A Level Mathematics 9709 Paper 53, May/June 2023

$[\text{P}(\text{HH})=\frac{1}{4}][\text{E}(X)=]4$

$[\text{P}(X=5)={\left(\frac{3}{4}\right)}^4\left(\frac{1}{4}\right)=]0.0791$

$[\text{P}(X<7)=]1-{\left(\frac{3}{4}\right)}^6$ or $\frac{1}{4}+\frac{3}{4}\times \frac{1}{4}+\frac{3^2}{4^2}\times \frac{1}{4}+\dots +\frac{3^5}{4^5}\times \frac{1}{4}$

$=\frac{3367}{4096},0.822$

$\text{Mean}=120\times 0.4=48$ $\text{Var}=120\times 0.4\times 0.6=28.8$

Substituting their $\mu$ and $\sigma$ into one $\pm$standardisation formula (any number for 35.5 or 54.5), condone ${\sigma }^2$ and $\sqrt{\sigma }$.

Using continuity correction 35.5, 36.5 or 53.5, 54.5 once in their standardisation formula.

$[=\text{P}(-2.3292<Z<1.211)=]0.8871+0.9900-1$

$=0.877$

$[\text{P}(X=4)=3\text{P}(X=2)]$ $4k(4+a)=3\times 2k(2+a)$ $16k+4ak=12k+6ak$

$a=2$

$3k+8k+15k+24k=1$

$k=\frac{1}{50}$

$\text{Var}(X)=\frac{3}{50}\times 1^2+\frac{8}{50}\times 2^2+\frac{15}{50}\times 3^2+\frac{24}{50}\times 4^2-{3.2}^2$

$[=11.08-{3.2}^2=]0.84[0],\frac{21}{25}$

Median = 99 [minutes]

$[\text{IQR}=]106-83$

23 [minutes]

The times for the Cheetahs are faster than the times for the Panthers

The times for the Cheetahs are more spread than the times for the Panthers

[Total time including Kenny = $99\times 20=$ ]1980

[Kenny's time =] $1980-1862$

= 118 [minutes]

$\text{P}(A)=\frac{10}{36}$ $\text{P}(B)=\frac{24}{36}$

$\text{P}(A\cap B)=\frac{8}{36}$

$\frac{10}{36}\times \frac{24}{36}$

$=\frac{5}{27},0.185[\mathrm{\ne }\frac{8}{36}]$ Events are not independent

$[\text{P}(B\cap A^{\mathrm{\prime }})=\frac{16}{36},0.4444]$ or their $\text{P}(B)$ – their $\text{P}(A\cap B)$ seen as numerator or denominator of conditional probability fraction.

$[\text{P}(A^{\mathrm{\prime }})=1-\frac{10}{36}]=\frac{26}{36},0.7222$ or $1-$ their $\text{P}(A)$ seen as denominator of conditional probability fraction.

$=\frac{8}{13}$

$\pm 1.282$ seen, cao – critical value.

Use of the $\pm$standardisation formula with 16, 28, $\sigma$ and a $z$-value (not 0.1, 0.9, 0.282, 0.5398, 0.8159) equated to a $z$-value.

$\sigma =9.36$

One term ${}^{12}{\text{C}}_x(p{)}^x(1-p{)}^{12-x},0<p<1.x\mathrm{\ne }0,1,2$.

Correct expression, accept unsimplified, no terms omitted leading to final answer.

0.111

Identifying at least one of $-1.3$ or $1.3$ as the appropriate $z$-values.

Calculating the appropriate probability area from 2 symmetrical $z$-values (leading to their final answer, expect > 0.5).

$=0.806,\frac{504}{625}$

$[\text{In 365 days }0.8064\times 365]$ $=294\text{ or }295$

$\frac{10!}{a!b!}-c,a\mathrm{\ne }b,a=1,2,b=1,4$, with $c$ being a positive integer.

$d-\frac{e!}{4!},e=8,9,10$, with $d$ being a positive integer.

= 60480

$\frac{6!}{2!}\times s$, with $s$ being a positive integer.

$\frac{t!}{r!}\times 4,r=1,2,3$ and $t=8,7,6$.

1440

Correct number of ways for identified scenarios of 2 or 3 As, accept unsimplified, www.

Add 3 values for 2, 3 and 4 As, no additional, incorrect or repeated scenarios. Accept unsimplified.

25