Numerical Measures of Central Tendency and Variability

The arithmetic mean is the most common measure of central tendency. It is simply the sum of the numbers divided by the number of numbers. The symbol "$\mu$" is used for the mean of a population. The symbol "$\mathrm{M}$" is used for the mean of a sample. The formula for $\mu$ is shown below:

where $\Sigma \mathrm{X}$ is the sum of all the numbers in the population and $N$ is the number of numbers in the population.

The formula for $M$ is essentially identical:

where $\Sigma \mathrm{X}$ is the sum of all the numbers in the sample and $\mathrm{N}$ is the number of numbers in the sample.

As an example, the mean of the numbers $1,2,3,6,8$ is $20 / 5=4$ regardless of whether the numbers constitute the entire population or just a sample from the population.

Table 1 shows the number of touchdown (TD) passes thrown by each of the 31 teams in the National Football League in the 2000 season. The mean number of touchdown passes thrown is 20.4516 as shown below.

Table 1. Number of touchdown passes.

Although the arithmetic mean is not the only "mean" (there is also a geometric mean), it is by far the most commonly used. Therefore, if the term "mean" is used without specifying whether it is the arithmetic mean, the geometric mean, or some other mean, it is assumed to refer to the arithmetic mean.

The median is also a frequently used measure of central tendency. The median is the midpoint of a distribution: the same number of scores is above the median as below it. For the data in Table 1, there are $31$ scores. The 16th highest score (which equals $20$) is the median because there are $15$ scores below the $\mathrm{16th}$ score and $15$ scores above the $\mathrm{16th}$ score. The median can also be thought of as the $\mathrm{50th}$ percentile.

Computation of the Median

When there is an odd number of numbers, the median is simply the middle number. For example, the median of $2$, $4$, and $7$ is $4$. When there is an even number of numbers, the median is the mean of the two middle numbers. Thus, the median of the numbers $2,4,7,12$ is $(4+7) / 2=5.5$. When there are numbers with the same values, then the formula for the third definition of the $\mathrm{50th}$ percentile should be used.

1. $(2+4+6+8) / 4=5$
   
   
2. 3
   Because there are 5 numbers, the median is the middle number when they are ranked from lowest to highest.
   
   
3. What is the mode of -2, 4, 0, 3, 0, 2, 4, 4, and 8?
   
   
4. Mean = 85, Median = 84, Mode = 75, so the mean of his scores is the highest.
   
   
5. If the teacher is going to use the median as the final grade, she should only argue the middle score (87). Changing the other scores by 2 points would not affect the median.
   

Learning Objectives

1. State when the mean and median are the same
2. State whether it is the mean or median that minimizes the mean absolute deviation
3. State whether it is the mean or median that minimizes the mean squared deviation
4. State whether it is the mean or median that is the balance point on a balance scale

In the section "What is central tendency," we saw that the center of a distribution could be defined three ways: (1) the point on which a distribution would balance, (2) the value whose average absolute deviation from all the other values is minimized, and (3) the value whose average squared difference from all the other values is minimized. From the simulation in this chapter, you discovered (we hope) that the mean is the point on which a distribution would balance, the median is the value that minimizes the sum of absolute deviations, and the mean is the value that minimizes the sum of the squared deviations.

Table 1 shows the absolute and squared deviations of the numbers $2, \, 3,  \, 4,  \, 9,$ and $16$ from their median of $\mathrm{4}$ and their mean of 6.8. You can see that the sum of absolute deviations from the median ($\mathrm{20}$) is smaller than the sum of absolute deviations from the mean ($\mathrm{22.8}$). On the other hand, the sum of squared deviations from the median ($\mathrm{174}$) is larger than the sum of squared deviations from the mean ($\mathrm{134.8}$).

Table 1. Absolute and squared deviations from the median of $\mathrm{4}$ and the mean of $\mathrm{6.8}$.

Figure 1 shows that the distribution balances at the mean of $\mathrm{6.8}$ and not at the median of $\mathrm{4}$. The relative advantages and disadvantages of the mean and median are discussed in the section "Comparing Measures" later in this chapter.

When a distribution is symmetric, then the mean and the median are the same. Consider the following distribution: $\mathrm{1,  \, 3,  \, 4,  \, 5, \,  6, \,  7, \, 9}$. The mean and median are both $\mathrm{5}$. The mean, median, and mode are identical in the bell-shaped normal distribution.

1. This is a definition of the median.
   
   
2. This is a definition of the mean
   
   
3. This is a definition of the mean.
   
   
4. The mean and the median are only the same when a distribution is symmetric. The mean and median are different when the distribution is skewed.
5. 15: The median minimizes the absolute deviations. To find the median, order the numbers from smallest to largest, and then find the middle number.
   
