A Level Mathematics 9709 Paper 62, Oct/Nov 2025

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9709/62 · Oct/Nov 2025

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The number, $X$ , of used computers donated to a charity has a constant average rate of 2.4 computers per week.

State a necessary condition for $X$ to have a Poisson distribution.

[1]

Now assume that $X$ has a Poisson distribution. Calculate the probability that the number of computers donated during a 4-week period is more than 6 and less than 9.

[3]

Use a suitable approximating distribution to calculate the probability that more than 50 computers are donated during a 20-week period.

[4]

The random variable $X$ has a normal distribution with mean 10 and standard deviation 3. The independent random variable $Y$ has a Poisson distribution with mean 4.

Find the standard deviation of $X + Y$ .

[3]

Find the standard deviation of $5 X - Y$ .

[3]

The times, in minutes, taken by students to complete a test have mean $μ$ and standard deviation $σ$ . The times taken by a random sample of 100 students are noted and are used to calculate a 95% confidence interval for $μ$ .

Given that the end points of the 95% confidence interval are 31.02 and 33.98, correct to 4 significant figures, calculate the value of $σ$ .

[3]

The calculation of the confidence interval required the use of the Central Limit theorem. Explain why it is valid to use the Central Limit theorem in this case.

[1]

A researcher calculates a number, $r$ , of 95% confidence interval for $μ$ . Find the largest value of $r$ such that the probability that all $r$ confidence intervals contain the true value of $μ$ is greater than 0.5.

[4]

An inspector believes that 18% of cups made at a certain factory contain flaws. The factory owner claims that the true percentage is less than 18%. The inspector examines a random sample of 40 cups and finds that 3 of them contain flaws.

Stating a necessary assumption, use a binomial distribution to test the factory owner’s claim at the 5% significance level.

[6]

Explain why it would not be appropriate to use the Poisson approximation to the binomial distribution to carry out the test in part (a).

[1]

A random variable $X$ has probability density function given by $f (x) = {\begin{cases} k (2 x^{2} - x^{3}) & 0 ⩽ x ⩽ 2, \\ 0 & otherwise. \end{cases}$

Show that $k = \frac{3}{4}$ .

[3]

b(i)

The median of $X$ is denoted by $m$ . Write down the value of $P (X ⩽ m)$ .

[1]

b(ii)

Hence find $P (E (X) ⩽ X ⩽ m)$ .

[5]

The weekly profit, in dollars, made by a certain firm has a normal distribution. In the past, the weekly profit had the distribution $N (736, 26^{2})$ . Following a change in management, the mean weekly profit for 35 randomly chosen weeks is $$ 725$ .

Stating a necessary assumption, test at the 2% significance level whether the mean weekly profit has decreased.

[6]

The mean weekly profit for another random sample of 35 weeks is found and a similar test is carried out at the 2% significance level. State the probability of a Type I error.

[1]

Given that the mean weekly profit is now in fact $$ 718$ , find the probability of a Type II error.

[5]

Mark scheme

9709/62 · Oct/Nov 2025

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1a · 1 mark

B1
Computers are donated independently or singly or randomly

1b · 3 marks

B1
λ = 9.6
M1
e−9.6(9.67/7! + 9.68/8!) = e−9.6 (1490.97 + 1789.16) = 0.10098 + 0.12118
A1
= 0.222 (3 sf)

1c · 4 marks

B1
N(48, 48)
M1
(50.5 − 48) / √48 [= 0.361]
M1
1 − Φ(‘0.361’)
A1
= 0.359 (3 sf)

2a · 3 marks

B1
Var(Y) = 4
M1
Var(X + Y) = 32 + ‘4’
A1
sd of X + Y = √13 = 3.61

2b · 3 marks

B1
Var(5X ) = 52 × 32
M1
Var(5X − Y) = Var(5X) + Var(Y) = “52 × 32” + “4”
A1
sd of 5X − Y = √229 or 15.1 (3 sf)

3a · 3 marks

B1
Sample mean = 32.5 soi or Width = 2.96
M1
32.5 + 1.96 × σ/√100 = 33.98 or 2.96 = 2 × 1.96 × σ/√100
A1
σ = 7.55 (3 sf)

3b · 1 mark

B1
Because the sample size is large.

3c · 4 marks

M1
0.95r > 0.5 Allow ‘=’
A1
r ln 0.95 > ln 0.5 Allow ‘=’ Correctly take logs and use log rule
A1
r < 13.5.... Allow ‘=’
A1
Largest value of r is 13

4a · 6 marks

B1
Assume flaws in cups are independent OR prob of containing a flaw is the same for all cups
B1
H0: P(contains flaw) = 0.18 H1: P(contains flaw) < 0.18
M1
[P(X ⩽ 3) =] 0.8240 + 40×0.8239×0.18 + 40C2×0.8238×0.182 + 40C3×0.8237×0.183 =0.0003569 + 0.0031338 + 0.013414 + 0.037298
A1
0.0542 (3 sf)
M1
0.0542>0.05
A1
[Accept H0] Insufficient evidence to accept factory owners claim. [Insufficient evidence that percentage less than 18%]

4b · 1 mark

B1
np = 7.2 which is too large, or > 5. Or n = 40 which is not greater than 50. Or p = 0.18 which is not small or > 0.1.

5a · 3 marks

M1
k ∫02 (2x2 − x3) dx = 1
A1
k [2x3/3 − x4/4]02 = 1
A1
k × 4/3 = 1 k = 3/4

5b(i) · 1 mark

B1
0.5

5b(ii) · 5 marks

M1
E(X) = 3/4 ∫02 (2x3 − x4) dx
A1
= 3/4 [2x4/4 − x5/5]02 = 6/5
M1
P(X < E(X)) = 3/4 ∫0'1.2' (2x2 − x3) dx
A1
= 3/4 [2x3/3 − x4/4]0'1.2' = 594/1250 or 297/625 or 0.4752
A1
P(E(X) ⩽ X ⩽ m) = 0.5 − ‘0.4752’ = 0.0248 or 31/1250

6a · 6 marks

B1
Assume sd is still 26 or is unchanged
B1
H0: Population mean = 736 H1: Population mean < 736
M1
± (725 − 736) / (26/√35)
A1
= −2.503
M1
−2.503 < −2.054(or 5 or 6) or 0.0062 < 0.02
A1
[Reject H0] There is sufficient evidence that (mean) weekly profit has decreased

6b · 1 mark

B1
0.02

6c · 5 marks

M1
(a − 736) / (26/√35) = −2.054 or -2.055
A1
a = 726.973
M1
('726.973' − 718) / (26/√35) (= 2.042)
M1
= 1 – ɸ('2.042')
A1
= 0.0203 to 0.0207

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