2020 H2 Mathematics Paper 2 Question 5

Discrete Random Variables (DRVs)

Answers

{0, 4, 10, 25}

\mathrm{Var}(S)=\frac{360 r^2 + 300 r - 340}{(3 r + 1)^2}

{r=3}

Tina's possible scores:

0, 4, 10, 25 \; \blacksquare

Let

{S}

denote the r.v. of Tina's score

\begin{align*} \mathrm{P}(S=0) &= \frac{1}{3r+1} \left( \frac{3r}{3r} \right) \times 2! \\ &= \frac{2}{3r+1} \\ \mathrm{P}(S=4) &= \frac{2r}{3r+1} \left( \frac{2r-1}{3r} \right) \\ &= \frac{2(2r-1)}{3(3r+1)} \\ \mathrm{P}(S=10) &= \frac{r}{3r+1} \left( \frac{2r}{3r} \right) \times 2! \\ &= \frac{4r}{3(3r+1)} \\ \mathrm{P}(S=25) &= \frac{r}{3r+1} \left( \frac{r-1}{3r} \right) \\ &= \frac{r-1}{3(3r+1)} \\ \end{align*}

${s}$	${0}$	${4}$	${10}$	${25}$
${\mathrm{P}(S=s)}$	${\frac{2}{3r+1}}$	${\frac{2(2r-1)}{3(3r+1)}}$	${\frac{4r}{3(3r+1)}}$	${\frac{r-1}{3(3r+1)}}$

\begin{align*} \mathrm{E}(S) &= \sum_s s \mathrm{P}(S=s) \\ &= \frac{8(2r-1)+40r+25(r-1)}{3(3r+1)} \\ &= \frac{81 r - 33}{3(3r+1)} \\ &= \frac{27 r - 11}{3r+1} \; \blacksquare \end{align*}

\begin{align*} \mathrm{E}(S^2) &= \sum_s s^2 \mathrm{P}(S=s) \\ &= \frac{32(2r-1)+400r+25^2(r-1)}{3(3r+1)} \\ &= \frac{1089 r - 657}{3(3r+1)} \\ &= \frac{363 r - 219}{3r+1} \\ \end{align*}

\begin{align*} & \mathrm{Var}(S) \\ &= \mathrm{E}(S^2) - \left(\mathrm{E}(S)\right)^2 \\ &= \frac{363 r - 219}{3r+1} - \frac{(27 r - 11)^2}{(3r+1)^2} \\ &= \frac{(363 r - 219)(3 r + 1)-(27 r - 11)^2}{(3 r + 1)^2} \\ &= \frac{360 r^2 + 300 r - 340}{(3 r + 1)^2} \; \blacksquare \end{align*}

\begin{gather*} \frac{360 r^2 + 300 r - 340}{(3 r + 1)^2} = 38 \\ 360 r^2 + 300 r - 340 = 38 (3 r + 1)^2 \\ 18 r^2 + 72 r - 378 = 0 \\ 18(r + 7)(r - 3) = 0 \\ \end{gather*}

Since

{r>0,}

r=3\;\blacksquare