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Agtuary
10-27-2002, 02:59 PM
I was attempting a problem where the expected number of items was either 7, 8, 9 or 10, with equal likelihood. The solution determined the Var(N) using E(N^2) - E(N)^2 = 73.5 - 8.5^2 = 1.25.

I am probably missing something very basic but why is this different than a uniform distribution from [7,10] with Var(N) = (10-7)^2/12 = 0.75?

bg23516
10-27-2002, 03:04 PM
The formula you are using is for the continuous uniform distribution, or "the Uniform Dist for a < x < b".

For "probabilities distributed uniformly with x= a, a+ 1, ..... , b" you have a discrete distribution and will have to memorize the formula for the variance of discrete uniform, or just derive it on the spot with E[X] and E[X^2].


HTH.
Marc