A Level Mathematics 9709 Paper 61, May/June 2026

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9709/61 · May/June 2026

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The masses, in grams, of apples of a certain type have the distribution $N (58.4, {6.3}^{2})$ . The apples are packed in bags, with each bag containing 10 randomly chosen apples. Find the probability that a randomly chosen bag contains apples whose total mass is between $550 g$ and $600 g$ .

[5]

In the past, $2 %$ of talking dolls made by a certain firm have been faulty. Following a change in the manufacturing process, a quality control officer suspects that the percentage of faulty talking dolls has increased. In order to test his suspicion, he selects 50 talking dolls at random. If the number of faulty talking dolls is more than 3, he will conclude that the percentage has increased.

State suitable null and alternative hypotheses for the test.

[1]

State what is meant by a Type II error in this context.

[1]

Given that the true percentage of faulty talking dolls is now $12 %$ , use a binomial distribution to find the probability of a Type II error.

[3]

In the work done by a typesetting department, incorrect characters appear randomly and independently at a constant average rate of 5 incorrect characters per 100 pages.

a(i)

A book that is typeset in the department contains 500 pages. The number of incorrect characters in the book is denoted by $X$ . State the distribution of $X$ , including the values of any parameters.

[1]

a(ii)

State an appropriate approximating distribution for $X$ , including the values of any parameters. Justify your choice of approximating distribution.

[2]

a(iii)

Use your approximating distribution to find $P (X > 35)$ .

[3]

Another book that is typeset in the department contains $n$ pages. The probability that the book contains fewer than 18 incorrect characters is less than 0.01. Use an approximating distribution to find the largest possible value of $n$ .

[5]

The score on one spin of a 5-sided spinner is denoted by the random variable $X$ with probability distribution as shown in the table.

$x$	$1$	$2$	$3$	$4$	$5$
$P (X = x)$	$0.1$	$0.3$	$0.4$	$0.1$	$0.1$

Show that $E (X) = 2.8$ .

[1]

It is given that $Var (X) = 1.16$ . The spinner is spun 300 times. The score on each spin is noted and the mean, $\overset{ˉ}{X}$ , of the 300 scores is found. Given that $P (2.8 - k < \overset{ˉ}{X} < 2.8 + k) = 0.2$ , find the value of $k$ .

[4]

Explain why it was necessary to use the Central Limit Theorem in your answer to 4(b).

[1]

For another similar spinner, it is claimed that $E (X) = 2.8$ . Fred suspects that this value is incorrect. The spinner is spun 300 times, and the mean of the 300 scores is found to be 2.92. Given that the variance of the score on one spin of this spinner is also 1.16, test Fred's suspicion at the $5 %$ significance level.

[5]

a(i)

$X$ and $Y$ are independent random variables such that $X \sim P o (2.5)$ and $Y \sim P o (λ)$ . Find $P (1 ⩽ X ⩽ 3)$ .

[2]

a(ii)

Given that $P (X = 5) = P (X + Y = 0)$ , find $P (X + Y = 1)$ .

[4]

The random variable $W$ has the distribution $B (200, 0.02)$ . It is given that, using the binomial distribution, $P (W = 2) = 0.146$ , correct to 3 decimal places. Show that a suitable approximating distribution gives the same value of $P (W = 2)$ correct to 2 decimal places, but not correct to 3 decimal places.

[4]

A random variable $X$ has probability density function given by $f (x) = {\begin{cases} \frac{8}{7} λ e^{- λ x} & 0 < x < \ln 2, \\ 0 & otherwise, \end{cases}$ where $λ$ is a constant.

Show that $λ = 3$ .

[4]

Use integration by parts to find $E (X)$ .

[4]

Mark scheme

9709/61 · May/June 2026

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1 · 5 marks

B1
N(58.4 × 10 = 584, …..)
B1
Variance = 6.3² × 10 = 396.9
M1
$\frac{550 -^{'} 584^{'}}{\sqrt{^{'} {396.9}^{'}}}$ , $\frac{600 -^{'} 584^{'}}{\sqrt{^{'} {396.9}^{'}}}$ [= −1.7066 and 0.8031]
M1
Φ('0.8031') − Φ(‘−1.7066’) = Φ('0.8031') − (1 − Φ(‘1.7066’))
A1
[= 0.7889 − 0.0440] = 0.745 (3 sf)

2a · 1 mark

B1
H₀: Percentage or p = 0.02 H₁: Percentage or p > 0.02

2b · 1 mark

B1
Concluding that the % of faulty talking dolls has not increased when it has

2c · 3 marks

B1
B(50, 0.12)
M1
0.88⁵⁰ + 50 × 0.88⁴⁹ × 0.12 + ⁵⁰C₂ × 0.88⁴⁸ × 0.12² + ⁵⁰C₃ × 0.88⁴⁷ × 0.12³ = 0.00167546 +0.011424 + 0.038165 + 0.083269
A1
0.135 (3 sf) (accept 0.134)

