Q44 / 90 · JEE Main 2024 (04 Apr Shift 2) · Mathematics

If the mean of the following probability distribution of a random variable $\mathrm{X}$ : $\begin{array}{|c|c|c|c|c|c|} \hline \mathrm{X} & 0 & 2 & 4 & 6 & 8 \\ \hline \mathrm{P}(\mathrm{X}) & a & 2 a & a+b & 2 b & 3 b \\ \hline \end{array}$ is $\frac{46}{9}$, then the variance of the distribution is

  • A
    $\frac{173}{27}$
  • Correct answer
    B
    $\frac{566}{81}$
  • C
    $\frac{151}{27}$
  • D
    $\frac{581}{81}$

Solution

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

Hence, the correct answer is option: 'B'.

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