Question

A notebook computer dealer mounts a new promotional campaign. Purchasers of new computers may, if dissatisfied for any reason, return them within 2 days of purchase and receive a full refund. The cost to the dealer of such a refund is $75. The dealer estimates that 14% of all purchasers will, indeed, return computers and obtain refunds. Suppose that 60 computers are purchased during the campaign period. Complete parts a and b. below. a. Find the mean and the variance of the number of these computers that will be returned for refunds. b. Find the mean and variance of the total refund cost that will accrue as a result of these 60 purchases. (NOTE: Number of computers returned is a binomial random variable. (There are only two outcomes - returned or not and the probability of a return is P=0.14). We can easily compute the mean and variance of a binomial random variable, and the variability of costs will depend on the variability of returned computers.) O A. Expected number of returns = 8.4 Variance of returns = 7.224 Variance of costs of returns for 60 computers = 541.8 B. Expected number of returns = 8.4 Variance of returns = (60-8.4)2 / (60-1) Variance of the costs of returns for 60 computers = 40635 O C. Expected number of returns = 8.4 Variance of returns = 7.224 Variance of the costs of returns for 60 computers = 5625 D. Expected number of returns = 8.4 Variance of returns = 7.224 Variance of the costs of returns for 60 computers = 40635

Answer 1

a)

Mean = np =    60*0.14=       8.4
Variance = np(1-p) =    60*0.14*(1-0.14)=       7.2240

b)

variance of total refund cost = Var(75X) = 75²*var(X) = 75^2*7.2240 =  40635

D. Expected number of returns = 8.4 Variance of returns = 7.224 Variance of the costs of returns for 60 computers = 40635

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answer: option d)