Question

The special case of the gamma distribution in which α is a positive integer n is called an Erlang distribution. If we replace β by 1 λ in the expression below, f(x; α, β) = 1 βαΓ(α) xα − 1e−x/β x ≥ 0 0 otherwise the Erlang pdf is as follows. f(x; λ, n) = λ(λx)n − 1e−λx (n − 1)! x ≥ 0 0 x < 0 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). (a) What is the expected value of X? E(X) = n λ​ Correct: Your answer is correct. If the time (in minutes) between arrivals of successive customers is exponentially distributed with λ = 0.5, how much time can be expected to elapse before the fifth customer arrives? min (b) If customer interarrival time is exponentially distributed with λ = 0.5, what is the probability that the fifth customer (after the one who has just arrived) will arrive within the next 16 min? (c) 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 occurring in an interval of length t has a Poisson distribution with parameter λt to write an expression (involving Poisson probabilities) for the Erlang cdf F(t; λ, n) = P(X ≤ t).

Answer 1