What is Chebyshev’s theorem?

Chebyshev’s theorem

Chebyshev’s theorem actually refers to several theorem, all proven by the Russian mathematician Pafnuty Chebyshev. They include: Chebyshev’s inequality, Bertrand’s postulate, Chebyshev’s sum inequality and Chebyshev’s equioscillation theorem.

Chebyshev’s inequality, also spelled Tchebysheff’s inequality, is the theorem most often used in statistics. It states that no more than 1/k² of the distribution’s values are more than “k” standard deviations away from the mean. With a normal distribution, standard deviations tell you how much of that distribution’s data are within k standard deviations from the mean. If you have a distribution isn’t normally distributed, you can apply Chebyshev’s theorem to help you find out what percentage of the data is clustered around the mean. You can read how to calculate Chebyshev’s inequality in this article. Although Chebyshev’s theorem usually means Chebyschev’s inequality in elementary statistics, three other theorems have an important place in mathematics.

Bertrand’s postulate
Bertand’s Postulate is used in number theory — it has very few applications to statistics and you probably won’t encounter it in an elementary statistics course. According to the University of Tennessee, the postulate states that if n is an integer greater than 3, then there is at least one prime between n and 2n-2. Itcan also be stated as “If n is a positive integer, then there is a prime p with n < p < 2n."

Chebyshev’s sum inequality
Chebyshev’s sum inequality is used in calculus. It states that:
If

Then

You won’t encounter this theorem in an elementary statistics course. However, if you take advanced statistics courses that include calculus, you might use this particular theorem.

Chebyshev’s equioscillation theorem
The Chebyshev equioscillation theorem demonstrates the pattern between a continuous function on a closed interval. You won’t come across this theory in statistics courses; It is used in numerical analysis courses at the graduate level and involves a somewhat complicated proof.