If the probability distribution of X is :

Then variance is equal to :

$\frac{14}{9}$

The correct answer is Option (2) → $\frac{14}{9}$

The expected Mean Value is,

$E(X)=∑X.P(X)$

$=2×\frac{1}{15}+3×\frac{2}{15}+4×\frac{3}{15}+5×\frac{4}{15}+6×\frac{5}{15}$

$=\frac{70}{15}$

$E(X^2)=∑X^2.P(X)$

$=2^2×\frac{1}{15}+3^2×\frac{2}{15}+4^2×\frac{3}{15}+5^2×\frac{4}{15}+6^2×\frac{5}{15}$

$=\frac{350}{15}$

Variance, $σ^2=E(X^2)-[E(X)]^2$

$=\frac{350}{15}-\frac{4900}{225}$

$=\frac{5250-4900}{225}=\frac{350}{225}=\frac{14}{9}$