Cambridge AS & A Level Mathematics 9709/62 is the May/June 2026 Paper 6 (Probability & Statistics 2), variant 2. It lasts 1 hour 15 minutes and is worth 50 marks. It is an A Level paper, taken with Papers 1, 3 and 5, because it builds on Paper 5. It had seven compulsory questions, answered on the question paper. The biggest were Question 4 (Poisson sums and a normal approximation) and Question 7 (a probability density function), at 10 marks each, followed by the z-test in Question 6 at 9 marks. This guide goes through the paper question by question with the final answers and what the official mark scheme gave marks for. The paper and its mark scheme are free on the Tutopiya Learning Portal.

For an overview of all six 9709 papers from this series, see A Level Maths 9709 May/June 2026 past papers: what came up and how it was marked.

9709/62 at a glance

Question	Topic	Marks
1	Unbiased estimates of the population mean and variance	3
2	Linear combination of independent normal variables	5
3	Poisson hypothesis test: hypotheses, rejection value, P(Type I error)	6
4	Sum of two Poisson variables, a conditional probability, a normal approximation	10
5	Approximate confidence intervals for a proportion	7
6	One-tailed z-test for a mean, the Central Limit Theorem, a range of sample sizes	9
7	Probability density function: E(X) and the exact median	10
	Total	50

Question-by-question walkthrough

The mark scheme uses three main types of mark. M marks are for a valid method applied to this problem, A marks are for accurate answers and need the M mark before them, and B marks are for a correct result or statement and stand alone. FT (follow through) lets a mark follow from your own earlier answer, dep marks need an earlier mark first, and SC is a special-case mark. CWO means correct working only. The cover asks for non-exact answers to 3 significant figures, and there are no marks for unsupported calculator answers.

Question 1: unbiased estimates (3 marks)

Ten values with Σx = 88.9 and Σx² = 794.47.

• Mean: 8.89. 88.9 ÷ 10.
• Variance: 0.461. s² = (1/9)(794.47 − 88.9²/10) = 4.149 ÷ 9.
• Marking: B1 for the mean, M1 for the unbiased variance formula with your sums, A1. The biased value 0.415 (dividing by 10) scored M0 A0.

Question 2: three train journeys (5 marks)

A ~ N(28, 15), B ~ N(23, 8) and C ~ N(99, 24), independent; find P(C > 2(A + B)).

• Answer: 0.390. Let D = C − 2(A + B). E(D) = 99 − 2 × 51 = −3. Var(D) = 24 + 4 × (15 + 8) = 116, because multiplying by 2 multiplies the variance by 4. Then P(D > 0) = 1 − Φ(3 ÷ √116) = 1 − Φ(0.2785) = 0.390 (0.39 accepted).
• Marking: B1 for the mean −3, M1 for the variance 24 + 4(15 + 8), M1 for standardising, M1 for the correct area, A1.

Question 3: Poisson hypothesis test (6 marks)

Vehicles in one minute at a junction had mean 2.4; after a change, one day’s count X is used to test at the 5% level whether the mean has increased.

• (a) Hypotheses (1 mark): H₀: population mean = 2.4; H₁: population mean > 2.4. B1. Writing “mean” without “population” (or λ or μ) did not score.
• (b) Smallest value that rejects H₀ (4 marks): under Po(2.4), P(X ≥ 5) = 1 − e⁻²·⁴(1 + 2.4 + 2.4²/2! + 2.4³/3! + 2.4⁴/4!) = 0.0959, more than 0.05; P(X ≥ 6) = 0.0959 − P(X = 5) = 0.0959 − 0.0602 = 0.0357, less than 0.05. So the smallest value is 6. M1 for a Poisson tail probability, M1 for P(X ≥ 5) or P(X ≥ 6), A1 for both values, then B1dep for 6. Working with P(X ≤ 4) = 0.9041 and P(X ≤ 5) = 0.9643 was equivalent. Point probabilities such as P(X = 6) scored nothing, even with the right answer.
• (c) P(Type I error) (1 mark): the probability of landing in the rejection region when H₀ is true, 0.0357. B1FT from your answer to (b), as long as it was below 0.05.

Question 4: two Poisson variables (10 marks)

D ~ Po(1.7) and E ~ Po(2.1), independent.

