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 August 23rd, 2017, 06:11 PM #1 Newbie   Joined: Aug 2017 From: Cincinnati, OH Posts: 1 Thanks: 0 Expected value of a function of a discrete random variable I'm new here, so perhaps this is beyond the intended purpose of the board. But, that doesn't mean it's beyond its members. Here is the question: An insurance company sells a one-year automobile policy with a deductible of 2. The probability that the insured will incur a loss is 0.05. If there is a loss, the probability of a loss of amount N is K/N for N = 1; ... ; 5 and K a constant. These are the only possible loss amounts and no more than one loss can occur. Determine the net premium for this policy. I have determined that K=60/137 Here is a chart of the probability distribution that I found: Pr(N=1)= p(1) = 60/137 Pr(N=2) = 30/137 Pr(N=3) = 20/137 Pr(N=4) = 15/137 Pr(N=5) = 12/137 Here's where I start to get unsure, but think it's still on the right track: E(K/N) = .05K*E(1/N) = K*(1(p(1)) + 1/2(p(2)) + 1/3(p(3)) + ... 1/5(p(5)) I'm not sure what to do with the deductible of 2. The answer is .0314 but I haven't found a way to get to it. Any help? Last edited by skipjack; August 24th, 2017 at 12:00 AM. August 24th, 2017, 12:12 AM #2 Global Moderator   Joined: Dec 2006 Posts: 21,028 Thanks: 2259 Tags discrete, expected, function, random, variable Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Keroro Advanced Statistics 2 August 17th, 2012 01:22 PM bureddogs44 Advanced Statistics 1 October 19th, 2011 01:39 PM hoyy1kolko Algebra 1 February 13th, 2011 05:32 AM hoyy1kolko Algebra 1 February 13th, 2011 05:08 AM adbroadband Algebra 1 January 31st, 2008 09:49 AM

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