Results 1 to 5 of 5

Math Help - 3-Sigma and Standard Deviation

  1. #1
    Newbie
    Joined
    Nov 2010
    Posts
    2

    3-Sigma and Standard Deviation

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Mar 2008
    Posts
    934
    Thanks
    33
    Awards
    1
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2010
    Posts
    2
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    chebyshev's is just a lower bound on the probabilities
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by awkward View Post
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: April 1st 2011, 03:22 PM
  2. Standard deviation
    Posted in the Statistics Forum
    Replies: 4
    Last Post: May 30th 2009, 05:40 AM
  3. Replies: 5
    Last Post: May 18th 2009, 04:14 AM
  4. standard deviation and mean deviation
    Posted in the Statistics Forum
    Replies: 1
    Last Post: March 16th 2008, 05:09 AM
  5. Standard deviation
    Posted in the Statistics Forum
    Replies: 9
    Last Post: September 13th 2007, 06:24 AM

Search Tags


/mathhelpforum @mathhelpforum