Chebyshev's inequality for Pareto distribution (3 sigma rule) According to the Chebyshev's inequality, if we take any distribution, we get >88.8889% of data in +-3  sigma interval.
For a normal distribution it is 99.97%.
How to calculate the interval for a Pareto or any other distribution?
 A: Let $X$ be a random variable with mean $\mu$ and standard deviation $\sigma$. By simple probability rules, we have
$$P\left(\mu-3\sigma \leq X \leq \mu+3\sigma\right) = P(X \leq \mu + 3\sigma) - P(X \leq \mu - 3\sigma).$$
The quantity $P(X \leq x) = \int_{-\infty}^xf(t)dt$ is called the cumulative distribution function (CDF), and is often written as $F(x)$. If the mean, standard deviation and CDF all exist in closed form, then this calculation is pretty straightforward.
For instance, if $X \sim \text{Pareto}(x_m=1, \alpha=3)$, then we have $\mu = 1.5$, $\sigma = \sqrt{0.75}$ and
$$F(x) = \begin{cases}
1 - x^{-3}, & x > 1 \\
0, & x \leq 1.
\end{cases}$$
So the density contained within three standard deviations of the mean can be computed as
\begin{align*}
\approx P(-1.1 \leq X \leq 4.1) &= P(X \leq 4.1) - P(X \leq -1.1) \\
&= 1 - 4.1^{-3} - 0 = 0.985
\end{align*}
The normal distribution yields the same answer for all parameter values, but the Pareto (and most distributions) depends on the values of the parameters. For instance, if $\alpha = 4$, we get $0.9824$.
It is also worth considering the case where $\alpha = 1$. In this case, the mean and variance of the Pareto distribution is infinite, and the question no longer makes sense.
