Suppose that the number of widgets sold in a given month (X) bythe Ajax Manufact

Question

Suppose that the number of widgets sold in a given month (X) bythe Ajax Manufacturing Company is well-approximated by aBinomial Distribution with a mean number of items sold of 2500 and(STD(X))2 = 1875. Find n and p for this random variable.Now suppose the cost associated with the sale of each item includesa fixed cost of $4.75 and a variable cost of $0.75 per unit sold,how much would you have to sell your widgets for, if you wished toset average monthly profits from the sale of widgets at$10,000?

Explanation / Answer

variance= 1875= np(1-p) mean = 2500 = np So there are 2 equations with 2 unknowns, so you can solvethis: 1) fill in np=2500 in the equation of the variance andobtain p: 1875= 2500(1-p) 0.75= 1-p -> p =0.25 2) fill in p in np=2500 to obtain n: np= 2500 n*0.25= 2500 --> n=10.000 I do not understand the second part of the question. The averageprofit has to be $ 10.000 and you sell on average 2500 widgets eachmonth. This means that you have to make $ 4.0 profit per sell. Ifeach atom costs 4.75 for the producer, the saleprice has tobe $ 8.75. I do not understand the variable cost of $ 0.75 .Can you specify this? Good Luck!

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