The probability of the queue having length $n$ is $p_{n} = (1-\rho)\rho^n$ so the expected mean is: $$ \begin{align} E(\bar{n}) &= \sum_{n=0}^\infty np_{n} \\ &= \sum_{n=0}^\infty n(1-\rho)\rho^n \\ &= (1-\rho) \rho \sum_{n=0}^\infty n\rho^{n-1} \\ \end{align} $$ Use this standard result $$ \boxed{ \begin{align} \sum_{n=0}^\infty n\rho^{n-1} = \frac{1}{(1-\rho)^2} \end{align} } $$ so $$ \begin{align} &= (1-\rho) \rho \sum_{n=0}^\infty n\rho^{n-1} \\ &= (1-\rho) \rho \frac{1}{(1-\rho)^2} \\ &= \frac{\rho}{1-\rho} \\ \end{align} $$ Finally we have the result: $$ \boxed{ \begin{align} E(\bar{n}) = \frac{\rho}{1-\rho} \end{align} } $$
To find the variance we first need to find this relation: $$ \begin{align} \sum_{n=0}^\infty n^2\rho^n &= \sum_{n=0}^\infty (n^2\rho^n-n\rho^n+n\rho^n) \\ &= \sum_{n=0}^\infty (n(n-1)\rho^n+n\rho^n) \\ &= \sum_{n=0}^\infty n(n-1)\rho^n+ \sum_{n=0}^\infty n\rho^n \\ &= 2\rho^2 \sum_{n=0}^\infty \frac{n(n-1)}{2} \rho^{n-2} + \rho \sum_{n=0}^\infty n\rho^{n-1} \\ \end{align} $$ Use these standard results $$ \boxed{ \begin{align} \sum_{n=0}^\infty \frac{n(n-1)}{2} \rho^{n-2} = \frac{1}{(1-\rho)^3} \\ \sum_{n=0}^\infty n\rho^{n-1} = \frac{1}{(1-\rho)^2} \end{align} } $$ so $$ \begin{align} & 2\rho^2 \sum_{n=0}^\infty \frac{n(n-1)}{2} \rho^{n-2} + \rho \sum_{n=0}^\infty n\rho^{n-1} \\ &= \frac{2\rho^2}{(1-\rho)^3} + \frac{\rho}{(1-\rho)^2} \\ &= \frac{\rho}{(1-\rho)^2} \left( \frac{2\rho}{1-\rho} + 1 \right) \\ &= \frac{\rho}{(1-\rho)^2} \left( \frac{2\rho + (1-\rho)}{1-\rho} \right) \\ &= \frac{\rho}{(1-\rho)^2} \left( \frac{1 + \rho}{1-\rho} \right) \\ &= \frac{\rho(1 + \rho)}{(1-\rho)^3} \end{align} $$ Giving the relation: $$ \boxed{ \begin{align} \sum_{n=0}^\infty n^2\rho^n &= \frac{\rho(1 + \rho)}{(1-\rho)^3} \\ \end{align} } $$
The variance then is $$ \begin{align} Var(\bar{n}) &= \sum_{n=0}^\infty n^2 p_{n} - E(\bar{n})^2 \\ &= \sum_{n=0}^\infty n^2(1-\rho)\rho^n - \left(\frac{\rho}{1-\rho}\right)^2 \\ &= (1-\rho) \sum_{n=0}^\infty n^2\rho^n - \left(\frac{\rho}{1-\rho}\right)^2 \\ &= (1-\rho) \frac{\rho(1 + \rho)}{(1-\rho)^3} - \frac{\rho^2}{(1-\rho)^2} \\ &= \frac{\rho + \rho^2}{(1-\rho)^2} - \frac{\rho^2}{(1-\rho)^2} \\ &= \frac{\rho}{(1-\rho)^2}\\ \end{align} $$ Finally we have the result: $$ \boxed{ \begin{align} Var(\bar{n}) &= \frac{\rho}{(1-\rho)^2} \end{align} } $$