Let X = {32, 32, 33, 34, 35, 38, 39, 41, 41, 41, 44, 44, 45, 46, 48} (Say the scores that students got on test one)

Let Y = {32, 32, 33, 34, 35, 38, 39, 41, 41, 41, 44, 44, 45, 46, 48} (say the scores that students got on test two. Coincidentally, each student got exactly the same score)

Let Z = X + Y

Let W = 2X

Let sigmax = standard deviation of X sigmay = standard deviation of Y sigmaz = standard deviation of Z sigmaw = standard deviation of W

By standard formulas, the standard deviation of Z should be sqrt(sigmax^2 + sigmay^2)

The standard deviation of W should be 2*sigma(X).

However, since W and Z are exactly the same distributions, they should have the same standard deviations. How is this possible?

By standard formulas I mean:

When you add distributions, you add the variances, not the standard deviation, but if you multiply a distribution by a constant, then you also times the standard deviation by the same constant.

• Your "standard formula" for $Z$ assumes independence between scores on the two tests. Is that true in your example? Commented Apr 20, 2016 at 15:11
• When you say "standard formulas" which formula are you using -- can you show something that gives it -- and why are you using that formula? Commented Apr 20, 2016 at 15:28

As implied by the comments, Z and W are not the "same." Z involves two independent distributions, while W involves two distributions that are perfectly correlated. You can see this yourself by simulation. Make (say $10^4$) draws from Z by drawing randomly from X and randomly from Y and adding the two each time. Make the same number of draws from W by drawing randomly from X and multiplying the draw by 2 each time. Now calculate the standard deviations for each group of $10^4$ draws. You will find they are not the same. The standard deviation for the draws from W is about $\sqrt{2}$ larger than that for Z.