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Is my conjecture correct that one standard deviation lies on the inflection points of the normal distribution curve (of the probability density function)? How can it be proved using the standard deviation formula?

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The question seems to be about the inflection points of the normal probability density function; the inflection point of the normal cumulative distribution function is at the mean. –  Henry Sep 2 '12 at 16:43

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We will find the inflection points of the density function of a general one-variable normal. This can be done in the usual calculus way, by examining the second derivative of the density function.

The density function $f(x)$ of the general one variable normal is given by $$f(x)=\frac{1}{\sqrt{2\pi}\sigma}\exp((x-\mu)^2/2\sigma^2).$$ Differentiate. We get a constant times $$(x-\mu)\exp((x-\mu)^2/2\sigma^2).$$ Differentiate the above expression. We get $$\exp((x-\mu)^2/2\sigma^2)-\frac{(x-\mu)^2}{\sigma^2}\exp((x-\mu)^2/2\sigma^2).$$ The second derivative is $0$ when $(x-\mu)^2=\sigma^2$, and changes sign there. That is the only place this happens, since $\exp(t)$ is never $0$. So the only inflection points are at $x=\mu\pm\sigma$, one "standard deviation unit" from the mean.

Remark: I do not know what you mean by the standard deviation formula. If it is $E(X-\mu)^2$, then the answer would be no, since there are plenty of density functions that do not even have an inflection point, and plenty that do but whose inflection point(s) are not $1$ standard deviation unit from the mean. So some special property of the normal needs to be used.

If you note that the density function of the general normal is just obtained by shifting and scaling the standard normal, it is enough to work with the simpler density function $$\frac{1}{\sqrt{2\pi}}\exp(-t^2/2),$$ and show that its inflection points are at $\pm 1$. The calculation is very similar to the one above, but looks simpler.

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Great answer. However the S.D. formula I had in mind was $\sqrt{E[(X-\mu)^2]}$. How is this related to $\sigma$ in the case of the normal distribution? –  chharvey Sep 9 '12 at 1:27
    
For the general normal, it is $\sigma$: just integrate. –  André Nicolas Sep 9 '12 at 1:45
    
So is it mere coincidence that the second derivative is zero at $\mu\pm\sigma$? –  chharvey Sep 9 '12 at 3:14

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