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(A) The random variable $X$ follows a discrete uniform distribution on the set $\{1, 2, \ldots, n\}$.The mean $E[X]$ is given by:$E[X] = \sum_{k=1}^{n

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

Vedclass · Answer

(A) The random variable $X$ follows a discrete uniform distribution on the set $\{1, 2, \ldots, n\}$.The mean $E[X]$ is given by:$E[X] = \sum_{k=1}^{n} k \cdot P(X=k) = \sum_{k=1}^{n} k \cdot \frac{1}{n} = \frac{1}{n} \cdot \frac{n(n+1)}{2} = \frac{n+1}{2}$.The expected value of $X^2$ is given by:$E[X^2] = \sum_{k=1}^{n} k^2 \cdot P(X=k) = \sum_{k=1}^{n} k^2 \cdot \frac{1}{n} = \frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6}$.The variance $Var(X)$ is given by:$Var(X) = E[X^2] - (E[X])^2 = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2$.Factoring out $\frac{n+1}{2}$:$Var(X) = \frac{n+1}{2} \left[ \frac{2n+1}{3} - \frac{n+1}{2} \right] = \frac{n+1}{2} \left[ \frac{4n+2 - 3n - 3}{6} \right] = \frac{n+1}{2} \cdot \frac{n-1}{6} = \frac{n^2-1}{12}$.

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X = x)$	$k$	$0$	$2k$	$5k$	$k$	$3k$

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

$Y$	$-1$	$0$	$1$
$P(Y)$	$0.6$	$0.1$	$0.2$

Let $X$ be the random variable taking values $1, 2, \ldots, n$ for a fixed positive integer $n$. If $P(X=k) = \frac{1}{n}$ for $1 \leq k \leq n$,then the variance of $X$ is

Similar Questions

Given below is the probability distribution of a discrete random variable $X$:
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$P(X = x)$ $k$ $0$ $2k$ $5k$ $k$ $3k$

Then $P(X \geq 4) = $

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