Negative and positive skewness: why are these names backwards? The skewness of a distribution is its third central moment,
$$
\mu_3 = \int (x - \mu)^3 f(x) \text{d}x \tag{1}
$$
where $\mu$ is the first raw moment, the mean. When I look at that equation, I think: if most of the distribution is to the left of the mean, then typically $x < \mu$ and $(x - \mu)$ is negative. Since cubing preserves signs, negative skewness means that more mass is to the left of the mean.
However, this is the opposite of how skewness is defined. A negatively skewed distribution leans right, and a positively skewed distribution leans left. Is there a good reason for this? Am I misunderstanding $(1)$?
 A: The total mass is one thing, but there's also the fact that the magnitude of the weight becomes larger as you move further from the mean. Take a Bernoulli distribution with $p<1/2,$ for example. The mean is $p$ and there is certainly more mass to the left of $p$. The third central moment is $$ E((X-p)^3)=p(1-p)^3 - (1-p)p^3.$$ Notice that smaller mass $p$ to the right is multiplied by a larger value than the larger mass $(1-p)$ to the left ($(1-p)^3$ vs $p^3$). The result is that the moment is actually positive... a little algebra gives $$ E((X-p)^3) = p(1-p)(1-2p) $$ and we can easily see that is positive if and only if  $p<1/2.$
So this shows that it is possible that a right-skewed distribution will have most mass to the left of the mean. And perhaps it is even a plausibility argument that this is the typical case. For more evidence, look at some of the most famous skewed distributions, like exponential, Poisson, and Pareto; all of them have this property, and it is often said that skewness is characterized in this way. However, this is actually a fallacy: it is possible, even in natural and non-pathological cases, for a right-skewed distribution to have most of its mass to the right of its mean.
