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5.2.q12 - Student Grady Simonton Course Math119 Elementary...

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Unformatted text preview: Student: Grady Simonton Course: Math119: Elementary Statistics - Spring 2.010 - CRN: 49239 Instructor: Shawn Parvini - 16 weeks Date: 3/18/10 Book: Triola: Elementary Statistics, 11e Time: 4:23 PM An insurance company charges a 21-year-old male a premium of $500 for a one-year $50,000 life insurance policy. A 21-year-old male has a 0.998 probability of living for a year. a. From the perspective of a 21-year-old male (or his estate), what are the values of the two different outcomes? The values are the total amount paid by the insurance company minus any premiums paid to the insurance company. The value if he lives is - 500 dollars. The value if he dies is 49,500 dollars. b. What is the expected value for a 21 -year-old male who buys the insurance? The expected value is the same as the mean of the random variable. 2 [x - P(x)] - (premium) - P(lives) + ( (policy value) - (premium)) - P(dies) — 500 - 0.998 + (50,000 — 500) - 0.002 — 400 dollars a c. What would be the cost of the insurance if the company just breaks even (in the long run with many such policies), instead of making a profit? For the company to break even, the expected value would need to be zero. This occurs when the cost of the premium weighted by the probability of survival is the same as the profit weighted by the probability of death. Let x be the premium. Since the policy is worth $50,000, the profit is 50,000 — x. 0.998x — 0.002(50,000 — x) = 0 0.993x— 100 +0.002x = 0 x = 100 dollars d. Given that the expected value is negative (so the insurance company can make a profit), why should a 21-year-old male or anyone else purchase life insurance? Someone should purchase life insurance if the opportunity to financially provide for loved ones in unlikely circumstances compensates for the expected value loss. Page 1 ...
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