1. From gamma distribution:
α (alpha) is a positive integer n
B (beta) is 1/λ

so we have:
f(x;y,n)=λ*(λx)^(n-1)*e^(-λx) / (n-1)!

NOTE: I USED y instead of λ in the math code because I get an error otherwise.

It can be shown that if the times between successive events are independent, each with an exponential distribution with parameter λ, then the total time X that elapses before all of the next n events occur has pdf f(x;λ,n).

1.
What is the expected value of X?
my work: E(X) = αB = n*1/λ = n/λ

If the time (in minutes) between arrivals of successive customers is exponentially distributed with λ = .5, how much time can be expected to elapse before the tenth customer arrives?
This is more than likely wrong, but I don't know how to approach this problem: $f(x;y,n)=f(n/y;y,n)=(y * (y*n/y)$^(n-1)*e^(-y*n/y)) $/(n-1)! = (y(n)^(n-1)e^(-n))/(n-1)! = (.5(10)^(10-1)e^(-10))/(10-1)! = .06$

2.
If customer inter arrival time is exponentially distributed with λ = .5, what is the probability that the tenth customer (after the one who has just arrived) will arrive within the next 30 min?
Think this is right: $f(x;y,n)=f(30;y,n)=$integral(0,30) $(.5(.5x)^(10-1)e^(-.5x))/(10-1)! = .93$

3.
The event [X ≤ t] occurs if at least n events occur in the next t units of time. Use the fact that the number of events occuring in an interval of lenth t has a Poisson distribution with parameter λt to write an expression (involving Poisson possibilities) for the Erlang cdf F(t; λ, n) = P(X ≤ t)
lost one this one..

2. Mods can lock this, I don't need help anymore. Thanks.