The p.m.f of a random variable X is given by | vedclass

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(B) The expected value $E(X)$ is defined as $\sum_{x=1}^{n} x \cdot P(X=x)$.Given $P(X=x) = \frac{2x}{n(n+1)}$ for $x = 1, 2, \ldots, n$.Thus,$E(X) =

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$\frac{2n+1}{3}$

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(B) The expected value $E(X)$ is defined as $\sum_{x=1}^{n} x \cdot P(X=x)$.Given $P(X=x) = \frac{2x}{n(n+1)}$ for $x = 1, 2, \ldots, n$.Thus,$E(X) = \sum_{x=1}^{n} x \cdot \frac{2x}{n(n+1)}$.$E(X) = \frac{2}{n(n+1)} \sum_{x=1}^{n} x^2$.Using the formula for the sum of squares of the first $n$ natural numbers,$\sum_{x=1}^{n} x^2 = \frac{n(n+1)(2n+1)}{6}$.Substituting this into the expression:$E(X) = \frac{2}{n(n+1)} \cdot \frac{n(n+1)(2n+1)}{6}$.$E(X) = \frac{2(2n+1)}{6} = \frac{2n+1}{3}$.

The p.m.f of a random variable $X$ is given by $P(X) = \frac{2x}{n(n+1)}$ for $x = 1, 2, 3, \ldots, n$,and $0$ otherwise. Then $E(X) = $

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