Results 1 to 1 of 1

Thread: How do you go from here?

  1. #1
    Sep 2009

    How do you go from here?

    N=2, using \sum_{i}^{N} \frac{f_{i} \left(x_{i}- \overline{x}\right)^2}{N}, prove \sigma = \sqrt{pq}

    Let p=\frac{f_1}{N}, and q=\frac{f_2}{N}

    The following is my algebraic work:

    \sigma = \sqrt{p \left( x_1-\left( px_1+qx_2 \right)\right)^2+q \left( x_2-\left( px_1+qx_2 \right)\right)^2}

    = \sqrt{p \left( x_1- px_1-qx_2 \right)^2+q \left( x_2- px_1-qx_2 \right)^2}

    = \sqrt{p \left( qx_1-qx_2 \right)^2+q \left( px_2-px_1 \right)^2}

    = \sqrt{pq^2 \left( x_1^2-2x_1x_2+x_2^2 \right)+p^2q \left( x_2^2- 2x_1x_2+x_1^2 \right)}

    = \sqrt{\left(pq^2+p^2q\right) \left( x_1-x_2 \right)^2 }

    = \sqrt{pq \left( x_1-x_2 \right)^2 }

    How do you go from here to \sqrt{pq}

    I posted the above yesterday morning, earnesty not knowing the solution, and I spent all night contemplating and discovered that the solution to the proof is only true in a special case of Binomial distribution, since the Standard Deviation in Binomial Distribution is \sqrt{npq}, where n \neq N of the given in the question.

    In this case n=1. Therefore, \left(x_1-x_2\right)^2=1 is required for all cases regardless, and for the solution, there is no other sample space but that of S={a, a+1}, 0\leq a, a\in R

    Now, case closed.
    Last edited by novice; Oct 11th 2009 at 11:16 AM. Reason: Found solution to share
    Follow Math Help Forum on Facebook and Google+

/mathhelpforum @mathhelpforum