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(B) The random variable $X$ follows a discrete uniform distribution on the set $\{1, 2, \ldots, m\}$.The mean $\bar{X} = E[X] = \sum_{n=1}^{m} n \cdot

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

Vedclass · Answer

(B) The random variable $X$ follows a discrete uniform distribution on the set $\{1, 2, \ldots, m\}$.The mean $\bar{X} = E[X] = \sum_{n=1}^{m} n \cdot P(X=n) = \frac{1}{m} \sum_{n=1}^{m} n = \frac{1}{m} \cdot \frac{m(m+1)}{2} = \frac{m+1}{2}$.The variance is given by $\operatorname{Var}(X) = E[X^2] - (E[X])^2$.$E[X^2] = \sum_{n=1}^{m} n^2 \cdot P(X=n) = \frac{1}{m} \sum_{n=1}^{m} n^2 = \frac{1}{m} \cdot \frac{m(m+1)(2m+1)}{6} = \frac{(m+1)(2m+1)}{6}$.Therefore,$\operatorname{Var}(X) = \frac{(m+1)(2m+1)}{6} - \left(\frac{m+1}{2}\right)^2$.$\operatorname{Var}(X) = \frac{m+1}{2} \left[ \frac{2(2m+1)}{3} - \frac{m+1}{2} \right] = \frac{m+1}{2} \left[ \frac{8m+4-3m-3}{6} \right] = \frac{m+1}{2} \cdot \frac{5m+1}{6}$ (Wait,let's re-calculate).$\operatorname{Var}(X) = \frac{(m+1)(2m+1)}{6} - \frac{(m+1)^2}{4} = \frac{m+1}{2} \left[ \frac{2(2m+1) - 3(m+1)}{6} \right] = \frac{m+1}{2} \left[ \frac{4m+2-3m-3}{6} \right] = \frac{m+1}{2} \cdot \frac{m-1}{6} = \frac{m^2-1}{12}$.

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

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