As is well known for the normal distribution, 68% of the probability mass is within one standard deviation of the mean, 95% within two standard deviations and 99.7% within 3 standard deviations.

However, I have some empirical distributions that are leptokurtic and negatively skewed. In such circumstances is there a formula based on their higher-order moments to calculate how much of the probability mass is within so many standard deviations of the mean?

I have a measurement and would like to give some sense of how far it is from the midpoint (mean or some other measure of central tendency).

Can this be done?

  • $\begingroup$ Without some additional assumptions you can't do all that much. See, for example, the many examples of distributions with the same skewness and kurtosis as the normal here; they have quite a variety of different quantile behaviours (without any attempt there to explore the possible extent of those behaviours). Similar effects occur at other values of skewness and kurtosis. $\endgroup$
    – Glen_b
    Sep 7, 2017 at 4:41

2 Answers 2


You can always calculate how many SDs values are from the mean by just plugging in sample values, (value $-$ mean)/SD, and then binning and counting.

Precise numerical facts such as you cite for the normal (Gaussian) in general depend on knowing one or more of the density, distribution or quantile functions, numerically if not analytically.

However, there aren't general relations available on just knowing the skewness or kurtosis. Skewness and kurtosis don't pin down the form of the distribution in general, as higher moments can vary too.


Here is a precise answer that shows that the median absolute deviation from the mean is not necessarily related to kurtosis.

Consider the family of distributions of $X = \mu + \sigma Z$, where $Z$ has the discrete distribution

$Z = -0.5$, with probability (wp) $.25$

$ = +0.5$, wp $.25$

$ = -1.2$, wp $.25 - \theta/2$

$ = +1.2$, wp $.25 - \theta/2$

$ = -\sqrt{0.155/\theta + 1.44}$, wp $\theta/2$

$ = +\sqrt{0.155/\theta + 1.44}$, wp $\theta/2$.

The family of distributions of $X$ is indexed by three parameters: $\mu$, $\sigma$, and $\theta$, with ranges $(-\infty, +\infty)$, $(0, +\infty)$ and $(0,.5)$.

In this family, $E(X) = \mu$, $Var(X) = \sigma^2$, and the median absolute deviation from the mean is $0.5\sigma$.

The kurtosis of $X$ is as follows:

kurtosis $= E(Z^4) = .5^4 * .5 + 1.2^4 * (.5 - \theta) + (0.155/\theta + 1.44)^2 * \theta$.

Within this family,

(i) kurtosis tends to infinity as $\theta \rightarrow 0$.

(ii) the distribution within the "shoulders" (i.e., within the $\mu \pm \sigma$ range) is constant for all values of kurtosis; it is simply the two points $\mu \pm \sigma/2$, wp $0.25$ each. This provides a counterexample to one interpretation of kurtosis, which states that larger kurtosis implies movement of mass away from the shoulders, simultaneously into the range between the shoulders and into the tails.

(iii) the "peak" of the distribution is also constant for all value of kurtosis; again, it is simply the two points $\mu \pm \sigma/2$, wp $0.25$ each. This provides a counterexample to the often given but obviously incorrect interpretation that larger kurtosis implies a more "peaked" distribution.

In this family, the central portion of the distribution actually becomes flatter as kurtosis increases, since the probabilities on $\mu \pm 1.2\sigma$ and $\mu \pm 0.5\sigma$ converge to the same value, $0.25$, as the kurtosis increases.

(iv) The median absolute deviation from the mean is constant, $0.5\sigma$, for all values of kurtosis.


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