For the following probability distribution determine standard deviation of the random variable X.

We have

${{\rm{X}}^{\rm{2}}}{\rm{P}}\left( {\rm{X}} \right)$ 0.8 4.5 4.8

We know that, standard deviation of $X = \sqrt {{\mathop{\rm Var}\nolimits} X}$

where, ${\mathop{\rm Var}\nolimits} X = E\left( {{X^2}} \right) - {[E(X)]^2}$

$= \sum\limits_{i = 1}^n {x_i^2} P\left( {{x_i}} \right) - {\left[ {\sum\limits_{i = 1}^n {{x_i}} {P_i}} \right]^2}$

$\therefore$ ${\mathop{\rm Var}\nolimits} X = [0.8 + 4.5 + 4.8] - {[0.4 + 1.5 + 1.2]^2}$

$= 10.1 - {(3.1)^2} = 10.1 - 9.61 = 0.49$
$\therefore$ Standard deviation of $X = \sqrt {{\mathop{\rm Var}\nolimits} X} = \sqrt {0.49} = 0.7$