Question

A discrete random variable $X$ can take all possible integer values from $1$ to $K$, each with a probability $\frac{1}{k}$. Its variance is

• A. $\frac{k^2}{4}$
• B. $\frac{(k+1{)}^2}{4}$
• C. $\frac{k^2-1}{12}$
• D. $\frac{k^2-1}{6}$

Answer 1

The correct option is C $\frac{k^2-1}{12}$

Variance = $\sum (X_i-\mu {)}^2.P_i$,
where $\mu$ is the mean of the given distribution.
$\mu =\sum X_i{P}_i=\frac{1}{k}(1+2+3+...k)=\frac{1}{k}.\frac{k(k+1)}{2}=\frac{k+1}{2}$
So, variance $V=\frac{1}{k}\sum {\left[X_i-\left(\frac{k+1}{2}\right)\right]}^2$
$=\frac{1}{k}[\sum X_i^2-\sum (k+1)X_i+\sum {\left(\frac{k+1}{2}\right)}^2]$
$=\frac{(k+1)(2k+1)}{6}-\frac{(k+1{)}^2}{2}+\frac{(k+1{)}^2}{4}$
$=\frac{(k+1)(2k+1)}{6}-\frac{(k+1{)}^2}{4}$
$=\frac{k+1}{12}(4k+2-3k-3)$
$=\frac{k^2-1}{12}$