Explanation

(c)
Given, distribution is
[table] $\mathbf{X}$ | 0 | 1 | 2 | 3 ; $\mathbf{P(X)}$ | $\frac{1}{3}$ | $\frac{1}{2}$ | 0 | $\frac{1}{6}$ [/table]
$\therefore\$Mean, $m = \sum_{i = 1}^{4}{p_{i}x_{i}}$
$= 0 \times \frac{1}{3} + 1 \times \frac{1}{2} + 2 \times 0 + 3 \times \frac{1}{6}$
$= 0 + \frac{1}{2} + 0 + \frac{1}{2} = 1$
Variance, $\sigma^{2} = \sum_{i = 1}^{4}{p_{i}\left( x_{1} - m \right)^{2}}$
$= \frac{1}{3}(0 - 1)^{2} + \frac{1}{2}(1 - 1)^{2} + 0(2 - 1)^{1} + \frac{1}{6}(3 - 1)^{2}$
$= \frac{1}{3} + 0 + 0 + \frac{2}{3} = 1$
$\therefore\ m = \sigma^{2} = 1$