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Chebychev's Inequality

  1. Apr 9, 2005 #1
    I thought that Chebychev's inequality is what would be used to solve this question, but the exceeding instead of the or more throws me off. Here is the question from Jim Pitman's book (I am studying for my final...)

    Suppose the IQ scores of million individuals have a mean of 100 and a SD of 10.

    Without making any further assumptions about the distribution of the scores, find an upper bound on the number of scores exceeding 130.

    Thank You. No need for the answer, just I need to know whose inequality/theorem should I use. Though if someone can kindly explain Chebychev's theorem in detail, it would be awesome. He makes no sense.
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  3. Apr 9, 2005 #2


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    Last edited: Apr 11, 2005
  4. Apr 10, 2005 #3
    No, it's the probability that X is _not_ within k standard deviations of the mean that is at most 1/k^2. I'm not sure if you can apply it directly here though--I guess if you assume that the probability that X is less than 70 could be 0, you could apply it, but I'm not sure that can happen.
  5. Apr 11, 2005 #4


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    The Chebyshev Inequality takes either of 2 Forms:

    [tex] 1: \ \ \ \ P(\mu - k \sigma \ \, \leq \ \, X \ \, \leq \ \, \mu + k \sigma) \ \ \, \color{red}\mathbf{\geq}\color{black} \ \, \ \left( 1 \ - \ \frac{1}{k^2} \right ) [/tex]

    [tex] 2: \ \ \ \ P\left(X \ \leq \ \mu - k \sigma \ \ \mathsf{\underline{\ or\ }}} \ \ X \ \geq \ \mu + k \sigma \right) \ \ \, \color{red}\mathbf{\leq}\color{black} \ \ \, \left(\frac{1}{k^2} \right ) [/tex]

    Therefore, Form #2 would apply to this problem with (k=3) and would supply the required upper bound of {(1/9) = 11.1% of Population}. This would be conservative since Form #2 is a double-tail upper-bound and only a single-tail upper bound is requested.

    Last edited: Apr 11, 2005
  6. Apr 11, 2005 #5
    Yes, but can it happen that NO observations lie more than 3 standard deviations below the mean AND the maximum number of them lie more than 3 standard deviations above the mean? Using the upper bound from Chebyshev's formula assumes this is possible. Is it?
  7. Apr 11, 2005 #6


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    The Chebyshev Inequality states that the TOTAL NUMBER of Population elements Less Than (μ - 3σ) or More Than (μ + 3σ) will have upper-bound of (1/9) of Population. The inequality itself makes no further statement about the specific distribution in those tails.

    Last edited: Apr 11, 2005
  8. Apr 11, 2005 #7
    True, by Chebyshev's theorem the upper bound is 1/9. However the question is whether there are other considerations that also do not depend on the particular distribution that would make the upper bound less than 1/9.
  9. Apr 11, 2005 #8
    Somewhat irrelevantly, I think that Chebyshev's theorem applies only to distributions and only approximately to populations, but I could be wrong. At least in my book it's only proved for distributions.
  10. Apr 11, 2005 #9


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    In previous msgs, "Population" (as opposed to "Sample") is used synonymously with "Distribution". Thus, you're correct in that the Inequality applies rigorously only to Distributions and Populations, but not to Samples from those Populations. For Samples, the Inequality can, of course, provide some approximate info.

  11. Apr 11, 2005 #10
    Well, I guess you could define a discrete distribution from a population as taking on each value with probability k/n where k is the number of elements taking on that value, and Chebyshev's theorem would work for that distribution. But what I'm wondering is if it's ever possible to have the full maximum 1/9 on the upper end of the distrubution.
    Last edited: Apr 12, 2005
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