3a(i) · 1 mark

B1
X ~ Po(25)

3a(ii) · 2 marks

B1
X ~ N(25, 25)
B1
(λ =) 25, which is > 15

3a(iii) · 3 marks

M1
$\frac{35.5 - 25}{\sqrt{25}}$ [= 2.1]
M1
1 − Φ(‘2.1’)
A1
= 0.0179 (3 sf)

3b · 5 marks

M1
mean $= \frac{n}{20}$ $\frac{17.5 - \frac{n}{20}}{\sqrt{\frac{n}{20}}} = z$
B1
z = −2.326
M1
$\frac{n}{20} - 2.326 \sqrt{\frac{n}{20}} - 17.5 = 0$
A1
$\sqrt{\frac{n}{20}} = 5.505$ or $\sqrt{n} = 24.619$
A1
[n = 606.1] Smallest value of n is 607

4a · 1 mark

B1
0.1 + 0.6 + 1.2 + 0.4 + 0.5 = 2.8

4b · 4 marks

B1
$\overset{ˉ}{X} \sim N (2.8$ , $\frac{1.16}{300})$
M1
$\frac{2.8 + k - 2.8}{\sqrt{\frac{1.16}{300}}} = z$
M1
z = Φ⁻¹(0.6) [= 0.253 or 0.254]
A1
k $[= z \times \sqrt{\frac{1.16}{300}}] = 0.0158$ (3 sf)

4c · 1 mark

B1
Population (X) is not normally distributed (accept ‘Population not known (to be Normal)’)

4d · 5 marks

B1
H₀: µ = 2.8 H₁: µ ≠ 2.8
M1
$\frac{2.92 - 2.8}{\sqrt{\frac{1.16}{300}}}$
A1
= 1.93
M1
‘1.93’ < 1.96
A1
There is insufficient evidence that Fred’s suspicion is true Or There is insufficient evidence that the mean score has changed [Not reject H₀]

5a(i) · 2 marks

M1
$e^{- 2.5} (2.5 + \frac{{2.5}^{2}}{2!} + \frac{{2.5}^{3}}{3!}) = e^{- 2.5} (2.5 + 3.125 + 2.6042) = 0.2052 + 0.2565 + 0.2138$
A1
= 0.675 (3sf) accept 0.676

5a(ii) · 4 marks

M1
$e^{- 2.5} \times \frac{{2.5}^{5}}{5!} = e^{- (2.5 + λ)}$ $[e^{λ} = 0.8138$ or $e^{- (2.5 + λ)} = 0.0668$ or $e^{- λ} = 1.229]$
M1
λ = −ln 0.8138 or 0.2060
M1
$P (X + Y = 1) = e^{- (2.5 + 0.2060)} \times (2.5 + 0.2060)$ or $e^{- 2.706} \times 2.706$
A1
= 0.181 (3 sf)

5b · 4 marks

B1
(λ=) 4
M1
$e^{- 4} \times \frac{4^{2}}{2!}$
A1
= 0.1465[25] or 0.1465[3]
A1
= 0.15 (2dp) and = 0.147 (3dp) [0.15 = 0.146 to 2dp, but 0.147 ≠ 0.146 to 3dp]

6a · 4 marks

M1
$\frac{8}{7}$ $\int_{0}^{\ln 2}$ $λ$ $e^{- λ x}$ $d x = 1$
A1
$\frac{8}{7}$ $[- e^{- λ x}]_{0}^{\ln 2} = 1$
M1
$\frac{8}{7}$ $[1 - e^{- λ \ln 2}] = 1$
M1
$[\frac{8}{7} [1 - \frac{1}{2^{λ}}] = 1]$
A1
$1 - \frac{1}{2^{λ}} = \frac{7}{8} \Rightarrow 2^{λ} = 8$ or $(λ = \frac{- \ln (\frac{1}{8})}{\ln 2}) \Rightarrow λ = 3$

6b · 4 marks

B1
$\frac{24}{7}$ $\int_{0}^{\ln 2}$ x $e^{- 3 x}$ $d x$
M1
$= \frac{24}{7}$ $[- \frac{x}{3} e^{- 3 x} - \int$ $(- \frac{e^{- 3 x}}{3}) d x]_{0}^{\ln 2}$
A1
$= \frac{24}{7}$ $[- \frac{x}{3} e^{- 3 x} - \frac{e^{- 3 x}}{9}]_{0}^{\ln 2}$
A1
$= \frac{1}{3} - \frac{1}{7} \ln$ 2 or 0.234 (3 sf)

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