JEE Main 2024 — Mathematics Probability & Statistics

$\displaystyle \begin{aligned} & \sum P_i=1 \\ & a+2 a+a+b+2 b+3 b=1\end{aligned}$​∑Pi​=1a+2a+a+b+2b+3b=1​ $4 a+6 b=1$4a+6b=1 ...(I) $\mathrm{E}(\mathrm{x})=$E(x)= mean $=\frac{46}{9}$=946​ $\displaystyle \begin{aligned} & \sum P_i X_i=\frac{46}{9} \Rightarrow 4 a+4 a+4 b+12 b+24 b=\frac{46}{9} \\ & 8 a+40 b=\frac{46}{9}\end{aligned}$​∑Pi​Xi​=946​⇒4a+4a+4b+12b+24b=946​8a+40b=946​​ $4 a+20 b=\frac{23}{9}$4a+20b=923​ ...(II) Subtract (I) from (II) we get $\begin{aligned}
& \mathrm{b}=\frac{1}{9} & \mathrm{a}=\frac{1}{12} \
& \text { Variance }=\mathrm{E}\left(\mathrm{x}{\mathrm{i}}{ }^2\right)-\mathrm{E}\left(\mathrm{x}{\mathrm{i}}\right)^2 \
& \mathrm{E}\left(\mathrm{x}_{\mathrm{i}}\right)^2=0^2 \times 9^2+2^2 \times 2 \mathrm{a}+4^2(\mathrm{a}+\mathrm{b})+6^2(2 \mathrm{b})+8^2(3 \mathrm{b}) \
& =24 \mathrm{a}+280 \mathrm{~b}
\end{aligned}$

Put $\mathrm{a}=\frac{1}{12} \quad \mathrm{~b}=\frac{1}{9}$a=121​ b=91​ $\begin{aligned}
& \mathrm{E}\left(\mathrm{x}{\mathrm{i}}^2\right)=2+\frac{280}{9}=\frac{298}{9} \
& \therefore \sigma^2=\mathrm{E}\left(\mathrm{x}{\mathrm{i}}^2\right)-\mathrm{E}\left(\mathrm{x}_{\mathrm{i}}\right)^2 \
& =\frac{298}{9}-\left(\frac{46}{9}\right)^2 \
& \sigma^2=\frac{298}{9}-\frac{2116}{81} \
& =\frac{566}{81}
\end{aligned}$