This preview shows page 1. Sign up to view the full content.
Unformatted text preview: In other words, we will have the same error above the mean value as we have below the mean value. How can we get around this problem? Well, instead of taking the absolute difference between each data point and the mean, let’s square this value instead. That is, let’s find ( 29 2 i x x. Since the square value of any negative number is positive, NOW we can add these points together, and get a legitimate nonzero value. We call this the “Variance”: ( 29 N x x V Variance N i i ∑ == = 1 2 Notice that this does not give us the mean value of each data point from the mean, but rather the square of this value. We are really not interested in the square distance from the mean, so we define the standard deviation as V Stddev = = s The smaller our standard deviation is, the less spread out our data is....
View
Full
Document
This note was uploaded on 01/27/2012 for the course CHM 2045 taught by Professor Josephwalter during the Fall '11 term at Dakota State.
 Fall '11
 JosephWalter
 Chemistry

Click to edit the document details