Question

A single-server waiting line system has an arrival pattern characterized by a Poisson distribution with 3 customers per hour. The average service time is 12 minutes. The service times are distributed according to the negative exponential distribution. The average time a customer can expect to wait in line is:

Answer 1

Answer:

18 minutes

Step-by-step explanation:

Given that:

The arrival time = 3 customers / hour

The avg. service rate (s) = 12 minutes per customer

To hour, we have:

$s = \dfrac{60}{12}$

s = 5 customers/ hour

Thus, the required average time for a customer needs to wait in line is:

$= \dfrac{3}{5}\times (5-3) hours \\ \\ = \dfrac{3}{5}\times 2 \\ \\ = \dfrac{3}{10} \ hours$

To minutes;

$= 3 \times \dfrac{60}{10} \ minutes$

= 18 minutes