Cambridge International AS & A Level Mathematics 9709: Probability & Statistics 2 – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover topic 6, Probability & Statistics 2, of the Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027 (Version 4): sections 6.1 to 6.5. This content is examined on Paper 6 (A Level only). A scientific calculator is allowed on every 9709 paper, so all questions are calculator allowed, but unsupported calculator answers earn no marks. Give non-exact answers to 3 significant figures.

Related: the study guide, the revision notes, the A Level Mathematics hub, the printable 9709 checklist and the free A Level diagnostic.

Questions

1. Faults occur at random in a roll of cloth at a mean rate of 0.8 per metre.

(a) Calculate the probability that a 3-metre length contains exactly 2 faults. [2] (b) Calculate the probability that a 1.5-metre length contains at least 1 fault. [2]

2. The random variable X has mean 12 and variance 5. The independent random variable Y has mean 7 and variance 2.

(a) Find E(3X − 4) and Var(3X − 4). [2] (b) Find E(2X − 3Y) and Var(2X − 3Y). [2]

3. 0.9% of the light bulbs made by a factory are faulty. A box contains 300 bulbs.

(a) Explain why a Poisson distribution is a suitable approximation for the number of faulty bulbs in a box. [1] (b) Use this approximation to find the probability that a box contains at most 2 faulty bulbs. [3]

4. The continuous random variable X has probability density function f(x) = kx² for 0 ≤ x ≤ 3, and f(x) = 0 otherwise.

(a) Show that k = 1/9. [2] (b) Find E(X). [2] (c) Find Var(X). [3] (d) Find the median of X. [3]

5. A machine fills packets of rice. The masses are normally distributed with standard deviation 4.5 g. The mean mass should be 250 g. A random sample of 30 packets has mean mass 248.4 g. Test, at the 2% significance level, whether the mean mass is less than 250 g. [5]

6. A road junction had a mean of 3.5 accidents per month. New warning signs are installed. In the next 2 months there are 2 accidents.

(a) State one condition needed for the number of accidents to be modelled by a Poisson distribution. [1] (b) Test, at the 5% significance level, whether the mean number of accidents has decreased. [4] (c) Find the probability of a Type I error in this test. [2] (d) The mean number of accidents is actually now 1.5 per month. Find the probability of a Type II error. [3]

7. The times, x minutes, taken by 60 randomly chosen customers to complete an online form are summarised by Σx = 1512 and Σx² = 39 000.

(a) Calculate unbiased estimates of the population mean and variance. [3] (b) Calculate a 98% confidence interval for the population mean time. [3] (c) Explain why it was not necessary to assume that the times are normally distributed. [1]

8. In a random sample of 150 voters, 45 support a new policy.

(a) Calculate an approximate 95% confidence interval for the proportion of all voters who support the policy. [3] (b) 50 such intervals are calculated from independent samples. State the expected number that contain the true proportion. [1]

9. Rods of type A have lengths A ~ N(42, 0.3²) cm. Rods of type B have lengths B ~ N(28, 0.4²) cm. All lengths are independent.

(a) Two type A rods and one type B rod are placed end to end. Find the probability that the total length is more than 113 cm. [4] (b) Find the probability that twice the length of a randomly chosen type B rod exceeds the length of a randomly chosen type A rod by more than 15 cm. [4]

10. To estimate the mean journey time of a school’s 1200 students, a council asks the first 40 students to arrive one morning.

