I did show that the set of modified means of a sample represents quite well its mean; its population being similar to the respective Population mean. To be clear: one has a size n random sample, the source, and from it we perform the sum from 1 to n: mmX= [W(i)*x(g)] where the parenthesis represent the sum of the products (Gauss notation) W(i) = i/[i], i=1, 2, ?, n, the arrival index, by the source item randomly, and exhaustively chosen from it. As mmX tends to the mean, I dare if ssP= [W(i)*x(g)^2] tends decently to the source sum of squares. If so we are able to find an artificial sum of squared difference from the mean, and to attain the variance of the sample. __1___Results Taking in account that ssP above, relative to a sample simulation, is a estimate of the IP (intra-permutation) sum of squares, then the sum of squared differences can be get by ssD= n*(ssP - mean*mean) where mean is from the source. Caution: every time a negative value is found we discard it, if not we use to calculate E(s)=[sqrt(ssD/(n-1)]. For practical proposes one should have an idea, for each n, what the ratio R = E(s)/s is.