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# 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=$\large\frac{105}{18}$-$\large(\frac{35}{18}$)$^2$
=$\large\frac{665}{324}$

edited Jun 4, 2013