• (a) P(D + E < 4) (3 marks): D + E ~ Po(3.8), so P(D + E ≤ 3) = e⁻³·⁸(1 + 3.8 + 3.8²/2! + 3.8³/3!) = 0.473 (0.474 accepted). B1 for the mean 3.8, M1 for the Poisson expression (one end error allowed), A1.
• (b) P(D = 1 given D + E < 4) (3 marks): if D = 1, then E must be 0, 1 or 2. P(D = 1) × P(E ≤ 2) = 0.31056 × 0.64963 = 0.20175, and 0.20175 ÷ 0.47348 = 0.426 (0.427 accepted). M1 for the numerator (one term missing or extra was allowed), M1 for dividing by your part (a), A1. An unjustified correct answer scored only 2.
• (c) P(T > 190), where T is the total of 100 values of D (4 marks): T ~ Po(170), approximated by N(170, 170). With a continuity correction, z = (190.5 − 170) ÷ √170 = 1.572, so P(T > 190) = 1 − Φ(1.572) = 0.0580. B1 for N(170, 170), M1 for standardising (a missing or wrong continuity correction still earned this M1), M1 for the correct area, A1. Working with the sample mean, N(1.7, 1.7/100), and 1.905 led to the same answer; mixing the two methods did not score.

Question 5: confidence intervals for a proportion (7 marks)

Eric hits the centre x times in 100 throws, and the width of his approximate 95% confidence interval for p is 0.165.

• (a) The two possible values of x (4 marks): the width is 2 × 1.96 × √(p(1 − p)/100) with p = x/100. Setting this equal to 0.165 gives p(1 − p) = 0.177172, so p² − p + 0.177172 = 0 (or x² − 100x + 1771.72 = 0). The roots give x = 77 or x = 23. M1 for the width equation (a missing factor of 2 was condoned), B1 for 1.96, M1 for forming the three-term quadratic, A1. The answers had to be whole numbers of throws; ending with 0.77 and 0.23 scored A0.
• (b) Find a (3 marks): each a% interval contains p with probability y = a/100, so the chance that exactly one of two independent intervals does is 2y(1 − y) = 0.1128. This gives y² − y + 0.0564 = 0, so y = 0.94 or 0.06, and since a > 90, a = 94. M1 for 2y(1 − y) (leaving out the 2 was allowed for this mark), M1 for the quadratic, A1.

Question 6: time spent in a shop (9 marks)

T ~ N(μ, 9.2); last year μ = 12.4. A sample of 75 customers has mean 11.9.

• (a) Test at the 5% level whether μ < 12.4 (5 marks): H₀: μ = 12.4, H₁: μ < 12.4. The test statistic is (11.9 − 12.4) ÷ √(9.2/75) = −1.428. Since −1.428 > −1.645 (or the p-value 0.0767 > 0.05), H₀ is not rejected: there is insufficient evidence that the mean time has decreased. B1 for the hypotheses (in terms of μ or the population mean), M1 for the standardisation using 75, A1 for −1.428, M1 for a valid comparison, A1FT for a conclusion in context that is not stated as certain and says “decreased”, not “changed”. The critical value method (11.82 < 11.9) earned the same marks. A two-tailed test was capped at 3 marks out of 5.
• (b) Was the Central Limit Theorem needed? (1 mark): No, because the population (T) is normally distributed, so the sample mean is normal whatever the sample size. B1 needed both “no” and the reason, referring to the population, not to the sample mean or the data.
• (c) Range of sample sizes (3 marks): with a sample mean of 11.8, rejecting H₀ needs (11.8 − 12.4) ÷ √(9.2/n) < −1.645, which gives √n > 1.645 × √9.2 ÷ 0.6, so n > 69.2 and n ≥ 70. M1 for the inequality, M1 for rearranging to make n (or √n) the subject using 1.645, A1 CWO. Writing n > 70 lost the A mark, because 70 itself works.

Question 7: probability density function (10 marks)

f(x) = a − x for 0 ≤ x ≤ a, and 0 otherwise.

• (a) Show E(X) = k√2 (6 marks): the area under f is ½a² = 1, so a = √2. Then E(X) = ∫₀^√2 x(√2 − x) dx = [√2x²/2 − x³/3] from 0 to √2 = √2 − 2√2/3 = ⅓√2, so k = ⅓. M1 for the area (as a triangle or an integral), M1 for setting it equal to 1 and solving, A1 for a = √2, M1 for integrating xf(x) with the correct limits, A1FT for the correct integration, A1 for the answer. Without the integration shown, a correct answer could earn only special-case marks.
• (b) Exact median (4 marks): the area from m up to √2 is a triangle, so ½(√2 − m)² = ½, which gives m² − 2√2m + 1 = 0 and m = √2 ± 1. m = √2 − 1, because √2 + 1 is greater than √2 and so outside the range of X. M1 for setting an area from 0 to m (or from m to √2) equal to ½, A1FT for the three-term quadratic, B1 for rejecting √2 + 1 with a reason, A1 for the exact answer.