(a) Explain why this sample may be unsatisfactory. [1] (b) Describe how random numbers could be used to choose a random sample of 40 students. [2] (c) Journey times have mean 24 minutes and standard deviation 9 minutes. A random sample of 40 students is taken. Find the probability that the sample mean is more than 26 minutes. [3]

11. Emails arrive at an office at random at a mean rate of 5 per hour. The office is open for 8 hours each day.

(a) Use a suitable approximation to find the probability that more than 45 emails arrive in one day. [4] (b) On one day, 52 emails arrive. Test, at the 5% significance level, whether the mean rate has increased. [4]

12. A seed company claims that 85% of its seeds germinate. A gardener plants 20 seeds and 14 germinate. Test, at the 5% significance level, whether the claim overstates the germination rate. [5]

Answers

1. (a) X ~ Po(2.4) [1]. P(X = 2) = e^(−2.4) × 2.4²/2 = 0.261 [1] (b) Y ~ Po(1.2); P(Y ≥ 1) = 1 − e^(−1.2) [1] = 0.699 [1] Examiner insight: the method mark depends on the correctly scaled mean; Po(0.8) for a 3-metre length loses both marks.

2. (a) E(3X − 4) = 3 × 12 − 4 = 32 [1]. Var(3X − 4) = 9 × 5 = 45 [1] (b) E(2X − 3Y) = 24 − 21 = 3 [1]. Var(2X − 3Y) = 4 × 5 + 9 × 2 = 38 [1] Examiner insight: a variance of 20 − 18 = 2 in (b) scores nothing, because the variances of independent variables add even when the combination is a difference.

3. (a) n = 300 is large (> 50) and np = 2.7 is small (< 5) [1] (b) X ~ Po(2.7) [1]. P(X ≤ 2) = e^(−2.7)(1 + 2.7 + 2.7²/2) [1] = 0.494 [1] Examiner insight: both conditions are needed, with numbers; “n is large” alone earns nothing.

4. (a) ∫₀³ kx² dx = k[x³/3]₀³ = 9k [1]. 9k = 1, so k = 1/9 [1] (b) E(X) = (1/9)∫₀³ x³ dx [1] = (1/9)(81/4) = 9/4 = 2.25 [1] (c) E(X²) = (1/9)∫₀³ x⁴ dx = (1/9)(243/5) = 5.4 [1]. Var(X) = 5.4 − 2.25² [1] = 0.338 (3 s.f.) [1] (d) (1/9)∫₀ᵐ x² dx = 1/2 [1], so m³/27 = 1/2 and m³ = 13.5 [1]. m = 2.38 (3 s.f.) [1] Examiner insight: (a) is a “show that”: the integral and 9k = 1 must both appear, or no marks are given.

5. H₀: μ = 250, H₁: μ < 250 [1]. z = (248.4 − 250)/(4.5/√30) [1] = −1.947 [1]. Critical value for a 2% lower tail is −2.054, and −1.947 > −2.054 [1]. Do not reject H₀: there is insufficient evidence at the 2% level that the mean mass is less than 250 g [1] Examiner insight: the final mark needs a non-definite conclusion in context; “the mean is 250 g” loses it.

6. (a) Accidents occur independently (or at random, or at a constant mean rate) [1] (b) H₀: λ = 7, H₁: λ < 7 (for 2 months) [1]. P(X ≤ 2) = e^(−7)(1 + 7 + 24.5) [1] = 0.0296 [1]. 0.0296 < 0.05, so reject H₀: there is evidence at the 5% level that the mean number of accidents has decreased [1] (c) P(X ≤ 3) = 0.0818 > 0.05, so the rejection region is X ≤ 2 [1]. P(Type I error) = 0.0296 [1] (d) New mean for 2 months = 3 [1]. P(Type II) = P(X ≥ 3 | λ = 3) = 1 − e^(−3)(1 + 3 + 4.5) [1] = 0.577 [1] Examiner insight: in (c), giving 0.05 scores zero; for a discrete test P(Type I) is the actual probability of the rejection region.

7. (a) x̄ = 1512/60 = 25.2 [1]. s² = (39 000 − 1512²/60)/59 [1] = 897.6/59 = 15.2 (3 s.f.) [1] (b) z = 2.326 [1]. 25.2 ± 2.326 × √(15.214/60) [1]. Interval (24.0, 26.4) [1] (c) The sample is large, so by the Central Limit Theorem the sample mean is approximately normally distributed [1] Examiner insight: dividing by 60 instead of 59 in (a) gives the biased estimate and loses the method and accuracy marks.

