Please, help me with section D?

Return to question Problem 10-16 10 points 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 5.0 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.) Answer is complete and correct. Number of 2.00 people a-2. How long would a customer expect to spend in the camera department (total time)? (Round your answer to 1 decimal place.) Answer is complete and correct. Average total time 15.0 minutes Return to question b. What is the utilization of the clerk? (Round your answer to 1 decimal place.) Answer is complete and correct. 66.7 % Utilization points 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.) Answer is complete and correct. Probability 29.6 % d. Another clerk has been hired for the camera department who also takes an average of 5 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 n/p in Exhibit 10.9 when determining Lq (.e., do not interpolate).) Check my work mode! This Snows what 15 Correct or incorrect for the Work you have completed 50 tal. It does not Inalcate completion. Return to question 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.) 10 points Answer is complete and correct. 29.6 % Probability d. Another clerk has been hired for the camera department who also takes an average of 5 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 lp in Exhibit 10.9 when determining La (.e., do not interpolate).) * Answer is complete but not entirely correct. Average total 0.6 minutes time Expected Number of Customer Waiting in Line (L ) for Various Values of S and Nu exhibit 10.9 9 10 11 12 0.1500 0.62650.000B 0.0500 0 1885 00111 0.0020 0.250 0. 500 0.5500 0.2 0 0940 0.5000010833 0672200450 0.9000 00603 0.0112 0.7000 0.7500 0.0000 0.850g 0.0000 0.250 1623300977 2.2500 0.1227 320000.1524 4.8167 0.1874 1000 02285 18.0500 00189 0.0240 0.0300 0.0371 00455 0.0032 0.0042 0.0064 0.0050 0.1171 0.3120 05321 0.0225 0.0005 0.102 00928 0.0048 1.4002 0.2772 0.0869 0.0159 1.0002 1.5283 2.3857 0.1010 02412 0.3542 0.5130 01042 CABO 00001 0.1453 0.0085 0.0111 0.0190 0.0282 00428 0.00780 0.01230 7.000 18.9370 1.0553 1.5187 2.2165 3.8200 0294 0.4116 05695 0.00650 0.01200 0.0010 0.1293 0.1801 0 2476 0.8365 4505 0.0230 0.04190 0.05000 0.08290 0.11430 0.155RO 0.00760 0.00800 005420 92999 14D 0.01290 001150 5.2000 2.9376 43009 6.6811 0.8104 10905 1.4444 0.27880 0.36620 0.48330 0.10060 0.03610 0.19460 0.0490 0.177000 08640 0.01260 0.01760 0.02430 QOCARON 0.32260 2 6482 3.6830 5.2981 030330 0.24200 0.50390 0.11650 0.15100 0.19560 0.04450 0.05010 0.07760 0.01680 0.00250 0.03010 1.39580 13.7701 31.1272 2.42000 224460 0.13030 0.16670 0.06240 0.06010 0.82490 1.06360 0.19473 1.72000 2.28250 2.91180 0.32280 0.40000 0.51740 0.65210 0.82040 1.03140 6.31390 0.11200 0.320000.14210 0.42120 0.17910 5.22660 1.63670 7.34440 2 07400 10.959702 54740 0.65320027970 0.81100 0.34710 1.00610 0.42910 4 48070 6 01 PO 8.38730 12.42040 20.6179 1.55260 1.06710 2.43010 3.07320 3.9021 5.1150 0.65020 0.7010 0.97000 120100 1.4754 1.817 2012 22489