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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

Answer :

akindelemf

Answer:

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 < p95 ) / P(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

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