# Can the normal pdf be rewritten to use mean absolute deviation as a parameter in place of standard deviation?

We know that the formula for the normal probability density function (pdf) is given by:

$$f(x;\mu,\sigma) = \frac{1}{\sigma\sqrt{2\pi}}e^{−(x−\mu)^2/ 2\sigma^2}$$

where $\mu$ is the mean and $\sigma$ is the standard deviation. But standard deviation doesn't have an intuitive meaning and it can be confusing for a beginning student of statistics.

Some educators have proposed that it would be better to use mean absolute deviation as a measure of dispersion, in part because it has a more intuitive meaning for students to understand.

Revisiting a 90-Year-Old Debate: The Advantages of the Mean Deviation
Stephen Gorard, British Journal of Educational Studies, Vol. 53, No. 4 (Dec., 2005), pp. 417-430.]

Resistance to using mean absolute deviation has primarily come from educators who point out that standard deviation is used in the normal probability function and is therefore tied to one of the most basic building blocks of modern statistics.

My question: Is it possible to rewrite the normal probability density function in terms of mean absolute deviation so that students who learn that measure of dispersion can make a seamless transition to the normal distribution? If so, what would the rewritten formula be?

• The fact that $\operatorname{var}(X_1+ \cdots + X_n) = \operatorname{var}(X_1) + \cdots + \operatorname{var}(X_n)$ if $X_1,\ldots,X_n$ are independent is why that measure of dispersion is used. If you toss a coin $1800$ times, what's the probability that the number of heads is at least $880$ but nor more than $908$? You couldn't answer that question easily without that identity. $\qquad$ – Michael Hardy Nov 22 '16 at 16:49

The mean absolute deviation for a normal random variable with standard deviation $\sigma$ is $m = \sigma \sqrt{2/\pi}$, so just replace $\sigma$ by $m \sqrt{\pi/2}$ in any formula for that random variable.
• Of course, you no longer enjoy nice properties when describing $m$ for the sum of two normals with identical means, and you can no longer use any of the ubiquitous tables of $P$-values. I would say this approach is a bad pedagogical idea, even though as R.I. says, it can be done fairly easily. – Mark Fischler Nov 22 '16 at 16:43
• $$f(x;\mu,\sigma) = \frac{1}{\sigma\sqrt{2\pi}}e^{−(x−\mu)^2/ (2\sigma^2)} = \underbrace{ \frac 1 {\pi m} e^{-(x-\mu)^2/(\pi m^2)} }_\text{This is it.}$$ – Michael Hardy Nov 22 '16 at 16:54