Let \(\left\{X_{n}ight\}_{n=1}^{\infty}\) be a sequence of independent and identically distributed random variables where \(X_{n}\) has a \(\operatorname{Exponential}(\theta)\) distribution. Prove that the standardized sample mean \(Z_{n}=n^{1 / 2} \theta^{-1}\left(\bar{X}_{n}-\thetaight)\) has a translated \(\operatorname{Gamma}\left(n, n^{-1 / 2}ight)\) distribution given by

\[\begin{aligned}P\left(Z_{n} \leq zight) & =P\left[n^{1 / 2} \theta^{-1}\left(\bar{X}_{n}-\thetaight) \leq zight] \\& =\int_{-n^{1 / 2}}^{z} \frac{n^{1 / 2}}{\Gamma(n)}\left(t+n^{1 / 2}ight)^{n-1} \exp \left[-n^{1 / 2}\left(t+n^{1 / 2}ight)ight] d t\end{aligned}\]