A manufacturer of car radios ships them to retailers in cartons of radios. The p

Question

A manufacturer of car radios ships them to retailers in cartons of radios. The profit per radio is $59.50, less shipping cost of $25 per carton, so the profit is $ per carton. To promote sales by assuring high quality, the manufacturer promises to pay the retailer if radios in the carton are defective. (The retailer is then responsible for repairing any defective radios.) Suppose radios are produced independently and that 5% of radios are defective. How many radios should be packed per carton to maximize expected net profit per carton?

Explanation / Answer

X has a binomial distribution of mean np and variance npq (where =0.05 and q=0.95). So E(X^2) = Variance + Mean^2 = npq + n^2 p^2 Therefore, the expected profit if the cartons have n radios is P = 59.5 n - 25 - 200 ( npq + n^2 p^2), so dP/dn = 59.5 - 200 (pq + 2 p^2 n) and if dP/dn=0 we must have n = (59.5 - 200 pq) / (400 p^2). Replacing p and q and calculating we get n=50 so that is the number of radios in the carton which will maximize net expected profit per carton

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