Question

Daily receipts for a department store are normally distributed with an average of $12,324 and a standard deviation of $2,345. The manager is being closed watched by the home office to determine his ability to generate revenues.

a. The home office declares that if revenues on this particular day are less than $12,000 the manager will be fired. How likely is it he will be looking for a new job?

b. Determine the amount at which the home office should set in order to keep the probability of firing the manager at 5%.

Answer 1

Let X be the random variable denoting the daily receipt of a department store.

Thus, X ~ N(12324, 2345) i.e. (X - 12324)/2345 ~ N(0,1).

a .The probability that revenue on any particular day is less than 12000 = P(X < 12000) = P[(X - 12324)/2345 < (12000 - 12324)/2345] = P[(X - 12324)/2345 < - 0.1382] = (-0.1382) = 0.4450. (Ans).

b. Let, P(Z < a) = 0.05, where, Z = (X - 12324)/2345

Thus, (a) = 0.05 i.e. a = (0.05) = -1.645.

Thus, P[(X - 12324)/2345 < a] = 0.05 i.e. P[(X - 12324)/2345 < - 1.645] = 0.05 i.e. P[X < 8466.475] = 0.05.

Hence, the required amount to be set = $8466.475. (Ans).