The variance of the following distribution is

$${x_i}$$	$$2$$	$$3$$	$$11$$
$$f\left( {{x_i}} \right)$$	$$\frac{1}{3}$$	$$\frac{1}{2}$$	$$\frac{1}{6}$$

Solution :
$${\text{We construct the following table :}}$$

$${x_i}$$	$$f\left( {{x_i}} \right)$$	$$x_i^2$$	$$f\left( {{x_i}} \right).{x_i}$$	$$f\left( {{x_i}} \right).x_i^2$$
2	$$\frac{1}{3}$$	$$4$$	$$\frac{2}{3}$$	$$\frac{4}{3}$$
3	$$\frac{1}{2}$$	$$9$$	$$\frac{3}{2}$$	$$\frac{9}{2}$$
11	$$\frac{1}{6}$$	$$121$$	$$\frac{11}{6}$$	$$\frac{121}{6}$$
	$$1$$		$$4$$	$$26$$

$$\eqalign{ & \therefore \,{\text{Variance}}\,{\sigma ^2} = \frac{{\sum {f\left( {{x_i}} \right).x_i^2} }}{{\sum {f\left( {{x_i}} \right)} }} - {\left( {\frac{{\sum {f\left( {{x_i}} \right).{x_i}} }}{{\sum {f\left( {{x_i}} \right)} }}} \right)^2} \cr & = \frac{{26}}{1} - {\left( {\frac{4}{1}} \right)^2} \cr & = 26 - 16 \cr & = 10 \cr} $$