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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: ______ / 80

Duration: 2 hours
Total Marks: 80

Instructions:

Answer all questions.
Write your working clearly and neatly.
The use of an approved graphing calculator (without symbolic algebra) is expected.
Show mathematical steps, not just calculator inputs.
Unless otherwise stated, give numerical answers to 3 decimal places.

Section A: Probability & Distributions (Questions 1 – 10)

1. A standard pack of 52 playing cards contains 4 suits (♠, ♥, ♦, ♣) each of 13 cards (Ace, 2, 3, …, 10, Jack, Queen, King).
(a) One card is drawn at random from the pack. Find the probability that it is either a diamond or a King. [2]

(b) Two cards are drawn at random without replacement. Find the probability that exactly one is a King and the other is a heart. [3]

2. Six different textbooks – three Mathematics, two Physics and one Chemistry – are to be arranged on a shelf.
(a) Find the number of possible arrangements if the three Mathematics books must be together. [2]

(b) A student selects any three of the six textbooks to take to school. Find the number of ways this can be done if at least one Physics book must be included. [3]

3. The events $A$ and $B$ are such that $P(A) = 0.4$ , $P(B) = 0.25$ and $P(A \cap B) = 0.1$ .
(a) Find $P(A \cup B)$ . [1]

(b) Find $P(A' \cap B)$ . [2]

(c) Determine whether $A$ and $B$ are independent, giving a reason for your answer. [2]

4. A discrete random variable $X$ has the probability distribution shown in the table.

$x$	1	2	3	4
$P(X=x)$	0.1	$p$	0.3	$q$

It is given that $E(X) = 2.8$ .
(a) Find the values of $p$ and $q$ . [3]

(b) Calculate $Var(X)$ . [2]

5. A biased coin is tossed repeatedly. The probability of obtaining a head on any toss is $0.35$ . Let $X$ be the number of heads obtained in 10 tosses.
(a) State the distribution of $X$ , including any parameters. [1]

(b) Find $P(X = 4)$ . [1]

(c) Find $P(2 \le X < 6)$ . [2]

(d) Calculate $E(3X - 2)$ and $Var(3X - 2)$ . [2]

6. The mass of a packet of chips, $M$ grams, follows a normal distribution with mean $52$ g and standard deviation $1.8$ g.
(a) Find the probability that a randomly chosen packet has a mass between $50.0$ g and $54.5$ g. [2]

(b) Packets are rejected if their mass is less than $50$ g. In a batch of $200$ packets, estimate the number rejected. [2]

7. The lifetime $T$ hours of a certain component is normally distributed with mean $\mu$ and variance $1600$ . It is known that $P(T > 580) = 0.2$ .
(a) Find the value of $\mu$ . [3]

(b) Find the value of $t$ such that $P(|T - \mu| < t) = 0.9$ . [3]

8. In a large population, $12\%$ of people have a certain allergy. A random sample of $200$ people is selected.
(a) Use a suitable approximation to find the probability that at most $30$ people in the sample have the allergy. [4]

(b) Explain why a normal approximation is valid in this case. [1]

9. A random sample of size $n$ is taken from a population with mean $75$ and variance $144$ . The sample mean is denoted by $\bar{X}$ .
(a) State the approximate distribution of $\bar{X}$ , quoting any theorem used. [2]

(b) Find the smallest value of $n$ such that $P(|\bar{X} - 75| < 1) \ge 0.95$ . [4]

10. A manufacturer claims that the mean breaking strength of a new fishing line is at least $18.0$ kg. A consumer group suspects the claim and tests a random sample of 50 lines. The population standard deviation is known to be $1.2$ kg. The sample mean is found to be $17.65$ kg.
(a) State appropriate hypotheses for the test. [1]

(b) Calculate the $p$ -value of the test. [2]

(c) State the conclusion of the test at the $1\%$ significance level, giving a reason. [2]

Section B: Sampling, Regression & Inference (Questions 11 – 20)

11. A health researcher measures the systolic blood pressure (mm Hg) of eight randomly chosen adults after a light exercise routine. The results are:

