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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