3-Sigma and Standard Deviation

**taik** · Nov 12th 2010, 04:23 AM

Hi all,

From this link: 68-95-99.7 rule - Wikipedia, the free encyclopedia, there is the relationship between sigma and stnadard deviation whereby 1-Sigma, 2-Sigma, 3-Sigma is 68-95-99.7 rule.

Can I confirm that this only applies to normal distribution? If so, then can I say that the term 1-Sigma, 2-Sigma or 3-Sigma would not have any meaning in non-normal distribution?

Thank you.

**awkward** · Nov 12th 2010, 09:30 AM

Yes, the 68-95-99.7 values are specific to the normal distribution and do not apply to other distributions.

About all that can be said in general is Chebyshev's inequality:

Chebyshev's inequality - Wikipedia, the free encyclopedia

**taik** · Nov 12th 2010, 03:38 PM

Hi awkard,

Thanks for the reply.

Actually there is someone who told me that this sigma can also be applied to non-normal distribution since it simply stands for area under the curve. However, I feel that this concept is too misleading.

Thanks for clarifying this.

**matheagle** · Nov 14th 2010, 10:23 PM

chebyshev's is just a lower bound on the probabilities

**CaptainBlack** · Nov 15th 2010, 12:17 AM

Originally Posted by awkward

Yes, the 68-95-99.7 values are specific to the normal distribution and do not apply to other distributions.

About all that can be said in general is Chebyshev's inequality:

Chebyshev's inequality - Wikipedia, the free encyclopedia

There is a variant of Chebyschev's inequality for RV with moments higher than the second.

If RV $$$ X$ has an $$$ n$ -th central moment $\mu_n$ then:

$Pr(|X-\overline{X}|\ge \lambda (\mu_n)^{\frac{1}{n}})\le \dfrac{1}{\lambda^n}$

This can be proven as can the standard Chebyshev's inequality using Markov's inequality:

$Pr(|X|\ge a) \le \dfrac{E(X)}{a}$

CB

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