What's new

# Hull RMFI EOC Q&A - Question 3.17

#### eldakrory

##### New Member
Subscriber
Hello David, i have a question regarding Q 3.17 page 38. ''The probability that losses will be greater than $200 million is the probability that a normally distributed variable is greater than one standard deviation above the mean. This is 0.1587.'' where did we come up with 0.1587? Edited by Nicole to include the text of the question Hull RMFI EOC Q&A - Question 3.17 An insurance company’s losses of a particular type per year are to a reasonable approximation normally distributed with a mean of$150 million and a standard deviation of $50 million. (Assume that the risks taken by the insurance company are entirely nonsystematic.) The one-year risk-free rate is 5% per annum with annual compounding. Estimate the cost of the following: (a) A contract that will pay in one-year’s time 60% of the insurance company’s costs on a pro rata basis. (b) A contract that pays$100 million in one-year’s time if losses exceed $200 million. Answers: a) The losses in millions of dollars are normally distributed with mean 150 and standard deviation 50. The payout from the reinsurance contract is therefore normally distributed with mean 90 and standard deviation 30. Assuming that the reinsurance company feels it can diversify away the risk, the minimum cost of reinsurance is 90/1.05 = 87.51 or$85.71 million.

b) The probability that losses will be greater than $200 million is the probability that a normally distributed variable is greater than one standard deviation above the mean. This is 0.1587. The expected payoff in millions of dollars is therefore 0.1587 × 100=15.87 and the value of the contract is 15.87/1.05 = 15.11 or$15.11 million.

Last edited by a moderator:

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
Hi @eldakrory I'm actually not sure to what question you refer exactly but 0.1587 = 1 - NORM.S.DIST(+1.0, true) = 1 - 84.13%. That is, when standard normal Z = +1.0, 84.13% of the area under the normal pdf is to the left. Put another way, the probability that a random standard normal will be less than or equal to 1.0 is 84.13%. Therefore the probability it will be greater than +1.0 is 15.87%. So on a Z-lookup table, you'd typically locate the p = 0.8413 under the Z = +1.0. Thanks,