A Level Further Mathematics 9231 Paper 42, Oct/Nov 2025

Data has outliers / is skewed / lacks symmetry, suggesting population/distribution may not be symmetrical.

H0: population median is [$]32,500. H1: population median is not [$]32,500.

Test statistic = 4 (or 11)

[$X\sim \text{B}(15,0.5)$] $\text{P}(X\le 4)=0.0592$ OR $\text{P}(X\ge 11)=0.0592$

'0.0592' > 0.05, accept ${\text{H}}_0$ / do not reject ${\text{H}}_0$ / not significant.

Insufficient evidence to reject [company’s] claim. OR Insufficient evidence to suggest that median [salary] is not [$]32,500.

$s^2=\frac{1}{11}{(}^{\mathrm{\prime }}{476117}^{\mathrm{\prime }}-\frac{{}^{\mathrm{\prime }}{2390}^{\mathrm{\prime }2}}{12})[=\frac{326}{33}=9.8788]$

3.106

$\frac{{}^{\mathrm{\prime }}{2390}^{\mathrm{\prime }}}{12}{\pm }^{\mathrm{\prime }}{3.106}^{\mathrm{\prime }}\sqrt{\frac{{}^{\mathrm{\prime }}{9.8788}^{\mathrm{\prime }}}{12}}$

$[196,202]$

$-1.796$

$\frac{{}^{\mathrm{\prime }}{199.17}^{\mathrm{\prime }}-k}{\sqrt{\frac{{}^{\mathrm{\prime }}{9.8788}^{\mathrm{\prime }}}{12}}}{>}^{\mathrm{\prime }}-{1.796}^{\mathrm{\prime }}$ or $\frac{{\mathrm{\mid }}^{\mathrm{\prime }}{199.17}^{\mathrm{\prime }}-k\mathrm{\mid }}{\sqrt{\frac{{}^{\mathrm{\prime }}{9.8788}^{\mathrm{\prime }}}{12}}}{<}^{\mathrm{\prime }}{1.796}^{\mathrm{\prime }}$

200.8

$m=69.522$

$n=0.156$ or $0.157$

${\text{H}}_0:\text{Po}(0.7)$ fits the data. ${\text{H}}_1:\text{Po}(0.7)$ does not fit the data.

Combining last 3 columns give 13 and 6.828 or 6.829.

$\frac{(88-99.317{)}^2}{99.317}+\frac{(73-69.552{)}^2}{69.552}+\cdots +\frac{(13-6.829{)}^2}{6.829}$ $[1.2896+0.1740+0.1142+5.5764]$ OR $\frac{{88}^2}{99.317}+\frac{{73}^2}{69.552}+\frac{{26}^2}{24.333}+\frac{{13}^2}{6.829}-200$

7.15 or 7.16

'7.15' < 7.815, accept ${\text{H}}_0$ / do not reject ${\text{H}}_0$ / not significant.

Insufficient evidence to reject expert’s claim. OR Insufficient evidence to suggest that $\text{Po}(0.7)$ is not a good fit/model [for the data].

$a=-\frac{1}{5}$

$b=\frac{7}{25}$

Attempt at differentiating both parts.

All correct, fully defined including domain. Domain must cover all reals, must not overlap at boundaries.

$\text{E}(X^2)=\frac{1}{5}{\int }_1^4{x}^2\text{d}x+\frac{1}{25}{\int }_4^6{x}^3\text{d}x=[\frac{1}{15}{x}^3{]}_1^4+[\frac{1}{100}{x}^4{]}_4^6$

$\text{Var}(X)=\frac{73}{5}-(\frac{529}{150}{)}^2$

2.16

10th percentile: $\frac{1}{5}x-\frac{1}{5}=\frac{1}{10}$ 90th percentile: $\frac{1}{50}{x}^2+\frac{7}{25}=\frac{9}{10}$

10th percentile = 1.5

90th percentile = $\sqrt{31}$

${\text{H}}_0:{\mu }_A={\mu }_B{\text{H}}_1:{\mu }_A\mathrm{\ne }{\mu }_B$

${}^{\mathrm{\prime }}{1331848}^{\mathrm{\prime }}-\frac{{}^{\mathrm{\prime }}{3264}^{\mathrm{\prime }2}}{8}[=136]$

${}^{\mathrm{\prime }}{964964}^{\mathrm{\prime }}-\frac{{}^{\mathrm{\prime }}{2406}^{\mathrm{\prime }2}}{6}[=158]$

$s_p^2=\frac{{}^{\mathrm{\prime }}{136}^{\mathrm{\prime }}{+}^{\mathrm{\prime }}{158}^{\mathrm{\prime }}}{8+6-2}=24.5$

24.5

t = \frac{\frac{'3264'}{8} - \frac{'2406'}{6}}{\sqrt{'s_p^2' (\frac{1}{8} + \frac{1}{6})}} = 2.62

2.62

'2.62' > 2.179, reject ${\text{H}}_0$ / significant.

Sufficient evidence to suggest that there is a difference in the mean tensile strengths of steel rods produced by the two machines.

$G_X^{\mathrm{\prime }}(t)=\frac{(3-2t{)}^2+4(3-2t)t}{(3-2t{)}^4}[=\frac{3+2t}{(3-2t{)}^3}]$

$\text{E}(X)=G_X^{\mathrm{\prime }}(1)=5$

$G_X^{\mathrm{\prime }\mathrm{\prime }}(t)=\frac{2(3-2t{)}^3+6(3+2t)(3-2t{)}^2}{(3-2t{)}^6}[=\frac{24+8t}{(3-2t{)}^4}]$

$\text{Var}(X)=G_X^{\mathrm{\prime }\mathrm{\prime }}(1)+G_X^{\mathrm{\prime }}(1)-(G_X^{\mathrm{\prime }}(1){)}^2=32+5-25$

$[\text{Var}(X)=]12$

$G_Z(t)=\frac{t^3}{(3-2t{)}^4}=\frac{1}{81}{t}^3(1+(-4)(-\frac{2}{3}t)+\dots )$

$\text{P}(Z=4)=\text{their coefficient of }{t}^4$. OR $\text{P}(Z=3)=\text{their coefficient of }{t}^3$.

$[\text{P}(Z=4)=]\frac{8}{243}$ and $[\text{P}(Z=3)=]\frac{1}{81}$

$\text{P}(Z>4)=1{-}^{\mathrm{\prime }}{\frac{8}{243}}^{\mathrm{\prime }}{-}^{\mathrm{\prime }}{\frac{1}{81}}^{\mathrm{\prime }}$

$\text{P}(Z>4)=\frac{232}{243}$