probability of arbitrary distribution that value is $> \mu+/-3\sigma$

Question

A normal distribution has the property: $P(X>\mu+/-3\sigma)=1-99.71$

What is the probability $P(X>\mu+/-3\sigma)$ for an arbitrary distribution? (Is it actually possible to have some general rule for an abitrary distribution? I read somewhere that $88.8\%$ of all values lie in the range $(\mu+/-3\sigma)$. Probably this is totally nonsense.)

Thanks!

Answer 1

It is true. This is a consequence of the Chebyshev Inequality.

This says that if $X$ is a random variable that has a mean $\mu$ and a variance $\sigma^2$, then $$\Pr(|X-\mu|\ge k\sigma)\le \frac{1}{k^2}.\tag{1}$$

Let $k=3$. Then from (1) we obtain that $$\Pr(|X-\mu|\le 3\sigma) \ge 1-\frac{1}{9}.$$ This is the $88.8\%$ that you quoted. Note that the Inequality says that the probability is at least $88.8\%$. In the case of the normal, it is far more than that, and one can come up with examples in which with probability $100\%$ a random variable lies within $3$ standard deviation units of the mean.

Note also that some distributions (and even useful ones) have a mean but not a variance, or have neither mean nor variance. (The requisite integrals, or sums, diverge.)