3 edited body edited Jul 28 '14 at 22:46 Alexis 17k44 gold badges4949 silver badges103103 bronze badges Squaring amplifies larger deviations. If your sample has values that are all over the chart then to bring the 68.2% within the first standard deviation your standard deviation needs to be a little wider. If your data tended to all fall around the mean then σ can be tighter. Some say that it is to simplify calculations. Using the positive square root of the square would have solved that so that argument doesn't float. $$|x| = \sqrt{x^{2}}$$ So if algebraic simplicity was the goal then it would have looked like this: $$\sigma = \text{E}\left[\sqrt{(x-\mu)^{2}}\right]$$ which yields the same results as $$\text{E}\left[|x-\mu|\right]$$. Obviously squaring this also has the affecteffect of amplifying outlying errors (doh!). Squaring amplifies larger deviations. If your sample has values that are all over the chart then to bring the 68.2% within the first standard deviation your standard deviation needs to be a little wider. If your data tended to all fall around the mean then σ can be tighter. Some say that it is to simplify calculations. Using the positive square root of the square would have solved that so that argument doesn't float. $$|x| = \sqrt{x^{2}}$$ So if algebraic simplicity was the goal then it would have looked like this: $$\sigma = \text{E}\left[\sqrt{(x-\mu)^{2}}\right]$$ which yields the same results as $$\text{E}\left[|x-\mu|\right]$$. Obviously squaring this also has the affect of amplifying outlying errors (doh!). Squaring amplifies larger deviations. If your sample has values that are all over the chart then to bring the 68.2% within the first standard deviation your standard deviation needs to be a little wider. If your data tended to all fall around the mean then σ can be tighter. Some say that it is to simplify calculations. Using the positive square root of the square would have solved that so that argument doesn't float. $$|x| = \sqrt{x^{2}}$$ So if algebraic simplicity was the goal then it would have looked like this: $$\sigma = \text{E}\left[\sqrt{(x-\mu)^{2}}\right]$$ which yields the same results as $$\text{E}\left[|x-\mu|\right]$$. Obviously squaring this also has the effect of amplifying outlying errors (doh!). 2 Edited for notation edited Jul 28 '14 at 21:03 Alexis 17k44 gold badges4949 silver badges103103 bronze badges Squaring amplifies larger deviations. If your sample has values that are all over the chart then to bring the 68.2% within the first standard deviation your standard deviation needs to be a little wider. If your data tended to all fall around the mean then σ can be tighter. Some say that it is to simplify calculations. Using the positive square root of the square would have solved that so that argument doesn't float. |x| = sqrt(x^2)$$|x| = \sqrt{x^{2}}$$ So if algebraic simplicity was the goal then it would have looked like this: σ = E[sqrt((x-μ)^2)]$$\sigma = \text{E}\left[\sqrt{(x-\mu)^{2}}\right]$$ which yields the same results as E[|x-μ|]$$\text{E}\left[|x-\mu|\right]$$. Obviously squaring this also has the affect of amplifying outlying errors (doh!). Squaring amplifies larger deviations. If your sample has values that are all over the chart then to bring the 68.2% within the first standard deviation your standard deviation needs to be a little wider. If your data tended to all fall around the mean then σ can be tighter. Some say that it is to simplify calculations. Using the positive square root of the square would have solved that so that argument doesn't float. |x| = sqrt(x^2) So if algebraic simplicity was the goal then it would have looked like this: σ = E[sqrt((x-μ)^2)] which yields the same results as E[|x-μ|]. Obviously squaring this also has the affect of amplifying outlying errors (doh!). Squaring amplifies larger deviations. If your sample has values that are all over the chart then to bring the 68.2% within the first standard deviation your standard deviation needs to be a little wider. If your data tended to all fall around the mean then σ can be tighter. Some say that it is to simplify calculations. Using the positive square root of the square would have solved that so that argument doesn't float. $$|x| = \sqrt{x^{2}}$$ So if algebraic simplicity was the goal then it would have looked like this: $$\sigma = \text{E}\left[\sqrt{(x-\mu)^{2}}\right]$$ which yields the same results as $$\text{E}\left[|x-\mu|\right]$$. Obviously squaring this also has the affect of amplifying outlying errors (doh!). 1 answered Jul 28 '14 at 20:57 Squaring amplifies larger deviations. If your sample has values that are all over the chart then to bring the 68.2% within the first standard deviation your standard deviation needs to be a little wider. If your data tended to all fall around the mean then σ can be tighter. Some say that it is to simplify calculations. Using the positive square root of the square would have solved that so that argument doesn't float. |x| = sqrt(x^2) So if algebraic simplicity was the goal then it would have looked like this: σ = E[sqrt((x-μ)^2)] which yields the same results as E[|x-μ|]. Obviously squaring this also has the affect of amplifying outlying errors (doh!).