Need Help with Mean and Variance Calculations for Distributions?

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Ok I am stuck up deriving the 'variance for Binomial Distribution' and mean for the 'Hypergeometric distribution '

For variance part , I first derived that variance can be written as =(second moment about origin) - (square of mean)

But I am having trouble calaculating the second moment about the origin .Please can sum1 tell me sum site which can help me?
 
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to get the "second moment about the mean" you need to sum i2P(i) for i= 0 to n. For the binomial distribution, with probabilities p, 1-p, P(i)= nCipi(1-p)n-i. That is, you are summing
[tex]\Sum_{i=0}^n _nC_i i^2 p^i (1-p)^{n-i}[/tex]
Can you relate that to the binomial theorem?
 
HallsofIvy said:
to get the "second moment about the mean" you need to sum i2P(i) for i= 0 to n. For the binomial distribution, with probabilities p, 1-p, P(i)= nCipi(1-p)n-i. That is, you are summing
[tex]\Sum_{i=0}^n _nC_i i^2 p^i (1-p)^{n-i}[/tex]
Can you relate that to the binomial theorem?

thats where I am stuck , I don't know how to solve this binomial further , its been a long time since I did Binomial, maybe lack of practice..
 
ok I read it , that's a good way to prove the mean of Binomial Distribution , but I want the proof for variance of Binomial , I want to solve it the same way as I told above.
 
Then keep reading! The first half of the page gives a very simple way of deriving the mean (Since one trial the value is either 0 or 1, the mean is 0*(1-p)+ 1(p)= p. Since trials are independent, the mean of n trials is the sum of the means of each: np) the second half of the page derives the variance in the same way.
 
ok thanks , I got hold of that idea.

One more thing , can u pls tell me how ot solve this thing:

[tex]\Sum_{i=0}^n _nC_i i^2 p^i (1-p)^{n-i}[/tex]

I am interested to know this.!
 
More detail please?

Hi, I think that web page is too trivial. Do you know a more detail page? Thanks!