test2-solutions

test2-solutions - UNIVERSITY OF VICTORIA MATHEMATICS 151...

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Unformatted text preview: UNIVERSITY OF VICTORIA MATHEMATICS 151, SECTION A02, FALL 2008 MIDTERM 2 SOLUTIONS, NOVEMBER 21 Question 1. An insurance company wishes to make an average profit of \$ 250 on each life insurance policy they sell. If there is a 0.0032 chance that a policy holder will pass away during the life of a policy, how much should a \$ 100,000 policy cost in dollars? (A) \$0 (B) \$100 (C) \$250 (D) \$500 (E) \$1000 (F) \$2000 (G) \$4000 (H) \$8000 (I) \$16000 (J) \$32000 Solution: Let X be the random variable corresponding to the profit in dollars that the company makes on a single policy. Let c be the cost of the policy in dollars. The probability distribution table for this random variable is: X = x c c- 100000 P ( X = x ) . 9968 . 0032 Here c- 100000 corresponds to the company taking in c dollars for the policy, and paying out \$ 100,000, which occurs 0 . 0032 of the time. Therefore we find 250 = E ( X ) = (0 . 9968)( c ) + (0 . 0032)( c- 100000) = c- 320 . Therefore c = 470, and so the answer is (D). Question 2. You roll a fair six sided die 12 times. What is the probability that you get at least 10 rolls that are strictly above 4 (either a 5 or a 6)? (A) . 0001 (B) . 0002 (C) . 0005 (D) . 001 (E) . 002 (F) . 005 (G) . 01 (H) . 02 (I) . 05 (J) . 1 Solution: This is a binomial distribution with n = 12 , p = 1 3 , and q = 2 3 . Thus we wish to find: P ( X ≥ 10) = P ( X = 10) + P ( X = 11) + P ( X = 12) = 12 10 1 3 10 2 3 2 + 12 11 1 3 11 2 3 1 + 12 12 1 3 12 2 3...
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test2-solutions - UNIVERSITY OF VICTORIA MATHEMATICS 151...

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