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

\(\begin{aligned} & \mathrm{X}=1,2,3, \ldots \mathrm{n} \\ & \begin{aligned} \mathrm{P}(\mathrm{X}) & =\frac{1}{\mathrm{n}} \\ \mathrm{E}(\mathrm{X}) & =\sum_{\mathrm{i}=1}^{\mathrm{n}} x_{\mathrm{i}} \mathrm{p}_{\mathrm{i}} \\ & =\frac{(1+2+3+\ldots+\mathrm{n})}{\mathrm{n}} \\ & =\frac{\mathrm{n}(\mathrm{n}+1)}{2 \mathrm{n}} \\ \mathrm{E}(\mathrm{X}) & =\frac{\mathrm{n}+1}{2}\end{aligned}\end{aligned}\)
\(\begin{aligned} & \operatorname{Var}(\mathrm{X})=\sum_{\mathrm{i}=1}^{\mathrm{n}} x_{\mathrm{i}}^2 \mathrm{p}_{\mathrm{i}}-[\mathrm{E}(\mathrm{X})]^2 \\ & =\frac{1^2+2^2+3^2+\ldots+\mathrm{n}^2}{\mathrm{n}}-\left(\frac{\mathrm{n}+1}{2}\right)^2 \\ & =\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6 \mathrm{n}}-\left(\frac{\mathrm{n}+1}{2}\right)^2 \\ & =\frac{(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}-\left(\frac{\mathrm{n}+1}{2}\right)^2 \\ & \operatorname{Var}(X)=E(X) \\ & \text {...[Given] } \\ & \frac{(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}-\left(\frac{\mathrm{n}+1}{2}\right)^2=\frac{\mathrm{n}+1}{2} \\ & \frac{2 \mathrm{n}^2+\mathrm{n}+2 \mathrm{n}+1}{6}-\left(\frac{\mathrm{n}^2+2 \mathrm{n}+1}{4}\right)=\frac{\mathrm{n}+1}{2} \\ & \frac{4 n^2+6 n+2-3 n^2-6 n-3}{12}=\frac{n+1}{2} \\ & \mathrm{n}^2-1=6(\mathrm{n}+1) \\ & \mathrm{n}^2-1=6 \mathrm{n}+6 \\ & \mathrm{n}^2-6 \mathrm{n}-7=0 \\ & \therefore \quad \mathrm{n}=-1 \text { or } \mathrm{n}=7 \\ & \text { But } \mathrm{n} \neq-1 \\ & \therefore \quad \mathrm{n}=7 \\ & \end{aligned}\)