# Expected time, exponential distribution

Suppose that you arrive at a single-teller bank to find five other customers in the bank, one being served and the other four waiting in line. You join the end of the line. If the service times are all exponential with rate $\mu$, what is the expected amount of time you will spend in the bank?

So, $\lambda=\mu$? Then the mean would equal $\frac{1}{\mu}$? That seems too simple, any hints?

Thank you.

The parameter called rate is indeed the one with the usual name $\lambda$. And the mean of the distribution with density function $\lambda e^{-\lambda t}$ (for $t\ge 0$) is $\frac{1}{\lambda}$.
This makes the choice of name $\mu$ for the rate surprising, since $\mu$ is a common name for the mean.
By the memorylessness property of the exponential, the time $X_0$ until the person currently being served is finished has exponential distribution with mean $\frac{1}{\mu}$. And the times of service $X_1$, $X_2$, $X_3$, and $X_4$ of the people waiting in line have mean $\frac{1}{\mu}$. And the time $X_5$ from the instant you get to the teller to the time you are finished has the same mean.
So your expected total time at the bank is $E(X_0+X_1+\cdots +X_4+X_5)$. By the linearity of expectation, this is $\frac{6}{\mu}$.
• I was a bit confused by the use of $\mu$ for $\lambda$, too. Thanks! – Alti Apr 8 '13 at 3:53