Statistics Question

Question

A certain insurance company sells an home and auto insurance policy that covers losses incurred by a policy holder, subject to a deductible of $202. Losses incurred follow an exponential distribution with a mean of $305. Find the 91 percentile of the actual losses that exceed the deductible.

Answer 1

Since losses follow an exponential distribution with mean 305, the probability that a random loss is less than or equal to x is given by the function:

CDF(x) = 1 - exp(-kx), where k = 1/305

and thus the probability that a random loss exceeds x is given by

1 - CDF(x) = exp(-kx)

The probability that a random loss exceeds the deductible (202) is thus given by:

exp(-k * 202) = exp(-202/305) = .5135...

To find the 91st percentile of those losses that exceed the deductible, we'll look for the cutoff x such that the probability of a loss being greater than x is equal to 9% (100 - 91 = 9) of the probability that a loss is greater than the deductible. We thus need to solve:

exp(-x/305) = .09 * .5135

This can easily be done using logarithms:

x = -305 * ln(.09 * .5135) = 936.42