For part a) I get $ E(X)=\alpha\beta=\frac{n}{\lambda}. $ Thus the answer is $\frac{10}{0.5}=20$ minutes. I am not sure how to do b). Any help?
The special case of the gamma distribution in which $\alpha$ is a positive integer $n$ is called an Erlang distribution. If we replace $\beta$ by $1/\lambda$ in Expression (4.7), the Erlang pdf is $ f(x;\lambda,n)=\frac{\lambda(\lambda{x})^{n-1}e^{-\lambda{x}}}{(n-1!)} $ for $ x\geq0$ and $ 0 $ otherwise.
It can be shown that if the times between successive events are independent, each with an exponential distribution with parameter $\lambda$, then the total time $X$ that elapses before all of the next $n$ events occur has pdf $f(x; l, n)$.
a. What is the expected value of $X$? If the time (in minutes) between arrivals of successive customers is exponentially distributed with $\lambda = .5$, how much time can be expected to elapse before the tenth customer arrives?
b. If customer interarrival time is exponentially distributed with $\lambda = .5$, what is the probability that the tenth customer (after the one who has just arrived) will arrive within the next 30 min?
c. The event ${X \leq t}$ occurs if and only if at least $n$ events occur in the next t units of time. Use the fact that the number of events occurring in an interval of length $t$ has a Poisson distribution with parameter $\lambda{t}$ to write an expression (involving Poisson probabilities) for the Erlang cumulative distribution function $F(t;\lambda,n)=O(X\leq t) $.