Mathematics - Statistics Question with Solution | TestHub

$\begin{array}{rl}& \sum P_i=1\\ & a+2a+a+b+2b+3b=1\end{array}$ $4a+6b=1$ ...(I) $\mathrm{E}(\mathrm{x})=$ mean $=\frac{46}{9}$ $\begin{array}{rl}& \sum P_i{X}_i=\frac{46}{9}\Rightarrow 4a+4a+4b+12b+24b=\frac{46}{9}\\ & 8a+40b=\frac{46}{9}\end{array}$ $4a+20b=\frac{23}{9}$ ...(II) Subtract (I) from (II) we get $\begin{array}{rl}& \mathrm{b}=\frac{1}{9}\&\mathrm{a}=\frac{1}{12}\\ & \text{ Variance }=\mathrm{E}\left({\mathrm{x}}_{\mathrm{i}}{}^2\right)-\mathrm{E}{\left({\mathrm{x}}_{\mathrm{i}}\right)}^2\\ & \mathrm{E}{\left({\mathrm{x}}_{\mathrm{i}}\right)}^2=0^2\times 9^2+2^2\times 2\mathrm{a}+4^2(\mathrm{a}+\mathrm{b})+6^2(2\mathrm{b})+8^2(3\mathrm{b})\\ & =24\mathrm{a}+280\mathrm{b}\end{array}$
Put $\mathrm{a}=\frac{1}{12}\mathrm{b}=\frac{1}{9}$ $\begin{array}{rl}& \mathrm{E}\left({\mathrm{x}}_{\mathrm{i}}^2\right)=2+\frac{280}{9}=\frac{298}{9}\\ & \therefore {\sigma }^2=\mathrm{E}\left({\mathrm{x}}_{\mathrm{i}}^2\right)-\mathrm{E}{\left({\mathrm{x}}_{\mathrm{i}}\right)}^2\\ & =\frac{298}{9}-{\left(\frac{46}{9}\right)}^2\\ & {\sigma }^2=\frac{298}{9}-\frac{2116}{81}\\ & =\frac{566}{81}\end{array}$