$\qquad 132,\; 141,\; 128,\; 136,\; 139,\; 134,\; 143,\; 131$

(a) Find unbiased estimates of the population mean and variance. [2]

(b) Assuming that systolic blood pressure is normally distributed, carry out a $t$ -test at the $5\%$ significance level to determine whether the population mean blood pressure after exercise exceeds $130$ mm Hg. State your hypotheses, test statistic, critical value and conclusion clearly. [5]

12. A factory has two machines, $X$ and $Y$ , producing metal rods. Random samples of $50$ rods from machine $X$ and $60$ rods from machine $Y$ are taken. The diameters (in mm) are measured, and the summarised results are:

$\qquad \bar{x} = 9.82,\; s_x^2 = 0.025$ ; $\qquad \bar{y} = 9.78,\; s_y^2 = 0.019$

(a) Obtain a $95\%$ confidence interval for the difference between the population mean diameters of the rods from the two machines, $\mu_X - \mu_Y$ . [4]

(b) Using the confidence interval in part (a), determine whether there is a significant difference between the two population means at the $5\%$ level, explaining your reasoning. [2]

13. A researcher records the number of hours of sunshine, $x$ , and the number of ice creams sold at a beach kiosk, $y$ , over eight days. The following summary statistics are obtained:

$\qquad n = 8,\;\; \sum x = 76,\;\; \sum y = 520,\;\; \sum x^2 = 780,\;\; \sum y^2 = 36\,400,\;\; \sum xy = 5\,180$

(a) Calculate the product moment correlation coefficient $r$ between $x$ and $y$ . [2]

(b) Comment on the strength of the linear relationship between hours of sunshine and ice cream sales. [1]

(c) Find the equation of the regression line of $y$ on $x$ , giving the coefficients to 3 decimal places. [3]

14. Using the regression line obtained in Question 13(c), estimate the number of ice creams sold on a day with $9.5$ hours of sunshine. [2]

Explain whether this estimate is likely to be reliable. [1]

15. The sales manager of a company believes that the weekly sales $y$ (in thousands of dollars) are related to the advertising expenditure $x$ (in thousands of dollars) by the model $y = a x^b$ , where $a$ and $b$ are constants.
(a) By taking logarithms, transform the model into a linear form suitable for regression. [1]

(b) The manager collects data over 10 weeks. The regression of $\ln y$ on $\ln x$ gives the equation $\ln y = 1.20 + 0.65 \ln x$ . Find the values of $a$ and $b$ , giving $a$ correct to 2 decimal places. [2]

16. A large population of students has a mean height of $165$ cm and a standard deviation of $9$ cm. No assumption is made about the distribution of heights. A random sample of $100$ students is selected.
(a) Write down the approximate distribution of the sample mean height, stating any theorem you have used. [2]

(b) Find the probability that the sample mean height is between $163.5$ cm and $166.5$ cm. [2]

17. The discrete random variable $X$ has the probability distribution $P(X = k) = c k^2$ , for $k = 1, 2, 3$ , where $c$ is a constant.
(a) Show that $c = \frac{1}{14}$ . [2]

(b) Find $E(X)$ and $Var(X)$ . [4]

(c) Find $E(2X + 5)$ and $Var(2X + 5)$ . [2]

18. A committee of 5 people is to be chosen from a group of 7 men and 5 women.
(a) Find the number of ways the committee can be chosen if it must contain exactly 3 men. [2]

(b) If the committee is chosen randomly, find the probability that it contains more women than men. [3]

19. In a large batch of light bulbs, $8\%$ are defective. A random sample of $180$ bulbs is selected. Let $X$ be the number of defective bulbs in the sample.
(a) State an approximate distribution for $X$ , justifying your choice. [2]

(b) Using this approximation, find the probability that the sample contains between $10$ and $20$ defective bulbs inclusive. [3]

(c) The batch is accepted if fewer than $12$ bulbs in the sample are defective. Find the approximate probability that the batch is accepted. [2]

20. A farmer claims that the mean mass of apples in his orchard is at least $150$ grams. To investigate, an inspector randomly selects $36$ apples and finds that their masses have a mean of $147.5$ g and a standard deviation of $12.0$ g.

