A Level Further Mathematics 9231 Paper 41, May/June 2023

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9231/41 · May/June 2023

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The lengths of the leaves of a particular type of tree are normally distributed with mean $μ cm$ . The lengths, $x cm$ , of a random sample of 12 leaves of this type are recorded. The results are summarised as follows. $Σ x = 91.2 Σ x^{2} = 695.8$ Find a 95% confidence interval for $μ$ .

[4]

The children at two large schools, $P$ and $Q$ , are all given the same puzzle to solve. A random sample of size 10 is taken from the children at school $P$ . Their individual times to complete the puzzle give a sample mean of 9.12 minutes and an unbiased variance estimate of $2.16 {minutes}^{2}$ . A random sample of size 12 is taken from the children at school $Q$ . Their individual times, $x$ minutes, to complete the puzzle are summarised by $Σ x = 99.6 Σ (x - \overset{ˉ}{x})^{2} = 21.5,$ where $\overset{ˉ}{x}$ is the sample mean. Times to complete the puzzle are assumed to be normally distributed with the same population variance. Test at the 5% significance level whether the population mean time taken to complete the puzzle by children at school $P$ is greater than the population mean time taken to complete the puzzle by children at school $Q$ .

[8]

A random sample of 50 values of the continuous random variable $X$ was taken. These values are summarised in the following table.

Interval	$1 ⩽ x < 1.5$	$1.5 ⩽ x < 2$	$2 ⩽ x < 2.5$	$2.5 ⩽ x < 3$	$3 ⩽ x < 3.5$	$3.5 ⩽ x ⩽ 4$
Observed frequency	$3$	$3$	$8$	$11$	$13$	$12$

It is required to test the goodness of fit of the distribution with probability density function $f$ given by $f (x) = {\begin{cases} \frac{1}{24} (\frac{4}{x^{2}} + x^{2}) & 1 ⩽ x ⩽ 4, \\ 0 & otherwise. \end{cases}$ The expected frequencies, correct to 4 decimal places, are given in the following table.

Interval	$1 ⩽ x < 1.5$	$1.5 ⩽ x < 2$	$2 ⩽ x < 2.5$	$2.5 ⩽ x < 3$	$3 ⩽ x < 3.5$	$3.5 ⩽ x ⩽ 4$
Expected frequency	$4.4271$	$a$	$6.1285$	$8.4549$	$b$	$14.9678$

Show that $a = 4.6007$ and find the value of $b$ .

[3]

Carry out a goodness of fit test, at the 10% significance level, to test whether $f$ is a satisfactory model for the data.

[6]

A random sample of 13 technology companies is chosen and the numbers of employees in 2018 and in 2022 are recorded.

Company	$A$	$B$	$C$	$D$	$E$	$F$	$G$	$H$	$I$	$J$	$K$	$L$	$M$
Number in 2018	$104$	$19$	$126$	$234$	$970$	$514$	$35$	$149$	$429$	$12$	$86$	$304$	$1104$
Number in 2022	$106$	$24$	$127$	$228$	$1012$	$525$	$32$	$156$	$449$	$24$	$78$	$294$	$1154$

A researcher claims that there has been an increase in the median number of employees at technology companies between 2018 and 2022.

Carry out a Wilcoxon matched-pairs signed-rank test, at the 5% significance level, to test whether the data supports this claim.

[7]

The researcher notices that the figures for company $G$ have been recorded incorrectly. In fact, the number of employees in 2018 was 32 and the number of employees in 2022 was 35. Explain, with numerical justification, whether or not the conclusion of the test in part (a) remains the same.

[2]

Harry has three coins. • One coin is biased so that, when it is thrown, the probability of obtaining a head is $\frac{1}{3}$ . • The second coin is biased so that, when it is thrown, the probability of obtaining a head is $\frac{1}{4}$ . • The third coin is biased so that, when it is thrown, the probability of obtaining a head is $\frac{1}{5}$ . The random variable $X$ is the number of heads that Harry obtains when he throws all three coins together.

Find the probability generating function of $X$ .

[3]

Isaac has two fair coins. The random variable $Y$ is the number of heads that Isaac obtains when he throws both of his coins together. The random variable $Z$ is the total number of heads obtained when Harry throws his three coins and Isaac throws his two coins. Find the probability generating function of $Z$ , expressing your answer as a polynomial in $t$ .

[4]

Use the probability generating function of $Z$ to find $E (Z)$ .

[2]

The continuous random variable $X$ has probability density function $f$ given by $f (x) = {\begin{cases} \frac{3}{28} (e^{\frac{1}{2} x} + 4 e^{- \frac{1}{2} x}) & 0 ⩽ x ⩽ 2 \ln 3, \\ 0 & otherwise. \end{cases}$

Find the cumulative distribution function of $X$ .

[3]

The random variable $Y$ is defined by $Y = e^{\frac{1}{2} X}$ . Find the probability density function of $Y$ .

[3]

Find the 30th percentile of $Y$ .

[3]

Find $E (Y^{4})$ .

