If X is a random variable with the following
probability distribution
 

X = x	–3	6	9
P(X = x)	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{1}{3}$

then the variance of X is 

 

Given 

Mean 

$\begin{array}{rcl}Mean \left(\overline{\mathrm{x}}\right)& =& \sum {\mathrm{x}}_{\mathrm{i}}\mathrm{P}(X={\mathrm{x}}_{\mathrm{i}})\\ & =& -3\left(\frac{1}{6}\right)+6\left(\frac{1}{2}\right)+9\left(\frac{1}{3}\right)\\ & =& -\frac{1}{2}+3+3\\ & =& \frac{11}{2}\end{array}$

Now variance $\left({\mathrm{o}}^2\right)=\sum {({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}})}^2 \mathrm{P}(X={\mathrm{x}}_{\mathrm{i}})$

                                 $$\begin{gathered} ={\left(-3-\frac{11}{2}\right)}^2\left(\frac{1}{6}\right)+{\left(6-\frac{11}{2}\right)}^2\left(\frac{1}{2}\right)+{\left(9-\frac{11}{2}\right)}^2\left(\frac{1}{3}\right) \\ =\frac{289}{24}+\frac{1}{8}+\frac{49}{12}=\frac{65}{4} \end{gathered}$$