Where the marks are on 9709/62

• Hypothesis testing carried 15 of the 50 marks: the Poisson test in Question 3 (6) and the z-test for a mean in Question 6 (9).
• The Poisson distribution and its normal approximation carried 10 marks (Question 4).
• Continuous random variables carried 10 marks (Question 7).
• Confidence intervals carried 7 marks (Question 5).
• Linear combinations of random variables carried 5 marks (Question 2).
• Unbiased estimates carried 3 marks (Question 1).

Questions 2, 4(c) and 6 all used the normal distribution, and the standard values 1.645 and 1.96 both appeared.

How to self-mark 9709/62 with the mark scheme

1. Check your hypotheses are about the population. Questions 3(a) and 6(a) needed “population mean”, λ or μ; “mean” on its own did not score.
2. Check your probabilities are tail probabilities. In Question 3(b), using P(X = 5) or P(X = 6) scored nothing, even with the right final answer.
3. Write the conclusion in context and without certainty. In Question 6(a), the final A1 needed “insufficient evidence that the mean time has decreased” or similar; a one-tailed comparison with −1.645 was needed for full marks.
4. Check inequality signs. In Question 6(c), n ≥ 70 was required; n > 70 lost the mark. In Question 4(a), “fewer than 4” means up to 3.
5. Give reasons where a value is rejected. In Question 7(b), the B1 was for rejecting √2 + 1 because it is outside 0 ≤ x ≤ √2.
6. Check accuracy and form. Probabilities to 3 significant figures (0.390, 0.0357, 0.473, 0.426, 0.0580), whole numbers in Question 5(a), an exact answer in Question 7(b) and an exact k in Question 7(a).
7. Log each lost mark as a wrong variance, a missing continuity correction, a wrong tail, a missing context, or an accuracy slip.

Sit 9709/62 as a timed mock

The paper gives 75 minutes for 50 marks, a minute and a half per mark. A workable plan:

Block	Marks	Time
Questions 1 and 2	8	11 minutes
Question 3 (Poisson test)	6	9 minutes
Question 4 (Poisson sums)	10	15 minutes
Question 5 (confidence intervals)	7	10 minutes
Question 6 (z-test)	9	13 minutes
Question 7 (probability density function)	10	14 minutes
Check, especially hypotheses, conclusions and inequality signs	–	3 minutes

Question 5 is short on marks but needs an unfamiliar set-up in both parts, so don’t let it run over; come back to it once Questions 6 and 7 are done if it stalls.

Other May/June 2026 9709 papers

• 9709 Paper 12 (Pure Mathematics 1)
• 9709 Paper 22 (Pure Mathematics 2)
• 9709 Paper 32 (Pure Mathematics 3)
• 9709 Paper 42 (Mechanics)
• 9709 Paper 52 (Probability & Statistics 1)

Practise and download

• Download 9709/62 and its mark scheme on the Learning Portal →
• All 9709 May/June 2026 papers: overview and marking patterns
• Try a free A Level Maths quiz

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Frequently asked questions

What topics were on 9709 Paper 62 May/June 2026? +

Seven questions: unbiased estimates of mean and variance, a linear combination of normal variables, a Poisson hypothesis test, sums of Poisson variables with a normal approximation, confidence intervals for a proportion, a z-test for a population mean, and a probability density function.

How many marks is 9709/62 and how long is it? +

It is out of 50 marks and lasts 1 hour 15 minutes. All seven questions are compulsory. Questions 4 (Poisson) and 7 (the probability density function) are the biggest at 10 marks each.

What was the smallest value of X that rejects H₀ in 9709/62 Question 3? +

6. Under Po(2.4), P(X ≥ 5) = 0.0959, which is more than 0.05, and P(X ≥ 6) = 0.0357, which is less than 0.05. So the rejection region is X ≥ 6, and P(Type I error) is 0.0357.

What was the conclusion of the z-test in 9709/62 Question 6? +

There was not enough evidence that the mean time had decreased. The test statistic was (11.9 − 12.4) ÷ √(9.2/75) = −1.428, which is above the critical value −1.645, so H₀ is not rejected. The conclusion had to be in context and not stated as certain.

What was the median in 9709/62 Question 7? +

√2 − 1. The density integrates to 1 only when a = √2, and the area from the median up to √2 is ½(√2 − m)² = ½, so m = √2 ± 1. The value √2 + 1 is rejected because it is larger than √2, outside the range of X.

Where can I download 9709/62 May/June 2026 and its mark scheme? +

Both are free to download and print on the Tutopiya Learning Portal, on the Cambridge AS & A Level Mathematics 9709 past papers page.

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