The random variable takes the values $1,2,3$, $\ldots, m$. If $P(X=n)=\frac{1}{m}$ to each $n$, then the…

The random variable takes the values $1,2,3$, $\ldots, m$. If $P(X=n)=\frac{1}{m}$ to each $n$, then the variance of $X$ is
  1. $\frac{(m+1)(2 m+1)}{6}$
  2. $\frac{m^2-1}{12}$
  3. $\frac{m+1}{2}$
  4. $\frac{m^2+1}{12}$

Solution

$\begin{aligned} & \operatorname{var}(X)=\sum_{i=2}^{\infty} P_i\left(X_i-\bar{X}\right)^2 \\ & \bar{X}=\frac{1+2+\ldots+m}{m}=\frac{m(m+1)}{2 \cdot m}=\frac{m+1}{2} \\ & \operatorname{var}(X)=\{P(X=1)+P(X=2)+\ldots+P(X=m)\} \\ & =\left\{\frac{1}{m}\left(1-\frac{m+1}{2}\right)^2+\frac{1}{m}\left(2-\frac{m+1}{2}\right)^2+\ldots+\frac{1}{m}\right. \\ & =\frac{1}{m}\left\{\left(1^2+\frac{(m+1)^2}{4}-2 \cdot 1\left(\frac{m+1}{2}\right)\right)\right. \\ & \quad+\left(2^2+\frac{(m+1)^2}{4}-2 \cdot 2 \frac{m+1}{2}\right)+\ldots \\ & \left.\quad+\left(m^2+\frac{(m+1)^2}{4}-m \cdot 2 \frac{(m+1)}{2}\right)\right\} \\ & =\frac{1}{m}\left\{\left(1^2+2^2+\ldots+m^2\right)+\frac{(m+1)^2}{4}(1+1+\ldots\right. \\ & \quad+m \text { times) }-(m+1)(1+2+3+\ldots+m)\} \\ & \left.=\frac{1}{12}\right\} \\ & =\frac{1}{12}\left\{\frac{m(m+1)(2 m+1)}{6}+\frac{m(m+1)^2}{4}\right. \\ & =\frac{(m+1)[2(2 m+1)+3(m+1)-6 m+1]}{2}\end{aligned}$

Asked in: AP EAMCET 2013

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