Chapter 8 Statistics

# 8.4 Range and Standard Deviation

Learning Objectives

By the end of this section, you will be able to:

• Calculate the range of a dataset
• Calculate the standard deviation of a dataset

Measures of centrality like the mean can give us only part of the picture that a dataset paints. For example, let’s say you’ve just gotten the results of a standardized test back, and your score was 138. The mean score on the test is 120. So your score is above average! But how good is it really? If all the scores were between 100 and 140, then you know your score must be among the best. But if the scores ranged from 0 to 200, then maybe 140 is good, but not great (though still above average). Knowing information about how the data are spread out can help us put a particular data value in better context. In this section, we’ll look at two numbers that help us describe the spread in the data: the range and the standard deviation. These numbers are called measures of dispersion.

# The Range

Our first measure of dispersion is the range, or the difference between the maximum and minimum values in the set. It’s the measure we used in the standardized test example above.

Let’s look at a couple of examples.

Example 1

You survey some of your friends to find out how many hours they work each week. Their responses are: 5, 20, 8, 10, 35, 12. What is the range?

The maximum value in the set is 35 and the minimum is 5, so the range is $35-5=30$.

Exercise 1

On your morning commute, you decide to record how long you have to wait each time you get caught at a red light. Here are the times in seconds: 12, 58, 35, 79, 21. What is the range?

Solution

67

The range is very easy to compute, but it depends only on two of the data values in the entire set. If there happens to be just one unusually high or low data value, then the range might give a distorted measure of dispersion. Our next measure takes every single data value into account, making it more reliable.

# The Standard Deviation

The standard deviation is a measure of dispersion that can be interpreted as approximately the average distance of every data value from the mean. (This distance from the mean is the “deviation” in “standard deviation.”)

### FORMULA

The standard deviation is computed as follows:

$s= \sqrt{\frac{ \sum (x- \bar{x})^{2}}{n-1}}$

Here, $x$ represents each data value, $\bar{x}$ is the mean of the data values, $n$ is the number of data values, and the capital sigma ($\sum$) indicates that we take a sum.

To compute the standard deviation using the formula, we follow the steps below:

1. Compute the mean of all the data values.
2. Subtract the mean from each data value.
3. Square those differences.
4. Add up the results in step 3.
5. Divide the result in step 4 by $n-1$.
6. Take the square root of the result in step 5.

Let’s see that process in action.

Example 2

You surveyed some of your friends to find out how many hours they work each week. Their responses were 5, 20, 8, 10, 35, 12. What is the standard deviation?

Let’s follow the six steps mentioned previously to compute the standard deviation.

Step 1: Find the mean: $\bar{x} = \frac{5+20+8+10+35+12}{6}=15$.

Step 2: Subtract the mean from each data value. To help keep track, let’s do this in a table. In the first row, we’ll list each of our data values (and we’ll label the row $x$); in the second, we’ll subtract $\bar{x}=15$ from each data value.

 $x$ 5 20 8 10 35 12 $x- \bar{x}$ -10 5 -7 -5 20 -3

Step 3: Square the differences. Let’s add a row to our table for those values:

 $x$ 5 20 8 10 35 12 $x- \bar{x}$ −10 5 –7 –5 20 –3 $(x- \bar{x} )^{2}$ 100 25 49 25 400 9

Step 4: Add up those squares: $100+25+49+58+400+9=608$

Step 5: Divide the sum by $n-1$. Since we have 6 data values, that gives us $\frac{608}{6-1} = 121.6$.

Step 6: Take the square root of the result: $\sqrt{121.6} \approx{11.027}$

Thus, the standard deviation is $s \approx{11.027}$.

Exercise 2

On your morning commute, you decide to record how long you have to wait each time you get caught at a red light. Here are the times in seconds: 12, 58, 35, 79, 21.

What is the standard deviation?

Solution

$s= \sqrt{752.5} \approx{27.432}$

definition