293. A company provides disability benefits to its employees. There are only two possible benefits: partial disability, costing the company 40, and total disability, costing the company 200. The company employs a number of married couples.
Let (X,Y) denote the company's disability costs for a randomly selected employed married couple. The joint probability function for (X,Y) is:

 Calculate the standard deviation of the total disability cost X + Y for the married couple.
(A) 12.3
(B) 13.8
(C) 15.7
(D) 16.6
(E) 19.8

251. Two independent estimates are to be made on a building damaged by fire. Each estimate is normally distributed with mean 10b and variance  b2 . Calculate the probability that the first estimate is at least 20 percent higher than the second. (A) 0.023 (B) 0.100 (C) 0.115 (D) 0.221 (E) 0.444

252. The independent random variables X and Y have the same mean. The coefficients of variation of X and Y are 3 and 4 respectively. Calculate the coefficient of variation of . (A) 5/4 (B) 7/4 (C) 5/2 (D) 7/2 (E) 7

253. Points scored by a game participant can be modeled by Z = 3X + 2Y – 5. X and Y are independent random variables with Var (X) = 3 and Var (Y) = 4. Calculate Var (Z). (A) 12 (B) 17 (C) 38 (D) 43 (E) 68

254. An actuary is studying hurricane models. A year is classified as a high, medium, or low hurricane year with probabilities 0.1, 0.3, and 0.6, respectively. The numbers of hurricanes in high, medium, and low years follow Poisson distributions with means 20, 15, and 10, respectively. Calculate the variance of the number of hurricanes in a randomly selected year. (A) 11.25 (B) 12.50 (C) 12.94 (D) 13.42 (E) 23.75

255. A dental insurance company pays 100% of the cost of fillings and 70% of the cost of root canals. Fillings and root canals cost 50 and 500 each, respectively. The tables below show the probability distributions of the annual number of fillings and annual number of root canals for each of the company’s policyholders.Calculate the expected annual payment per policyholder for fillings and root canals. (A) 90.00 (B) 102.50 (C) 132.50 (D) 250.00 (E) 400.00

256. A loss under a liability policy is modeled by an exponential distribution. The insurance company will cover the amount of that loss in excess of a deductible of 2000. The probability that the reimbursement is less than 6000, given that the loss exceeds the deductible, is 0.50. Calculate the probability that the reimbursement is greater than 3000 but less than 9000, given that the loss exceeds the deductible. (A) 0.28 (B) 0.35 (C) 0.50 (D) 0.65 (E) 0.72

257. Let X be the percentage score on a college-entrance exam for students who did not participate in an exam-preparation seminar. X is modeled by a uniform distribution on [a, 100]. Let Y be the percentage score on a college-entrance exam for students who did participate in an exam-preparation seminar. Y is modeled by a uniform distribution on [1.25a, 100]. It is given that . Calculate the  percentile of Y. (A) 80 (B) 85 (C) 90 (D) 92 (E) 95

258. In a study of driver safety, drivers were categorized according to three risk factors. Exactly 1000 drivers exhibited each individual risk factor. Also, for each of the risk factors, there were exactly 400 drivers exhibiting that risk factor and neither of the other two risk factors. Finally, there were exactly 300 drivers who exhibited all three risk factors and 500 who exhibited none of the three risk factors. Calculate the number of drivers in the study. (A) 2000 (B) 2300 (C) 2450 (D) 2750 (E) 3500

259. An insurance company examines its pool of auto insurance customers and gathers the following information:i) All customers insure at least one car.ii) 64% of the customers insure more than one car.iii) 20% of the customers insure a sports car.iv) Of those customers who insure more than one car, 15% insure a sports car.Calculate the probability that a randomly selected customer insures exactly one car, and that the car is not a sports car.(A) 0.16(B) 0.19(C) 0.26(D) 0.29(E) 0.31

260. An insurance company has found that 1% of all applicants for life insurance have diabetes. Calculate the probability that five or fewer of 200 randomly selected applicants have diabetes. (A) 0.85 (B) 0.88 (C) 0.91 (D) 0.95 (E) 0.98

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