Thorem stating how much of a population is within n standard deviations from the mean












2














In a statistics class I took in college, I remember learning about a theorem stating that 50% of the population must be within 1σ from μ, 75% within 1σ, and so on, regardless of the distribution. The distribution can be tighter, but this much is a guarantee.



People don't seem to be familiar with this, I'm not sure I remembered the numbers right, and I'm not sure this even exists. Is this a thing?










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    2














    In a statistics class I took in college, I remember learning about a theorem stating that 50% of the population must be within 1σ from μ, 75% within 1σ, and so on, regardless of the distribution. The distribution can be tighter, but this much is a guarantee.



    People don't seem to be familiar with this, I'm not sure I remembered the numbers right, and I'm not sure this even exists. Is this a thing?










    share|cite|improve this question







    New contributor




    David Ehrmann is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.























      2












      2








      2







      In a statistics class I took in college, I remember learning about a theorem stating that 50% of the population must be within 1σ from μ, 75% within 1σ, and so on, regardless of the distribution. The distribution can be tighter, but this much is a guarantee.



      People don't seem to be familiar with this, I'm not sure I remembered the numbers right, and I'm not sure this even exists. Is this a thing?










      share|cite|improve this question







      New contributor




      David Ehrmann is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      In a statistics class I took in college, I remember learning about a theorem stating that 50% of the population must be within 1σ from μ, 75% within 1σ, and so on, regardless of the distribution. The distribution can be tighter, but this much is a guarantee.



      People don't seem to be familiar with this, I'm not sure I remembered the numbers right, and I'm not sure this even exists. Is this a thing?







      probability-distributions standard-deviation






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      asked 1 hour ago









      David Ehrmann

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          Yes, some of it is true, and comes from Tschebychev inequality. It says that
          $$P(|X-mu|le nsigma)ge 1-frac1{n^2}.$$
          This gives the mentioned $0.75$ for $n=2$, but the $0.5$ actually appears if you take $n=sqrt2=1.41ldots$. It says nothing for $n=1$.



          This is valid for any distribution, provided it has a well defined finite mean and a finite variance/s.d. For other distributions exact values can be determined, which are necessarily equal or greater (quite bigger, for most usual distributions) than those given by this theorem.






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            You're thinking of Chebyshev's inequality.



            For any $k$, in any distribution with a mean and standard deviation, the probability of being more than $k$ standard deviations away from the mean is no more than $frac1{k^2}$.



            Most distributions, of course, are much tighter than this; the theorem is the worst case.






            share|cite|improve this answer





















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              2 Answers
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              2 Answers
              2






              active

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              active

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              active

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              3














              Yes, some of it is true, and comes from Tschebychev inequality. It says that
              $$P(|X-mu|le nsigma)ge 1-frac1{n^2}.$$
              This gives the mentioned $0.75$ for $n=2$, but the $0.5$ actually appears if you take $n=sqrt2=1.41ldots$. It says nothing for $n=1$.



              This is valid for any distribution, provided it has a well defined finite mean and a finite variance/s.d. For other distributions exact values can be determined, which are necessarily equal or greater (quite bigger, for most usual distributions) than those given by this theorem.






              share|cite|improve this answer


























                3














                Yes, some of it is true, and comes from Tschebychev inequality. It says that
                $$P(|X-mu|le nsigma)ge 1-frac1{n^2}.$$
                This gives the mentioned $0.75$ for $n=2$, but the $0.5$ actually appears if you take $n=sqrt2=1.41ldots$. It says nothing for $n=1$.



                This is valid for any distribution, provided it has a well defined finite mean and a finite variance/s.d. For other distributions exact values can be determined, which are necessarily equal or greater (quite bigger, for most usual distributions) than those given by this theorem.






                share|cite|improve this answer
























                  3












                  3








                  3






                  Yes, some of it is true, and comes from Tschebychev inequality. It says that
                  $$P(|X-mu|le nsigma)ge 1-frac1{n^2}.$$
                  This gives the mentioned $0.75$ for $n=2$, but the $0.5$ actually appears if you take $n=sqrt2=1.41ldots$. It says nothing for $n=1$.



                  This is valid for any distribution, provided it has a well defined finite mean and a finite variance/s.d. For other distributions exact values can be determined, which are necessarily equal or greater (quite bigger, for most usual distributions) than those given by this theorem.






                  share|cite|improve this answer












                  Yes, some of it is true, and comes from Tschebychev inequality. It says that
                  $$P(|X-mu|le nsigma)ge 1-frac1{n^2}.$$
                  This gives the mentioned $0.75$ for $n=2$, but the $0.5$ actually appears if you take $n=sqrt2=1.41ldots$. It says nothing for $n=1$.



                  This is valid for any distribution, provided it has a well defined finite mean and a finite variance/s.d. For other distributions exact values can be determined, which are necessarily equal or greater (quite bigger, for most usual distributions) than those given by this theorem.







                  share|cite|improve this answer












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                  share|cite|improve this answer










                  answered 1 hour ago









                  Alejandro Nasif Salum

                  4,294118




                  4,294118























                      4














                      You're thinking of Chebyshev's inequality.



                      For any $k$, in any distribution with a mean and standard deviation, the probability of being more than $k$ standard deviations away from the mean is no more than $frac1{k^2}$.



                      Most distributions, of course, are much tighter than this; the theorem is the worst case.






                      share|cite|improve this answer


























                        4














                        You're thinking of Chebyshev's inequality.



                        For any $k$, in any distribution with a mean and standard deviation, the probability of being more than $k$ standard deviations away from the mean is no more than $frac1{k^2}$.



                        Most distributions, of course, are much tighter than this; the theorem is the worst case.






                        share|cite|improve this answer
























                          4












                          4








                          4






                          You're thinking of Chebyshev's inequality.



                          For any $k$, in any distribution with a mean and standard deviation, the probability of being more than $k$ standard deviations away from the mean is no more than $frac1{k^2}$.



                          Most distributions, of course, are much tighter than this; the theorem is the worst case.






                          share|cite|improve this answer












                          You're thinking of Chebyshev's inequality.



                          For any $k$, in any distribution with a mean and standard deviation, the probability of being more than $k$ standard deviations away from the mean is no more than $frac1{k^2}$.



                          Most distributions, of course, are much tighter than this; the theorem is the worst case.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 1 hour ago









                          jmerry

                          1,26916




                          1,26916






















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