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

According to the given condition, probability distribution function is

\(\therefore E(X)=\frac{1+2+\ldots+n}{n}=\frac{n(n+1)}{2 n}=\frac{n+1}{2}\)
\(E\left(X^2\right)=\frac{1^2+2^2+\ldots+n^2}{n} =\frac{n(n+1)(2 n+1)}{6 n} \)
\( =\frac{(n+1)(2 n+1)}{6}\)
\(\therefore \operatorname{Var}(X) =E\left(X^2\right)-[E(X)]^2 \)
\( =\frac{(n+1)(2 n+1)}{6}-\frac{(n+1)^2}{4} \)
\( =\frac{2 n^2+3 n+1}{6}-\frac{n^2+2 n+1}{4} \)
\( =\frac{4 n^2+6 n+2-3 n^2-6 n-3}{12}=\frac{n^2-1}{12}\)
Given that \(\frac{\operatorname{Var}(X)}{E(X)}=\frac{4}{1}\)
\(\frac{\frac{(n+1)(n-1)}{12}}{\frac{(n+1)}{2}}=\frac{4}{1} \)
\( \therefore \frac{n-1}{6}=\frac{4}{1} \)
\( \therefore n=1+24 \)
\( \therefore n=25\)