A Level Further Mathematics 9231 Paper 42, Oct/Nov 2024

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9231/42 · Oct/Nov 2024

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A scientist is investigating the lengths of the leaves of a certain type of plant. The scientist assumes that the lengths of the leaves of this type of plant are normally distributed. He measures the lengths, $x cm$ , of the leaves of a random sample of 8 plants of this type. His results are as follows.

3.5

4.2

3.8

5.2

2.9

3.7

4.1

3.2

Find a 90% confidence interval for the population mean length of leaves of this type of plant. [4]

[4]

The random variable $X$ has probability generating function $G_{X} (t)$ given by $G_{X} (t) = \frac{1}{5} + p t + q t^{2},$ where $p$ and $q$ are constants.

Given that $E (X) = 1.1$ , find the numerical value of $Var (X)$ .

[4]

The random variable $Y$ has probability generating function $G_{Y} (t)$ given by $G_{Y} (t) = \frac{2}{3} t (1 + \frac{1}{2} t^{2}) .$ The random variable $Z$ is the sum of independent observations of $X$ and $Y$ . Find the probability generating function of $Z$ .

[2]

Find $P (Z > 2)$ .

[1]

State the most probable value of $Z$ .

[1]

Rosie sows 5 seeds in each of 150 plant pots. The number of seeds that germinate is recorded for each pot. The results are summarised in the following table.

Number of seeds that germinate	0	1	2	3	4	5
Number of pots	12	40	43	35	16	4

Rosie suggests that the number of seeds that germinate follows the binomial distribution $B (5, p)$ .

Use Rosie's results to show that $p = 0.42$ .

[1]

Carry out a goodness of fit test, at the 10% significance level, to test whether the distribution $B (5, 0.42)$ is a good fit for the data.

[9]

The random variable $X$ has probability density function $f$ given by $f (x) = {\begin{cases} \frac{1}{21} (x - 1)^{2} & 2 ⩽ x ⩽ 5, \\ 0 & otherwise. \end{cases}$

Find the cumulative distribution function of $X$ .

[3]

The random variable $Y$ is defined by $Y = (X - 1)^{4}$ . Find the probability density function of $Y$ .

[3]

Find the median value of $Y$ .

[2]

Find $E (Y)$ .

[2]

Dev owns a small company which produces bottles of juice. He uses two machines, $X$ and $Y$ , to fill empty bottles with juice. Dev is investigating the volumes of juice in the bottles. He chooses a random sample of 35 bottles filled by machine $X$ and a random sample of 60 bottles filled by machine $Y$ . The volumes of juice, $x$ and $y$ respectively, measured in suitable units, are summarised by $\sum x = 30.8, \sum x^{2} = 29.0, \sum y = 62.4, \sum y^{2} = 76.8.$ Dev claims that the mean volume of juice in bottles filled by machine $Y$ is greater than the mean volume of juice in bottles filled by machine $X$ . A test at the $α %$ significance level suggests that there is sufficient evidence to support Dev's claim. Find the set of possible values of $α$ . [9]

[9]

A sports college keeps records of the times taken by students to run one lap of a running track. The population median time taken is 51.0 seconds. After a month of intensive training, a random sample of 22 new students run one lap of the track, giving times, in seconds, as follows.

$51.3$	$52.0$	$53.4$	$49.2$	$49.3$	$51.1$	$52.2$	$47.2$
$53.0$	$48.5$	$49.4$	$50.3$	$50.8$	$51.6$	$49.1$	$52.3$
$51.8$	$52.4$	$47.9$	$48.9$	$50.6$	$51.9$

It is claimed that the intensive training has led to a decrease in the median time taken to run one lap of the track. Carry out a Wilcoxon signed-rank test, at the 5% significance level, to test whether there is sufficient evidence to support the claim. [9]

[9]

Mark scheme

9231/42 · Oct/Nov 2024

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1 · 4 marks

M1
$Σ x = 30.6 Σ x^{2} = 120.52$ $s^{2} = \frac{1}{7} [120.52 - \frac{{30.6}^{2}}{8}] [= 0.49643]$
M1
$CI = \frac{30.6}{8} \pm 1.895 \sqrt{\frac{s^{2}}{8}}$
B1
1.895 seen
A1
$[3.35, 4.30]$

2a · 4 marks

B1
$[G_{X} (t) = \frac{1}{5} + p t + q t^{2}, G_{X}^{'} (t) = p + 2 q t,] p + 2 q = 1.1$
M1
$\frac{1}{5} + p + q = 1, p + q = \frac{4}{5} and solve$
A1
$[p = \frac{1}{2},] q = \frac{3}{10}$
A1
$G_{X}^{''} (t) = 2 q = \frac{3}{5}, Var (X) = 0.6 + 1.1 - {1.1}^{2} = 0.49$

