0
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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 $$\Pr[S_{10} \ge 22] = \frac{369915925}{18144}e^{-11} \approx 0.340511.$$

$\endgroup$
  • $\begingroup$ Hi, thank u! Also, how did u obtain 369915925 on the numerator? And how did u get 18144 as your denominater? $\endgroup$ – Bobby Aug 3 '17 at 21:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.