   
6. The mean minimizes the squared deviations. To find the mean, find the sum of the values (140) and divide by number of values in your data set (5). $140/5 = 28$
   
   
7. The mean (in this case, 6.5) is the point at which a distribution balances.

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

Quiz 1

Quiz 2

Figure 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.

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: $\mathrm{10, \, 2,  \, 5,  \, 6,  \, 7,  \, 3,  \, 4}$? Well, the highest number is $\mathrm{10}$, and the lowest number is $\mathrm{2}$, so $\mathrm{10 - 2 = 8}$. The range is $\mathrm{8}$. Let's take another example. Here's a dataset with $\mathrm{10}$ numbers: $\mathrm{99,  \, 45,  \, 23,  \, 67,  \, 45,  \, 91,  \, 82,  \, 78,  \, 62,  \, 51}$. What is the range? The highest number is $\mathrm{99}$ and the lowest number is $\mathrm{23}$, so $\mathrm{99 - 23}$ equals $\mathrm{76}$; the range is $\mathrm{76}$. Now consider the two quizzes shown in Figure 1. On Quiz 1, the lowest score is $\mathrm{5}$ and the highest score is $\mathrm{9}$. Therefore, the range is $\mathrm{4}$. The range on Quiz 2 was larger: the lowest score was $\mathrm{4}$ and the highest score was $10$. Therefore the range is $\mathrm{6}$.

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

For Quiz 1, the $\mathrm{75th}$ percentile is $\mathrm{8}$ and the $\mathrm{25th}$ percentile is $\mathrm{6}$. The interquartile range is therefore $\mathrm{2}$. For Quiz 2, which has greater spread, the $\mathrm{75th}$ percentile is $\mathrm{9}$, the $\mathrm{25th}$ percentile is $\mathrm{5}$, and the interquartile range is $\mathrm{4}$. Recall that in the discussion of box plots, the $\mathrm{75th}$ percentile was called the upper hinge and the $\mathrm{25th}$ 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 $\mathrm{2}$. If a distribution is symmetric, the median plus or minus the semi-interquartile range contains half the scores in the distribution.

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 Quiz 1 are shown in Table 1. The mean score is $\mathrm{7.0}$. Therefore, the column "Deviation from Mean" contains the score minus $\mathrm{7}$. The column "Squared Deviation" is simply the previous column squared.

Table 1. Calculation of Variance for Quiz 1 scores.

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 $\mathrm{1.5}$. Therefore, the variance is $\mathrm{1.5}$. Analogous calculations with Quiz 2 show that its variance is $\mathrm{6.7}$. The formula for the variance is:

$\sigma^{2}=\frac{\sum(X-\mu)^{2}}{N}$

where $\sigma^{2}$ is the variance, $\mu$ is the mean, and $N$ is the number of numbers. For 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:

$\begin{aligned} M &=(1+2+4+5) / 4=12 / 4=3 \\ S^{2} &=\left[(1-3)^{2}+(2-3)^{2}+(4-3)^{2}+(5-3)^{2}\right] /(4-1) \\ &=(4+1+1+4) / 3=10 / 3=3.333 \end{aligned}$

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.

$\sigma^{2}=\frac{\sum X^{2}-\frac{\left(\sum X\right)^{2}}{N}}{N}$

and

$s^{2}=\frac{\sum X^{2}-\frac{\left(\sum X\right)^{2}}{N}}{N-1}$

For this example,

The standard deviation is simply the square root of the variance. This makes the standard deviations of the two quiz distributions $\mathrm{1.257}$ and $\mathrm{2.203}$. 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, $\mathrm{68\%}$ of the distribution is within one standard deviation of the mean, and approximately $\mathrm{95\%}$ of the distribution is within two standard deviations of the mean. Therefore, if you had a normal distribution with a mean of $\mathrm{50}$ and a standard deviation of $\mathrm{10}$, then $\mathrm{68\%}$  of the distribution would be between $50 - 10 = 40$ and $50 +10 =60$. Similarly, about $\mathrm{95\%}$ 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 σ; the symbol for an estimate computed in a sample is $\mathrm{s}$. Figure 2 shows two normal distributions. The red distribution has a mean of $\mathrm{40}$ and a standard deviation of $\mathrm{5}$; the blue distribution has a mean of $\mathrm{60}$ and a standard deviation of $\mathrm{10}$. For the red distribution, $\mathrm{68\%}$ of the distribution is between $\mathrm{35}$ and $\mathrm{45}$; for the blue distribution, $\mathrm{68\%}$ is between $\mathrm{50}$ and $\mathrm{70}$.

Figure 2. Normal distributions with standard deviations of $\mathrm{5}$ and $\mathrm{10}$.

R code

1. $8 - 2 = 6$
   
   
2. The variance would be larger if these numbers represented a sample because you would divide by $\mathrm{N-1}$ (instead of just $\mathrm{N}$).
   
   
3. 3.2842
   
   
4. 25th% = 10, 75th% = 19, 19 - 10 = 9