Mean absolute deviation (MAD) analogy to 68-95-99 rule With MAD, 50% of all values fall within one absolute deviation. How many within two, and three?

Assuming Normally distributed data (since the $68-95-99$ rule comes from the Normal distribution) I estimate that about $82\%$ of samples lie within $\pm 2\text{MAD}$ of the median and around $96\%$ of samples lie within $\pm 3 \text{MAD}$ of the median. This is based upon three assumptions

*

*Your estimate of central tendency, $m$, is the sample median

*You estimate MAD by $\text{MAD} = \frac{1}{n} \sum_{i = 1}^n |x_i - m|$

*your data are Normally distributed

R code:
set.seed(123)
x <- rnorm(10^6) 
m = median(x)
mad = median(abs(x - m))
mean(abs(x - m) < 1*mad)
[1] 0.5
mean(abs(x - m) < 2*mad)
[1] 0.822464
mean(abs(x - m) < 3*mad)
[1] 0.956922