2b · 2 marks

M1
$G_{Z} (t) = (\frac{1}{5} + \frac{1}{2} t + \frac{3}{10} t^{2}) \times \frac{2}{3} t (1 + \frac{1}{2} t^{2})$
A1
$\frac{1}{30} (4 t + 10 t^{2} + 8 t^{3} + 5 t^{4} + 3 t^{5})$

2c · 1 mark

B1
$\frac{16}{30} = \frac{8}{15} oe$

2d · 1 mark

B1
$(Z =) 2$

3a · 1 mark

B1
$\overset{ˉ}{x} = \frac{40 + 86 + 105 + 64 + 20}{150} = \frac{315}{150} = 2.1, p = \frac{2.1}{5} = 0.42$

3b · 9 marks

B1
Calculate expected frequencies (must be seen) at least 2 correct to at least 2 decimal places.
B1
At least 4 correct to at least 2 decimal places.
M1
Combine last two columns
M1
Chi-squared contributions: 0.4712 0.5314 1.4416 0.1521 1.3088
A1
Test statistic = 3.905
B1
$H_{0}$ : $Binomial B (5, 0.42)$ fits the data $H_{1}$ : $Binomial B (5, 0.42)$ does not fit the data
B1
Critical value is 6.251
M1
‘3.905’ < ‘6.251’ Accept $H_{0}$
A1
Insufficient evidence to suggest that $B (5, 0.42)$ is a not a good fit (to the data)

4a · 3 marks

M1
Integrate $\frac{1}{63} (x^{3} - 3 x^{2} + 3 x) [+ c]$ oe.
A1
Correct $F (x)$ (with $c = - \frac{2}{63}$ evaluated).
A1
0 and 1 correct, inequalities correct.

4b · 3 marks

M1
$G (y) = \frac{1}{63} (y^{0.75} - 1)$
M1
Differentiate and change variable in domain.
A1
$g (y) = {\begin{cases} \frac{1}{84} y^{- 0.25} & 1 ⩽ y ⩽ 256, \\ 0 & otherwise. \end{cases}$

4c · 2 marks

M1
$\frac{1}{63} (y^{0.75} - 1) = 0.5$
A1
$y = 104$

4d · 2 marks

M1
$\int_{1}^{256} \frac{1}{84} y (y^{- 0.25}) d y = \int_{1}^{256} \frac{1}{84} (y^{0.75}) d y = [\frac{1}{84} \times \frac{4}{7} y^{1.75}]$
A1
$111$

5 · 9 marks

B1
$s_{x}^{2} = \frac{1}{34} [29.0 - \frac{{30.8}^{2}}{35}] [= 0.055765]$ $s_{y}^{2} = \frac{1}{59} [76.8 - \frac{{62.4}^{2}}{60}] [= 0.201763]$
M1
$s^{2} = \frac{0.055765}{35} + \frac{0.201763}{60} [= 0.004956]$
A1
$s^{2} = \frac{0.055765}{35} + \frac{0.201763}{60} [= 0.004956]$
B1
$H_{0} : μ_{x} = μ_{y}$ $H_{1} : μ_{x} < μ_{y}$
M1
$z = \frac{\frac{30.8}{35} - \frac{62.4}{60}}{s}$
A1
$\frac{- 0.25}{\sqrt{0.004956}} = - 2.27$
M1
$Corresponding probability is 0.9885$
M1
$One tail: 100 \times (1 - ’0.9885’) = 1.15$
A1
$α > 1.15$

6 · 9 marks

M1
Signed differences from 51.0
0.3 1.0 2.4 - 1.8 - 1.7 0.1 1.2 - 3.8
2.0 - 2.5 - 1.6 - 0.7 - 0.2 0.6 - 1.9 1.3
0.8 1.4 - 3.1 - 2.1 - 0.4 0.9
M1
3 9 19 -15 -14 1 10 -22
17 -20 -13 -6 -2 5 -16 11
7 12 -21 -18 -4 8
A1
Smaller sum is 102
B1
$H_{0}$ : population median is 51.0 (seconds) $H_{1}$ : population median is less than 51.0 (seconds)
B1
Normal approximation: $mean = 126.5$ $variance = 948.75$
M1
$\frac{’102’ + 0.5 - 126.5}{\sqrt{948.75}}$
A1
$- 0.779$
M1
‘-0.779’ > -1.645 accept $H_{0}$
A1
Insufficient evidence to suggest that (population) median is less than 51.0 seconds/ Insufficient evidence to support the claim

0.3	1.0	2.4	- 1.8	- 1.7	0.1	1.2	- 3.8
2.0	- 2.5	- 1.6	- 0.7	- 0.2	0.6	- 1.9	1.3
0.8	1.4	- 3.1	- 2.1	- 0.4	0.9

3	9	19	-15	-14	1	10	-22
17	-20	-13	-6	-2	5	-16	11
7	12	-21	-18	-4	8

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