Question

A dry-cleaning business offers a pick-up and delivery service for a 15 percent surcharge. Management believes 70 percent of the existing customers will take advantage of this service. They are also considering offering customers the option of opening an account and receiving monthly bills. They believe 40 percent of customers (independent of whether or not they use the pick-up service) will use the account service. If the two services are introduced to the market, what is the probability that a customer uses only one of these services, but not the other?

From historical data, the Probabilities of 0, 1, 2, 3, 4, or 5 calls in a 5 minute- period are 0.1, 0.15, 0.25, 0.20, 0.15, and 0.15, respectively. If X is a random variable for the number of calls arriving in a five-minute period, what is the Mean of X? Secondly, what is the Variance of X?

Please show work and Thanks

P

Answer 1

(1) A -> using delivery service, B -> using account service. Here, P(A) = 70% = 0.7 and P(B) = 40% = 0.4. Prob.(uses only one of these services, but not the other) = P(A^c) * P(B) + P(A) * P(B^c) = ((0.3 * 0.4) + (0.7 * 0.6) = 0.12 + 0.42 = 0.54.

(2) Given, x = (0, 1, 2, 3, 4, 5) and P(x) = (0.1, 0.15, 0.25, 0.20, 0.15, 0.15). Hence, Mean of X = E(X) = sum(x * P(x)) = (0 * 0.1) + (1 * 0.15) + ..... + (5 * 0.15) = 2.6. Now, E(X²) = sum(x * x * P(x)) = (0² * 0.1) + ... + (5² * 0.15) = 9.1. Hence, Variance of X = Var(X) = E(X²) - E²(X) = 9.1 - (2.6²) = 2.34.