PDA

View Full Version : Hypergeometric Distribution



Jonas
June 28th 2006, 10:17 AM
I am having a bit of trouble understanding the mean of the hypergeometric distribution and the variance of the hypergeometric distribution. I will continue with the rest of this message in the context of the defects problem where there are 20 defective parts and 30 parts that are not defective. According to the book that I am using, the probability of choosing a defective on a choice is 20/50 and the expected number of defectives is n(20/50) where n is the number of choices. I am uncertain that this is true and I don't know how to prove it.

I thought, there are many possible ratios that are above and below 20/50.

ratios=p:

20/50

ratios<p:

19/49, 18/48, 17/47,

ratios>p:

20/49, 20/48, 20/47

Is there a way to prove that the average ratio is equal to 20/50?

I don't know where to start deriving the variance of the hypergeometric distribution. Maybe I could start with the formula npq(20-5)/(50-1). The first portion makes sense for the variance of the binomial distribution but I don't understand where (20-5)/(50-1) comes from.

Your advice is appreciated.