In English: "The probability that the outcome of an experiment with the random variable X will fall more than k standard deviations beyond the mean of X , \mu , is less than \frac1{k^2} ."

Or: "The proportion of the total area under the probability distribution function of X outside of k standard deviations from the mean \mu is at most \frac1{k^2} ."


Let S be the sample space for a random variable, X , and let f_X(x) stand for the pdf of X . Let R_1,R_2,R_3 partition S , such that for every sample point x\in S

R_2&|x-\mu|\le k\sigma\\





since the term that evaluates to the variance in R_2 has been subtracted on the right-hand side.

For any sample point x\in R_1


Notice that the direction of the inequality changes since squaring causes the right-hand expression to become positive.

And for any sample point x\in R_3


So, for any sample point x\in R_1 or x\in R_3 , it can be said that (x-\mu)^2>k^2\sigma^2 , and so


Dividing each side of the inequality by (k\sigma)^2 results in


Or, in other terms


And thus the original claim is proven.


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