(a) Stating your hypotheses, carry out a hypothesis test at the $5\%$ significance level to determine whether the farmer’s claim should be rejected. Use the critical region approach. [5]

(b) Explain what is meant by a Type I error in the context of this test and state the probability of making a Type I error. [2]

--- END OF QUIZ ---

Answers

A-Level Maths H2 Quiz – Statistics Probability

ANSWER KEY & MARKING SCHEME

Section A: Probability & Distributions

1.
(a) P(diamond or King) = P(diamond) + P(King) – P(diamond and King)
= 13/52 + 4/52 – 1/52 = 16/52 = 4/13.
[2 marks] – 1 for correct addition rule, 1 for correct answer.

(b) Exactly one King and one heart, but note the King of hearts is both a King and a heart.
Case 1: King (not hearts) and heart (not King): 3 non-♥ Kings, 12 non-King hearts → 3×12 = 36 ways.
Case 2: King of hearts and a non-King, non-heart card: 1 × 36 = 36 ways.
Total favourable = 72. Total ways = C(52,2) = 1326. Probability = 72/1326 = 12/221.
[3 marks] – 1 for identifying overlapping card, 1 for correct case analysis, 1 for correct probability.

2.
(a) Treat the three Maths books as a block. That gives 4 items (block + 2 Physics + 1 Chem) → 4! = 24 arrangements. Inside block, 3! = 6. Total = 24 × 6 = 144.
[2 marks] – 1 for block idea, 1 for multiplication.

(b) Total ways to choose any 3 from 6 = C(6,3) = 20.
Ways with no Physics (i.e., choose 3 from 4 non-Physics books) = C(4,3) = 4.
Number with at least one Physics = 20 – 4 = 16.
[3 marks] – 1 for total, 1 for complement, 1 for answer.

3.
(a) P(A ∪ B) = P(A) + P(B) – P(A∩B) = 0.4 + 0.25 – 0.1 = 0.55.
[1 mark]

(b) P(A′ ∩ B) = P(B) – P(A∩B) = 0.25 – 0.1 = 0.15.
[2 marks] – 1 for method, 1 for correct.

(c) If independent, P(A)P(B) = 0.4×0.25 = 0.1 = P(A∩B). Yes, they are independent.
[2 marks] – 1 for product, 1 for conclusion with reason.

4.
(a) From probabilities summing to 1: 0.1 + p + 0.3 + q = 1 → p + q = 0.6.
E(X) = 1(0.1) + 2p + 3(0.3) + 4q = 0.1 + 2p + 0.9 + 4q = 1.0 + 2p + 4q = 2.8 → 2p + 4q = 1.8.
Solve: 2(0.6 – q) + 4q = 1.8 → 1.2 – 2q + 4q = 1.8 → 2q = 0.6 → q = 0.3, p = 0.3.
[3 marks] – 1 for sum equation, 1 for expectation equation, 1 for solving.

(b) E(X²) = 1²(0.1) + 2²(0.3) + 3²(0.3) + 4²(0.3) = 0.1 + 1.2 + 2.7 + 4.8 = 8.8.
Var(X) = E(X²) – [E(X)]² = 8.8 – 2.8² = 8.8 – 7.84 = 0.96.
[2 marks] – 1 for E(X²), 1 for Var.

5.
(a) X ~ B(10, 0.35).
[1 mark]

(b) P(X = 4) = C(10,4) * (0.35)⁴ * (0.65)⁶ ≈ 0.2376. (Accept 0.238)
[1 mark]

(c) P(2 ≤ X < 6) = P(X=2,3,4,5). Sum ≈ 0.1081 + 0.2668 + 0.2376 + 0.1580 = 0.7705. (Accept 0.771)
[2 marks] – 1 for setting up, 1 for correct.

