A Level Mathematics 9709 Paper 62, May/June 2025

Question paper and mark scheme50 marks8 questions

Solve it on ExamPal

Practise

Untimed. Write on the paper and check as you go.

Start

Exam

Timed, with a clock you set, like sitting the real paper.

Start

Opens full screen.

Download PDF Mark scheme

100%

Opening the question paper

Read as textThe questions and the mark scheme as plain text. Easier to read on a phone.

Question paper

9709/62 · May/June 2025

Download PDF

One of a group of three students is to be chosen at random. Explain how a single throw of a fair six-sided dice could be used to make the choice. [1]

[1]

The times, in minutes, taken by students to complete a test are normally distributed with mean 125 and variance 50. Two students are chosen at random. Find the probability that the difference between the times taken by these two students to complete the test is more than 12 minutes. [5]

[5]

The height of a certain species of plant is denoted by $H cm$ . The heights of a random sample of 100 plants were measured, and the following results were found. • The mean, $\overset{ˉ}{h}$ , for the sample was 80.2. • An unbiased estimate of the population variance of $H$ was 15.6. Calculate the value of $Σ h^{2}$ . [3]

[3]

The random variable $X$ has the distribution $Po (15)$ .

Write down an expression in terms of $e$ for $P (X = 12)$ . [1]

[1]

It is given that $P (X = n) = P (X = n + 1)$ . Write down an equation in $n$ , and hence find the value of $n$ . [3]

[3]

A biased spinner has four sides. Each side is of a different colour: yellow, red, green or black. The probability, $p$ , that the spinner will land on red is unknown. The spinner was spun 200 times, and the proportion, $a$ , of times that it landed on red was noted. This proportion was used to calculate an approximate 90% confidence interval for $p$ . The width of this confidence interval was 0.1066 correct to 4 significant figures. Find the two possible values of $a$ . [4]

[4]

The amount of time, in minutes, spent by a customer on one visit to a certain shop is modelled by the random variable $X \sim N (μ, σ^{2})$ . In the past, the values of $μ$ and $σ$ were 10.5 and 3.8 respectively. The shop has recently moved to a new location, and the manager hopes that the new value of $μ$ will be greater than 10.5. He takes a random sample of 10 customers and notes the time they each spend in the shop. He then calculates the sample mean $\overset{ˉ}{x}$ for these 10 times. Using a hypothesis test at the 5% significance level, the manager finds that there is sufficient evidence to conclude that the new value of $μ$ is greater than 10.5. Stating a necessary assumption, find the smallest possible value of $\overset{ˉ}{x}$ . [4]

[4]

Use suitable approximating distributions to answer the following.

a(i)

The random variable $W$ has the distribution $B (700, 0.005)$ . Find $P (W ⩾ 4)$ . [3]

[3]

a(ii)

Two values of $W$ are chosen at random. Find the probability that the sum of these two values is less than 3. [3]

[3]

The random variable $X$ has the distribution $Po (200)$ . Use a suitable approximating distribution to find $P (X > 205)$ . [4]

[4]

The random variable $X$ has probability density function given by $f (x) = {\begin{cases} \frac{k x^{2}}{a^{2}} & 0 ⩽ x ⩽ a, \\ 0 & otherwise, \end{cases}$ where $k$ and $a$ are positive constants.

Show that $k = \frac{3}{a}$ . [3]

[3]

It is given that $E (X) = 1$ . Find the value of $a$ . [3]

[3]

Find the median of $X$ . [3]

[3]

Birgitte has a six-sided dice. She suspects that the dice is biased so that the probability, $p$ , that it will show a six on one throw is less than $\frac{1}{6}$ . She throws the dice 30 times and finds that it shows a six on exactly 2 throws.

Use a binomial distribution with a 5% significance level to test Birgitte's suspicion. [5]

[5]

Later, Birgitte carries out a similar test at the 5% significance level, using another 30 throws of the dice. Calculate the probability of a Type I error. [2]

[2]

Given that the value of $p$ is actually 0.02, calculate the probability of a Type II error. [3]

[3]

Mark scheme

9709/62 · May/June 2025

Download PDF

1a · 1 mark

B1
E.g. 1–2: choose student 1; 3–4: choose student 2; 5–6: choose student 3. Must be un-ambiguous i.e. if no example, need to say 2 different numbers per person AND different to the other two people so that all 6 numbers on dice used.

