Some mean/variance problems

[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)= _nC_ipⁱ(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]

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)= _nC_ipⁱ(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 don't know how to solve this binomial further , its been a long time since I did Binomial, maybe lack of practice..

 

[Dr.Brain]

please sum1 help.

 

[HallsofIvy]

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

 

[Dr.Brain]

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.

 

[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 thanks , 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]

More detail please?

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