(d) E(3X – 2) = 3E(X) – 2 = 3(10×0.35) – 2 = 3(3.5) – 2 = 10.5 – 2 = 8.5.
Var(3X – 2) = 9 Var(X) = 9 × (10×0.35×0.65) = 9 × 2.275 = 20.475.
[2 marks] – 1 for E, 1 for Var.

6.
(a) X ~ N(52, 1.8²). P(50 < X < 54.5) = P((50–52)/1.8 < Z < (54.5–52)/1.8) = P(–1.111… < Z < 1.3889) ≈ 0.8666 + 0.8665? Use normal: Φ(1.389) – Φ(–1.111) = 0.9177 – 0.1335 = 0.7842. (Accept 0.784–0.785)
[2 marks] – 1 for standardising, 1 for answer.

(b) P(reject) = P(X < 50) = Φ(–1.111) ≈ 0.1335. Expected number = 200 × 0.1335 ≈ 26.7, so about 27 packets.
[2 marks] – 1 for prob, 1 for estimate.

7.
(a) T ~ N(μ, 40²) since variance = 1600. P(T > 580) = 0.2 → P(Z > (580–μ)/40) = 0.8. (580–μ)/40 = Φ⁻¹(0.8) ≈ 0.8416. μ = 580 – 40×0.8416 = 580 – 33.664 = 546.336 ≈ 546.3.
[3 marks] – 1 for sd, 1 for inverse, 1 for μ.

(b) P(|T – μ| < t) = 0.9 → P(–t/40 < Z < t/40) = 0.9 → 2Φ(t/40) – 1 = 0.9 → Φ(t/40) = 0.95 → t/40 = 1.6449 → t ≈ 65.796.
[3 marks] – 1 for setting up, 1 for normal quantile, 1 for t.

8.
(a) X ~ B(200, 0.12). np = 24, nq = 176 both > 5, so X ≈ N(24, 200×0.12×0.88 = 21.12). Continuity correction: P(X ≤ 30) ≈ P(Z ≤ (30.5 – 24)/√21.12) = P(Z ≤ 6.5/4.595) = P(Z ≤ 1.415) ≈ 0.9214.
[4 marks] – 1 for np, nq, 1 for continuity correction, 1 for standardisation, 1 for answer.

(b) Valid because np = 24 > 5 and n(1–p) = 176 > 5, so normal approximation is appropriate.
[1 mark]

9.
(a) By the Central Limit Theorem, for large n, $\bar{X} \sim N(75, 144/n)$ approximately.
[2 marks] – 1 for CLT statement, 1 for distribution.

(b) P(| $\bar{X}$ – 75| < 1) = P(–1 < $\bar{X}$ – 75 < 1) = P(–1/(12/√n) < Z < 1/(12/√n)) = 2Φ(√n/12) – 1 ≥ 0.95 → Φ(√n/12) ≥ 0.975 → √n/12 ≥ 1.96 → √n ≥ 23.52 → n ≥ 553.19. Smallest n = 554.
[4 marks] – 1 for standardising, 1 for inequality, 1 for finding quantile, 1 for final n.

10.
(a) H₀: μ = 18.0 (or μ ≥ 18.0 could be used; two-tailed? Actually suspicion of lower, so one-tailed)
H₁: μ < 18.0. (Accept H₀: μ = 18, H₁: μ < 18).
[1 mark]

(b) Test statistic: z = (17.65 – 18) / (1.2/√50) = –0.35 / 0.1697 ≈ –2.062.
p-value = P(Z < –2.062) ≈ 0.0196.
[2 marks] – 1 for test statistic, 1 for p-value.

(c) p-value = 0.0196 > 0.01, so do not reject H₀. There is insufficient evidence at 1% level to reject the manufacturer’s claim.
[2 marks] – 1 for comparison, 1 for conclusion in context.

