A notebook computer dealer mounts a new promotional campaign. Purchasers of new computers may, if dissatised 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 15% of all purchasers will, indeed, return computers and obtain refunds. Suppose that 90 computers are purchased during the campaign period. Complete parts a. and b. below. a. Find the mean and the standard deviation of the number of these computers that will be returned for refunds. The mean, u, is . (Round to one decimal place as needed.) The standard deviation. 6, is . (Round to one decimal place as needed.) b. Find the mean and standard deviation of the total refund cost that will accrue as a result of these 90 purchases. The mean, u, is $ . (Round to two decimal places as needed.) The standard deviation, 6, is $ . (Round to two decimal places as needed.)