A Level Mathematics 9709 Paper 62, Oct/Nov 2024

$\lambda =4.5$

$1-{\text{e}}^{-4.5}(1+4.5+\frac{{4.5}^2}{2}+\frac{{4.5}^3}{3!})=1-{\text{e}}^{-4.5}(1+4.5+10.125+15.1875)=1-(0.011109+0.049999+0.11248+0.16872)$

0.658 (3 sf)

254 [m]

$263=\text{‘}254\text{’}+1.96\times \frac{\sigma }{\sqrt{50}}$ oe or $2\times 1.96\times \frac{\sigma }{\sqrt{50}}=18$

$[\sigma =\frac{9\sqrt{50}}{1.96}=]$. s.d. = 32.5 [m] (3 sf)

No Because the sample mean is approximately normally distributed [for large n]

${\text{H}}_0:p=0.8{\text{H}}_1:p>0.8$

$[\text{Assuming }{\text{H}}_0,\text{P}(X⩾23)=]{}^{25}{\text{C}}_{23}\times {0.2}^2\times {0.8}^{23}+{}^{25}{\text{C}}_{24}\times 0.2\times {0.8}^{24}+{0.8}^{25}=0.070835+0.0236118+0.0037779$

= 0.0982

0.0982 < 0.1

$[\text{There is evidence to reject }{\text{H}}_0]$ There is sufficient evidence to suggest that $p$ has increased

Not suitable as model does not allow for more than 25 employees to use the canteen/Not suitable as uses a sample instead of all employees/Not suitable doesn’t include all employees /Not suitable as 30 is only just bigger than 25 should have used 30 OR Suitable as owner knows that not all employees use the canteen, or similar

$\frac{36-35}{8.1÷\sqrt{140}}$ $[=1.461]$

$1-\mathrm{\Phi }(\text{‘}1.461\text{’})$

= 0.0720 (3 sf)

$[{\mathrm{\Phi }}^{-1}(0.986)]=2.197\text{ to }2.198$

$\pm \frac{a-35}{8.1÷\sqrt{140}}=\pm \text{‘}2.198\text{’}$

a = 36.5 (3 sf)

${\text{e}}^{-3.7}(\frac{{3.7}^3}{3!}+\frac{{3.7}^4}{4!}+\frac{{3.7}^5}{5!})={\text{e}}^{-3.7}(8.44217+7.80900+5.77866)=0.20872+0.19307+0.14287$

= 0.545 (3 sf)

$[\lambda ]=6.3$

${\text{e}}^{-6.3}(1+6.3+\frac{{6.3}^2}{2!}+\frac{{6.3}^3}{3!})={\text{e}}^{-6.3}(1+6.3+19.845+41.6745)=0.0018363+0.011569+0.0364415+0.076527$

= 0.126 (3 sf)

$\text{L}\sim \text{N}(37,37),\text{O}\sim \text{N}(26,26)$

$(\text{O}-\text{L})\sim \text{N}(-11,\dots )$

$\text{var}=37+26$

$\frac{0-\text{‘-}11\text{’}}{\sqrt{\text{‘}63\text{’}}}[=1.386]\text{ or }\frac{0+0.5-\text{‘-}11\text{’}}{\sqrt{\text{‘}63\text{’}}}[=1.449]$

$1-\mathrm{\Phi }(\text{‘}1.386\text{’})\text{ or }1-\mathrm{\Phi }(\text{‘}1.449\text{’})$

= 0.0828 or 0.0829 (3 sf) or = 0.0737 or 0.0736 (3 sf)

Min and max times [to complete challenge]

${\int }_a^b\frac{1}{x^2}\text{d}x=1$

$[{[-\frac{1}{x}]}_a^b=1]-\frac{1}{b}+\frac{1}{a}=1$

$-a+b=ab\text{ or }b=a(b+1)$ $\text{OE}\Rightarrow a=\frac{b}{b+1}\text{ AG.}$

$\text{E}(X)={\int }_a^b\frac{1}{x}\text{d}x$

$=\ln$ $b-\ln$ a or $\ln$ $b\text{–}\ln$ (b/(b+1)

$[=\ln b-(\ln b-\ln (b+1))]=\ln b\text{–}\ln (b\mathrm{/}(b+1)=\ln 3$ $=\ln (b+1)=\ln 3\text{ or }b+1=3\text{ or }{b}^2+b=3b\text{ or }\frac{b}{\frac{b}{b+1}}=3$

b = 2 (AG) $a=\frac{2}{3}$

${\int }_{\text{‘}\frac{2}{3}\text{’}}^m\frac{1}{x^2}\text{d}x=0.5\text{ or }{\int }_m^2\frac{1}{x^2}\text{d}x=0.5$

$[-\frac{1}{x}{]}_{\text{‘}\frac{2}{3}\text{’}}^m=0.5\text{ or }[-\frac{1}{x}{]}_m^2=0.5$

$[-\frac{1}{m}+\frac{3}{2}=0.5]\text{ or }[-\frac{1}{2}+\frac{1}{m}=0.5]$ $m=1$

0.02 or 2%

${\text{H}}_0:\mu =2.3{\text{H}}_1:\mu >2.3$

$s^2=\frac{100}{99}(\frac{580}{100}-(2.38{)}^2)\text{ or }1\mathrm{/}99(580-{238}^2\mathrm{/}100)$

= 0.137 = 113/825 or s = 0.370 (3 sf) and $\bar{x}$ = 238/100 [= 2.38]

$\frac{2.38-2.3}{\sqrt{\frac{\text{‘}0.137\text{’}}{100}}}[=2.161\text{ or }2.162]$

= 2.16 (3 sf) OR 0.0153/0.0154 if area comparison used

‘2.16’ > 2.054 (or 2.055) OR ‘0.0153 or 0.0154’<0.02

[There is evidence to reject Ho.] There is sufficient evidence to suggest that the [mean] height [in scientist’s region] is greater than 2.3 [m] OR there is sufficient evidence to suggest that the scientist’s claim is justified.

Not possible since ${\text{H}}_0$ was rejected.