# Thread: How do you go from here?

1. ## 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.