Cambridge AS & A Level Further Mathematics 9231/42 is the May/June 2026 Paper 4 (Further Probability & Statistics), variant 2. It lasts 1 hour 30 minutes, is worth 50 marks, and you answer all six questions. It counts towards the full A Level and is one of the two applied options for AS Further Maths. Three of the six questions were hypothesis tests: a sign test, a chi-squared test and a paired t-test. The biggest questions were Question 2 on the chi-squared test and Question 6 on a probability generating function (10 marks each). This guide goes through the paper question by question with the 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 four 9231 papers from this series, see A Level Further Maths 9231 May/June 2026 past papers: what came up and how it was marked.

9231/42 at a glance

Question	Topic	Marks
1	Items bought per customer: why a Wilcoxon signed-rank test may not suit; a sign test for a median at 5%	7
2	Chi-squared test for independence of bus company and reliability; which company looks least reliable	10
3	95% confidence interval for a difference in means with a pooled variance	6
4	Piecewise cumulative distribution function: median, constants a and b, the PDF, E(1/X)	9
5	Paired t-test at 10% on a claim of more than 5 marks’ improvement, and its assumption	8
6	Probability generating function: a and k, Var(X), P(Y = 0) for a sum of three observations	10
	Total	50

Question-by-question walkthrough

The mark scheme uses the standard Cambridge mark types. M marks are for a correct method applied to this problem, A marks are for accurate answers and need the M mark before them, and B marks stand alone. A DM (or DB) mark depends on an earlier mark flagged with an asterisk. FT (follow through) lets you score with your own earlier value, and AG means the answer is given on the paper, so every step must be shown. Non-exact answers go to 3 significant figures unless a question says otherwise.

On this paper, every hypothesis test ended with two marks of the same kind: an M1 for comparing your statistic or probability with the right critical value, and an A1 for a conclusion in context that uses uncertain language (“sufficient evidence to suggest…”, never “proves”).

Question 1: sign test for a median (7 marks)

The median number of items bought used to be 8. A sample of 15 customers had values from 6 up to 74, shown in a stem-and-leaf diagram.

• (a) Why not a Wilcoxon signed-rank test (1 mark): the data have outliers and look skewed, so the population may not be symmetrical, which the Wilcoxon test needs. A comment about tied ranks was also accepted.
• (b) Sign test at 5% (6 marks):
  • H₀: population median = 8; H₁: population median > 8 (B1). It has to be the median, not the mean.
  • three values are below 8 and twelve above, and under H₀ the number below 8 follows X ~ B(15, 0.5) (B1)
  • P(X ≤ 3) = (1 + 15 + 105 + 455) × 0.5¹⁵ = 0.0176 (M1 A1)
  • 0.0176 < 0.05, so reject H₀ (M1)
  • there is sufficient evidence to support the manager’s belief that the median has increased (A1)
• Alternative: a normal approximation, N(7.5, 3.75) with a continuity correction, gives z = ±2.07, compared with 1.645. This also earned full marks.

Question 2: chi-squared test for independence (10 marks)

One hundred journeys from companies A, B and C were classed as on time, late or cancelled.

• (a) Test at 5% (8 marks):
  • H₀: reliability of service is independent of bus company (B1). Both “reliability” and “company” had to appear.
  • expected frequencies (row total × column total ÷ 100), for example 12.25, 19.11 and 17.64 for on time (M1 A1, at least 3 significant figures)
  • the cancelled row has expected values 3.50, 5.46 and 5.04, and 3.50 is below 5, so cells must be combined (M1). Either combine the late and cancelled rows, or combine columns A and B.
  • work out Σ(O − E)²/E (M1): 5.89 with rows combined, or 1.03 with columns combined (A1)
  • both are below the critical value 5.991 for 2 degrees of freedom, so do not reject H₀ (M1)
  • there is insufficient evidence to suggest reliability is not independent of company (A1). “No evidence” and “sufficient evidence” wordings were not accepted.
  • Cap: with no cells combined the statistic is 6.73, compared with 9.488, and the most this route could score was 5 out of 8.
• (b) The least reliable company (2 marks): company A (B1). Its observed late and cancelled journeys are above the expected frequencies, which is not the case for B and C (DB1). With rows combined, A had 18 observed against 12.75 expected.

Question 3: pooled confidence interval (6 marks)

Seven human geography students (Σx = 445, Σx² = 29144) and nine physical geography students (Σy = 310, Σy² = 11622). Normal distributions with equal variances are assumed.

