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Untensity of Ponda forogramare pe week or to come an Anonduce posing and loyal Gatorant of whom today buy the program follow tribution for each ind promises to a recycling chat pays only 10 per program. The standard deviation 6.000 grams and meet regram 52 Rotero the stand somalt for a) What is the cost of understag demand wat tog Casad your response.place for University football programs are printed 1 week prior to each home game. Attendance averages 60,000 screaming and loyal fans, of whom two-thirds usually buy the program following a normal distribution, for $4 mal each. Unsold programs are sent to a recycling center that pays only 20 cents per program. The standard deviation is 5,000 programs, and the cost to print each program is $1. a) What is the cost of underestimating demand for each program? C = Cost of shortage (underestimated) = Sales price/unit – Cost/unit = $4 – $1 = S 3 b) What is the overage cost per program? Co = Cost of overage (overestimated) = Cost/unit – Salvage value/unit = $1 – $ 20 = $ 0.80 c) How many programs should be ordered per game? Determine the mean for the number of sold programs using that only two-thirds of fans usually buy the program HE – ().60.0 (60,000) = 40.000 (Enter your response as a whole number.) Now compute the service level, that is, the probability of not stocking out Service level cs 3 C + C 3 +0.80 0.7895 (Enter your response as a real number rounded to four decimal places.) decimal cs 3 Service level = 0.7895 (Enter your response as a real number rounded to four decimal C+C 3 + 0.80 places.) Therefore, find the Z score for the normal distribution that yields a probability of 0.7895. Hence, 78,95% of the area under the normal curve must be to the left of the optimal stocking level. a H = 40,000 Stocking 6 = 5,000 level 78.95% For an area of 0.7895, the Z value is 0.805 (enter your response as a real number rounded to three decima/ places.) Determine the correct formula to find the optimal stocking level. Choose the correct answer below. A. Optimal stocking level = (H + 0).Z B. Optimal stocking level = (u – 0).Z c. Optimal stocking level = + 2.0 D. Optimal stocking level = H- Given u = 40,000, 6=5,000, and Z = 0.805, compute the optimal stocking level. Optimal stocking level = 40,000 + (0.805)(5,000) = 44,025 (Enter your response as a whole number.) d) What is the stockout risk for this order size? Recall that the service level is the complement of the probability of a stockout. Stockout risk = 1 – Service level = 1 -0.7895 = 0.2105 (Enter your response as a real number rounded to four decimal places.) Therfore, 44,025 programs should be printed to each home game. 1f 44.025 programs are printed, the stockout risk is 0.2105 or 21.05%.