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The random variable takes the value $1,2,3,..........,m$. If $P(X=n)=\large\frac{1}{m}$ to each n, then the variance of $X$ is

\[(a)\;\frac{(m+1)(2m+1)}{6} \quad (b)\;\frac{m^2-1}{12} \quad (c)\;\frac{m+1}{2} \quad (d)\;\frac{m^2+1}{12} \]
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answered Nov 7, 2013 by pady_1

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