Question about calculating Variance
Here's the problem I'm bugged about: An insurance company has 5000 policies and assumes these policies are all independent. Each policy is governed by the same distribution with a mean of $495 and a variance of $30,000. What is the probability that the total claims for the year will be less than $2,500,000?
So I calculate the expected value of the 5000 policies as E(5000X) = 5000*E(X) = 5000*495 = 2,475,000.
Then, the next step is to calculate the variance V(5000X) = 5000^2 * V(X) = 25,000,000 * 30,000 = 750,000,000,000. However, I am told that this is not the correct way to calculate the variance in this case. The property of V(aX) = a^2 * V(X) is one I've used for awhile and gotten comfortable with. Why does it not apply in this case?
Re: Question about calculating Variance
Quote:
Originally Posted by
arcketer 
Here's the problem I'm bugged about: An insurance company has 5000 policies and assumes these policies are all independent. Each policy is governed by the same distribution with a mean of $495 and a variance of $30,000. What is the probability that the total claims for the year will be less than $2,500,000?
So I calculate the expected value of the 5000 policies as E(5000X) = 5000*E(X) = 5000*495 = 2,475,000.
Then, the next step is to calculate the variance V(5000X) = 5000^2 * V(X) = 25,000,000 * 30,000 = 750,000,000,000. However, I am told that this is not the correct way to calculate the variance in this case. The property of V(aX) = a^2 * V(X) is one I've used for awhile and gotten comfortable with. Why does it not apply in this case?
The mean is 5000*495
The variance is 5000*30000
CB
Re: Question about calculating Variance
I know, the question however, is why does the variance identity V(aX) = a^2 * V(X) apply?
Re: Question about calculating Variance
Quote:
Originally Posted by
arcketer I know, the question however, is why does the variance identity V(aX) = a^2 * V(X) apply?
=\int_D (ax-\overline{ax})^2f(x)\; dx=\int_D (ax-a\overline{x})^2f(x)\; dx)
........... ^2f(x)\; dx=a^2\int_D (x-\overline{x})^2f(x)\; dx=a^2V(X))
But this does not apply to this problem because you do not have a RV multiplied by 5000. You have the sum of 5000 independent identically distributed random variables and as the variance of the sum of two independednt random is the sum of their variances the variance of the sum of 5000 iid RVs is 5000 times the sum of the variance of one of them.
CB
