A Level Mathematics 9709 Paper 63, May/June 2023

Attempt to integrate $x\text{f}(x)$; ignore limits.

Correct integration and limits.

$=\frac{3}{8}$

One correct. Ignore incorrect numbers.

Both correct and no extra numbers seen. (Allow other correct use of list of digits).

These numbers are not independent of the previous numbers OR Only a finite number of digits used

$z=1.645$

$$z\times \frac{\sqrt{\frac{x}{100}\times (1-\frac{x}{100})}}{100}=0.07896$$

$$[x(100-x)={100}^3\times {0.07896}^2÷{1.645}^2]$$ $$x^2-100x+2304=0$$

$x=36\text{ or }64$

$${0.1}^2=0.01$$

Mean $=7\times 65.2=456.4$ (or Mean $=7\times 65.2\times 50=22820$)

Var $=7\times {3.6}^2[=90.72]$ (or Var $=7\times {3.6}^2$)

$22000\mathrm{/}50=440$ used in standardising equation (or Var $={50}^2\times \text{’90.72’ }[=226800]$)

For standardising with their values. No mixed methods.

For correct probability area consistent with their values.

$=0.0425\text{ or }0.0426$

$$\bar{x}=1700\mathrm{/}50=34$$

$$\text{Est}({\sigma }^2)=\frac{50}{49}(\frac{59050}{50}-{34}^2)\text{ or }\frac{1}{49}(59050-\frac{{1700}^2}{50})$$

$$=25.5\text{ (3 sf) or }\frac{1250}{49}$$

${\text{H}}_0$: Population mean time $=32.4$, ${\text{H}}_1$: Population mean time $\mathrm{\ne }32.4$

$$\frac{34-32.4}{\frac{\sqrt{\text{’25.5’}}}{\sqrt{50}}}$$

$=2.24\text{ (3 sf)}$

$\text{’2.24’}<2.326$

[Not reject ${\text{H}}_0$] Insufficient evidence that (mean) time has changed

Distribution of times in the population is normal

$X\sim \text{Po}(2.5)$

$${\text{e}}^{-2.5}(1+2.5+\frac{{2.5}^2}{2}+\frac{{2.5}^3}{3!})$$

$=0.758\text{ (3 sf)}$

$$\text{E}(X)=\frac{5}{2}\text{ or }2.5,\text{ Var}(X)=\frac{4999}{2000}\text{ or }2.4995$$

These are almost equal

$$\frac{1}{2}\pi {\left(\sqrt{\frac{2}{\pi }}\right)}^2$$

$=1$, which is the area under a PDF [and $\text{f}(x)\ge 0$]

$${\cos }^{-1}\left(\frac{\sqrt{\frac{1}{\pi }}}{\sqrt{\frac{2}{\pi }}}\right)=\frac{\pi }{4}$$ (or Use of correct limits $\sqrt{\frac{1}{\pi }}$ to $\sqrt{\frac{2}{\pi }}$)

Area of sector $=\frac{1}{4}$ (or Find equation of curve $x^2+y^2=\frac{2}{\pi }$, M1 in alt method)

Area of triangle $AOB=\frac{1}{2}OA\times OB=\frac{1}{2}\times \sqrt{\frac{1}{\pi }}\times \sqrt{\frac{2}{\pi }-\frac{1}{\pi }}$ or $\text{Area of triangle }AOB=\frac{1}{2}OA\times OB\times \sin (AOB)=\frac{1}{2}\times \sqrt{\frac{1}{\pi }}\times \sqrt{\frac{2}{\pi }}\sin \frac{\pi }{4}$ (or $y=\sqrt{\frac{2}{\pi }-x^2}$, A1 in alt method)

$\frac{1}{2\pi }\text{ or }0.1592$ (or Attempt to integrate (any limits), M1 in alt method)

‘$\frac{1}{4}$’ – ‘$\frac{1}{2\pi }$’ or ‘0.25’ – ‘0.1592’ (or Correct integration with correct limits, A1 in alt method)

$=\frac{1}{4}-\frac{1}{2\pi }\text{ or }0.0908\text{ (3sf)}$

${\text{H}}_0$: Pop mean no. people $=3.03$ or $1.01$ (per 20 min), ${\text{H}}_1$: Pop mean no. people $>3.03$ or $1.01$ (per 20 min)

Use of $\text{Po}(3.03)$

$$=1-{\text{e}}^{-3.03}(1+3.03+\frac{{3.03}^2}{2}+\frac{{3.03}^3}{3!}+\frac{{3.03}^4}{4!}+\frac{{3.03}^5}{5!})$$

$=0.0870\text{ (3sf) }[0.0869727]$

$0.0870>0.05$

(Do not reject ${\text{H}}_0$) Insufficient evidence to believe (mean) number of people has increased

"0.0869727" – ${\text{e}}^{-3.03}\times \frac{{3.03}^6}{6!}$ or $0.869727-{\text{e}}^{-3.03}(1.0748)$ or $0.869727-0.05193$ or $1-{\text{e}}^{-3.03}(1+3.03+\frac{{3.03}^2}{2}+\frac{{3.03}^3}{3!}+\frac{{3.03}^4}{4!}+\frac{{3.03}^5}{5!}+\frac{{3.03}^6}{6!})$

$0.0350\text{ or }0.0351$

Concluding that the (mean) number of people (using the path per 20 mins in the evening) has increased when it has not

A value for the true mean

Number of people using the path per 20 mins in the evening.