Probability question involving exponential distribution.

[Hodgey8806]

Homework Statement

The total claim for a health insurance policy follows a distribution with density function

f(x) = (1/1000)e[itex]^{-x/1000}[/itex], x>0

The premium for the policy is set at 100 over the expected total claim amount. If 100 policies are sold, what is the approximate probability that the insurance company will have some claims exceeding the premiums collected?

Homework Equations

I want to understand their method a bit. Would this binomial expression be approximate?
They use the expected value of claim, expected value of premium, and a normal approximation.

The Attempt at a Solution

My solution:
The probability of a claim being greater than 110000 = e^(-110)
Thus the probability of being less is 1-e^(-110)
P(N≥1) = 1-P(N=0) = 1 - (1)(1-e^(-110)[itex]^{100}[/itex](e^(-110))[itex]^{0}[/itex]
I don't have a program to evaluate this, so would they be approximate?

Their solution:
100*E(X) = 100*1000 = 100000
100*E(X+100) = 100*1100 = 110000
σ=10000

P(X>110000) = P(z>1) = .1587

Thanks!

 

[Ray Vickson]

Homework Statement

The total claim for a health insurance policy follows a distribution with density function

f(x) = (1/1000)e[itex]^{-x/1000}[/itex], x>0

The premium for the policy is set at 100 over the expected total claim amount. If 100 policies are sold, what is the approximate probability that the insurance company will have some claims exceeding the premiums collected?

Homework Equations

I want to understand their method a bit. Would this binomial expression be approximate?
They use the expected value of claim, expected value of premium, and a normal approximation.

The Attempt at a Solution

My solution:
The probability of a claim being greater than 110000 = e^(-110)
Thus the probability of being less is 1-e^(-110)
P(N≥1) = 1-P(N=0) = 1 - (1)(1-e^(-110)[itex]^{100}[/itex](e^(-110))[itex]^{0}[/itex]
I don't have a program to evaluate this, so would they be approximate?

Their solution:
100*E(X) = 100*1000 = 100000
100*E(X+100) = 100*1100 = 110000
σ=10000

P(X>110000) = P(z>1) = .1587

Thanks!

Your solution is incorrect, but so it the one you say "they" give.

First, for a single policy, what is the premium? The expected claim = 1000, so the premium is 1100 ($). What is the probability that a single claim exceeds the premium? That is p = Pr{X > 1100} = exp(-1100/1000) = exp(-11/10) ≈ 0.3328710837.

So, for 100 independence policies sold, the probability that NONE of them exceed the premium is the probability that NO 'successes' occur in 100 independent trials with success probability p per trial.

 

[Hodgey8806]

Thats the way I tried it the first time. But I was thinking they might have been asking for individual claims that exceed the total expected premium.

So I wantes to verify if their solution ia actually an approximation of what I did.

Thanks!