For the following probability distribution.
X 1 2 3 4
P(X) $\frac{1}{{10}}$ $\frac{1}{5}$ $\frac{3}{{10}}$ $\frac{2}{5}$
$E\left( {{X^2}} \right)$ is equal to

$E\left( {{X^2}} \right) = \Sigma {X^2}P(X) = 1 \cdot \frac{1}{{10}} + 4 \cdot \frac{1}{5} + 9 \cdot \frac{3}{{10}} + 16 \cdot \frac{2}{5}$

$= \frac{1}{{10}} + \frac{4}{5} + \frac{{27}}{{10}} + \frac{{32}}{5}$

$= \frac{{1 + 8 + 27 + 64}}{{10}} = 10$