A Level Further Mathematics 9231 Paper 43, Oct/Nov 2023

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9231/43 · Oct/Nov 2023

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Maya is an athlete who competes in 1500-metre races. Last summer her practice run times had mean 4.22 minutes. Over the winter she has done some intense training to try to improve her times. A random sample of 10 of her practice run times, $x$ minutes, this summer are summarised as follows. $Σ x = 42.05 Σ x^{2} = 176.83$ Maya's new practice run times are normally distributed. She believes that on average her times have improved as a result of her training. Test, at the 5% significance level, whether Maya's belief is supported by the data. [6]

[6]

A town council has published its plans for redeveloping the town centre and residents are being asked whether they approve or disapprove. A random sample of 250 responses has been selected from residents in the four main streets in the town: North, East, South and West Streets. The results are shown in the table.

	North Street	East Street	South Street	West Street
Approve	$33$	$54$	$42$	$26$
Disapprove	$19$	$39$	$28$	$9$

Test, at the 5% significance level, whether the opinions of the residents are independent of the streets on which they live. [7]

[7]

Scientists are studying the effects of exercise on LDL blood cholesterol levels. Over a three-month period, a large group of people exercised for 20 minutes each day. For a randomly chosen sample of 10 of these people, the LDL blood cholesterol levels were measured at the beginning and the end of the three-month period. The results, measured in suitable units, are as follows.

Person		$A$	$B$	$C$	$D$	$E$	$F$	$G$	$H$	$I$	$J$
Cholesterol	Beginning	$72$	$84$	$120$	$90$	$102$	$135$	$64$	$75$	$80$	$88$
level	End	$64$	$76$	$105$	$92$	$105$	$115$	$67$	$75$	$75$	$84$

Test, at the 2.5% significance level, whether there is evidence that the population mean LDL blood cholesterol level has reduced by more than 2 units after the three-month period. [7]

[7]

State any assumption that you have made in part (a). [1]

[1]

As shown in the diagram, the continuous random variable $X$ has probability density function $f$ given by $f (x) = {\begin{cases} m x & 0 ⩽ x ⩽ 2, \\ \frac{k}{x^{2}} + c & 2 ⩽ x ⩽ 6, \\ 0 & otherwise, \end{cases}$ where $m$ , $k$ and $c$ are constants.

Given that $P (X ⩽ 2) = \frac{1}{3}$ , show that $m = \frac{1}{6}$ and find the values of $k$ and $c$ . [4]

[4]

Find the exact numerical value of the interquartile range of $X$ . [5]

[5]

The random variable $X$ has the geometric distribution $G e o (p)$ .

Show that the probability generating function of $X$ is $\frac{p t}{1 - q t}$ , where $q = 1 - p$ . [3]

[3]

Use the probability generating function of $X$ to show that $V a r (X) = \frac{q}{p^{2}}$ . [5]

[5]

Kenny throws an ordinary fair 6-sided dice repeatedly. The random variable $X$ is the number of throws that Kenny takes in order to obtain a 6. The random variable $Z$ denotes the sum of two independent values of $X$ . Find the probability generating function of $Z$ . [2]

[2]

A school is conducting an experiment to see whether the distance that children can throw a ball increases in hot weather. On a cold day, all the children at the school were asked to throw a ball as far as possible. The distances thrown were measured and recorded. The median distance thrown by a random sample of 25 of the children was 22.0 m. The children were asked to throw the ball again on a hot day. The distances thrown by the same 25 children were measured and recorded and these distances, in m, are shown below.

$21.2$	$23.5$	$22.9$	$18.6$	$19.4$
$22.1$	$26.5$	$20.2$	$25.7$	$20.6$
$22.3$	$17.4$	$22.2$	$27.0$	$23.9$
$28.2$	$22.6$	$27.2$	$23.0$	$23.7$
$19.8$	$22.7$	$23.3$	$21.5$	$24.3$

The teacher claims that on average the distances thrown will be further when it is hot. Carry out a Wilcoxon signed-rank test, at the 5% significance level, to test whether the data supports the teacher's claim. [10]

[10]

Mark scheme

9231/43 · Oct/Nov 2023

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1 · 6 marks

B1
$H_{0} : μ = 4.22 H_{1} : μ < 4.22$
M1
$\overset{ˉ}{x} = \frac{42.05}{10} [= 4.205], s^{2} = \frac{1}{9} [176.83 - \frac{{42.05}^{2}}{10}] [= \frac{13}{12000} = 0.0010833]$
M1
$t = \frac{4.205 - 4.22}{\sqrt{\frac{s^{2}}{10}}}$
A1
$t = - 1.44$
M1
$Tabular value = 1.833 : ‘ 1.44 ’ < 1.833, accept H_{0}$
A1
There is insufficient evidence to support Maya’s belief.

