A Level Mathematics 9709 Paper 61, May/June 2025

($\lambda$) = 4

${\text{e}}^{-4}(1+4+\frac{4^2}{2!}+\frac{4^3}{3!})={\text{e}}^{-4}(1+4+8+10.6666)=0.018316+0.073263+0.146525+0.19537$

= 0.433

N(6, ....)

${\sigma }^2=\frac{3}{2}=1.5$ (or $\sigma =\sqrt{1.5}$ or $\frac{\sqrt{6}}{2}$ or 1.22 (3sf))

$\frac{6.2-6}{\sqrt{\frac{1.5}{100}}}$

= 1.633

$\text{P}(\bar{X}>6.2)=1-\mathrm{\Phi }(\text{‘}1.633\text{’})$

= 0.0512 or 0.0513 (3sf)

Population ($X$) is not normally distributed / Population ($X$) is Binomial

$\text{Est}(\mu )=\frac{2750}{60}$ or $\frac{275}{6}$ or $45.8\text{ (3 sf)}$

$\text{Est}({\sigma }^2)=\frac{60}{59}(\frac{127000}{60}-{\left(\frac{275}{6}\right)}^2)$

= 16.2 (3 sf) or $\frac{2875}{177}$

${\text{H}}_0:\text{Population mean time}=45{\text{H}}_1:\text{Population mean time}>45$

$\frac{\text{‘}\frac{275}{6}\text{’}-45}{\sqrt{\frac{\text{‘}16.24294\text{’}}{60}}}$

= 1.602 to 1.595

‘1.602’ < 1.645

[Accept ${\text{H}}_0$] There is insufficient evidence [at 5% level] to reject the company’s claim OR There is insufficient evidence to accept the passenger’s belief OR There is insufficient evidence that the mean time is more than 45 minutes

Mean = 6 [$]

Variance = ${0.40}^2\times 9$

= 1.44 [${\mathrm{\$}}^2$]

Mean = $35\times \text{‘6’}=210$ [$]

Variance = $35\times \text{‘1.44’}$

= 50.4 [${\mathrm{\$}}^2$]

${\text{e}}^{-3.5}(\frac{{3.5}^3}{3!}+\frac{{3.5}^4}{4!}+\frac{{3.5}^5}{5!})={\text{e}}^{-3.5}(7.1458+6.2526+4.37682)=0.21579+0.18881+0.13217$

= 0.537

Use of $\text{N}(1.2\times 100+2.3\times 200,\dots )=\text{N}(580,\dots )$

$\text{Var}(S)=1.2\times 100+2.3\times 200=580$

$\frac{600.5-\text{‘580’}}{\sqrt{\text{‘580’}}}$

= 0.851

$1-\mathrm{\Phi }(\text{‘0.581’})$

= 0.197 (3sf)

$\frac{5}{2}\times \frac{{\lambda }^3}{3!}+\frac{{\lambda }^4}{4!}=\frac{{\lambda }^5}{5!}$ Or $\left(\frac{5}{2}\right){\text{e}}^{-\lambda }\left(\frac{{\lambda }^3}{3!}\right)+{\text{e}}^{-\lambda }\left(\frac{{\lambda }^4}{4!}\right)={\text{e}}^{-\lambda }\left(\frac{{\lambda }^5}{5!}\right)$

${\lambda }^2-5\lambda -50=0$

$(\lambda -10)(\lambda +5)=0$ or $\lambda =\frac{5\pm \sqrt{25+4\times 50}}{2}$ $\lambda =10$

${\text{H}}_0:\text{Proportion (at college) owning Pumpkin phone}=0.25$ ${\text{H}}_1:\text{Proportion (at college) owning Pumpkin phone}<0.25$

$\text{B}(30,0.1)$ and $\text{P}(X⩾5)$ attempted

$1-({0.9}^{30}+30\times {0.9}^{29}\times 0.1+{}^{30}{\text{C}}_2\times {0.9}^{28}\times {0.1}^2+{}^{30}{\text{C}}_3\times {0.9}^{27}\times {0.1}^3+{}^{30}{\text{C}}_4\times {0.9}^{26}\times {0.1}^4)=1-(0.042391+0.141304+0.22766+0.236088+0.177066)$

= 0.175 (3 sf) accept 0.176

$\frac{1}{8}\pm z\sqrt{\frac{\frac{1}{8}\times \frac{7}{8}}{40}}$

$z=1.96$

0.0225 to 0.227 (3 sf)

${\int }_0^{\frac{1}{2}}(1+\cos \pi x)\text{d}x$

$={[x+\frac{\sin \pi x}{\pi }]}_0^{\frac{1}{2}}$

$=\frac{1}{2}+\frac{\sin \frac{\pi }{2}}{\pi }=\frac{1}{2}+\frac{1}{\pi }$

${\int }_0^1(x+x\cos \pi x)\text{d}x$

$={[\frac{x^2}{2}+x\frac{\sin \pi x}{\pi }-\int \frac{\sin \pi x}{\pi }\text{d}x]}_0^1$

$={[\frac{x^2}{2}+x\frac{\sin \pi x}{\pi }+\frac{\cos \pi x}{{\pi }^2}]}_0^1$

$=\frac{1}{2}+\frac{\sin \pi }{\pi }+\frac{\cos \pi }{{\pi }^2}-\frac{\cos 0}{{\pi }^2}$

$=\frac{1}{2}-\frac{2}{{\pi }^2}$