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Thread: How do you go from here?

  1. #1
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    How do you go from here?

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

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

    The following is my algebraic work:

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

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

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

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

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

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

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

    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 $\displaystyle \sqrt{npq}$, where $\displaystyle n \neq N$ of the given in the question.

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

    Now, case closed.
    Last edited by novice; Oct 11th 2009 at 10:16 AM. Reason: Found solution to share
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