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Thank you so much, I appreciate it! Question 8 (1 point) Patients arrive randomly at an eye care clinic for eye exam. Suppose that there
Thank you so much, I appreciate it!
Question 8 (1 point) Patients arrive randomly at an eye care clinic for eye exam. Suppose that there is only one optometrist. The time required for the exam varies from patient to patient. Arrival rates have been found to follow the Poisson distribution (i.e., exponentially distributed inter-arrival times), and the service times follow the exponential distribution. The average arrival rate is 12 patients per hour, and the average service rate is 15 patients per hour. Patients stay in the waiting room until the optometrist is ready to see them. How many patients, on the average, will be in the waiting room? 1 patient 3.2 patients 4 patients 0.8 patients Question 9 (1 point) Customers arrive at a Service Canada centre with inter-arrival times that are on average 9 minutes. What is the hourly arrival rate customers at this centre? Note: Round your answer to 2 decimal places. Your Answer: Answer Question 10 (1 point) At a fashion retailer, there are 3 cashiers providing checkout service simultaneously. On average, customers arrive at the checkout area every 6 minutes. It has been estimated that the arrival process is a Poisson process. The average checkout time for each customer is 12 minutes, with its standard deviation equal to 18 minutes. Suppose that customers form a single line. What is the average length of the line (i.e., average number of customers in the waiting line)? Note: 1. Keep 2 decimal places for your final answer. Either use Excel for your calculation, or keep at least 4 decimal places for your intermediate numbers. 2. The Poisson arrival process has exponentially distributed inter-arrival times. YourStep by Step Solution
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