It can be calculated by $\frac{3\,(\text{mean}-\text{median})}{\text{standard deviation}}$.
Does the 3 make the result easier to work with somehow, or what?
$\begingroup$
$\endgroup$
2
-
3$\begingroup$ I think the idea (or ideal) is to match Pearson's other measure of skewness based on mean and mode as closely as possible. Pearson didn't seem to realise that (without the 3), the "second" measure falls in $[-1,1]$, which was proved formally later. It now can seem strange that Pearson emphasised using mean and mode, but that measure is well determined in his system of distributions. A more common modern stance is that mode can be difficult to estimate, not that there aren't many ways to do it (which, conversely, is the problem). $\endgroup$– Nick CoxCommented Oct 28, 2015 at 20:24
-
1$\begingroup$ See the question and some of the discussion here which explains where the "3" arises. $\endgroup$– Glen_bCommented Oct 28, 2015 at 22:47
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
This is a very interesting question.
The answer lies in history. There was one skewness measure, which was very popular initially, say in the beginning of XX-th century: $$\frac{mean-mode}{s.d.}$$. Since the mode is difficult to estimate the next best thing was to use the (approximate) relationship $$mean - mode \approx 3 (mean - median)$$ This relationship was discovered empirically for near symmetric distributions, as Yule explained in the text below.
Here's the reference: Yule, Introduction to the theory of Statistics (1922), see p.150 and p.121
