# Find the variance of the distribution.

 x 0 1 2 3 4 5 P(x) $\Large \frac{1}{6}$ $\Large \frac{5}{18}$ $\Large \frac{2}{9}$ $\Large \frac{1}{6}$ $\Large \frac{1}{9}$ $\Large \frac{1}{18}$

Toolbox:
• variance=$$\sum(P_iX^2_i$$)-($$\sum(P_iX_i)$$$$^2$$
$$\sum(P_iX^2_i)=\Large\;o\;\times\;\frac{1}{0}\;+1\times\;\frac{5}{18}\;+2\times\;\frac{2}{9}\;+3\times\;\frac{1}{6}\;+4\times\;\frac{1}{9}\;+5\times\;\frac{1}{18}$$
=$$\large\frac{5}{18}\;+\;\frac{4}{9}\;+\;\frac{3}{6}\;+\;\frac{4}{9}\;+\;\frac{5}{18}$$
=$$\large\frac{35}{18}$$
($$\sum(P_iX_i)$$$$^2$$=$$\large\frac{1}{6}$$$$\times$$0+$$\large\frac{5}{18}$$$$\times$$1$$^2$$+$$\large\frac{2}{9}$$$$\times$$2$$^2$$+$$\large\frac{1}{6}$$$$\times$$3$$^2$$+$$\large\frac{1}{9}$$$$\times$$4$$^2$$+$$\large\frac{1}{18}$$$$\times$$5$$^2$$
=$$\frac{5}{18}$$+$$\large\frac{8}{9}$$+$$\large\frac{6}{9}$$+$$\large\frac{16}{9}$$+$$\large\frac{25}{18}$$
=$$\large\frac{105}{18}$$
variance=$$\large\frac{105}{18}$$-$$\large(\frac{35}{18}$$)$$^2$$
=$$\large\frac{665}{324}$$

answered Mar 1, 2013
edited Jun 4, 2013