CHEM 1&amp;2 Lab Manual &amp; worksheets pg 228

# CHEM 1&2 Lab Manual & worksheets pg 228 - In other...

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Basic Laboratory Statistics General Chemistry I and II Lab Manual Dakota State University page 228 of 232 be 101.33, and the median would be (99.46+101.33)/2=100.40. Notice that this would also change the mean to 100.27; why? The closer the mean, the median and the mode are to one another, the more “normalized” the data is (that is, the closer the data would fit to a curve created with no systematic errors at all). Variance and Standard Deviation We would like to have a measure of how close our data points are to one another, or, better still, how close they are to the mean. For any given data point, we can simply do a subtraction, i x x - , but the problem is, if we try to take the average distance from the mean, or ( 29 N x x N i i = - 1 , we will find that this gives us a value of zero, because of the way that the mean is defined.
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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 non-zero 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....
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## 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.

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