A Level Further Mathematics 9231 Paper 43, Oct/Nov 2025

$s^2=\frac{1}{9}({}^{\mathrm{\prime }}{71314}^{\mathrm{\prime }}-\frac{{}^{\mathrm{\prime }}{844}^{\mathrm{\prime }2}}{10})[=\frac{134}{15}=8.933]$

2.262

$\frac{{}^{\mathrm{\prime }}{844}^{\mathrm{\prime }}}{10}{\pm }^{\mathrm{\prime }}{2.262}^{\mathrm{\prime }}\sqrt{\frac{{}^{\mathrm{\prime }}{8.933}^{\mathrm{\prime }}}{10}}$

$[82.3,86.5]$

The distribution of estimates of angles is normal. OR The estimates are a random sample from some population. OR The estimates are independent. OR Underlying distribution is normal.

Population unlikely to be normal as $\theta$ is close to right-angle / data is skewed. OR Estimates may not be independent, for example due to collusion. OR No indication that sample is random.

$p=\frac{{\mathrm{e}}^{-2.8}(2.8{)}^3}{3!}\times 100=22.248$

$q=\frac{(29-22.25{)}^2}{22.25}=2.05$

${\mathrm{H}}_0$: Po(2.8) fits the data. ${\mathrm{H}}_1$: Po(2.8) does not fit the data. OR ${\mathrm{H}}_0$: Po(2.8) is a satisfactory model for the number of cars (entering the car park). ${\mathrm{H}}_1$: Po(2.8) is not a satisfactory model for the number of cars (entering the car park).

$2.739+0.241+2.152{+}^{\mathrm{\prime }}{2.049}^{\mathrm{\prime }}+0.425+3.753+0.037[=11.396]$

${}^{\mathrm{\prime }}{11.4}^{\mathrm{\prime }}<12.59$, accept ${\mathrm{H}}_0$ / do not reject ${\mathrm{H}}_0$ / not significant.

Insufficient evidence to suggest that Po(2.8) is not a good fit to the data. OR Insufficient evidence to suggest that the manager’s claim is false.

$\sum (x-\bar{x}{)}^2=9\times 0.36=3.24$

${}^{\mathrm{\prime }}{55.8}^{\mathrm{\prime }}-\frac{{}^{\mathrm{\prime }}{16.6}^{\mathrm{\prime }2}}{5}$

$\frac{3.24+0.688}{10+5-2}(=0.302)$

$s=0.550$

$\frac{1}{4}\times 28\times (28+1)[=203]$ $\frac{1}{24}\times 28\times (28+1)\times (56+1)[=1928.5]$

$[z=]\frac{T+0.5{-}^{\mathrm{\prime }}{203}^{\mathrm{\prime }}}{\sqrt{{}^{\mathrm{\prime }}{1928.5}^{\mathrm{\prime }}}}$

$\frac{T+0.5{-}^{\mathrm{\prime }}{203}^{\mathrm{\prime }}}{\sqrt{{}^{\mathrm{\prime }}{1928.5}^{\mathrm{\prime }}}}>-2.576$

$T>89.4$

$T=90$

$\frac{1}{16}{\left[\frac{2}{3}{x}^{\frac{3}{2}}\right]}_0^4+\frac{1}{k}{\left[2\sqrt{x}\right]}_4^9=1$ $\frac{1}{24}(8-0)+\frac{2}{k}(3-2)=1$ OR $\frac{1}{3}+\frac{2}{k}=1$

$\frac{2}{k}=\frac{2}{3},k=3$

$\frac{1}{16}{\int }_0^4\sqrt{x}\mathrm{d}x+\frac{1}{3}{\int }_4^m{x}^{-\frac{1}{2}}\mathrm{d}x=\frac{1}{2}$ OR $\frac{1}{3}{\int }_m^9{x}^{-\frac{1}{2}}\mathrm{d}x=0.5$

$\frac{1}{3}+\frac{2}{3}(\sqrt{m}-2)=\frac{1}{2}$ OR $\frac{2}{3}(3-\sqrt{m})=\frac{1}{2}$

$m=\frac{81}{16}[=5.0625]$

Attempt to integrate both parts, limits not required. May be seen in part 5(a).

For constant –1 obtained in expression for $4⩽x⩽9$. May be seen in part 5(a).

$\mathrm{G}(y)=\{\begin{array}{ll}\frac{1}{24}{y}^3& 0⩽y<2\\ \frac{2}{3}y-1& 2⩽y⩽3\end{array}$

For differentiation, limits not required.

$\mathrm{g}(y)=\{\begin{array}{ll}\frac{1}{8}{y}^2& 0⩽y<2\\ \frac{2}{3}& 2⩽y⩽3\\ 0& \text{otherwise}\end{array}$

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

Differences: $5,0,-1,6,1,-2,8,7,15$

$s_d^2=\frac{1}{8}({}^{\mathrm{\prime }}{405}^{\mathrm{\prime }}-\frac{{}^{\mathrm{\prime }}{39}^{\mathrm{\prime }2}}{9})[=29.5]$

$\frac{\frac{{}^{\mathrm{\prime }}{39}^{\mathrm{\prime }}}{9}}{\sqrt{\frac{{}^{\mathrm{\prime }}{29.5}^{\mathrm{\prime }}}{9}}}$

$=2.393$

${}^{\mathrm{\prime }}{2.393}^{\mathrm{\prime }}>1.860$, reject ${\mathrm{H}}_0$ / significant.

Sufficient evidence to suggest new training programme results in reduced times. OR Sufficient evidence to support the coach’s belief.

The population of differences may not be normally distributed. The population of differences may not be symmetrical (and hence not normal).

A Wilcoxon matched-pairs signed-rank test. OR A paired-sample sign test.

${\mathrm{G}}_X(t)=a+2at+b{t}^2$

${\mathrm{G}}_X^{\mathrm{\prime }}(t)=2a+2bt$ $\mathrm{E}(X)={\mathrm{G}}_X^{\mathrm{\prime }}(1)=2a+2b$

${\mathrm{G}}_X^{\mathrm{\prime }\mathrm{\prime }}(t)=2b$

$\mathrm{V}\mathrm{a}\mathrm{r}(X)=2b+(2a+2b)-(2a+2b{)}^2$

$2b+(2a+2b)-(2a+2b{)}^2=2b+(2a+2b)(1-(2a+2b))=2b+2(a+b)(1-2a-2b)$

${\mathrm{G}}_Y(t)=(a+2at+b{t}^2{)}^{10}$

${\mathrm{G}}_Y^{\mathrm{\prime }}(t)=20(a+bt)(a+2at+b{t}^2{)}^9$ $\mathrm{E}(Y)=20(a+b)(3a+b{)}^9$

$3a+b=1$, hence $\mathrm{E}(Y)=20(a+b)=10\mathrm{E}(X)$.

Binomial

$\mathrm{B}(10,\frac{2}{3})$