# 3.12: Measures of Variability

Skills to Develop

- Determine the relative variability of two distributions
- Compute the range
- Compute the inter-quartile range
- Compute the variance in the population
- Estimate the variance from a sample
- Compute the standard deviation from the variance

### What is Variability?

Variability refers to how "spread out" a group of scores is. To see what we mean by spread out, consider graphs in Figure \(\PageIndex{1}\). These graphs represent the scores on two quizzes. The mean score for each quiz is \(7.0\). Despite the equality of means, you can see that the distributions are quite different. Specifically, the scores on \(\text{Quiz 1}\) are more densely packed and those on \(\text{Quiz 2}\) are more spread out. The differences among students were much greater on \(\text{Quiz 2}\) than on \(\text{Quiz 1}\).

\(\text{Quiz 1}\)

\(\text{Quiz 1}\)

* Figure *\(\PageIndex{1}\)

*: Bar charts of two quizzes*

The terms variability, spread, and dispersion are synonyms, and refer to how spread out a distribution is. Just as in the section on central tendency where we discussed measures of the center of a distribution of scores, in this chapter we will discuss measures of the variability of a distribution. There are four frequently used measures of variability: the range, interquartile range, variance, and standard deviation. In the next few paragraphs, we will look at each of these four measures of variability in more detail.

### Range

The range is the simplest measure of variability to calculate, and one you have probably encountered many times in your life. The range is simply the highest score minus the lowest score. Let’s take a few examples. What is the range of the following group of numbers: \({10, 2, 5, 6, 7, 3, 4}\)? Well, the highest number is \(10\), and the lowest number is \(2\), so \(10 - 2 = 8\). The range is \(8\). Let’s take another example. Here’s a dataset with \(10\) numbers: \({99, 45, 23, 67, 45, 91, 82, 78, 62, 51}\). What is the range? The highest number is \(99\) and the lowest number is \(23\), so \(99 - 23\) equals \(76\); the range is \(76\). Now consider the two quizzes shown in Figure \(\PageIndex{1}\). On \(\text{Quiz 1}\), the lowest score is \(5\) and the highest score is \(9\). Therefore, the range is \(4\). The range on \(\text{Quiz 2}\) was larger: the lowest score was \(4\) and the highest score was \(10\). Therefore the range is \(6\).

### Interquartile Range

The interquartile range (IQR) is the range of the middle \(50\%\) of the scores in a distribution. It is computed as follows:

\[IQR = 75^{th} percentile - 25^{th} percentile\]

For \(\text{Quiz 1}\), the \(75^{th}\) percentile is \(8\) and the \(25^{th}\) percentile is \(6\). The interquartile range is therefore \(2\). For \(\text{Quiz 2}\), which has greater spread, the \(75^{th}\) percentile is \(9\), the \(25^{th}\) percentile is \(5\), and the interquartile range is \(4\). Recall that in the discussion of box plots, the \(75^{th}\) percentile was called the upper hinge and the \(25^{th}\) percentile was called the lower hinge. Using this terminology, the interquartile range is referred to as the \(H\)-spread.

A related measure of variability is called the semi-interquartile range. The semi-interquartile range is defined simply as the interquartile range divided by \(2\). If a distribution is symmetric, the median plus or minus the semi-interquartile range contains half the scores in the distribution.

### Variance

Variability can also be defined in terms of how close the scores in the distribution are to the middle of the distribution. Using the mean as the measure of the middle of the distribution, the variance is defined as the average squared difference of the scores from the mean. The data from \(\text{Quiz 1}\) are shown in Table \(\PageIndex{1}\). The mean score is \(7.0\). Therefore, the column "Deviation from Mean" contains the score minus \(7\). The column "Squared Deviation" is simply the previous column squared.

