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Pillow
05-16-2002, 09:55 AM
This is problem #6 from the November 2000 exam.

PROBLEM: An insurance company issues 1250 vision care insurance policies. The number of claims filed by a policyholder under a vision care insurance policy during one year is a Poisson random variable with mean 2. Assume the numbers of claims filed by distinct policyholders are independent of one another.

What is the approximate probability that there is a total of between 2450 and 2600 claims during a one year period.

MY PROBLEM - I understand how to find the "new" mean for all the policies. It's just 1250*E(X) = 1250 * 2 = 2500.

What I can never seem to quite understand completely is how they go about figuring the "new" variance/standard deviation. Anyone have any tricks or better explanations?

Thanks for you help.

Gandalf
05-16-2002, 10:34 AM
General rule: If you have the sum of n independent random variables each with the same distribution (e.g., the 1250 policies here), then Var(sum of n) = n * Var(each one).

So here, with 1250 policies each with variance 2, Var(sum) = 1250 * 2 = 2,500.

Std Dev = square root of variance = 50.

[To get that the variance of each one was 2, you need to know that the mean and variance of any Poisson distribution is the same.]
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