2 · 7 marks

B1
$H_{0} : opinion is independent of street H_{1} : opinion is not independent of street$
M1
Expected frequencies: at least 2 correct values or expressions.
A1
Expected frequencies: 6 correct values or expressions.
M1
Calculate chi-squared values: $0.0179 + 0.2323 + 0.0452 + 0.8521 + 0.0292 + 0.3790 + 0.0737 + 1.3902$
A1
3.02
M1
$Tabular value, 3 degrees of freedom 7.815 : ‘ 3.02 ’ < 7.815, accept H_{0}$
A1
Insufficient evidence to suggest that opinion depends on street.

3a · 7 marks

B1
$H_{0} : μ_{B} - μ_{E} = 2 H_{1} : μ_{B} - μ_{E} > 2$
M1
| 8 | 8 | 15 | –2 | –3 | 20 | –3 | 0 | 5 | 4 |
M1
$\sum d = 52, \sum d^{2} = 816, s^{2} = \frac{1}{9} [816 - \frac{52^{2}}{10}] [= \frac{2728}{45} = 60.62]$
M1
$t = \frac{\frac{52}{10} - 2}{\sqrt{\frac{s^{2}}{10}}}$
A1
$t = 1.29967, 1.30$
M1
$Tabular value = 2.262. ‘ 1.30 ’ < 2.262, accept H_{0}$
A1
Insufficient evidence to suggest that cholesterol level has reduced by more than 2.

3b · 1 mark

B1
Population differences are normally distributed

4a · 4 marks

B1
$P (X ⩽ 2) = \frac{1}{3}, so {[\frac{m x^{2}}{2}]}_{0}^{2} = \frac{1}{3} ⟹ m (\frac{4}{2}) = \frac{1}{3}, m = \frac{1}{6}$
M1
$\int_{2}^{6} (\frac{k}{x^{2}} + c) d x = \frac{2}{3} so {[- \frac{k}{x} + c x]}_{2}^{6} = 4 c + \frac{1}{3} k = \frac{2}{3} or k + 12 c = 2$
M1
$Also, 2 m = \frac{k}{4} + c or \frac{1}{3} = \frac{k}{4} + c or 3 k + 12 c = 4$
A1
$Solve, k = 1, c = \frac{1}{12}$

4b · 5 marks

B1
$LQ : \int_{0}^{L} \frac{1}{6} x d x = \frac{1}{4}, L = \sqrt{3}$
M1
$UQ : \frac{1}{3} + \int_{2}^{U} (\frac{k}{x^{2}} + c) d x = \frac{3}{4}$
M1
$- \frac{k}{U} + U c + \frac{k}{2} - 2 c = \frac{5}{12}$ $U^{2} - U - 12 = 0$
A1
$U = 4$
A1
$IQR = 4 - \sqrt{3}$

5a · 3 marks

B1
$P (X = r) = p (1 - p)^{r - 1} = p q^{r - 1}$
M1
$G (t) = \sum p t (q t)^{r - 1}$
A1
$G (t) = p t (\frac{1}{1 - q t}) = \frac{p t}{1 - q t}$

5b · 5 marks

M1
$G^{'} (t) = p q t (1 - q t)^{- 2} + p (1 - q t)^{- 1} [= p (1 - q t)^{- 2}]$
M1
$G^{''} (t) = 2 p q^{2} t (1 - q t)^{- 3} + 2 p q (1 - q t)^{- 2} [= 2 q p (1 - q t)^{- 3}]$
M1
$G^{'} (1) = p (1 - q)^{- 2} = \frac{1}{p}, G^{''} (1) = \frac{2 q}{p^{2}}$
M1
$Var (X) = \frac{2 q}{p^{2}} + \frac{1}{p} - {(\frac{1}{p})}^{2}$
A1
$\frac{q}{p^{2}}$

5c · 2 marks

B1
$G_{X} (t) = \frac{\frac{1}{6} t}{1 - \frac{5}{6} t} [= \frac{t}{6 - 5 t}]$
B1
$G_{Z} (t) = (G_{X} (t))^{2} = {(\frac{t}{6 - 5 t})}^{2}$

6 · 10 marks

B1
$H_{0} : population medians are equal$ $H_{1} : population median after > population median before$
M1
Signed differences table
M1
Attempt at ranks (ignore signs)
A1
$(W_{+} = 214) W_{-} = 111$
B1
$Normal: mean = \frac{1}{4} n (n + 1) = \frac{1}{4} \times 25 \times 26 [= 162.5]$
B1
$Variance = \frac{1}{24} n (n + 1) (2 n + 1) = \frac{1}{24} \times 25 \times 26 \times 51 [= 1381.25]$
M1
$z - value: \frac{111.5 - 162.5}{\sqrt{1381.25}}$
A1
$- 1.37$
M1
$Tabular value is - 1.645 : ‘ - 1.37 ’ > - 1.645, or 0.915 < 0.95 oe, accept H_{0},$
A1
Insufficient evidence to support the teacher’s claim Insufficient evidence to suggest that the distances thrown are further when it is hot

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