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A cafeteria serving line has a coffee urn from which customers serve themselves Arrivals at the urn follow a Poisson distribution at the rate of three per minute. Customer service times are exponentially distributed with a mean of 15 seconds.


a) How many customers on average would you expect to see at the coffee station?


b) How long would you expect it to take to get a cup of coffee?


c) What percentage of time is the urn being used?


d) What is the probability that there are three or more people at the coffee station?

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Answer:

3 ; 1 minute ;75%

Step-by-step explanation:

Given :

Service time = 15 seconds = 15/60 = 0.25 minute per customer

μ = number of customers per minute = 1/0.25 = 4

Arrival rate, λ= 3 persons per minute

L = λ ÷ (μ - λ)

L = 3 ÷ (4 - 3)

L = 3 ÷ 1

L = 3

b) How long would you expect it to take to get a cup of coffee?

L / arrival rate (λ)

3 / 3

= 1 minute

c) What percentage of time is the urn being used?

Utilization rate = λ/ μ

= 3 / 4

= 0.75 = 0.75 * 100% = 75%

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