A Level Mathematics 9709 Paper 63, May/June 2024

[$\lambda =$] 4

${\text{e}}^{-4}(\frac{4^2}{2!}+\frac{4^3}{3!}+\frac{4^4}{4!})$ or ${\text{e}}^{-4}(8+10.67+10.67)$ or $0.1465+0.19537+0.19537$

$=0.537$ (3sf)

$n=4000>50$ and either $np=4<5$ or $p=0.001<0.1$

$3.12+z\times \frac{\sigma }{\sqrt{150}}=3.23$

$z=1.96$

$\sigma =0.687$ (3sf) [cm]

Yes, because population [of widths] not given to be normally distributed

For N and mean $=\pm (5\times 50-10\times 26)$ SOI

For var $=5\times 2.4+10\times 1.8$ SOI

$\frac{0-(\text{’-10’})}{\sqrt{\text{’30’}}}[=1.826]$

$1-\mathrm{\Phi }(\text{’1.826’})$

$=0.0339$ or $0.034[0]$ (3sf)

${\text{H}}_0:p=0.24$ ${\text{H}}_1:p<0.24$

$\text{P}(X\le 2)={0.76}^{25}+25\times {0.76}^{24}\times 0.24+{}^{25}{\text{C}}_2\times {0.76}^{23}\times {0.24}^2$ or $0.0010479+0.0082732+0.0313513$

$=0.0407$

$0.0407<0.05$

[Evidence to reject ${\text{H}}_0$.] There is sufficient evidence to suggest that the support for the Today Party has decreased.

Attempted integration of $\text{f}(x)$ and equated to 1.

Correct integration and substitute correct limits: ${[a\frac{x^2}{2}-\frac{x^4}{4}]}_0^{\sqrt{2}}=1⟹a-\frac{4}{4}=1$

$a=2$

${\int }_0^m(2x-x^3)\text{d}x=\frac{1}{2}$

$m^2-\frac{m^4}{4}=\frac{1}{2}$

$m^4-4m^2+2=0⟹m^2=\frac{4\pm \sqrt{16-8}}{2}[=2\pm \sqrt{2}]$

$m=\sqrt{2-\sqrt{2}}$ or $0.765$ (3sf)

${\int }_0^{\sqrt{2}}(2x^2-x^4)\text{d}x$

${[\frac{2x^3}{3}-\frac{x^5}{5}]}_0^{\sqrt{2}}$

$[=\frac{4\sqrt{2}}{3}-\frac{4\sqrt{2}}{5}]=\frac{8}{15}\sqrt{2}$

$\widehat{\mu }=\frac{307}{200}$ or $1.535$

$\mathrm{\Sigma }{x}^{2}f=627$, $\text{Est}({\sigma }^2)=\frac{200}{199}(\frac{\text{’627’}}{200}-{\text{’1.535’}}^2)$ or $\frac{1}{199}(\text{’627’}-\frac{{\text{’307’}}^2}{200})$

$=0.783$

${\text{H}}_0:\mu =1.65$ ${\text{H}}_1:\mu <1.65$

$\frac{\text{’1.535’}-1.65}{\sqrt{0.783÷200}}$

$=-1.838$ or $-1.84$

$\mathrm{\Phi }(0.05)$ and $\mathrm{\Phi }(0.01)$ attempted

$-1.645>-1.838>-2.326$ [Hence significant at 5% but not 1% level]

At the 1% level ${\text{H}}_0$ is not rejected Or a Type I error can only occur if ${\text{H}}_0$ is rejected.

$\lambda =1.9+2.2[=4.1]$

${\text{e}}^{-4.1}(1+4.1+\frac{{4.1}^2}{2!}+\frac{{4.1}^3}{3!})$ or ${\text{e}}^{-4.1}(1+4.1+8.405+11.487)$ or $0.01657+0.06795+0.13929+0.19037$

$=0.414$ (3sf)

$\text{P}(X+Y<4\text{ and }X=2)=\text{P}(2,0\text{ or }2,1)$

$={\text{e}}^{-1.9}\times \frac{{1.9}^2}{2}({\text{e}}^{-2.2}+{\text{e}}^{-2.2}\times 2.2)[=0.0957]$

$\text{P}(X=2\mid X+Y<4)=\frac{\text{’0.0957’}}{\text{’0.414’}}$

$0.231$ (3sf)

$\text{E}(X+Y)=4.1\text{Var}(X+Y)=4.1$ or $\text{Po}(246)$

Normal and var $=\frac{4.1}{60}$ Or normal and var $=246$

$\frac{4.0-4.1}{\sqrt{4.1÷60}}$ or totals method $\frac{240-246}{\sqrt{246}}$ or use of continuity correction

$=-0.383$ (3sf)

$\mathrm{\Phi }(\text{’-0.383’})=1-\mathrm{\Phi }(\text{’0.383’})$

$=0.351$ (3sf)