A Level Mathematics 9709 Paper 63, May/June 2026

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9709/63 · May/June 2026

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A five-sided spinner has edges marked 1, 2, 3, 4, 5. Ayesha believes that the spinner is biased so that it lands on five more often than it should. She spins the spinner 200 times, and it lands on five on 60 spins. Use Ayesha’s result to calculate an approximate 94% confidence interval for the probability that the spinner lands on five on one spin.

[3]

Bettie believes that 30% of students at her school are vegetarian. Amir claims that the proportion is less than 30%. In order to test his claim, Amir takes a random sample of 20 students at the school. He finds that exactly 2 students in the sample say that they are vegetarian.

Use a binomial distribution to test Amir’s claim at the 2.5% significance level.

[5]

State the smallest significance level at which Amir’s claim would be rejected. Give your answer correct to 3 significant figures.

[1]

A random variable $X$ takes values between 0 and 6 only and has probability density function $f (x)$ as shown in the diagram, where $a$ is a constant. Given that $E (X) = 2$ , find $Var (X)$ .

[6]

Cars arrive at a junction at a constant average rate of 3.2 per minute.

State a condition for the number of cars arriving per minute to have a Poisson distribution.

[1]

It is now given that the number of cars arriving per minute has the distribution $P o (3.2)$ . Find the probability that at least 4 cars arrive during a randomly chosen 3-minute period.

[3]

Given that at least 4 cars arrive during a particular randomly chosen 3-minute period, find the probability that exactly 5 cars arrive during this period.

[2]

Use a suitable approximating distribution to find the probability that more than 60 cars arrive during a randomly chosen 20-minute period.

[5]

The lengths, $X cm$ , of Wamba snakes have mean $μ cm$ . Research has shown that, for the whole population of Wamba snakes, $μ = 41.5$ . A researcher claims that, for the Wamba snakes in a particular region, the mean length is less than 41.5. She measures the lengths of a random sample of 150 Wamba snakes in this region. The results are summarised below. $n = 150 Σ x = 6030 Σ x^{2} = 252014$

Find unbiased estimates of the population mean and variance of $X$ for Wamba snakes in this region, showing that the unbiased estimate of variance is 64.483 correct to 3 decimal places.

[3]

Assuming that $Var (X) = 64.483$ for the whole population of Wamba snakes, test the researcher’s claim at the 2% significance level.

[5]

Use your answer to 5(b) to explain whether it is possible for the researcher to make a Type I error.

[1]

Later the researcher carries out a similar test at the 2% significance level, using the same hypotheses and another random sample of 150 Wamba snakes from this region. Given that the population mean length of Wamba snakes for this region is $38.7 cm$ , find the probability of a Type II error.

[5]

One batch of a cake mixture contains $X litres$ of ingredient $A$ and $Y litres$ of ingredient $B$ , where $X$ and $Y$ have the independent distributions $X \sim N (25.5, 0.07)$ and $Y \sim N (26.0, 0.08)$ .

Find the probability that, in a randomly chosen batch of the mixture, the number of litres of ingredient $A$ is greater than the number of litres of ingredient $B$ .

[5]

The mass of 1 litre of ingredient $A$ is $m kg$ and the mass of 1 litre of ingredient $B$ is $1.5 m kg$ , where $m$ is a constant. Given that the probability that the total mass of a randomly chosen batch of the mixture is less than $70 kg$ is 0.3, find the value of $m$ .

