Here is the summation (from i=0 to n) of i^2 * (n!/(n-i)!(i!)) * p^i * q^(n-i)

I know that the answer is equal to np[(n-1)p+1].

I am not really sure how to get what is in the bracket of the answer.

Please help me out.

Thanks.

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- May 8th 2011, 01:28 AMohm713Finding the expected value of x^2 in binomial distribution
Here is the summation (from i=0 to n) of i^2 * (n!/(n-i)!(i!)) * p^i * q^(n-i)

I know that the answer is equal to np[(n-1)p+1].

I am not really sure how to get what is in the bracket of the answer.

Please help me out.

Thanks. - May 8th 2011, 02:17 AMSpringFan25
There are easier ways to do this than evaluating that sum

You know that

$\displaystyle E(X^2) = Var(X) + E^2(X)$

You also know that*by definition*the binomial variable X is the sum of n independant bernoulli trials. Its not difficult to compute the mean and variance of the sum of n independant bernoulli trials, then you can use the result above. - May 8th 2011, 02:25 AMohm713
THank you for your quick reply.

However, I would actually like to know how to get that answer, which is np[(n-1)p+1] using the summation and binomial distribution properties.

Could you please show me the steps? - May 8th 2011, 04:53 AMSpringFan25
http://quicklatex.com/cache3/ql_8468...b2406a5_l3.png

Instead of recognising the left hand sum as an expectation, you can use the algebra in the following link to show explicitly that it is (n-1)p.

Mean or Expectation Value

__Spoiler__: - May 8th 2011, 06:30 AMmatheagle
It's easier to find E(X(X-1)) first and then use $\displaystyle E(X(X-1))= E(X^2)-np$