Let $X$ be the total service time for $10$ customers. Estimate the probability that $X > 22$ minutes.

Assume that the service time for a customer at a bank is exponentially distributed with mean service time $2$ minutes. Let $X$ be the total service time for $10$ customers. Estimate the probability that $X > 22$ minutes.

Attempt: I tried to set up and take the integral $$\int_{22}^{z} λe^{10x}\,\mathrm dx$$ then I eventually did the integral to get $2-2e^{-λ(z-x)}$. unfortunately I was unable to figure this out

The sum of $n$ IID exponential random variables with rate $\lambda$ is a gamma random variable with shape $n$ and rate $\lambda$; i.e., if $$X \sim \operatorname{Exponential}(\lambda), \quad f_X(x) = \lambda e^{-\lambda x}, \quad x > 0,$$ then for an IID sample drawn from this distribution $$S_n = X_1 + \cdots + X_n \sim \operatorname{Gamma}(n, \lambda), \quad f_S(s) = \frac{\lambda^n s^{n-1} e^{-\lambda s}}{\Gamma(n)}, \quad s > 0.$$ It follows that the precise probability is given by the integral $$\Pr[S_{10} \ge 22] = \int_{s=22}^\infty \frac{(1/2)^{10} s^9 e^{-s/2}}{9!} \, ds.$$ We can perform the computation by a tedious repeated integration by parts (aka tabular integration). However, we can also approximate the probability using a normal distribution. The mean and variance of a gamma distribution is $$\operatorname{E}[S_n] = n/\lambda = \mu, \quad \operatorname{Var}[S_n] = n/\lambda^2 = \sigma^2$$ again where the parametrization is by rate (and in your case, $\lambda = 1/2$). Thus $$\Pr[S_{10} \ge 22] = \Pr\left[\frac{S_{10} - \mu}{\sigma} \ge \frac{22 - 20}{\sqrt{40}}\right] \approx \Pr[Z \ge 0.316228] \approx 0.375915.$$ The exact probability is $$\Pr[S_{10} \ge 22] = \frac{369915925}{18144}e^{-11} \approx 0.340511.$$