[5]

Mark scheme

9709/63 · May/June 2026

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1 · 3 marks

M1
Use of correct formula. Any z-value. Must be a z value. Either side (+ or – not necessarily ±).
B1
z = 1.881 or 1.882
A1
0.239 to 0.361 (3 sf)

2a · 5 marks

B1
H₀: P(vegetarian) or p = 0.3 H₁: P(vegetarian) or p < 0.3
M1
P(≤ 2 vegetarian in 20 | H₀) = 0.7²⁰ + 20×0.7¹⁹×0.3 + ²⁰C₂×0.7¹⁸×0.3² OR = 0.0007979 + 0.0068393 + 0.027846
A1
= 0.0355 (3 sf)
M1
0.0355 > 0.025
1 mark
Insufficient evidence to suggest that the proportion of vegetarians is less than 30% OR Insufficient evidence [at 2.5% level] to support Amir’s claim Oe [Do not reject H₀]

2b · 1 mark

1 mark
3.55% or 3.56%

3 · 6 marks

M1
½ × 6 × a = 1
A1
a = ⅓
1 mark
y = -1/18 x + ⅓ or y = -a/6 x + a
M1
∫₀⁶ (-1/18 x³ + ⅓ x²) dx or ∫₀⁶ (-a/6 x³ + a x²) dx
1 mark
= [-1/72 x⁴ + 1/9 x³]₀⁶ [= 6] or [-a/24 x⁴ + a/3 x³]₀⁶ [= 18a]
A1
Var(X) = 6 − 2² = 2

4a · 1 mark

B1
Cars arrive independently or singly or at random

4b · 3 marks

B1
λ = 9.6
M1
1 − e⁻⁹·⁶(1 + 9.6 + 9.6²/2! + 9.6³/3!) OR 1 – e⁻⁹·⁶(1 + 9.6 + 46.08 + 147.456) OR 1 – (0.0000677 + 0.0006502 + 0.003121 + 0.009987)
A1
= 0.986 (3 sf)

4c · 2 marks

M1
(e⁻⁹·⁶ × 9.6⁵/5!) / '0.986' OR 0.04602 / ‘0.986’
A1
= 0.0467 (3 sf)

4d · 5 marks

B1
N(64, 64)
M1
(60.5 − 64) / √64
A1
= −0.4375
M1
1 − Φ(‘−0.4375’) = Φ(‘0.4375’)
A1
= 0.669 (3 sf)

5a · 3 marks

B1
est(μ) = 6030/150 or 40.2 or 201/5
M1
est(σ²) = 150/149 (252014/150 − (6030/150)²) or 1/149 (252014 − 6030²/150) [= 9608/149]
A1
= 64.483

5b · 5 marks

B1
H₀: μ = 41.5 H₁: μ < 41.5
M1
(6030/150 − 41.5) / √(64.483/150)
A1
= −1.983
M1
−1.983 > −2.054 or 1.983 < 2.054 or 0.0237 > 0.02
1 mark
There is insufficient evidence that the (mean) lengths (of Wamba snakes) are less than 41.5 cm or there is insufficient evidence to support researcher’s claim [Do not reject H₀]

5c · 1 mark

B1
No because H₀ not rejected

5d · 5 marks

M1
(x̄ − 41.5) / √(64.483/150) = −2.054 or −2.055
A1
Critical value of x̄ is 40.15(3)
M1
(40.153 − 38.7) / √(64.483/150) [= 2.216 or 2.217]
M1
1 − Φ(“2.216”)
A1
= 0.0134 or 0.0133 or 0.0135 (3 sf)

6a · 5 marks

B1
E(X − Y) = 25.5 – 26.0 (= −0.5)
B1
Var( X − Y) = 0.07 + 0.08 (= 0.15)
M1
(0 − (−0.5)) / √'0.15' [= 1.29099…]
M1
1 − Φ(‘1.29099…’)
A1
= 0.0983 or 0.0984 (3 sf)

6b · 5 marks

B1
E(total mass) = m(25.5 + 1.5 × 26) ( = 64.5m)
B1
Var(total mass) = m²(0.07 + 1.5² × 0.08) (= 0.25m²)
M1
(70 − '64.5m') / '0.5m' = z
B1
z [= −Φ⁻¹(0.7)] = −0.524
A1
[70 = m(−0.262 + 64.5)] m = 1.09 (3 sf)

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