A Level Further Mathematics 9231 Paper 42, May/June 2025

${\text{H}}_0:{\mu }_x={\mu }_y{\text{H}}_1:{\mu }_x\mathrm{\ne }{\mu }_y$

$s_x^2=\mathrm{103.218...}$ $s_y^2=\mathrm{171.139...}$

$[\frac{s_x^2}{120}+\frac{s_y^2}{80}=2.999]$

Value of $z$ obtained using correct form with their $s_x^2$ and their $s_y^2$

$[z=]\frac{\frac{10470}{120}-\frac{6560}{80}}{\sqrt{\frac{s_x^2}{120}+\frac{s_y^2}{80}}}=3.03$

'3.03' > 2.576, reject ${\text{H}}_0$ (significant). OR $\text{P}(Z>\text{’3.03’})=0.00122<0.005$, reject ${\text{H}}_0$ (significant).

There is sufficient evidence to suggest the mean spend of customers is different on weekends compared to weekdays.

$5a+\frac{3}{2}a=1$, leading to $a=\frac{2}{13}$.

Attempt to solve 2 independent equations in $b$ and $c$ or 3 independent equations in $a$, $b$ and $c$.

$b=\frac{16}{39},c=\frac{2}{39}$

$[\text{E}(X)=]{\int }_0^5\frac{2}{13}x\text{d}x+{\int }_5^{8}x(\frac{16}{39}-\frac{2}{39}x)\text{d}x$

$[\frac{1}{13}{x}^2{]}_0^5+[\frac{8}{39}{x}^2-\frac{2}{117}{x}^3{]}_5^8$

$\frac{43}{13}=3.31$

$\frac{2}{13}m=0.5$ OR ${\int }_0^m\frac{2}{13}\text{d}x=0.5$

$m=\frac{13}{4}=3.25$

$\frac{2}{13}x,\frac{16}{39}x-\frac{1}{39}{x}^2(+k)$

$k=-\frac{25}{39}$

Both parts with $\sqrt{y}$ substituted (condone missing $k$), ignore limits.

$\text{G}(y)=\{\begin{array}{ll}0& y<0\\ \frac{2}{13}\sqrt{y}& 0⩽y⩽25\\ \frac{16}{39}\sqrt{y}-\frac{1}{39}y-\frac{25}{39}& 25<y⩽64\\ 1& y>64\end{array}$

$[\frac{0\times 1844+1\times 143+2\times 11+3\times 0+4\times 1+5\times 0+6\times 1}{2000\times 6}=]\frac{7}{480}$

$a=2000-(1831.3+6.016+0.119+0.001)=162.6$ OR $a=\left(\genfrac{}{}{0px}{}{6}{1}\right)\times \frac{7}{480}\times (1-\frac{7}{480}{)}^5\times 2000=162.6$

${\text{H}}_0:\text{A binomial distribution is a satisfactory model for the data.}$ ${\text{H}}_1:\text{A binomial distribution is not a satisfactory model for the data.}$

$\frac{(1844-1831.3{)}^2}{1831.3}+\frac{(143-162.6{)}^2}{162.6}+\frac{(13-6.136{)}^2}{6.136}[=0.08807+2.3626+7.6784]$ OR $\frac{{1844}^2}{1831.3}+\frac{{143}^2}{162.6}+\frac{{13}^2}{6.136}-2000$

10.1

[1 degree of freedom so critical value is 7.879.] '10.1' > 7.879, reject ${\text{H}}_0$ (significant).

Sufficient evidence to suggest that the manager’s model is not satisfactory.

Not independent or probability of “success” may not be constant or plausible reason in context (for example, higher number of boxes with more than 1 broken egg due to boxes being dropped).

$\mathrm{\Sigma }(x-\bar{x}{)}^2=40104-\frac{1}{10}\times {612}^2[=2649.6]$ $\mathrm{\Sigma }(y-\bar{y}{)}^2=27460-\frac{1}{8}\times {444}^2[=2818]$

$s_p^2=\frac{2649.6+2818}{10+8-2}[=341.725]$

2.120, SOI.

Correct form, must be a $t$ value.

Correct un-simplified confidence interval, SOI.

$[-12.9,24.3]$

Because zero lies inside the confidence interval.

${\text{H}}_0:m_s=m_c$ ${\text{H}}_1:m_s>m_c$

Attempt at differences.

Attempt at (signed) ranks.

$(P=51,Q=4)T=4$

'4' $⩽10$, reject ${\text{H}}_0$ (significant).

Sufficient evidence to suggest that children, on average, complete the cartoon puzzle more quickly than the seaside puzzle. OR Sufficient evidence to support the researcher’s belief.

$[X\sim \text{B}(10,0.5)]\text{P}(X⩽2)=0.0547$

'0.0547' > 0.05, accept ${\text{H}}_0$ (not significant).

In part 5(a) ${\text{H}}_0$ is rejected, in part 5(b) ${\text{H}}_0$ is accepted, so opposite conclusion reached.

The order of the tests means that children may do the second test quicker (or slower) because, for example, they have had practice (or are tired).

$k=1-a$

$(1-a{t}^2{)}^{-1}=1+a{t}^2(+\dots )$ OR $\text{P}(Y=0)={\text{G}}_Y(0)=k,\text{P}(Y=2)=\frac{1}{2}{\text{G}}_Y^{\mathrm{\prime }\mathrm{\prime }}(0)=\frac{1}{2}(2ak)$

$\text{P}(Y>2)=1-k(1+a)$ OR $\text{P}(Y>2)=1-k-ak$

$\text{P}(Y>2)=1-(1-a)(1+a)=a^2$ OR $\text{P}(Y>2)=1-(1-a)-a(1-a)=a^2$

${\text{G}}_Y^{\mathrm{\prime }}(t)=\frac{0.32t}{(1-0.2t^2{)}^2}$

$\text{E}(Y)={\text{G}}_Y^{\mathrm{\prime }}(1)=\frac{0.32}{(1-0.2{)}^2}=0.5$