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MathMan's derivation is correct. If you use

E(|x|)=(2/(sqrt(2pi)*sigma)*intergral from 0 to infinity of x*exp(-x^2/2sigma^2)dx, then change variable (for example, y = x^2/2sigma^2), you will get his formula.

Since the question was about normal distribution with zero mean, the answer is correct. If distributional assumption is different, the answer would be different as well. If you'd like to come up with a range of values for a wide range of symmetric distributions, you can try to minimize and maximize sqrt((1/n)*sum(x_k^2)) given 1/n*(sum(x_k^2))=1% and all x_k>=0. That might a fun problem to think about.