Chebyshev’s inequality says that at least 1 -1/*K*^{2} of data from a sample must fall within *K* standard deviations from the mean, where *K* is any positive real number greater than one. This means that we don’t need to know the shape of the distribution of our data. With only the mean and standard deviation, we can determine the amount of data a certain number of standard deviations from the mean.

The following are some problems to practice using the inequality.

### Example #1

A class of second graders has a mean height of five feet with a standard deviation of one inch. At least what percent of the class must be between 4’10” and 5’2”?

### Solution

The heights that are given in the range above are within two standard deviations from the mean height of five feet. Chebyshev’s inequality says that at least 1 – 1/2^{2} = 3/4 = 75% of the class is in the given height range.

### Example #2

Computers from a particular company are found to last on average for three years without any hardware malfunction, with a standard deviation of two months. At least what percent of the computers last between 31 months and 41 months?

### Solution

The mean lifetime of three years corresponds to 36 months. The times of 31 months to 41 months are each 5/2 = 2.5 standard deviations from the mean. By Chebyshev’s inequality, at least 1 – 1/(2.5)6^{2} = 84% of the computers last from 31 months to 41 months.

### Example #3

Bacteria in a culture live for an average time of three hours with a standard deviation of 10 minutes. At least what fraction of the bacteria live between two and four hours?

### Solution

Two and four hours are each one hour away from the mean. One hour corresponds to six standard deviations. So at least 1 – 1/6^{2} = 35/36 =97% of the bacteria live between two and four hours.

### Example #4

What is the smallest number of standard deviations from the mean that we must go if we want to ensure that we have at least 50% of the data of a distribution?

### Solution

Here we use Chebyshev’s inequality and work backward. We want 50% = 0.50 = 1/2 = 1 – 1/*K*^{2}. The goal is to use algebra to solve for *K*.

We see that 1/2 = 1/*K*^{2}. Cross multiply and see that 2 =*K*^{2}. We take the square root of both sides, and since *K* is a number of standard deviations, we ignore the negative solution to the equation. This shows that *K* is equal to the square root of two. So at least 50% of the data is within approximately 1.4 standard deviations from the mean.

### Example #5

Bus route #25 takes a mean time of 50 minutes with a standard deviation of 2 minutes. A promotional poster for this bus system states that “95% of the time bus route #25 lasts from ____ to _____ minutes.” What numbers would you fill in the blanks with?

### Solution

This question is similar to the last one in that we need to solve for *K*, the number of standard deviations from the mean. Start by setting 95% = 0.95 = 1 – 1/*K*^{2}. This shows that 1 - 0.95 = 1/*K*^{2}. Simplify to see that 1/0.05 = 20 = *K*^{2}. So *K* = 4.47.

Now express this in the terms above. At least 95% of all rides are 4.47 standard deviations from the mean time of 50 minutes. Multiply 4.47 by the standard deviation of 2 to end up with nine minutes. So 95% of the time, bus route #25 takes between 41 and 59 minutes.