Question

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

• A. $\frac{(m+1)(2m+1)}{6}$
• B. $\frac{m^2-1}{12}$
• C. $\frac{m+1}{2}$
• D. $\frac{m^2+1}{12}$

Answer 1

Answer: (B)

$\frac{m^2-1}{12}$

X	1	2	3	...	m
P(X)	1/m	1/m	1/m	...	1/m

$Var\left(X\right)=E\left(X^2\right)-\left(E\right(X){)}^2$

$Var\left(X\right)=\sum X^{2}P\left(X\right)-(\sum XP(X){)}^2$

$Var\left(X\right)=\sum X^2\left(\frac{1}{m}\right)-(\sum X(\frac{1}{m}){)}^2$

$Var\left(X\right)=\frac{1}{m}\sum X^2-\frac{1}{m^2}(\sum X{)}^2$

$Var\left(X\right)=\frac{1}{m}\left(\frac{m(m+1)(2m+1)}{6}\right)-(\frac{m(m+1)}{2m}{)}^2$

$Var\left(X\right)=(m+1)(\frac{(2m+1}{6}-\frac{m+1}{4})$

$Var\left(X\right)=(m+1)\left(\frac{(8m+4-6m-6}{24}\right)$

$Var\left(X\right)=(m+1)\left(\frac{(2m-2}{24}\right)$

$Var\left(X\right)=(m+1)\left(\frac{m-1}{12}\right)$

$Var\left(X\right)=\left(\frac{m^2-1}{12}\right)$

option B is correct.