176
Critical value at 5% (one-tail): z_crit = −1.6449
Since −1.3176 > −1.6449, we do NOT reject H₀. [4 marks]

(c) There is insufficient evidence at the 5% significance level to reject the company's claim that the mean lifetime is at least 120 hours. [1 mark]

Question 13

(a) x̄ = (502 + 498 + 501 + 497 + 503 + 499 + 500 + 496 + 501 + 498)/10 = 4995/10 = 499.5 g [1 mark]

(b) H₀: μ = 500
H₁: μ ≠ 500 (two-tail test)
σ = 2.5, n = 10
Test statistic: z = (499.5 − 500)/(2.5/√10) = −0.5/0.7906 = −0.6325
Critical values at 5% (two-tail): ±1.96
Since |−0.6325| < 1.96, we do NOT reject H₀.
There is insufficient evidence at the 5% significance level to reject the manufacturer's claim that μ = 500 g. [5 marks]

Question 14

(a) E(X̄) = μ
Var(X̄) = σ²/n [2 marks]

(b) X̄ is an unbiased estimator of μ because its expected value equals the population parameter μ, i.e., E(X̄) = μ. [1 mark]

Question 15

(a) Unbiased estimate of p: p̂ = 112/200 = 0.56 [1 mark]

(b) Approximate 95% CI: p̂ ± z₀.₀₂₅ × √[p̂(1 − p̂)/n]
z₀.₀₂₅ = 1.96
SE = √(0.56 × 0.44 / 200) = √(0.2464/200) = √0.001232 = 0.03510
Margin of error = 1.96 × 0.03510 = 0.06880
CI: 0.56 ± 0.0688 → (0.491, 0.629) (3 d.p.) [3 marks]

Section D: Correlation and Regression (Questions 16–20)

Question 16

(a) r = [nΣxy − (Σx)(Σy)] / √{[nΣx² − (Σx)²][nΣy² − (Σy)²]}
= [8(7728) − (96)(624)] / √{[8(1248) − 96²][8(49536) − 624²]}
= [61824 − 59904] / √{[9984 − 9216][396288 − 389376]}
= 1920 / √(768 × 6912)
= 1920 / √5308416
= 1920 / 2304 = 0.8333... ≈ 0.833 [2 marks]

(b) r = 0.833 indicates a strong positive linear correlation between hours spent studying and test score. As study hours increase, test scores tend to increase. [1 mark]

Question 17

(a) b = [nΣxy − (Σx)(Σy)] / [nΣx² − (Σx)²] = 1920/768 = 2.5
a = ȳ − bx̄ = (624/8) − 2.5(96/8) = 78 − 2.5(12) = 78 − 30 = 48
Regression line: y = 48 + 2.5x [3 marks]

(b) When x = 14: y = 48 + 2.5(14) = 48 + 35 = 83
This is interpolation since x = 14 lies within the range of the original data (x-values from the data range approximately 8 to 16 based on mean 12 and spread). The estimate is reliable because it falls within the observed range of x-values. [2 marks]

Question 18

(a) The gradient −0.8 means that for each unit increase in x, y is expected to decrease by 0.8 units, on average. [1 mark]

(b) x = 50 is far outside the range of the original data (5 to 30). This would be extrapolation, and the linear relationship observed may not hold beyond the range of the data. The prediction would be unreliable. [1 mark]

Question 19

(a) When x = 10: y = 3.2(10) + 1.5 = 32 + 1.5 = 33.5 = ȳ.
Thus (10, 33.5) lies on the line. [1 mark]

(b) The regression line of x on y minimises the sum of squared horizontal deviations, while the regression line of y on x minimises vertical deviations. The two lines are different unless r = ±1. Simply rearranging the y-on-x equation does not give the x-on-y equation. [1 mark]

Question 20

(a) r = 0.92 indicates a very strong positive linear correlation between advertising expenditure and sales revenue. Higher advertising expenditure is strongly associated with higher sales revenue. [1 mark]

(b) Correlation does not imply causation. While there is a strong association, other factors (e.g., market conditions, product quality, seasonality) may influence sales revenue. The claim of a definite increase cannot be justified solely by the correlation. [1 mark]

END OF ANSWER KEY