Customers enter the camera department of a store at the average rate of eight per hour....

Customers enter the camera department of a store at the average rate of eight per hour. The department is staffed by one employee, who takes an average of 3.0 minutes to serve each arrival. Assume this is a simple Poisson arrival, exponentially distributed service time situation. Use Exhibit 10.9.

a-1. As a casual observer, how many people would you expect to see in the camera department (excluding the clerk)? (Round your answer to 2 decimal places.)

a-2. How long would a customer expect to spend in the camera department (total time)? (Round your answer to 1 decimal place.)

c. What is the probability that there are more than two people in the camera department (excluding the clerk)? (Round your intermediate calculations to 3 decimal places and final answer to 1 decimal place.)

d. Another clerk has been hired for the camera department who also takes an average of 3 minutes to serve each arrival. How long would a customer expect to spend in the department now? (Round your answer to 1 decimal place. Use the closest value of λ/μ in Exhibit 10.9 when determining Lq (i.e., do not interpolate).)

Solutions

Expert Solution

Given,

= Mean number of arrivals per time period = 8 arrivals per hour

= Mean number of people served per time period = 3 minutes per customer = 20 customers per hour

Average number of customers in the system = = 8 / 20 - 8 = 8 / 12 = 2/3 = 0.67

Average time a customer spends in the system = Waiting Time + Service time = = 1 / (20 - 8)

= 1/12 hours = 5 minutes

Probability that there are more than k customers in the system =

Probability that there are more than 2 customers in the system = = 0.064

Probability that there are no students in the system : (In the below image)

Hence, average time that a customer spends in the system = 3.126 minutes

keosha answered 2 hours ago

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