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

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(C) Given that $X$ takes values $1, 2, \ldots, n$ with equal probability $P(X) = \frac{1}{n}$.$E(X) = \sum_{i=1}^{n} x_i p_i = \frac{1}{n} \sum_{i=1}^

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$7$

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(C) Given that $X$ takes values $1, 2, \ldots, n$ with equal probability $P(X) = \frac{1}{n}$.$E(X) = \sum_{i=1}^{n} x_i p_i = \frac{1}{n} \sum_{i=1}^{n} i = \frac{1}{n} \cdot \frac{n(n+1)}{2} = \frac{n+1}{2}$.$E(X^2) = \sum_{i=1}^{n} x_i^2 p_i = \frac{1}{n} \sum_{i=1}^{n} i^2 = \frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6}$.$\operatorname{Var}(X) = E(X^2) - [E(X)]^2 = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}ight)^2$.Given $\operatorname{Var}(X) = E(X)$,we have:$\frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4} = \frac{n+1}{2}$.Dividing by $(n+1)$ (since $n 
eq -1$):$\frac{2n+1}{6} - \frac{n+1}{4} = \frac{1}{2}$.Multiply by $12$:$2(2n+1) - 3(n+1) = 6$.$4n + 2 - 3n - 3 = 6$.$n - 1 = 6 \implies n = 7$.

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

$X = x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$k$	$\frac{1}{5}$	$2k$	$\frac{3}{10}$	$k$

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

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