A random variable X assumes values 1, 2, 3, l | vedclass

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(C) The probability distribution is given by the table:| $X=x$ | $1$ | $2$ | $3$ | $\ldots$ | $n$ ||---|---|---|---|---|---|| $P(X=x)$ | $\frac{1}{n}$

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(C) The probability distribution is given by the table:| $X=x$ | $1$ | $2$ | $3$ | $\ldots$ | $n$ ||---|---|---|---|---|---|| $P(X=x)$ | $\frac{1}{n}$ | $\frac{1}{n}$ | $\frac{1}{n}$ | $\ldots$ | $\frac{1}{n}$ |$E(X) = \sum x P(X=x) = \frac{1}{n} (1 + 2 + \ldots + n) = \frac{1}{n} \cdot \frac{n(n+1)}{2} = \frac{n+1}{2}$$E(X^2) = \sum x^2 P(X=x) = \frac{1}{n} (1^2 + 2^2 + \ldots + n^2) = \frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6}$$\operatorname{Var}(X) = E(X^2) - [E(X)]^2 = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2$$= \frac{(n+1)}{2} \left[ \frac{2n+1}{3} - \frac{n+1}{2} \right] = \frac{(n+1)}{2} \left[ \frac{4n+2 - 3n - 3}{6} \right] = \frac{(n+1)(n-1)}{12} = \frac{n^2-1}{12}$Given $\frac{\operatorname{Var}(X)}{E(X)} = \frac{4}{1}$,we have:$\frac{(n^2-1)/12}{(n+1)/2} = 4$$\frac{(n+1)(n-1)}{12} \cdot \frac{2}{n+1} = 4$$\frac{n-1}{6} = 4$$n-1 = 24$$n = 25$

$X = x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X = x)$	$0$	$k$	$2k$	$2k$	$3k$	$k^2$	$2k^2$	$7k^2 + k$

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

$A$ random variable $X$ assumes values $1, 2, 3, \ldots, n$ with equal probabilities. If $\operatorname{var}(X) : E(X) = 4 : 1$,then $n$ is equal to

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