The random variable X takes the values 1, 2, | vedclass

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(B) The mean of $X$ is given by $\bar{X} = E(X) = \sum_{n=1}^{m} n \cdot P(X=n) = \sum_{n=1}^{m} n \cdot \frac{1}{m} = \frac{1}{m} \cdot \frac{m(m+1)}

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

Vedclass · Answer

(B) The mean of $X$ is given by $\bar{X} = E(X) = \sum_{n=1}^{m} n \cdot P(X=n) = \sum_{n=1}^{m} n \cdot \frac{1}{m} = \frac{1}{m} \cdot \frac{m(m+1)}{2} = \frac{m+1}{2}$.The variance of $X$ is given by $	ext{Var}(X) = E(X^2) - [E(X)]^2$.First,calculate $E(X^2) = \sum_{n=1}^{m} n^2 \cdot P(X=n) = \frac{1}{m} \sum_{n=1}^{m} n^2 = \frac{1}{m} \cdot \frac{m(m+1)(2m+1)}{6} = \frac{(m+1)(2m+1)}{6}$.Now,$	ext{Var}(X) = \frac{(m+1)(2m+1)}{6} - \left( \frac{m+1}{2} ight)^2 = \frac{(m+1)(2m+1)}{6} - \frac{(m+1)^2}{4}$.Taking $\frac{m+1}{2}$ as a common factor: $	ext{Var}(X) = \frac{m+1}{2} \left[ \frac{2m+1}{3} - \frac{m+1}{2} ight] = \frac{m+1}{2} \left[ \frac{4m+2 - 3m - 3}{6} ight] = \frac{m+1}{2} \cdot \frac{m-1}{6} = \frac{m^2-1}{12}$.

$X=x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X=x)$	$0.1$	$k$	$0.2$	$2k$	$0.3$	$k$

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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