# verification on formula

• May 29th 2008, 09:23 PM
Judi
verification on formula
I don't know how to write sigma notation here, so I am going to word it instead

sum of square of the difference of the x and mean = sum of the square of the x minus square of the sum of x divided by n

I have been using this formula, and I know it works.
But now that I think about it...n has not been squared. I know x bar is equal to sum of x divided by n

I don't know whether you can understand what I am trying to say

I thank you all for your precious time,

Judi
• May 29th 2008, 10:15 PM
mr fantastic
Quote:

Originally Posted by Judi
I don't know how to write sigma notation here, so I am going to word it instead

sum of square of the difference of the x and mean = sum of the square of the x minus square of the sum of x divided by n

I have been using this formula, and I know it works.
But now that I think about it...n has not been squared. I know x bar is equal to sum of x divided by n

I don't know whether you can understand what I am trying to say

I thank you all for your precious time,

Judi

$\sum_{i=1}^{n} (x_i - \mu)^2 = \sum_{i=1}^{n} (x_i)^2 - 2 \sum_{i=1}^{n} x_i \mu + \sum_{i=1}^{n} \mu^2$

$\sum_{i=1}^{n} (x_i)^2 - 2 \mu \sum_{i=1}^{n} x_i + n \mu^2$

$\sum_{i=1}^{n} (x_i)^2 - 2 \mu (n \mu) + n \mu^2$

$\sum_{i=1}^{n} (x_i)^2 - n \mu^2$

$\sum_{i=1}^{n} (x_i)^2 - n \left( \frac{\sum_{i=1}^n x_i}{n}\right)^2$

$\sum_{i=1}^{n} (x_i)^2 - \frac{\left(\sum_{i=1}^n x_i\right)^2}{n}$