Question

Repair calls are handled by one repairman st a photocopy shop. Repair time, including travel time, is exponentisly distributed, with a mean of 2.0 hours per call. Requests for oopier repairs oome in at a mean rste of 2.1 per eight-hour day (assume Poisson) a. Determine the average number of oustomers awaiting repairs (Round your answer to 2 decimal places.) Number of customers 0.58 b. Determine system utilization. (Round your answer to the nearest whole percent. Omit the "%" sign in your response.) System utlization 53 % c. Determine the amount of time c. Detemine the amount of time during an eight-hour day that the repairman is not out on a call (Use your Round your answer to 2 decimal places) Amount of time 3.76 hours d. Determine the probability of two or more customers in the system. (Do not round calculations. Round your answer to 4 decimal places) Probability 0.2494 Hints References eBook& Resources

Please help somewhat confused on all parts

Answer 1

R 2.1 per day μ-22-4 per day a.The average number of customers awaiting repairs. = ( 2.1 )2 / 4 ( 4-2.1-4.41/4 (1.9) = 4.41 / 7.6-0.5802 = 0.58 (rounded off to 2 decimals) b.System utilization -R x 100 / μ-2.1 x 100 / 4-0.525 x 100-52.5% 53 (rounded off to whole percent) c.The amount of time during an eight-hour day that the repairman is not out on a call. System not utilized-100%-system utilized-100%-53% 47% Amount of time during 8 hour day that repairman is not out on a call-47% x 8-3.76 hours d.The probability of two or more customers in the system (N 22) Prob (N-1) 0.525 x (1 - 0.525)- 0.525 x 0.475-0.249375-0.2494 (rounded off to 2 decimals) Note : If it is Prob (N >- 2)1 [0.475 + 0.249]-1 [0.724]0.276