A Level Further Mathematics 9231 Paper 41, May/June 2024

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9231/41 · May/June 2024

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The times taken by members of a large cycling club to complete a cross-country circuit have a normal distribution with mean $μ$ minutes. The times taken, $x$ minutes, are recorded for a random sample of 14 members of the club. The results are summarised as follows, where $\overset{ˉ}{x}$ is the sample mean. $\overset{ˉ}{x} = 42.8 Σ (x - \overset{ˉ}{x})^{2} = 941.5$ Find a 95% confidence interval for $μ$ .

[4]

A large number of students are taking a Physics course. They are assessed by a practical examination and a written examination. The marks out of 100 obtained by a random sample of 15 students in each of the examinations are as follows.

Student	$A$	$B$	$C$	$D$	$E$	$F$	$G$	$H$	$I$	$J$	$K$	$L$	$M$	$N$	$O$
Practical examination	$66$	$63$	$24$	$52$	$59$	$76$	$88$	$51$	$48$	$36$	$91$	$72$	$68$	$67$	$60$
Written examination	$63$	$57$	$39$	$50$	$47$	$71$	$87$	$65$	$56$	$39$	$78$	$70$	$61$	$62$	$70$

Use a sign test, at the 10% significance level, to test whether, on average, the practical examination marks are higher than the written examination marks.

[5]

A factory produces metal discs. The manager claims that the diameters of these discs have a median of $22.0 mm$ . The diameters, in mm, of a random sample of 12 discs produced by this factory are as follows.

22.4

20.9

22.8

21.5

23.2

22.9

23.9

21.7

19.8

23.6

22.6

23.0

Carry out a Wilcoxon signed-rank test, at the 10% significance level, to test whether there is any evidence against the manager's claim.

[7]

State an assumption that is necessary for this test to be valid.

[1]

The random variable $Y$ is the sum of two independent observations of the random variable $X$ . The probability generating function $G_{Y} (t)$ of $Y$ is given by $G_{Y} (t) = \frac{t^{2}}{(4 - 3 t)^{4}} .$

Find $E (Y)$ .

[3]

Write down an expression for the probability generating function of $X$ .

[1]

Find $P (X = 4)$ .

[3]

Two companies, $P$ and $Q$ , produce a certain type of paint brush. An independent examiner rates the quality of the brushes produced as poor, satisfactory or good. He takes a random sample of brushes from each company. The examiner's ratings are summarised in the table.

Company	Poor	Satisfactory	Good
$P$	$18$	$43$	$64$
$Q$	$22$	$22$	$31$

Test, at the 5% significance level, whether quality of brushes is independent of company.

[7]

Compare the quality of the brushes produced by the two companies.

[1]

Jade is a swimming instructor at a sports college. She claims that, as a result of an intensive training course, the mean time taken by students to swim 50 metres has reduced by more than 1 second. She chooses a random sample of 10 students. The times taken, in seconds, before and after the training course are recorded in the table.

Student	$A$	$B$	$C$	$D$	$E$	$F$	$G$	$H$	$I$	$J$
Time before course	$54.2$	$47.4$	$52.1$	$59.0$	$55.3$	$51.0$	$48.9$	$52.2$	$58.4$	$51.4$
Time after course	$50.1$	$46.3$	$52.5$	$58.8$	$51.4$	$48.4$	$49.5$	$48.7$	$58.3$	$51.4$

Test, at the 10% significance level, whether Jade's claim is justified.

[7]

State an assumption that is necessary for this test to be valid.

[1]

The continuous random variable $X$ has probability density function $f$ given by $f (x) = {\begin{cases} \frac{x}{4} (4 - x^{2}) & 0 ⩽ x ⩽ 2, \\ 0 & otherwise. \end{cases}$

Find $V a r (\sqrt{X})$ .

[4]

The continuous random variable $Y$ is defined by $Y = X^{2}$ . Find the probability density function of $Y$ .

[4]

Find the exact value of the median of $Y$ .

[2]

Mark scheme

9231/41 · May/June 2024

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1 · 4 marks

B1
$s^{2} = \frac{941.5}{13}$
M1
CI: $42.8 \pm 2.160 \sqrt{\frac{s^{2}}{14}}$
A1
2.160 seen in a CI formula
A1
$[37.9, 47.7]$

2 · 5 marks

B1
$H_{0}$ : practical results and written results are equal $H_{1}$ : practical results are greater than written results
M1
There are 10 + signs; $B (15, 0.5)$ soi
A1
$(P (X ⩾ 10) = 1 - P (X ⩽ 9) =) 1 - 0.849 = 0.151$
M1
Compare with 0.1: '0.151' > 0.1 accept $H_{0}$
A1
Insufficient evidence to suggest that practical results are greater than written results