Section B

11.
(a) n = 8. Σx = 132+141+128+136+139+134+143+131 = 1084.
Sample mean $\bar{x}$ = 1084/8 = 135.5.
Σx² = 132²+141²+128²+136²+139²+134²+143²+131² = 17424+19881+16384+18496+19321+17956+20449+17161 = 147072.
Unbiased estimate of mean = 135.5.
Unbiased variance s² = (Σx² – n $\bar{x}$ ²)/(n–1) = (147072 – 8×135.5²)/7 = (147072 – 8×18360.25)/7 = (147072 – 146882)/7 = 190/7 ≈ 27.14 (3 s.f.).
[2 marks] – 1 for mean, 1 for variance.

(b) H₀: μ = 130; H₁: μ > 130 (one‑tailed).
s = √27.14 ≈ 5.209.
t = (135.5 – 130) / (5.209/√8) = 5.5 / 1.842 ≈ 2.986.
Critical value at 5% with 7 df: t₀.₀₅,₇ = 1.895.
Since 2.986 > 1.895, reject H₀. There is sufficient evidence that the mean blood pressure after exercise exceeds 130 mm Hg.
[5 marks] – 1 for hypotheses, 1 for test statistic, 1 for critical value, 1 for comparison, 1 for conclusion in context.

12.
(a) Pooled variance: s²_p = ((50–1)×0.025 + (60–1)×0.019)/(50+60–2) = (49×0.025 + 59×0.019)/108 = (1.225 + 1.121)/108 = 2.346/108 ≈ 0.021722.
SE = √(s²_p(1/50 + 1/60)) = √(0.021722 × 0.03667) ≈ √0.0007966 ≈ 0.02822.
Difference $\bar{x} – \bar{y} = 0.04$ .
Critical value t₀.₀₂₅,₁₀₈ ≈ 1.984 (or use z 1.96 if using normal approx).
95% CI: 0.04 ± 1.984×0.02822 → 0.04 ± 0.0559 → (–0.0159, 0.0959).
[4 marks] – 1 for pooled variance, 1 for SE, 1 for t, 1 for interval.

(b) The 95% CI contains 0, so we cannot reject the null hypothesis that μ_X = μ_Y at the 5% level. There is no significant difference.
[2 marks]

13.
(a) r = (nΣxy – Σx Σy) / √( [nΣx² – (Σx)²] [nΣy² – (Σy)²] )
= (8×5180 – 76×520) / √( (8×780 – 76²)(8×36400 – 520²) )
= (41440 – 39520) / √( (6240 – 5776)(291200 – 270400) ) = 1920 / √(464 × 20800)
√(464×20800) = √(9,651,200?) compute: 464×20800=9,651,200, sqrt≈3106.64. r = 1920/3106.64 ≈ 0.6180. So r ≈ 0.618.
[2 marks] – 1 for formula, 1 for correct.

(b) Moderately strong positive linear correlation between hours of sunshine and ice cream sales.
[1 mark]

(c) b = (nΣxy – Σx Σy)/(nΣx² – (Σx)²) = 1920/464 ≈ 4.137931.
a = $\bar{y}$ – b $\bar{x}$ = 520/8 – 4.1379×(76/8) = 65 – 4.1379×9.5 = 65 – 39.3103 = 25.6897.
Regression line: y = 25.690 + 4.138x (3 decimal places).
[3 marks] – 1 for b, 1 for a, 1 for equation.

14.
(a) y = 25.690 + 4.138(9.5) = 25.690 + 39.311 = 65.001 ≈ 65.0 ice creams.
[2 marks] – 1 for substitution, 1 for value.

(b) The estimate is likely to be reliable because 9.5 hours lies within the range of observed sunshine hours (assuming data range includes 9.5) and the correlation is moderately strong. However extrapolation risks might exist but here it’s probably interpolation. (Accept any valid comment.)
[1 mark]

15.
(a) Take natural logs: ln y = ln a + b ln x → Y = A + b X where Y = ln y, X = ln x, A = ln a.
[1 mark]

(b) From ln y = 1.20 + 0.65 ln x, so A = 1.20, b = 0.65. Thus a = e^1.20 ≈ 3.32 (2 d.p.). So a = 3.32, b = 0.65.
[2 marks] – 1 for b, 1 for a.

