Let X be a random variable whose probability distribution is given by…

X	1	3	5	7
$X^2$	1	9	25	49
P(X)	$\frac{1}{3}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{3}$

mean = $E(X) = ∑pixi = 1×\frac{1}{3}+3×\frac{1}{6}+5×\frac{1}{6}+7×\frac{1}{3}$

$E(X) =\frac{2+3+5+7}{6}=\frac{24}{6}=4=E(X)$

so $E(X^2) =∑pi{x_i}^2 =1×\frac{1}{3}+9×\frac{1}{6}+25×\frac{1}{6}+49×\frac{1}{3}$

$E(X^2) =\frac{2+9+25+98}{6}=\frac{134}{6}$

$σ^2$ (variance) = $E(X^2)-(E(X))^2$

$=\frac{134}{6}-4^2=\frac{67}{3}-16$

$=\frac{67-48}{3}=\frac{19}{3}$