[2]

Mark scheme

9231/41 · May/June 2023

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1 · 4 marks

M1
$\overset{ˉ}{x} = \frac{91.2}{12} [= 7.6] s_{x}^{2} = \frac{1}{11} (695.8 - \frac{{91.2}^{2}}{12}) [= 0.2436 (\frac{67}{275})]$
M1
CI: $7.6 \pm t_{11} (0.975) \times \sqrt{\frac{0.2436}{12}}$
B1
2.201 seen
A1
$7.6 \pm 0.3136 = [7.29, 7.91]$

2 · 8 marks

M1
$\overset{ˉ}{x} = \frac{99.6}{12} [= 8.3] s_{x}^{2} = \frac{21.5}{11} [= 1.9545 or \frac{43}{22}]$
B1
$H_{0} : μ_{P} = μ_{Q} H_{1} : μ_{P} > μ_{Q}$
M1
$Pooled estimate = \frac{9 \times 2.16 + 11 \times 1.9545}{10 + 12 - 2}$
A1
2.047 (2.05)
M1
$t = \frac{9.12 - 8.3}{\sqrt{2.047 \times (\frac{1}{10} + \frac{1}{12})}}$
A1
1.339 (or 1.338)
M1
'1.339' < 1.725 Accept $H_{0}$ / not significant
A1
Insufficient evidence to suggest that the (mean) time taken at $P$ is greater than the (mean) time taken at $Q$ .

3a · 3 marks

M1
$\int_{1.5}^{2} \frac{1}{24} (\frac{4}{x^{2}} + x^{2}) d x = \frac{1}{24} [- \frac{4}{x} + \frac{x^{3}}{3}]$
A1
$a = 50 \times 0.092014 = 4.6007$
B1
$b = 11.4211$

3b · 6 marks

M1
Combine first two columns: 6, 9.0278
M1
Calculate value of chi-squared: $1.0155 + 0.5715 + 0.7661 + 0.2183 + 0.5885$
A1
3.16
B1
$H_{0}$ : f is a good fit for the data $H_{1}$ : f is not a good fit for the data
M1
Tabular value of chi-squared: 7.78 ‘3.16’ < 7.78 and accept $H_{0}$
A1
Insufficient evidence to suggest that f is not a good fit for the data Condone: sufficient evidence to suggest that f is a good fit for the data.

4a · 7 marks

M1
2 5 1 –6 42 11 –3 7 20 12 –8 –10 50
A1
2 4 1 –5 12 9 –3 6 11 10 –7 –8 13
A1
[ $P = 68$ ,] $Q = 23$
B1
$H_{0}$ : difference in population medians = 0 $H_{1}$ : population median in 2022 > population median in 2018
B1
Critical value 21
M1
‘23’ > ‘21’ and accept $H_{0}$
A1
Insufficient evidence to support researcher’s belief / insufficient evidence that the (median) number of employees in 2022 is greater than the (median) number of employees in 2018

4b · 2 marks

M1
Rank for $G$ will now be +3, giving $Q = 20$ which is < 21
A1
Change in conclusion

5a · 3 marks

B1
$P (3 H) = \frac{1}{60} P (2 H) = \frac{9}{60} P (1 H) = \frac{26}{60} P (0 H) = \frac{24}{60}$
M1
$G_{X} (t) = \frac{24}{60} + \frac{26}{60} t + \frac{9}{60} t^{2} + \frac{1}{60} t^{3}$
A1
$G_{X} (t) = \frac{24}{60} + \frac{26}{60} t + \frac{9}{60} t^{2} + \frac{1}{60} t^{3}$

5b · 4 marks

B1
$G_{Y} (t) = \frac{1}{4} + \frac{1}{2} t + \frac{1}{4} t^{2}$
M1
$G_{Z} (t) = (\frac{24}{60} + \frac{26}{60} t + \frac{9}{60} t^{2} + \frac{1}{60} t^{3}) (\frac{1}{4} + \frac{1}{2} t + \frac{1}{4} t^{2})$
M1
$\frac{1}{240} (24 + 74 t + 85 t^{2} + 45 t^{3} + 11 t^{4} + t^{5})$
A1
$\frac{1}{240} (24 + 74 t + 85 t^{2} + 45 t^{3} + 11 t^{4} + t^{5})$

5c · 2 marks

M1
Differentiate: $G_{Z}^{'} (t) = \frac{1}{240} (74 + 170 t + 135 t^{2} + 44 t^{3} + 5 t^{4})$
A1
$G_{Z}^{'} (1) = \frac{428}{240} = \frac{107}{60} = 1.78$

6a · 3 marks

M1
$F (x) = \frac{3}{28} (2 e^{\frac{1}{2} x} - 8 e^{- \frac{1}{2} x} (+ c))$
A1
$F (x) = \frac{3}{28} (2 e^{\frac{1}{2} x} - 8 e^{- \frac{1}{2} x}) + \frac{9}{14}$
B1
Correct ranges including $0 ⩽ x ⩽ 2 \ln 3$ associated with their $F (x)$ $F (x) = 0 for x < 0 and 1 for x > 2 \ln 3$

6b · 3 marks

M1
$G (y) = \frac{3}{28} (2 y - \frac{8}{y} + 6)$
B1
$1 ⩽ y ⩽ 3$
A1
$g (y) = \frac{3}{28} (2 + \frac{8}{y^{2}})$

6c · 3 marks

M1
$\frac{3}{28} (2 y - \frac{8}{y} + 6) = \frac{3}{10}$
M1
$5 y^{2} + 8 y - 20 = 0$
A1
y = 1.35

6d · 2 marks

M1
$E (Y^{4}) = \int_{1}^{3} \frac{3}{28} (2 + \frac{8}{y^{2}}) y^{4} d y = \int_{1}^{3} \frac{3}{28} (2 y^{4} + 8 y^{2}) d y = \frac{3}{28} [\frac{2}{5} y^{5} + \frac{8}{3} y^{3}]$
A1
17.8

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