Question

Consider the probability distribution of a random variable $x$ :
$\begin{array}{cccccc}X& 0& 1& 2& 3& 4\\ P\left(X\right)& 0.1& 0.25& 0.3& 0.2& 0.15\end{array}$
Calculate
(ii) Variance of $X$

Answer 1

we have,
$\begin{array}{cccccc}X& 0& 1& 2& 3& 4\\ P\left(X\right)& 0.1& 0.25& 0.3& 0.2& 0.15\\ XP\left(X\right)& 0& 0.25& 0.6& 0.6& 0.60\\ X^{2}P\left(X\right)& 0& 0.25& 1.2& 1.8& 2.40\end{array}$

$\mathrm{Var}\left(X\right)=E\left(X^2\right)-\left[E\right(X){]}^2$

Where, $E\left(X\right)=\mu ={\sum }_{i=1}^n{x}_i{P}_i\left(xi\right)$

And $E\left(X^2\right)={\sum }_{i=1}^n{x}_i^{2}P\left(x_i\right)$

$\therefore E\left(X\right)=0+0.25+0.6+0.6+0.60=2.05$

$E\left(X^2\right)=0+0.25+1.2+1.8+2.40=5.65$

(ii) $V\left(X\right)=1.44475$