11.4 Measures of Central Tendency

Let's begin by trying to find the most "typical" value of a data set.

Note that we just used the word "typical" although in many cases you might think of using the word "average." We need to be careful with the word "average" as it means different things to different people in different contexts. One of the most common uses of the word "average" is what mathematicians and statisticians call the arithmetic mean, or just plain old mean for short. "Arithmetic mean" sounds rather fancy, but you have likely calculated a mean many times without realizing it; the mean is what most people think of when they use the word "average".

Example 3

The one hundred families in a particular neighborhood are asked their annual household income, to the nearest $5 thousand dollars. The results are summarized in a frequency table below.

Income (thousands of dollars)	Frequency
15	6
20	8
25	11
30	17
35	19
40	20
45	12
50	7

Show solution

Calculating the mean by hand could get tricky if we try to type in all 100 values:

$\frac{15 + \dots + 15 + 20 + \dots + 20 + 25 + \dots + 25 + \dots}{100}$

We could calculate this more easily by noticing that adding 15 to itself six times is the same as $15 \cdot 6 = 90$ . Using this simplification, we get

$\frac{15 \cdot 6 + 20 \cdot 8 + 25 \cdot 11 + 30 \cdot 17 + 35 \cdot 19 + 40 \cdot 20 + 45 \cdot 12 + 50 \cdot 7}{100} = \frac{3390}{100} = 33.9$

The mean household income of our sample is 33.9 thousand dollars ($33,900).

Example 4

Extending off the last example, suppose a new family moves into the neighborhood example that has a household income of $5 million ($5000 thousand). Adding this to our sample, our mean is now:

import statistics
import matplotlib.pyplot as plt

# The book's 100 households, as a frequency table (income in $thousands).
# CHANGE a frequency, or add a row, and every number below moves.
table = [(15, 6), (20, 8), (25, 11), (30, 17), (35, 19), (40, 20), (45, 12), (50, 7)]
incomes = [value for value, freq in table for _ in range(freq)]

NEW_NEIGHBOR = 5000     # <-- CHANGE ME. The book's millionaire is 5000 ($5 million).
                        #     Try 60 (a well-off neighbor) and then 50000.

def describe(label, data):
    print("%-28s n=%-5d mean=%8.2f  median=%7.2f  mode=%5d"
          % (label, len(data), statistics.mean(data), statistics.median(data),
             statistics.mode(data)))

describe("original neighborhood", incomes)
describe("+ the new neighbor", incomes + [NEW_NEIGHBOR])
print()
print("One household out of %d moved the MEAN by %+.1f thousand"
      % (len(incomes) + 1,
         statistics.mean(incomes + [NEW_NEIGHBOR]) - statistics.mean(incomes)))
print("and the MEDIAN by %+.1f thousand."
      % (statistics.median(incomes + [NEW_NEIGHBOR]) - statistics.median(incomes)))
print()
print("The mean is a balance point, so one weight far out on the see-saw drags it.")
print("The median only counts POSITIONS, so it never notices how far out the outlier is.")

# Sweep the outlier from ordinary to absurd and watch the two curves separate.
xs = list(range(20, 5001, 20))
means   = [statistics.mean(incomes + [x]) for x in xs]
medians = [statistics.median(incomes + [x]) for x in xs]

plt.figure(figsize=(7, 3.8))
plt.plot(xs, means, label="mean", color="#c00000")
plt.plot(xs, medians, label="median", color="#2e75b6")
plt.axvline(NEW_NEIGHBOR, ls=":", color="grey", label="your NEW_NEIGHBOR")
plt.xscale("log")
plt.xlabel("income of the one new household ($ thousands, log scale)")
plt.ylabel("centre of the 101 households ($ thousands)")
plt.title("The mean chases the outlier; the median shrugs")
plt.legend()
plt.tight_layout()
plt.show()

Show solution

$\frac{15 \cdot 6 + 20 \cdot 8 + 25 \cdot 11 + 30 \cdot 17 + 35 \cdot 19 + 40 \cdot 20 + 45 \cdot 12 + 50 \cdot 7 + 5000 \cdot 1}{101} = \frac{8390}{101} = 83.069$

While 83.1 thousand dollars ($83,069) is the correct mean household income, it no longer represents a “typical” value.

Imagine the data values on a see-saw or balance scale. The mean is the value that keeps the data in balance, like in the picture below.

If we graph our household data, the $5 million data value is so far out to the right that the mean has to adjust up to keep things in balance

The same balance picture for the household income data once the 5 million dollar family is added. Three blocks, one wide and two narrow, are bunched together just to the left of the fulcrum, while a single small block sits at the far right end of a very long plank. The fulcrum, which marks the mean, has had to slide right to the edge of the cluster to balance that one distant value.

For this reason, when working with data that have outliers – values far outside the primary grouping – it is common to use a different measure of center, the median.

Example 7

Let us return now to our original household income data

Income (thousands of dollars)	Frequency
15	6
20	8
25	11
30	17
35	19
40	20
45	12
50	7

Show solution

Here we have 100 data values. If we didn’t already know that, we could find it by adding the frequencies. Since 100 is an even number, we need to find the mean of the middle two data values - the 50^th and 51^st data values. To find these, we start counting up from the bottom:

$\begin{matrix} There are 6 data values of $15, so & Values 1 to 6 are $15 thousand \\ The next 8 data values are $20, so & Values 7 to (6 + 8) = 14 are $20 thousand \\ The next 11 data values are $25, so & Values 15 to (14 + 11) = 25 are $25 thousand \\ The next 17 data values are $30, so & Values 26 to (25 + 17) = 42 are $30 thousand \\ The next 19 data values are $35, so & Values 43 to (42 + 19) = 61 are $35 thousand \end{matrix}$

From this we can tell that values 50 and 51 will be $35 thousand, and the mean of these two values is $35 thousand. The median income in this neighborhood is $35 thousand.

In addition to the mean and the median, there is one other common measurement of the "typical" value of a data set: the mode.

The mode is fairly useless with data like weights or heights where there are a large number of possible values. The mode is most commonly used for categorical data, for which median and mean cannot be computed.

Example 9

In our vehicle color survey, we collected the data

Color	Frequency
Blue	3
Green	5
Red	4
White	3
Black	2
Grey	3

For this data, Green is the mode, since it is the data value that occurred the most frequently.

It is possible for a data set to have more than one mode if several categories have the same frequency, or no modes if each every category occurs only once.

Your Turn

Try it Now 6

Reviewers were asked to rate a product on a scale of 1 to 5. Find

The mean rating
The median rating
The mode rating

Rating	Frequency
1	4
2	8
3	7
4	3
5	1

Answer

The mean is $\frac{1 \cdot 4 + 2 \cdot 8 + 3 \cdot 7 + 4 \cdot 3 + 5 \cdot 1}{23} \approx 2.5$
There are 23 data values, so the median will be the 12^th data value. Ratings of 1 are the first 4 values, while a rating of 2 are the next 8 values, so the 12^th value will be a rating of 2. The median is 2.
The mode is the most frequent rating. The mode rating is 2.

Adapted from Math in Society by David Lippman, hosted on LibreTexts (math.libretexts.org) and licensed under CC BY-SA 3.0. Changes were made. License: CC-BY-SA-3.0.