# 2.9 Chevyshev’s theorem

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This module explains Chevyshev’s Theorem as it pertains to the spread of non-normal data. Given an data set, Chevyshev’s Theorem gives a worst case scenario for the percentage of data within a given number of standard deviations from the mean.

Chevyshev’s Theorem

The proportion (or fraction) of any data set lying within K standard deviations of the mean is always at least 1 - $\frac{1}{{K}^{2}}$ , where K is any positive number greater then 1. Why is this?

For K = 2, the proportion is 1 - $\frac{1}{{2}^{2}}$ = 1 - $\frac{1}{4}$ = $\frac{3}{4}$ , hence $\frac{3}{4}$ ths or 75% of the data falls within 2 standard deviations of the mean.

For K = 3, the proportion is 1 - $\frac{1}{{3}^{2}}$ = 1 - $\frac{1}{9}$ = $\frac{8}{9}$ , hence $\frac{8}{9}$ ths or approximately 89% of the data falls within 3 standard deviations of the mean.

Using the data from the pre-calculus class exams and K = 2, this means that at least 75% of the scores fall between 73.5 - 2(17.9) and 73.5 + 2(17.9), or between 37.7 and 109.3.

In actual fact all but one data value falls in this range, however Chevyshev's Theorem gives the worst case scenario.

Using the pre-calculus class exams, what would the range of values be for at least 89% of the data according to Chevyshev’s Theorem?

73.5 – 3(17.9) to 73.5 + 3(17.9) or 19.8 to 127.2

Using Chevyshev’s Theorem, what percent of the data would fall between 46.65 and 100.35?

Step 1: Find how far the maximum (or minimum) value is from the mean. 100.35 – 73.5 = 26.85

Step 2: How many standard deviations does 26.85 represent? 26.85/17.9 = 1.5. Hence K = 1.5

Step 3: If K = 1.5, then the percentage is $1-\frac{1}{{1.5}^{2}}\approx 0.55556$ , or approximately 56%

Given a data set with a mean of 56.3 and a standard deviation of 8.2, use this information and Chevyshev’s Theorem to answer the following questions.

What percent of the data lies within 2.2 standard deviation from the mean?

$1-\frac{1}{{2.2}^{2}}\approx 0.793$ or 79%

For the given sent of data, about 79% of the data falls between which two values?

56.3 – 2.2(8.2) = 38.26 and 56.3 + 2.2(8.2) = 74.34

What percent of the data lies between the values 45.64 and 66.96?

Step 1: Find how far the maximum value is from the mean: 66.96 – 56.3 = 10.66

Step 2: How many standard deviations does 10.66 represent? 10.66/8.2 = 1.3. Hence K = 1.3

Step 3: If K = 1.3, then the percentage is 1 - $\frac{1}{1\text{.}{3}^{2}}$ .408 or approximately 41%

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