A Level Further Mathematics 9231 Paper 43, Oct/Nov 2024

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9231/43 · Oct/Nov 2024

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Ellie is investigating the heights of two types of beech tree, $A$ and $B$ , in a certain region. She has chosen a random sample of 60 beech trees of type $A$ in the region, recorded their heights, $x m$ , and calculated unbiased estimates for the population mean and population variance as $35.6 m$ and $4.95 m^{2}$ respectively. Ellie also chooses a random sample of 50 beech trees of type $B$ in the region and records their heights, $y m$ . Her results are summarised as follows. $Σ y = 1654 Σ y^{2} = 54 850$ Find a 95% confidence interval for the difference between the population mean heights of type $A$ and type $B$ beech trees in the region.

[6]

A school with a large number of students is updating its logo. Each student has designed a new logo and two teachers have each awarded a mark out of 50 for each logo. The marks awarded to a random sample of 12 students are shown in the following table.

Student	$A$	$B$	$C$	$D$	$E$	$F$	$G$	$H$	$I$	$J$	$K$	$L$
Teacher 1	$36$	$38$	$40$	$36$	$22$	$34$	$45$	$44$	$48$	$35$	$28$	$30$
Teacher 2	$38$	$42$	$32$	$41$	$32$	$41$	$42$	$50$	$36$	$44$	$42$	$41$

One of the students claims that Teacher 2 is awarding higher marks than Teacher 1.

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

[7]

It was later discovered that Teacher 1 had entered her mark for student $C$ incorrectly. Her intended mark was 24 not 40. This was corrected. Determine whether this correction affects the conclusion of the test carried out in part (a).

[2]

A statistician believes that the number of telephone calls received by an advice centre in a 10-minute interval can be modelled by the Poisson distribution $P o (1.9)$ . The number of calls received in a randomly chosen 10-minute interval was recorded on each of 100 days. The results are summarised in the table, together with some of the expected frequencies corresponding to the distribution $P o (1.9)$ .

Complete the table.

Number of calls	$0$	$1$	$2$	$3$	$4$	$5$	$6 ormore$
$Observedfrequency$	$10$	$18$	$35$	$21$	$11$	$4$	$1$
$Expectedfrequency$	$14.957$	$28.418$	$26.997$				$1.322$

[2]

Carry out a goodness of fit test, at the 10% significance level, to determine whether the statistician's belief is reasonable.

[6]

The continuous random variable $X$ has probability density function $f$ given by $f (x) = {\begin{cases} k x^{3} & 0 ⩽ x < 1, \\ k (5 - x) & 1 ⩽ x ⩽ 5, \\ 0 & otherwise, \end{cases}$ where $k$ is a constant.

Sketch the graph of $f$ .

[1]

Show that $k = \frac{4}{33}$ .

[2]

Find the cumulative distribution function of $X$ .

[3]

Find the median value of $X$ .

[4]

Nikita has three coins. One coin is fair, one coin is biased so that the probability of obtaining a head is $\frac{1}{3}$ and the third coin is biased so that the probability of obtaining a head is $\frac{1}{5}$ . The random variable $X$ is the number of heads that Nikita obtains when he throws all three coins at the same time.

Find the probability generating function of $X$ .

[3]

Rajesh has two fair six-sided dice with faces labelled 1, 2, 3, 4, 5, 6. The random variable $Y$ is the number of 4s that Rajesh obtains when he throws the two dice. The random variable $Z$ is the sum of the number of heads obtained by Nikita and the number of 4s obtained by Rajesh. Find the probability generating function of $Z$ , expressing your answer as a polynomial.

[4]

Use your answer to part (b) to find $E (Z)$ .

[2]

Ansal is investigating the wingspans of Monarch butterflies in two different regions, $X$ and $Y$ . He takes a random sample of 8 Monarch butterflies from region $X$ and records their wingspans, $x cm$ . His results are as follows. $8.2 7.0 7.3 8.8 7.8 8.5 9.2 7.4$ Ansal also takes a random sample of 9 Monarch butterflies from region $Y$ and records their wingspans, $y cm$ . His results are summarised as follows. $Σ y = 71.10 Σ y^{2} = 567.13$ Ansal suspects that the mean wingspan of Monarch butterflies from region $X$ is greater than the mean wingspan of Monarch butterflies from region $Y$ . It is known that the wingspans of Monarch butterflies in regions $X$ and $Y$ are normally distributed with equal population variances. Test, at the 10% significance level, whether Ansal's suspicion is supported by the data.

