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The special case of the gamma distribution in which is a positive integer n is called an Erlang distribution. If we replace by 1 in the expression below, f(x; , ) = 1 Γ() x − 1e−x/ x ≥ 0 0 otherwise the Erlang pdf is as follows. f(x; , n) = (x)n − 1e−x (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 , then the total time X that elapses before all of the next n events occur has pdf f(x; , n). (a) What is the expected value of X?

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To find the expected value of \( X \) for an Erlang distribution with parameters \( \lambda \) and \( n \), we use the formula:

Expected value of \( X \), \( E[X] = n ÷ \lambda \)

Where:

- \( n \) is the shape parameter of the Erlang distribution.

- \( \lambda \) is the rate parameter (reciprocal of the scale parameter).

In the Erlang distribution, \( n \) represents the number of events occurring in a given interval of time, and \( \lambda \) represents the average rate of events per unit time.

So, the expected value of \( X \) is \( \frac{n}{\lambda} \).

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