What is Chebyshev's inequality?

Chebyshev’s inequality says that at least ${\displaystyle 1-\frac{1}{K^2}}$ of data from a sample must fall within K standard deviations from the mean, where K is any positive real number greater than one.
Explanation:
Let play with a few value of K:
1. K=2 we have ${\displaystyle 1-\frac{1}{K^2}=1-\frac{1}{4}=\frac{3}{4}=75\mathrm{\%}}$. So Chebyshev’s would tell us that 75% of the data values of any distribution must be within two standard deviations of the mean.
2. K=3 we have ${\displaystyle 1-\frac{1}{K^2}=1-\frac{1}{9}=\frac{8}{9}=89\mathrm{\%}}$. This time we have 89% of the data values within three standard deviations of the mean.
3. K=4 we have ${\displaystyle 1-\frac{1}{K^2}=1-\frac{1}{16}=\frac{15}{16}=93.75\mathrm{\%}}$. Now we have 93.75% of the data within four standard deviations of the mean.
This is consistent to saying that in Normal distribution 68% of the data is one standard deviation from the mean, 95% is two standard deviations from the mean, and approximately 99% is within three standard deviations from the mean. The difference is Chebyshev's theorem extends this principle to any distribution.

Chebyshev’s inequality is a probability theory that guarantees that within a specified range or distance from the mean, for a large range of probability distributions, no more than a specific fraction of values will be present. In other words, only a definite fraction of values will be found within a specific distance from the mean of a distribution.
The formula for the fraction for which no more than a certain number of values can exceed is ${\displaystyle \frac{1}{K^2}}$; in other words, ${\displaystyle \frac{1}{K^2}}$ of a distribution’s values can be more than or equal to K standard deviations away from the mean of the distribution. Further, it also holds that ${\displaystyle 1-\left(\frac{1}{K^2}\right)}$ of a distribution’s values must be within, but not including, K standard deviations away from the mean of the distribution.