An insurance company sells an auto insurance policy that covers losses incurred by a policyholder, subject to a deductible of 100. Losses incurred follow an exponential distribution with a mean of 300.

Calculate the 95th percentile of actual losses that exceed the deductible. Give your answer rounded to the nearest whole number. Hint: You can either do this one 'directly' or you can use the 'memoryless' property of the exponential distribution.

A) 600 B) 700 C) 800 D) 900 E) 1000

Option E - 1000

Step-by-step explanation:

Let X stand for actual losses incurred.

Given that X follows an exponential distribution with mean 300,

To find the 95-th percentile of all claims that exceed 100.

In other words,

0.95 = Pr (100 < x 100)

= Fx (P95) - Fx (100) / 1 - Fx (100)

, where Fx is the cumulative distribution function of X

since, Fx (x) = 1 - e^ (-x/300)

0.95 = 1 - e^ (-P95/300) - [ 1 - e^ (-100/300) ] / 1 - [ 1 - e^ (-100/300) ]

= e^ (-1/3) - e^ ( - P95//300) / e^ (-1/3)

= 1 - e^1/3 e^ (-P95/300)

The solution is given by, e^ ( - P95/300) = 0.05e^ (-1/3)

P95 = - 300 ln (0.05e^ (-1/3))

= 999

= 1000