• Working:
  • Σ(x − x̄)² = 29144 − 445²/7 = 854.7 and Σ(y − ȳ)² = 11622 − 310²/9 = 944.2 (M1)
  • pooled variance s² = (854.7 + 944.2)/(7 + 9 − 2) = 128.5 (M1 A1)
  • t with 14 degrees of freedom at 97.5%: 2.145 (B1)
  • interval: (445/7 − 310/9) ± 2.145 × √(128.5 × (1/7 + 1/9)) (M1)
• Answer: [16.9, 41.4] minutes (A1). The interval for ȳ − x̄, [−41.4, −16.9], was also accepted.
• Marking: the last M1 needs a t-value, not a z-value, and follows through your s. The final answer had to be written as an interval, not as a centre plus or minus a width.

Question 4: piecewise cumulative distribution function (9 marks)

F(x) is −x³/108 + x²/12 for 0 ≤ x ≤ 3, then −x²/32 + 7x/16 + a for 3 < x ≤ b.

• (a) Median is 3 (1 mark, AG): F(3) = 9/12 − 27/108 = 0.75 − 0.25 = 0.5. Using the second piece with a = −17/32 scored B0, because a is not yet known. If you solved F(x) = ½ instead, you had to reject the roots 3 ± 3√3 as outside [0, 3].
• (b) a and b (3 marks):
  • continuity at 3: −9/32 + 21/16 + a = 0.5, so a = −17/32 (B1, AG)
  • F(b) = 1 gives b² − 14b + 49 = 0 (M1), so b = 7 (A1)
• (c) The PDF (2 marks): differentiate each piece.
  • f(x) = −x²/36 + x/6 for 0 ≤ x ≤ 3
  • f(x) = −x/16 + 7/16 for 3 < x ≤ 7
  • f(x) = 0 otherwise
  • Marking: B1 for both pieces differentiated, B1FT for the full definition with a domain covering all real numbers.
• (d) E(1/X) (3 marks):
  • ∫₀³ (−x/36 + 1/6) dx + ∫₃⁷ (−1/16 + 7/(16x)) dx
  • = [x/6 − x²/72] from 0 to 3 + [−x/16 + (7/16) ln x] from 3 to 7
  • = 1/8 + (7/16) ln(7/3) ≈ 0.496
  • Marking: B1FT for the integral set up from your f(x), M1 for integrating both parts with a log term and limits, A1 for 0.496.

Question 5: paired t-test (8 marks)

Eight students sat a test before and after a revision course. The claim is that the course raises scores by more than 5 marks on average.

• (a) t-test at 10% (7 marks):
  • H₀: μ_d = 5; H₁: μ_d > 5, where d is the population mean increase (B1)
  • differences: 6, 24, 12, 8, 5, 26, 13, 2 (M1, at most two errors). Σd = 96, Σd² = 1694, mean 12
  • unbiased variance s² = (1694 − 96²/8)/7 = 542/7 = 77.43 (M1)
  • t = (12 − 5) ÷ √(77.43/8) = 2.25 (M1 A1). Leaving out the 5 scored M0.
  • critical value for 7 degrees of freedom, one tail at 10%: 1.415. Since 2.25 > 1.415, reject H₀ (M1)
  • there is sufficient evidence to support Margaret’s claim (A1)
  • Caps: a biased variance scored at most 4 out of 7, and a two-sample test instead of a paired one at most 2 out of 7.
• (b) Assumption (1 mark): the population of differences is normally distributed. “The population mean is normal” and “the data are normal” both scored B0.

Question 6: probability generating function (10 marks)

G_X(t) = (k/t)(at + 1)², with E(X) = 1/3.

• (a) Show a = 2 and find k (5 marks):
  • G(1) = 1 gives k(a + 1)² = 1 (B1*)
  • G′(1) = 1/3 gives ka² − k = 1/3 (M1*)
  • dividing: (a² − 1)/(a + 1)² = 1/3, so (a − 1)/(a + 1) = 1/3 and a = 2 (DM1, A1 AG)
  • then k = 1/9 (B1)
• (b) Var(X) (2 marks): G_X(t) = (1/9)(4t + 4 + 1/t), so X is −1, 0 or 1 with probabilities 1/9, 4/9 and 4/9. Either use E(X²) = 5/9, or G″(1) = 2/9 with Var(X) = G″(1) + G′(1) − [G′(1)]². Both give Var(X) = 4/9. B1FT for each step.
• (c) P(Y = 0) for the sum of three observations (3 marks):
  • G_Y(t) = [G_X(t)]³ = (2t + 1)⁶/(729t³) (B1)
  • P(Y = 0) is the coefficient of t⁰: (1/729) × ⁶C₃ × 2³ (M1)
  • P(Y = 0) = 160/729 ≈ 0.219 (A1)
  • Alternative: list the cases. (−1, 0, 1) in 6 orders gives 6 × (1/9)(4/9)(4/9), and (0, 0, 0) gives (4/9)³. This earned the same 3 marks.

