A Level Mathematics 9709 Paper 62, May/June 2026

$\text{Est}(\mu )=\left[\frac{88.9}{10}\right]=\frac{889}{100}=8.89$

$\text{Est}({\sigma }^2)=\frac{10}{9}(\frac{794.47}{10}-{\text{’8.89’}}^2)\text{ or }\frac{1}{9}(794.47-{88.9}^2\mathrm{/}10)$

$=0.461$

$C-2(A+B)\sim \text{N}(-3,\dots )$

$\text{Var}(C-2(A+B))=24+4(15+8)[=116]$

$\frac{0-(-3)}{\sqrt{\text{’116’}}}[=0.27854\dots ]$

$1-\mathrm{\Phi }(\text{’0.27854’})=1-(0.6096\text{ to }0.6099)$

$=0.390\text{ (3 sf)}$

${\text{H}}_0:\text{Population mean}=2.4{\text{H}}_1:\text{Population mean}>2.4$

$\text{P}(X⩾a\mid X\sim \text{Po}(2.4))$ attempted, where $a$ is any integer 3 or more (or $\text{P}(X⩽b)$ attempted where $b$ is any integer 2 or more)

$\text{P}(X⩾5\text{ or }6\mid X\sim \text{Po}(2.4))$ attempted (or $\text{P}(X⩽4\text{ or }5)$ attempted)

$[\text{P}(X⩾6)]=0.0357$ (or $[\text{P}(X⩽5)]=0.9643$)

Smallest value is 6

0.0357 (3 s.f.) Accept 0.036 (2 s.f.)

$[D+E\sim \text{Po}](3.8)$

$\text{P}(D+E<4)={\text{e}}^{-3.8}(1+3.8+\frac{{3.8}^2}{2!}+\frac{{3.8}^3}{3!})={\text{e}}^{-3.8}(1+3.8+7.22+9.1453)=0.02237+0.08501+0.1615+0.2046$

$[=0.47348]=0.473\text{ (3 sf) accept }0.474$

$\text{P}(D=1)\times \text{P}(E=0,1,2)={\text{e}}^{-1.7}\times 1.7\times {\text{e}}^{-2.1}(1+2.1+\frac{{2.1}^2}{2})=0.03803+0.079864+0.083856[=0.20175]$

$\frac{\text{P}(D=1)\times \text{P}(E=0,1,2)}{\text{P}(D+E<4)}=\frac{{\text{e}}^{-1.7}\times 1.7\times {\text{e}}^{-2.1}(1+2.1+\frac{{2.1}^2}{2})}{\text{Their }0.47348}[=\frac{0.20175}{0.47348}]$

$=0.426\text{ (3 sf) accept }0.427$

$T\sim \text{N}(170,170)$

$\frac{190.5-170}{\sqrt{170}}[=1.572]$

$1-\mathrm{\Phi }(\text{’1.572’})$

$=0.0580\text{ Allow }0.058\text{ Allow }0.0579$

$2\times 1.96\times \sqrt{\frac{x(100-x)}{{100}^3}}=0.165\text{ or in terms of }p\left(\text{or }\frac{x}{100}\right)$

$1.96\text{ seen}$

$x^2-100x+1771.72=0\text{ or }{p}^2-p+0.177172=0$

$=77\text{ or }23$

$2\times y(1-y)=0.1128$

$y^2-y+0.0564=0[y=0.94[\text{or }0.06]]$

$\alpha =94$

${\text{H}}_0:\mu =12.4{\text{H}}_1:\mu <12.4$

$\frac{11.9-12.4}{\sqrt{\frac{9.2}{75}}}$

$=-1.428\text{ or }-1.427\text{OR }0.0767\text{ or }0.0768\text{ or }0.0766\text{ (3sf)}$

$-1.428>-1.645\text{ or }1.428<1.645\text{ or }0.0767>0.05$

Insufficient evidence to suggest that the (mean) time (spent by customers) has decreased / Insufficient evidence to suggest that $\mu <12.4$ [Do not reject ${\text{H}}_0$]

No because population normal

$\frac{11.8-12.4}{\sqrt{\frac{9.2}{n}}}<-1.645$

$\sqrt{n}>1.645÷0.6\times \sqrt{9.2}\text{ or }n>{1.645}^2÷{0.6}^2\times 9.2$

$[n>69.2]n⩾70$

$\text{Area}=\frac{1}{2}\times a\times a\text{ or }{\int }_0^a(a-x)\text{d}x$

'$\frac{1}{2}{a}^2$' $=1$

$a=\sqrt{2}$

$\text{E}(X)={\int }_0^{\sqrt{2}}(\sqrt{2}x-x^2)\text{d}x$

$={[\frac{\sqrt{2}{x}^2}{2}-\frac{x^3}{3}]}_0^{\sqrt{2}}$

$[=\sqrt{2}-\frac{2}{3}\sqrt{2}]=\frac{1}{3}\sqrt{2}$

$\frac{1}{2}\times (\sqrt{2}-m)\times (\sqrt{2}-m)=\frac{1}{2}\text{ or }{\int }_0^m(\sqrt{2}-x)\text{d}x=\frac{1}{2}\text{ or }{\int }_m^{\sqrt{2}}(\sqrt{2}-x)\text{d}x=\frac{1}{2}$ Or in terms of $a$ $\int (a-x)\text{d}x$ limits 0 and $m$

$m^2-2\sqrt{2}m+1=0(m^2-2am+1=0)[m=\frac{2\sqrt{2}\pm \sqrt{8-4}}{2}=\sqrt{2}\pm 1]$

$\sqrt{2}+1\text{ is not in the domain of }X$

$m=\sqrt{2}-1$