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

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According to the given condition, probability distribution function is

$\therefore \quad E(X)=\frac{1+2+\ldots+n}{n}=\frac{n(n+1)}{2 n}=\frac{n+1}{2}$ $\begin{aligned} E\left(X^2\right)=\frac{1^2+2^2+\ldots+n^2}{n} & =\frac{n(n+1)(2 n+1)}{6 n} \\ & =\frac{(n+1)(2 n+1)}{6}\end{aligned}$ $\begin{aligned} \therefore \quad \operatorname{Var}(X) & =E\left(X^2\right)-[E(X)]^2 \\ & =\frac{(n+1)(2 n+1)}{6}-\frac{(n+1)^2}{4} \\ & =\frac{2 n^2+3 n+1}{6}-\frac{n^2+2 n+1}{4} \\ & =\frac{4 n^2+6 n+2-3 n^2-6 n-3}{12}=\frac{n^2-1}{12}\end{aligned}$ Given that $\frac{\operatorname{Var}(X)}{E(X)}=\frac{4}{1}$ $\begin{array}{ll} & \frac{\frac{(n+1)(n-1)}{12}}{\frac{(n+1)}{2}}=\frac{4}{1} \\ \therefore & \frac{n-1}{6}=\frac{4}{1} \\ \therefore & n=1+24 \\ \therefore & n=25 \end{array}$

A random variable X assumes values $1,2,3, \ldots \ldots ., \mathrm{n}$ with equal probabilities. If…

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