3a · 7 marks

M1
Signed differences 0.4 -1.1 0.8 -0.5 1.2 0.9 1.9 -0.3 -2.2 1.6 0.6 1.0
M1
Ranks: 2 -8 5 -3 9 6 11 -1 -12 10 4 7
A1
$(P_{+} = 54) P_{-} = 24$
B1
$H_{0}$ : population median = 22.0 $H_{1}$ : population median $\neq$ 22.0
B1
Critical value from table = 17
M1
'24' > '17' Accept $H_{0}$
A1
Insufficient evidence to support (population) median not being 22.0 Insufficient evidence against the manager’s claim

3b · 1 mark

B1
Distribution is symmetric about the population median Underlying/population distribution is symmetric about the median

4a · 3 marks

M1
$G_{Y} (t) = \frac{t^{2}}{(4 - 3 t)^{4}}$ , $G_{Y}^{'} (t) = \frac{(4 - 3 t)^{4} (2 t) - - 12 t^{2} (4 - 3 t)^{3}}{(4 - 3 t)^{8}}$ or $\frac{2 t}{(4 - 3 t)^{4}} + \frac{12 t^{2}}{(4 - 3 t)^{5}}$
M1
$E (Y) = G_{Y}^{'} (1) = 14$
A1
14

4b · 1 mark

B1
$G_{X} (t) = \frac{t}{(4 - 3 t)^{2}}$

4c · 3 marks

M1
$G_{X} (t) = t (4 - 3 t)^{- 2} = \frac{t}{16} {(1 - \frac{3}{4} t)}^{- 2} = \frac{t}{16} (1 + \frac{3}{2} t - 3 \times \frac{9}{16} t^{2} + 4 \times \frac{27}{64} t^{3} + \dots)$
M1
Use $P (X = 4) = their coefficient of t^{4}$
A1
$\frac{27}{256}$

5a · 7 marks

M1
Calculate E values
A1
Expected values table
M1
Chi-squared contributions
A1
Test statistic = 6.56
B1
$H_{0}$ : Quality (of items) is independent of company $H_{1}$ : Quality (of items) is not independent of company
M1
6.56 > 5.991 Reject $H_{0}$
A1
There is sufficient evidence to suggest that the quality of brushes is not independent of company.

5b · 1 mark

B1
$P$ produces better quality brushes (than $Q$ ) or $Q$ produces worse quality brushes (than $P$ ) $P$ has fewer poor quality brushes than expected AND $Q$ has more poor quality brushes than expected

6a · 7 marks

M1
Signed differences: 4.1 1.1 -0.4 0.2 3.9 2.6 -0.6 3.5 0.1 0
M1
$\sum d = 14.5$ , $\sum d^{2} = 52.81$ $s^{2} = \frac{1}{9} (52.81 - \frac{{14.5}^{2}}{10}) = 3.532$
B1
$H_{0} : μ_{B} - μ_{A} = 1$ $H_{1} : μ_{B} - μ_{A} > 1$
M1
$t = \frac{1.45 - 1}{\sqrt{\frac{s^{2}}{10}}}$
A1
0.757
M1
Critical value is 1.383: ‘0.757’ < 1.383 Accept $H_{0}$
A1
Insufficient evidence to support Jade’s claim

6b · 1 mark

B1
Distribution of population differences is normal

7a · 4 marks

B1
$f (x) = {\begin{cases} \frac{x}{4} (4 - x^{2}) & 0 ⩽ x ⩽ 2, \\ 0 & otherwise. \end{cases}$ $(E (X) =) \int_{0}^{2} \frac{1}{4} (4 x^{2} - x^{4}) d x = \frac{1}{4} {[\frac{4}{3} x^{3} - \frac{1}{5} x^{5}]}_{0}^{2} = \frac{16}{15} (= 1.067)$
M1
$(E (\sqrt{X}) =) \int_{0}^{2} \frac{1}{4} (4 x^{1.5} - x^{3.5}) d x = \frac{1}{4} {[\frac{8}{5} x^{2.5} - \frac{2}{9} x^{4.5}]}_{0}^{2} [= \frac{32}{45} \sqrt{2} = 1.005 (7)]$
M1
$Var (\sqrt{X}) = E (X) - (E (\sqrt{X}))^{2} used$
A1
$[\frac{16}{15} - {(\frac{32}{45} \sqrt{2})}^{2}] = 0.0553$

7b · 4 marks

B1
$F (x) = \frac{1}{16} (8 x^{2} - x^{4}) [0 ⩽ x ⩽ 2]$
M1
$G (y) = \frac{1}{16} (8 y - y^{2}) [0 ⩽ y ⩽ 4]$
M1
$g (y) = \frac{1}{16} (8 - 2 y) [0 ⩽ y ⩽ 4]$
A1
$g (y) = {\begin{cases} \frac{1}{16} (8 - 2 y) & 0 ⩽ y ⩽ 4, \\ 0 & otherwise. \end{cases}$

7c · 2 marks

M1
$\frac{1}{16} (8 m - m^{2}) = \frac{1}{2}$
A1
$[m^{2} - 8 m + 8 = 0,] m = 4 - 2 \sqrt{2}$

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