If X is a random variable with probability distribution as given below.

Then, the variance of the distribution is

$\frac{62}{49}$

The correct answer is Option (3) → $\frac{62}{49}$ **

Given probabilities:

$k,\;2k,\;k,\;3k$

Total probability:

$k + 2k + k + 3k = 7k = 1$

$k = \frac{1}{7}$

Now compute the mean:

$E(X) = 0\cdot\frac{1}{7} + 1\cdot\frac{2}{7} + 2\cdot\frac{1}{7} + 3\cdot\frac{3}{7}$

$= 0 + \frac{2}{7} + \frac{2}{7} + \frac{9}{7} = \frac{13}{7}$

Compute $E(X^2)$:

$E(X^2) = 0 + 1^2\cdot\frac{2}{7} + 2^2\cdot\frac{1}{7} + 3^2\cdot\frac{3}{7}$

$= \frac{2}{7} + \frac{4}{7} + \frac{27}{7} = \frac{33}{7}$

Variance:

$\text{Var}(X) = E(X^2) - [E(X)]^2$

$= \frac{33}{7} - \left(\frac{13}{7}\right)^2$

$= \frac{33}{7} - \frac{169}{49}$

$= \frac{231}{49} - \frac{169}{49} = \frac{62}{49}$

Variance = $\frac{62}{49}$