- #1

Hodgey8806

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## 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!