# Some mean/variance problems (1 Viewer)

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#### Dr.Brain

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?

#### HallsofIvy

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
$$\Sum_{i=0}^n _nC_i i^2 p^i (1-p)^{n-i}$$
Can you relate that to the binomial theorem?

#### Dr.Brain

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
$$\Sum_{i=0}^n _nC_i i^2 p^i (1-p)^{n-i}$$
Can you relate that to the binomial theorem?
thats where I am stuck , I dont know how to solve this binomial further , its been a long time since I did Binomial, maybe lack of practice..

#### HallsofIvy

Look at this:
http://www.bbc.co.uk/education/asguru/maths/14statistics/03binomialdistribution/12meanandvariance/index.shtml [Broken]

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#### Dr.Brain

ok I read it , thats 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.

#### HallsofIvy

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.

#### Dr.Brain

ok thanx , I got hold of that idea.

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

$$\Sum_{i=0}^n _nC_i i^2 p^i (1-p)^{n-i}$$

I am interested to know this.!

#### firestar

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

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