JEE Main 2024 — Statistics Question with Solution

$\begin{array}{rl}& \sum P_i=1\\ & a+2a+a+b+2b+3b=1\end{array}$ $4a+6b=1$ ...(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 $\begin{aligned} & \mathrm{b}=\frac{1}{9} \& \mathrm{a}=\frac{1}{12} \\ & \text { Variance }=\mathrm{E}\left(\mathrm{x}_{\mathrm{i}}{ }^2\right)-\mathrm{E}\left(\mathrm{x}_{\mathrm{i}}\right)^2 \\ & \mathrm{E}\left(\mathrm{x}_{\mathrm{i}}\right)^2=0^2 \times 9^2+2^2 \times 2 \mathrm{a}+4^2(\mathrm{a}+\mathrm{b})+6^2(2 \mathrm{~b})+8^2(3 \mathrm{~b}) \\ & =24 \mathrm{a}+280 \mathrm{~b} \end{aligned}$<br/>Put$\mathrm{a}=\frac{1}{12} \quad \mathrm{~b}=\frac{1}{9}$ $\begin{aligned} & \mathrm{E}\left(\mathrm{x}_{\mathrm{i}}^2\right)=2+\frac{280}{9}=\frac{298}{9} \\ & \therefore \sigma^2=\mathrm{E}\left(\mathrm{x}_{\mathrm{i}}^2\right)-\mathrm{E}\left(\mathrm{x}_{\mathrm{i}}\right)^2 \\ & =\frac{298}{9}-\left(\frac{46}{9}\right)^2 \\ & \sigma^2=\frac{298}{9}-\frac{2116}{81} \\ & =\frac{566}{81} \end{aligned}$