Statistics - Chebyshev's Theorem



The fraction of any set of numbers lying within k standard deviations of those numbers of the mean of those numbers is at least

${1-\frac{1}{k^2}}$

Where −

  • ${k = \frac{the\ within\ number}{the\ standard\ deviation}}$

and ${k}$ must be greater than 1

Example

Problem Statement

Use Chebyshev's theorem to find what percent of the values will fall between 123 and 179 for a data set with mean of 151 and standard deviation of 14.

Solution

  • We subtract 151-123 and get 28, which tells us that 123 is 28 units below the mean.

  • We subtract 179-151 and also get 28, which tells us that 151 is 28 units above the mean.

  • Those two together tell us that the values between 123 and 179 are all within 28 units of the mean. Therefore the "within number" is 28.

  • So we find the number of standard deviations, k, which the "within number", 28, amounts to by dividing it by the standard deviation −

${k = \frac{the\ within\ number}{the\ standard\ deviation} = \frac{28}{14} = 2}$

So now we know that the values between 123 and 179 are all within 28 units of the mean, which is the same as within k=2 standard deviations of the mean. Now, since k > 1 we can use Chebyshev's formula to find the fraction of the data that are within k=2 standard deviations of the mean. Substituting k=2 we have −

${1-\frac{1}{k^2} = 1-\frac{1}{2^2} = 1-\frac{1}{4} = \frac{3}{4}}$

So ${\frac{3}{4}}$ of the data lie between 123 and 179. And since ${\frac{3}{4} = 75}$% that implies that 75% of the data values are between 123 and 179.

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