Where the marks are on 9231/42

• Hypothesis tests carried 21 of the 50 marks: the sign test in Question 1(b) (6), the chi-squared test in Question 2(a) (8) and the paired t-test in Question 5(a) (7).
• Probability generating functions carried 10 marks: Question 6.
• Continuous random variables carried 9 marks: Question 4.
• Confidence intervals carried 6 marks: Question 3.
• Interpreting and choosing tests carried 4 marks: Question 1(a), Question 2(b) and Question 5(b).

By topic, the t-distribution (Questions 3 and 5) carried 14 marks, non-parametric tests (Question 1) 7 and the chi-squared test (Question 2) 10.

How to self-mark 9231/42 with the mark scheme

1. Check your hypotheses word by word. Question 1 needed “population median”, not mean. Question 2 needed both “reliability” and “company”. Question 5 needed a population mean difference of 5, not 0.
2. Mark the test statistic with its method marks. The probability 0.0176, the statistic 5.89 or 1.03, and t = 2.25 each earn an A1, but only after the M1 for the method behind them.
3. Apply the caps before anything else. No cells combined in Question 2 caps you at 5/8. A biased variance in Question 5 caps you at 4/7, and a two-sample test at 2/7.
4. Check the comparison step. Each test has an M1 for comparing with the right critical value: 0.05 (or 1.645), 5.991 for 2 degrees of freedom, and 1.415 for a one-tailed t on 7 degrees of freedom.
5. Judge the conclusion strictly. It must be in context, follow from correct working and use uncertain language. In Question 2, “no evidence” was not accepted.
6. Treat AG parts as all-or-nothing. In Question 4(a) and (b) and Question 6(a), the printed result scores only if the working really gets there, and in 4(a) not by using the value of a you are asked to find later.
7. Log each lost mark as wrong hypotheses, wrong test or distribution, an arithmetic slip, or a weak conclusion, and revise in that order.

Sit 9231/42 as a timed mock

The paper gives 90 minutes for 50 marks, about 1.8 minutes per mark.

Question	Marks	Time
Question 1 (sign test)	7	12 minutes
Question 2 (chi-squared)	10	18 minutes
Question 3 (confidence interval)	6	11 minutes
Question 4 (CDF and PDF)	9	16 minutes
Question 5 (paired t-test)	8	14 minutes
Question 6 (PGF)	10	15 minutes
Check hypotheses and conclusions	–	4 minutes

Question 2 takes the longest per mark: nine expected frequencies, a decision on which cells to combine, then a second table and the sum.

Other May/June 2026 9231 papers

• 9231 Paper 12 (Further Pure Mathematics 1)
• 9231 Paper 22 (Further Pure Mathematics 2)
• 9231 Paper 32 (Further Mechanics)

Practise and download

• Download 9231/42 and its mark scheme on the Learning Portal →
• All 9231 May/June 2026 papers: overview and marking patterns
• Try a free A Level Further Maths quiz

Need help with a 9231 topic? Book a free trial lesson with an A Level Further Maths tutor →

Frequently asked questions

What topics were on 9231 Paper 42 May/June 2026? +

Six questions: a sign test for a median (and why a Wilcoxon test may not suit the data), a chi-squared test for independence of bus company and reliability, a pooled two-sample confidence interval, a piecewise cumulative distribution function, a paired t-test, and a probability generating function.

How long is 9231/42 and how many marks is it? +

It lasts 1 hour 30 minutes and is out of 50 marks, with six compulsory questions. Questions 2 and 6 are the biggest at 10 marks each.

Was the chi-squared result in 9231/42 Question 2 significant? +

No. The cancelled row had expected frequencies below 5, so you combine rows (late and cancelled) or columns (A and B). The statistic is then 5.89 or 1.03, both below the critical value of 5.991 with 2 degrees of freedom, so there is not enough evidence that reliability depends on the company.

What happens if you don't combine cells in 9231/42 Question 2? +

The mark scheme capped an answer with no rows or columns combined at 5 out of 8. That route gives a statistic of 6.73, compared with 9.488 on 4 degrees of freedom.

What was P(Y = 0) in 9231/42 Question 6? +

160/729, about 0.219. With a = 2 and k = 1/9, X takes the values −1, 0 and 1 with probabilities 1/9, 4/9 and 4/9. The sum of three observations is 0 either when all three are 0 or when they are −1, 0 and 1 in any of 6 orders.

Where can I download 9231/42 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 Further Mathematics 9231 past papers page.

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