Answered

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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:

Sagot :

ajeigbeibraheem

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:

[tex]s = \dfrac{60}{12}[/tex]

s = 5 customers/ hour

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

[tex]= \dfrac{3}{5}\times (5-3) hours \\ \\ = \dfrac{3}{5}\times 2 \\ \\ = \dfrac{3}{10} \ hours[/tex]

To minutes;

[tex]= 3 \times \dfrac{60}{10} \ minutes[/tex]

= 18 minutes

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