[8]

Mark scheme

9231/43 · Oct/Nov 2024

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1 · 6 marks

B1
$s_{x}^{2} = \frac{1}{49} (54850 - \frac{1654^{2}}{50}) [= 2.769]$
M1
$s^{2} = \frac{4.95}{60} + \frac{2.769}{50}$
A1
$[= 0.1379]$
M1
CI: $35.6 - \frac{1654}{50} \pm z s$
B1
1.96 seen.
A1
$[1.79, 3.25]$

2a · 7 marks

M1
$2 4 - 8 5 10 7 - 3 6 - 12 9 14 11$
M1
$1 3 - 7 4 9 6 - 2 5 - 11 8 12 10$
A1
(Sum of + ranks = 58) Test statistic = 20
B1
$H_{0}$ : population median for teacher 2 = population median for teacher 1 $H_{1}$ : population median for teacher 2 > population median for teacher 1
B1
Critical value, from tables, is 17
M1
‘20’ > 17, accept $H_{0}$
A1
Insufficient evidence to support the claim Insufficient evidence that the scores of Teacher 2 are higher than those of Teacher 1

2b · 2 marks

B1
New test statistic is ‘13’
B1
$13 < 17$ , so conclusion is now ‘reject $H_{0}$ ’, (sufficient evidence to support claim)

3a · 2 marks

B1
One correct.
B1
All correct.

3b · 6 marks

M1
Table with last two or three columns combined.
M1
Contributions to test statistic are: 1.6428 3.8192 2.3724 0.8905 0.9609(7)
A1
Test statistic is 9.69
B1
$H_{0}$ : $Po (1.9)$ is a good fit for the data $H_{1}$ : $Po (1.9)$ is not a good fit for the data
M1
Critical value is 7.779, compare ‘9.69’ > 7.779 reject $H_{0}$
A1
Sufficient evidence to suggest that $Po (1.9)$ is a not a good fit for the data/ Sufficient evidence to reject/not support the statistician’s claim

4a · 1 mark

B1
Correct sketch.

4b · 2 marks

B1
$\frac{k x^{4}}{4}$ with limits 0 and 1 gives $\frac{k}{4}$ $k (5 x - \frac{x^{2}}{2})$ with limits 1 and 5 gives $8 k$
B1
$\frac{k}{4} + 8 k = 1, k = \frac{4}{33}$

4c · 3 marks

M1
Integrate both parts.
A1
Middle two parts correct AEF.
A1
0 and 1 correct.

4d · 4 marks

M1
$\int_{0}^{1} k x^{3} d x + \int_{1}^{m} k (5 - x) d x = \frac{1}{2}$
M1
$\frac{4}{33} [\frac{1}{4} + 5 m - \frac{1}{2} m^{2} - 5 + \frac{1}{2}] = \frac{1}{2}$ or $\frac{4}{33} [\frac{25}{2} - 5 m + \frac{1}{2} m^{2}] = \frac{1}{2}$ or $\frac{4}{33} [5 m - \frac{1}{2} m^{2} - \frac{17}{4}] = \frac{1}{2}$
1 mark
$4 m^{2} - 40 m + 67 = 0$
A1
$m = \frac{1}{2} (10 - \sqrt{33}) = 2.13$

5a · 3 marks

B1
$P (0 heads) = \frac{8}{30}, P (1 head) = \frac{14}{30}, P (2 heads) = \frac{7}{30}, P (3 heads) = \frac{1}{30}$
M1
$G_{X} (t) = \frac{8}{30} + \frac{14}{30} t + \frac{7}{30} t^{2} + \frac{1}{30} t^{3}$
A1
FT their probabilities that sum to one.

5b · 4 marks

B1
$G_{Y} (t) = \frac{25}{36} + \frac{10}{36} t + \frac{1}{36} t^{2}$
M1
$G_{Z} (t) = \frac{1}{30 \times 36} (8 + 14 t + 7 t^{2} + t^{3}) (25 + 10 t + t^{2})$
M1
Multiply to obtain quintic polynomial.
A1
$\frac{1}{1080} (200 + 430 t + 323 t^{2} + 109 t^{3} + 17 t^{4} + t^{5})$

5c · 2 marks

M1
$G_{Z}^{'} (1) = \frac{1}{1080} (430 + 646 + 327 + 68 + 5)$
A1
$\frac{1476}{1080} = \frac{41}{30} = 1.37$

6 · 8 marks

M1
$s_{x}^{2} = \frac{1}{7} (519.46 - \frac{{64.2}^{2}}{8}) [= 0.607857]$ $s_{y}^{2} = \frac{1}{8} (567.13 - \frac{{71.1}^{2}}{9}) [= 0.680]$
M1
$s^{2} = \frac{8 \times 0.680 + 7 \times 0.607857}{9 + 8 - 2}$
A1
$[= 0.64633]$
B1
$H_{0}$ : $μ_{X} - μ_{Y} = 0$ $H_{1}$ : $μ_{X} - μ_{Y} > 0$
M1
$t = \frac{8.025 - 7.9}{s \sqrt{\frac{1}{8} + \frac{1}{9}}}$
A1
$= \frac{0.125}{0.3906} = 0.320$
M1
‘0.320’ < 1.341 accept $H_{0}$
A1
Insufficient evidence to support Ansal’s suspicion Insufficient evidence to suggest that wingspan from region $X$ is greater than wingspan from region $Y$ .

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