I'm trying to solve a problem where all arriving items (arrival exponential $\lambda = 1/5$) are divided into into groups, those who are served within 5 units of time and those who have their service time $>5$. The service of items is also exponential $\mu = 1/5$
So if I want to calculate the mean service time for each group:
$$E[S_1] = \int_0^{5}{s (1/5) e^{-(1/5) s} ds}=(5e-10)/e = 1.32$$ $$E[S_2] = \int_5^{\infty}{s (1/5) e^{-(1/5) s} ds}=10/e = 3.6$$
So before going further, I don't understand why my mean service time for the second group of items is 3.6 when only items with service time $>5$ are considered. Do I have an incorrect approach? Do I need to condition it somehow?
Your $\mu$ = $\frac{1}{5}$ tells you that the mean service time is 5 minutes from 0 to $\infty$.
If you add $E(s_1)$ and $E(s_2)$, it is 5. One in the interval $ 0-5$ and the other in the interval (5 - $\infty$).
So I am not surprised that the $E(s_2)$ $= \frac{10}{e}$
For truncated exponential, read the above article, http://en.wikipedia.org/wiki/Truncated_distribution#Expectation_of_truncated_random_variable
Accordingly, $E(S/S>5) = \dfrac{\int_5^{\infty} sf(s)ds}{1-F(5)}$
The top part you already calculated it to be $\frac{10}{e}$
$F(5) = 1-\frac{1}{e}$
$1-F(5) = \frac{1}{e}$
Now $E(S/S>5) = \dfrac{\frac{10}{e}}{\frac{1}{e}} = 10$ which now makes sense