**Table \(\PageIndex{1}\):** Calculation of Variance for \(\text{Quiz 1}\) scores

Scores | Deviation from Mean | Squared Deviation |
---|---|---|

9 | 2 | 4 |

9 | 2 | 4 |

9 | 2 | 4 |

8 | 1 | 1 |

8 | 1 | 1 |

8 | 1 | 1 |

8 | 1 | 1 |

7 | 0 | 0 |

7 | 0 | 0 |

7 | 0 | 0 |

7 | 0 | 0 |

7 | 0 | 0 |

6 | -1 | 1 |

6 | -1 | 1 |

6 | -1 | 1 |

6 | -1 | 1 |

6 | -1 | 1 |

6 | -1 | 1 |

5 | -2 | 4 |

5 | -2 | 4 |

Means | ||

7 | 0 | 1.5 |

One thing that is important to notice is that the mean deviation from the mean is \(0\). This will always be the case. The mean of the squared deviations is \(1.5\). Therefore, the variance is \(1.5\). Analogous calculations with \(\text{Quiz 2}\) show that its variance is \(6.7\). The formula for the variance is:

\[s^2=\frac{\sum (X-\mu )^2}{N}\]

where \(\sigma ^2\) is the variance, \(\mu\) is the mean, and \(N\) is the number of numbers. For \(\text{Quiz 1}\), \(\mu = 7\) and \(N = 20\).

If the variance in a sample is used to estimate the variance in a population, then the previous formula underestimates the variance and the following formula should be used:

\[s^2=\frac{\sum (X-M)^2}{N-1}\]

where \(s^2\) is the estimate of the variance and \(M\) is the sample mean. Note that \(M\) is the mean of a sample taken from a population with a mean of \(\mu\). Since, in practice, the variance is usually computed in a sample, this formula is most often used. The simulation "estimating variance" illustrates the bias in the formula with \(N\) in the denominator.

Let's take a concrete example. Assume the scores \({1, 2, 4,\: and\; 5 }\) were sampled from a larger population. To estimate the variance in the population you would compute \(s^2\) as follows:

\[M = \dfrac {1 + 2 + 4 + 5}{4} = \dfrac {12}{4} = 3\]

\[\begin{align*} s^2 &= \dfrac{[(1-3)^2 + (2-3)^2 + (4-3)^2 + (5-3)^2]}{(4-1)}\\ &= \dfrac{(4 + 1 + 1 + 4)}{3}\\ &= \dfrac{10}{3}\\ &= 3.333 \end{align*}\]

There are alternate formulas that can be easier to use if you are doing your calculations with a hand calculator. You should note that these formulas are subject to rounding error if your values are very large and/or you have an extremely large number of observations.

\[\sigma ^2=\cfrac{\sum X^2-\cfrac{(\sum X)^2}{N}}{N}\]

and

\[s^2=\cfrac{\sum X^2-\cfrac{(\sum X)^2}{N}}{N-1}\]

For this example,

\[\sum X^2=1^2+2^2+4^2+5^2=46\]

\[\dfrac {(\sum X)^2}{N}=\dfrac {(1+2+4+5)^2}{4}=\dfrac {144}{4}=36\]

\[\sigma ^2=\dfrac {(46-36)}{4}=2.5\]

\[s^2=\dfrac {(46-36)}{3}=3.333\; \; \text{as with the other formula}\]

### Standard Deviation

The standard deviation is simply the square root of the variance. This makes the standard deviations of the two quiz distributions \(1.225\) and \(2.588\). The standard deviation is an especially useful measure of variability when the distribution is normal or approximately normal (see Chapter on Normal Distributions) because the proportion of the distribution within a given number of standard deviations from the mean can be calculated. For example, \(68\%\) of the distribution is within one standard deviation of the mean and approximately \(95\%\) of the distribution is within two standard deviations of the mean. Therefore, if you had a normal distribution with a mean of \(50\) and a standard deviation of \(10\), then \(68\%\) of the distribution would be between \(50 - 10 = 40\) and \(50 +10 =60\). Similarly, about \(95\%\) of the distribution would be between \(50 - 2 \times 10 = 30\) and \(50 + 2 \times 10 = 70\). The symbol for the population standard deviation is \(\sigma\); the symbol for an estimate computed in a sample is \(s\). Figure \(\PageIndex{2}\) shows two normal distributions. The red distribution has a mean of \(40\) and a standard deviation of \(5\); the blue distribution has a mean of \(60\) and a standard deviation of \(10\). For the red distribution, \(68\%\) of the distribution is between \(35\) and \(45\); for the blue distribution, \(68\%\) is between \(50\) and \(70\).

* Figure *\(\PageIndex{2}\)

**:**Normal distributions with standard deviations of 5 and 10### Contributor

Online Statistics Education: A Multimedia Course of Study (http://onlinestatbook.com/). Project Leader: David M. Lane, Rice University.