8. (a) p̂ = 45/150 = 0.3 [1]. 0.3 ± 1.96 × √(0.3 × 0.7/150) [1]. Interval (0.227, 0.373) [1] (b) 0.95 × 50 = 47.5 [1] Examiner insight: using 1.645 instead of 1.96 in (a) loses the method mark and the accuracy mark after it.

9. (a) T = A₁ + A₂ + B, E(T) = 112 [1]. Var(T) = 2 × 0.09 + 0.16 = 0.34 [1]. z = (113 − 112)/√0.34 = 1.715 [1]. P(T > 113) = 1 − Φ(1.715) = 0.0432 [1] (b) D = 2B − A, E(D) = 56 − 42 = 14 [1]. Var(D) = 4 × 0.16 + 0.09 = 0.73 [1]. z = (15 − 14)/√0.73 = 1.170 [1]. P(D > 15) = 0.121 [1] Examiner insight: (b) needs 2² × 0.16; writing 2 × 0.16 treats 2B as two separate rods and loses the variance mark.

10. (a) Early arrivals (e.g. those living nearby) may have shorter journeys; not every student has an equal chance of selection [1] (b) Number the students 1 to 1200 [1]. Generate random numbers in this range and choose the matching students, ignoring repeats, until 40 are chosen [1] (c) By the CLT, X̄ ≈ N(24, 81/40) [1]. z = (26 − 24)/(9/√40) = 1.4055 [1]. P(X̄ > 26) = 1 − Φ(1.4055) = 0.0799 [1] Examiner insight: in (c), using 9 instead of 9/√40 is a method error and loses all later marks.

11. (a) X ~ Po(40), and λ > 15, so use N(40, 40) [1]. P(X > 45) ≈ P(Y > 45.5) [1]. z = 5.5/√40 = 0.870 [1]. P = 1 − Φ(0.870) = 0.192 [1] (b) H₀: λ = 40, H₁: λ > 40 [1]. P(X ≥ 52) ≈ P(Y > 51.5), z = 11.5/√40 = 1.818 [1]. P = 0.0345 [1]. 0.0345 < 0.05, so reject H₀: there is evidence at the 5% level that the mean rate has increased [1] Examiner insight: “more than 45” means X ≥ 46, so use 45.5; 44.5 or 46.5 loses the correction and accuracy marks.

12. H₀: p = 0.85, H₁: p < 0.85 [1]. Under H₀, X ~ B(20, 0.85) [1]. P(X ≤ 14) = 0.0673 [1]. 0.0673 > 0.05 [1]. Do not reject H₀: there is insufficient evidence at the 5% level that the claim overstates the germination rate [1] Examiner insight: compare the tail P(X ≤ 14), not P(X = 14); a single-value probability loses the method mark and all later marks.

Where marks are usually lost

• The Poisson mean not rescaled for a new interval.
• Variances subtracted, or 2B treated as B₁ + B₂.
• A “show that” for k with the key equation missing.
• Hypotheses written with x̄ or the observed count instead of μ, p or λ.
• The significance level quoted as the Type I error probability for a discrete test.
• The Type II probability calculated under H₀ instead of the stated true value.
• Conclusions stated as certain, or with no context.

Next steps

• Probability & Statistics 2 revision notes
• Probability & Statistics 2 study guide
• A Level Mathematics hub
• Printable 9709 checklist
• A Level diagnostic, or all free 10-minute diagnostics
• Book a free trial class

Official syllabus

Cambridge International AS & A Level Mathematics 9709 syllabus, for exams in 2026 and 2027 (Version 4), Cambridge University Press & Assessment. Topic 6, Probability & Statistics 2 (for Paper 6): sections 6.1 to 6.5.