Merlyn, sorry for the incorrectness.. tired I guess. OK, where did you get the 9826 for (Xi)sq ?

by Merlyn


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The standard deviation you have posted is not correct.





Let X1, X2, X3, ... , Xn be a random sample. The mean, xbar is found by:





xbar =





n


鈭?Xi / n


i = 1





the sample variance is:





n


鈭?(Xi - xbar)虏 / (n - 1)


i = 1





This can be written in another form that is easier to compute





1/(n-1) * { [ 鈭?(Xi)虏 ] - n * xbar虏 }








the variance is divided by n -1, not n. this is done because if you divide by n you have a biased estimator for the population variance. using n - 1 yields an unbiased estimator.





the sample standard deviation is found by taking the square root of the variance.





== -- == -- == -- ==





We have in this small example





67, 43, 28, 52





xBar = 1/4 * (67 + 43 + 28 + 52) = 47.5





the variance of the sample is:


1/(n-1) * { [ 鈭?(Xi)虏 ] - n * xbar虏 }





1/3 * { 9826 - 4 * 2256.25} = 267





the stddev of the sample is the sqrt of the variance





sqrt(267) = 16.34013|||9826 for (Xi)sq ... it is found by summing all the squared values from the sample. See below.











We have in this small example





67, 43, 28, 52





xBar = 1/4 * (67 + 43 + 28 + 52) = 47.5





the variance of the sample is:


1/(n-1) * { [ 鈭?(Xi)虏 ] - n * xbar虏 }





1/3 * { [ 67^2 + 43^2 + 28^2 + 52^2 ] - 4 * 2256.25}





1/3 * { 9826 - 4 * 2256.25} = 267





the stddev of the sample is the sqrt of the variance





sqrt(267) = 16.34013