Calculate Standard Deviation From N Sets of Numbers - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-24T00:45:17Z http://mathoverflow.net/feeds/question/16373 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16373/calculate-standard-deviation-from-n-sets-of-numbers Calculate Standard Deviation From N Sets of Numbers Yan Cheng CHEOK 2010-02-25T06:18:12Z 2010-02-25T12:37:37Z <p>Currently, I have the following information</p> <ol> <li>Standard deviation = 1.6667</li> <li>Size of set = 4</li> <li>Average = 2.5</li> </ol> <p>From the following number set A : {1, 2, 3, 4}</p> <ol> <li>Standard deviation = 1.6667</li> <li>Size of set = 4</li> <li>Average = 1.5</li> </ol> <p>From the following number set B : {0, 1, 2, 3}</p> <p>Now, I do not have the detailed element information in both set A and set B. I do not know set A is having {1, 2, 3, 4}. I also do not know set B is having {0, 1, 2, 3}.</p> <p>What I only know is, the Standard deviation, Size of set and Average for set A and B.</p> <p>Now, I would like to compute the Standard deviation for the combination of set A and set B. (Of course, there should be no problem in computing Size of set and Average of set A and set B). Is it possible to do so?</p> http://mathoverflow.net/questions/16373/calculate-standard-deviation-from-n-sets-of-numbers/16390#16390 Answer by TonyK for Calculate Standard Deviation From N Sets of Numbers TonyK 2010-02-25T12:37:37Z 2010-02-25T12:37:37Z <ol> <li>Do you mean variance? The standard deviation is just the square root of the variance.</li> <li>The variance of these sets is 5/4, not 5/3.</li> <li>The <em>unbiased sample estimate</em> of these sets is 5/3.</li> </ol> <p>Anyway, if you have the mean and variance (or unbiased sample estimate) of a set, then you can get the sum of squares just by multiplying by n (resp. n-1). So you can get the sum of squares over A and B. Then just use the fact that the variance is equal to the mean of the squares minus the square of the mean.</p>