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Home  >>  CBSE XII  >>  Math  >>  Probability

Find the variance of the distribution.

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

 

1 Answer

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 by poojasapani_1
edited Jun 4, 2013 by poojasapani_1
 

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