
From gamma distribution:
α (alpha) is a positive integer n
B (beta) is 1/λ
so we have:
f(x;y,n)=λ*(λx)^(n1)*e^(λx) / (n1)!
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: $\displaystyle f(x;y,n)=f(n/y;y,n)=(y * (y*n/y)$^(n1)*e^(y*n/y))$\displaystyle /(n1)! = (y(n)^(n1)e^(n))/(n1)! = (.5(10)^(101)e^(10))/(101)! = .06$
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: $\displaystyle f(x;y,n)=f(30;y,n)=$integral(0,30) $\displaystyle (.5(.5x)^(101)e^(.5x))/(101)! = .93$
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..

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