1b · 5 marks

B1
$E (D) = 0$ , $Var (D) = 100$
M1
$\frac{12 - 0}{\sqrt{^{'} 100^{'}}} [= 1.2]$
M1
$1 - Φ (^{'} {1.2}^{'})$
A1
$= 0.115 (3sf)$
1 mark
$P (Difference > 12 minutes) = 0.23 [0]$

2 · 3 marks

M1
For attempt biased or unbiased = 15.6.
A1
$15.6 = \frac{100}{99} (\frac{Σ h^{2}}{100} - {80.2}^{2})$ OR $15.6 = 1 / 99 (Σ h^{2} - 8020^{2} / 100)$
A1
$Σ h^{2} = 644748.4 or 645000 (3 sf) or 3223742 / 5$

3a · 1 mark

B1
$e^{- 15} \times \frac{15^{12}}{12!}$

3b · 3 marks

B1
$e^{- 15} \times \frac{15^{n}}{n!} = e^{- 15} \times \frac{15^{n + 1}}{(n + 1)!}$
M1
$1 = \frac{15}{n + 1}$
A1
n = 14

4 · 4 marks

M1
$2 \times 1.645 \times \sqrt{\frac{a (1 - a)}{200}} = 0.1066$
B1
$z = 1.645$
M1
$a^{2} - a + 0.20997 = 0$
A1
$a = 0.3 (00) 3sf or 0.7 (00) 3sf$

5 · 4 marks

B1
sd ( $σ$ ) remains at 3.8, or is unchanged
M1
$= 1.645$
B1
$z = \frac{\overset{ˉ}{x} - 10.5}{\frac{3.8}{\sqrt{10}}} = \pm 1.645$
A1
[Smallest value of] $\overset{ˉ}{x} = 12.5 (3 sf)$

6a(i) · 3 marks

B1
$(λ) = 3.5$
M1
$1 - e^{-3.5}(1 + 3.5 + \frac{3.5^2}{2!} + \frac{3.5^3}{3!}) = 1 - e^{-3.5}(1 + 3.5 + 6.125 + 7.1458) = 1 - (0.030197 + 0.105691 + 0.1849589 + 0.215785)$
A1
$= 0.463 (3 sf)$

6a(ii) · 3 marks

B1
$(λ =) 7$
M1
$e^{- 7} (1 + 7 + \frac{7^{2}}{2!}) = e^{- 7} (1 + 7 + 24.5) = 0.0009119 + 0.006383 + 0.0223411$
A1
$= 0.0296 (3 sf)$

6b · 4 marks

M1
$N (200, 200)$
M1
$\frac{205.5 - 200}{\sqrt{200}} [= 0.38891]$
M1
$1 - Φ (“ 0.38891 ”)$
A1
$= 0.349 (3 sf)$

7a · 3 marks

M1
$\int_{0}^{a} \frac{k}{a^{2}} x^{2} d x = 1$
A1
${[\frac{k x^{3}}{3 a^{2}}]}_{0}^{a} = 1$
A1
$\Rightarrow \frac{k a}{3} = 1 \Rightarrow k = \frac{3}{a}$

7b · 3 marks

M1
$\int_{0}^{a} \frac{3}{a^{3}} x^{3} d x$
A1
${[\frac{3}{a^{3}} \frac{x^{4}}{4}]}_{0}^{a} = 1 [\Rightarrow \frac{3}{4} a = 1]$
A1
$a = \frac{4}{3} or 1.33 (3 sf)$

7c · 3 marks

M1
$\frac{81}{64} \int_{0}^{m} x^{2} d x = 0.5$
1 mark
$\frac{81}{64} {[\frac{x^{3}}{3}]}_{0}^{m} = 0.5 \Rightarrow \frac{27}{64} m^{3} = 0.5$
A1
$m = \sqrt[3]{\frac{32}{27}} or 1.06 (3 sf)$

8a · 5 marks

B1
$H_{0} : p = \frac{1}{6} H_{1} : p < \frac{1}{6}$
M1
$(\frac{5}{6})^{30} + 30 (\frac{5}{6})^{29} (\frac{1}{6}) +^{30} C_{2} (\frac{5}{6})^{28} (\frac{1}{6})^{2} = 0.00421272 + 0.0252763 + 0.07330$
A1
$= 0.103$
M1
‘0.103’ > 0.05
1 mark
[Accept $H_{0}$ ] Insufficient evidence to suggest that the probability is less than $\frac{1}{6}$ Or Insufficient evidence to support Birgitte’s suspicion

8b · 2 marks

M1
$P (X ⩽ 1) = (\frac{5}{6})^{30} + 30 (\frac{5}{6})^{29} (\frac{1}{6}) (= 0.029489...)$
A1
$P (Type I error) = 0.0295 (3 sf)$

8c · 3 marks

M1
$P (Type II error) = 1 - P (X ⩽ 1 ∣ p = 0.02)$
M1
$= 1 - ({0.98}^{30} + 30 \times {0.98}^{29} \times 0.02) = 1 - (0.54548 + 0.33397)$
A1
$= 0.121 (3 sf)$

Questions

1
Question 16 marks
Mark scheme ›
2
Question 23 marks
Mark scheme ›
3
Question 34 marks
Mark scheme ›
4
Question 44 marks
Mark scheme ›
5
Question 54 marks
Mark scheme ›
6
Question 610 marks
Mark scheme ›
7
Question 79 marks
Mark scheme ›
8
Question 810 marks
Mark scheme ›