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

The correct option is B $\frac{m^2-1}{12}$

$E\left(x\right)=\sum _{i=1}^m{x}_1{P}_i=(1\cdot \frac{1}{m}+2\cdot \frac{2}{m}+3\cdot \frac{1}{m}+....m\cdot \frac{1}{m})$

$=(1+2+3+...m)\cdot \frac{1}{m}$

$=\frac{m(m+1)}{2}\cdot \frac{1}{m}$

$=\frac{(m+1)}{2}$

$E\left(x^2\right)=\sum _{i=1}^m{x}_i^2{p}_i=(1^2\cdot \frac{1}{m}+2^2\cdot \frac{2}{m}+....m^2\cdot \frac{1}{m})$

$=(1^2+2^2+....m^2)\cdot \frac{1}{m}$

$=\frac{m(m+1)(2m+1)}{6}\cdot \frac{1}{m}$

$=\frac{(m+1)(2m+1)}{6}$

$\therefore var\left(x\right)=E\left(x^2\right)-\left[E\right(x){]}^2$

$=\frac{(m+1)(2m+1)}{6}-\frac{(m+1{)}^2}{4}$

$=\frac{(m+1)(4m+2-3m-3)}{12}$

$=\frac{(m+1)(m-1)}{12}$

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