The discrete random variable X can take all p | vedclass

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Question

(A) The probability distribution is given by:$E(X) = \sum_{i=1}^{k} x_i p_i = \frac{1}{k} + \frac{2}{k} + \dots + \frac{k}{k} = \frac{1}{k} \sum_{i=1}

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

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(A) The probability distribution is given by:$E(X) = \sum_{i=1}^{k} x_i p_i = \frac{1}{k} + \frac{2}{k} + \dots + \frac{k}{k} = \frac{1}{k} \sum_{i=1}^{k} i = \frac{1}{k} \cdot \frac{k(k+1)}{2} = \frac{k+1}{2}$$E(X^2) = \sum_{i=1}^{k} x_i^2 p_i = \frac{1^2 + 2^2 + \dots + k^2}{k} = \frac{1}{k} \cdot \frac{k(k+1)(2k+1)}{6} = \frac{(k+1)(2k+1)}{6}$Variance $= E(X^2) - [E(X)]^2$$= \frac{(k+1)(2k+1)}{6} - \left( \frac{k+1}{2} \right)^2$$= \frac{2k^2 + 3k + 1}{6} - \frac{k^2 + 2k + 1}{4}$$= \frac{2(2k^2 + 3k + 1) - 3(k^2 + 2k + 1)}{12}$$= \frac{4k^2 + 6k + 2 - 3k^2 - 6k - 3}{12}$$= \frac{k^2 - 1}{12}$

$X = x$	$1$	$2$	$3$	$\dots$	$n$
$P(X = x)$	$\frac{1}{n}$	$\frac{1}{n}$	$\frac{1}{n}$	$\dots$	$\frac{1}{n}$

$X$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X)$	$0.15$	$0.23$	$0.12$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

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

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