16.
(a) By the Central Limit Theorem, $\bar{X} \sim N(165, 9^2/100) = N(165, 0.81)$ approximately.
[2 marks]

(b) P(163.5 < $\bar{X}$ < 166.5) = P((163.5–165)/0.9 < Z < (166.5–165)/0.9) = P(–1.667 < Z < 1.667) = 2Φ(1.667)–1 ≈ 2×0.9522 – 1 = 0.9044.
[2 marks] – 1 for standardisation, 1 for answer.

17.
(a) Sum of probabilities = c(1²+2²+3²) = c(1+4+9) = 14c = 1 → c = 1/14.
[2 marks]

(b) E(X) = Σ k×P(X=k) = (1×1 + 2×4 + 3×9)/14 = (1+8+27)/14 = 36/14 = 18/7 ≈ 2.571.
E(X²) = Σ k²×P(X=k) = (1×1 + 4×4 + 9×9)/14 = (1+16+81)/14 = 98/14 = 7.
Var(X) = 7 – (18/7)² = 7 – 324/49 = (343–324)/49 = 19/49 ≈ 0.3878.
[4 marks] – 2 for E(X), 2 for Var.

(c) E(2X+5) = 2E(X)+5 = 2×(18/7)+5 = 36/7+5 = 71/7 ≈ 10.143.
Var(2X+5) = 4 Var(X) = 4×19/49 = 76/49 ≈ 1.551.
[2 marks] – 1 for E, 1 for Var.

18.
(a) Choose 3 men from 7: C(7,3)=35; choose 2 women from 5: C(5,2)=10; total = 35×10 = 350.
[2 marks]

(b) More women than men means 3 women, 2 men OR 4 women, 1 man OR 5 women, 0 men.
C(5,3)×C(7,2) = 10×21 = 210; C(5,4)×C(7,1) = 5×7 = 35; C(5,5)×C(7,0) = 1×1 = 1; total favourable = 246.
Total ways = C(12,5) = 792. Probability = 246/792 = 41/132 ≈ 0.3106.
[3 marks] – 1 for enumerating cases, 1 for counts, 1 for probability.

19.
(a) X ~ B(180, 0.08) approx N(μ=14.4, σ²=180×0.08×0.92=13.248). Normal approximation valid because np=14.4>5 and n(1-p)=165.6>5.
[2 marks]

(b) P(10 ≤ X ≤ 20) with continuity correction: P(9.5 ≤ X ≤ 20.5) ≈ P( (9.5–14.4)/√13.248 ≤ Z ≤ (20.5–14.4)/√13.248 ) = P(–1.346 ≤ Z ≤ 1.676) ≈ Φ(1.676) – Φ(–1.346) = 0.9529 – 0.0891 = 0.8638.
[3 marks] – 1 for continuity, 1 for standardisation, 1 for answer.

(c) P(X < 12) = P(X ≤ 11) cont.corr.: P(X ≤ 11.5) ≈ P(Z ≤ (11.5–14.4)/√13.248 = –2.9/3.639 = –0.797) ≈ Φ(–0.797) ≈ 0.2127.
[2 marks] – 1 for continuity, 1 for value.

20.
(a) H₀: μ = 150; H₁: μ < 150 (one‑tailed).
Since population variance unknown, use t‑test. s = 12.0, n = 36, SE = 12/6 = 2.
Test statistic t = (147.5 – 150)/2 = –2.5/2 = –1.25.
Degrees of freedom = 35. Critical value at 5% one‑tailed: t₀.₀₅,₃₅ ≈ –1.690 (or use table: 1.690).
Since –1.25 > –1.690, do not reject H₀. There is insufficient evidence to reject the farmer’s claim.
[5 marks] – 1 hypotheses, 1 test statistic, 1 df/critical value, 1 comparison, 1 conclusion.

(b) A Type I error is rejecting H₀ when it is actually true. In context, that would mean concluding the mean mass is less than 150 g when it really is at least 150 g. The probability of a Type I error is the significance level, 0.05.
[2 marks] – 1 for definition, 1 for probability.

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