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

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(D) The random variable $X$ takes values $1, 2, \ldots, n$ with probability $p_i = \frac{1}{n}$ for each $i$.The expected value $E(X) = \sum p_i x_i =

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$25$

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(D) The random variable $X$ takes values $1, 2, \ldots, n$ with probability $p_i = \frac{1}{n}$ for each $i$.The expected value $E(X) = \sum p_i x_i = \frac{1}{n} \sum_{i=1}^{n} i = \frac{1}{n} \cdot \frac{n(n+1)}{2} = \frac{n+1}{2}$.The variance $V(X) = E(X^2) - [E(X)]^2 = \sum p_i x_i^2 - [E(X)]^2$.$E(X^2) = \frac{1}{n} \sum_{i=1}^{n} i^2 = \frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6}$.$V(X) = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2 = \frac{(n+1)}{2} \left[ \frac{2n+1}{3} - \frac{n+1}{2} \right] = \frac{(n+1)}{2} \left[ \frac{4n+2-3n-3}{6} \right] = \frac{(n+1)(n-1)}{12} = \frac{n^2-1}{12}$.Given the ratio $\frac{V(X)}{E(X)} = 4$,we have $\frac{(n^2-1)/12}{(n+1)/2} = 4$.$\frac{(n-1)(n+1)}{12} \cdot \frac{2}{n+1} = 4$.$\frac{n-1}{6} = 4 \Rightarrow n-1 = 24 \Rightarrow n = 25$.

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

Outcome	Probability
$\omega_{1}$	$0.1$
$\omega_{2}$	$0.01$
$\omega_{3}$	$0.05$
$\omega_{4}$	$0.03$
$\omega_{5}$	$0.01$
$\omega_{6}$	$0.2$
$\omega_{7}$	$0.6$

$A$ random variable $X$ assumes values $1, 2, 3, \ldots, n$ with equal probabilities. If the ratio of the variance of $X$ to the expected value of $X$ is equal to $4$,then the value of $n$ is:

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