MAD in relation to 95% confidence MAD (Median Absolute Deviation) is:
$\text{MAD} =  M_i(|x_i-M_j(x_j)|)$
where $M()$ is the median operator ($M_i(x_i) = \text{median}(x_1,...,x_n)$).
I'd like to scale the MAD in such a way as to include (say) 95% of a distribution around the median, the way that that 95% of a normal distribution is within $1.96\sigma$ of the mean.
That is, if $m = M_i(x_i)$ and $d = \text{MAD}_i(x_i)$, make an interval like $m \pm b\cdot d$ (where $b$ depends on the distribution you are dealing with) that includes 95% of the distribution. 
Can this be done?
 A: I know the original post is over a year old, but I would like some more information on this topic. I currently run a proficiency program for manure testing and soil testing laboratories. A colleague, who knows much more about statistics than I do, suggested the following to get a 95% confidence interval using the MAD and median.


*

*Calculate the median and MAD values.

*Remove results exceeding plus or minus 4.0 MAD units from the median as outliers.

*Recalculate the median and MAD values on the reduced data set. 

*Results exceeding plus or minus 2.9 MAD units from the second median are outside the 95% confidence interval.


There is one other kicker. I use the statistical program R. When calculating MAD I use the following:
mad(x, constant = 1)

The default in R is: constant = 1.4826. 
Typically, we have from 140 to 200 datapoints for each analysis. Often, the results are right skewed, occasionally left skewed, and rarely normally distributed. After removing the 4.0 MAD outliers, we have a much more normally distributed histogram. I suspect at that point we might be able to use mean and SD to calculate the confidence interval. 
For a number of years, we just ran the data one time. Labs were flagged for accuracy if their results deviated by more than 2.5 MAD units from the median. I have compared both methods, and usually 2.5 MAD units from the median (just one calculation) is quite close to the two-step method using 2.9 MAD units from the median after removing the 4.0 outliers.
I hope this method gives us a 95% confidence interval. But, if anyone has a better suggestion, I'd like to hear it.
A: Yes it can! In this rather recent preprint authors defined asymptotic CI of Median absolute deviation (MAD) as
$$
\left[L,U\right]_{MAD} = \left[ \widehat{MAD}_X ± \textit{z}_{1 - \alpha / 2} \frac{ \widehat{ASD}(\mathcal{MAD}, F_n)) }{ \sqrt{n} } \right]
$$
where ASD is a squared root of asymptotic variation, the $z_{1 − \alpha/2}$ is the $(1 − \alpha/2)×100$ percentile of the standard normal distribution. But then getting into how to calculate asymptotic variation< I got a bit lost, but all the math is in the preprint.
Futhermore, this CI was also implemented in R in DescTools package.
