# What purpose does multiplying by 3 serve in Pearson's second coefficient of skewness?

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?

• 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). – Nick Cox Oct 28 '15 at 20:24
• See the question and some of the discussion here which explains where the "3" arises. – Glen_b Oct 28 '15 at 22:47

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.