Section 6.8: Measures of Center

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Timothy Cooley
Orange Coast College

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Learning Objectives

Compute the mean of a data set
Compute the median of a data set
Compute the mode of a data set
Compute the midrange of a data set
Compare the four measures of central tendency

Measures of central tendency (sometimes referred to as Measures of Center) are statistical values that describe the center or typical value of a quantitative data set. They give us a single number that represents the "middle" or "most common" point of the data.

Measures of Central Tendency

The four measures of central tendency are called the mean, median, mode, and midrange.

Note: Most statistical textbooks indicate that there are only three measures of central tendency which are mean, median, and mode.

The first measure of central tendency, mean, is a type of average and is probably the measure of central tendency most people are familiar with. However, there are other measures of average, such as the median, mode, and midrange. The term “average” is more general and can refer to different central values depending on the context. In everyday language, “average” usually means the arithmetic mean (the same as the mean). In statistics, however, “average” could also refer to the mean, median, or mode, all of which are measures of central tendency.

For example, if someone says “the average household income,” they are referring to the mean, not a median or a mode. If they say “the average student score was the middle score,” they’re using median as the average. All means are averages, but not all averages are means. The mean is a precise, mathematical type of average.

Definition: Mean

The mean is a type of average. It's a way to describe the center or typical value of a set of numbers.

How to find the mean:

Step 1 — Add up all the values in the set.
Step 2 — Divide the total by how many values there are.

The mathematical formula for sample mean denoted by $\large\bar{x}$ is the following:

For a data set $\large x_1, x_2, x_3, \dots, x_n$, the mean is calculated by the formula

\[\large\bar{x} = \frac{\sum x_i}{n} = \frac{x_1 + x_2 + x_3 + \dots + x_n}{n},\nonumber\]

where $\large n$ is the sample size.

In statistics, the population mean $\mu$ is the average of all values in a population, while the sample mean $\large\bar{x}$ is the average of a sample and is used to estimate the population mean. For all intense purposes for this text, mean in general and sample mean are the same thing.

Example #6.8.1 🤔

Find the mean of the data set: 4, 8, 6, 2, 10

✅ Solution:

Step 1 — Add the numbers: 4 + 8 + 6 + 2 + 10 = 30
Step 2 — Divide by how many values there are which is 5: 30 ÷ 5 = 6

So, the mean is 6.

Using the mathematical formula for mean, we have

\[\large\bar{x} = \frac{\sum x_i}{5} = \frac{x_1 + x_2 + x_3 + x_4 + x_5}{5}=\frac{4 + 8 + 6 + 2 + 10}{5}=\frac{30}{5}=6\nonumber\]

Example #6.8.2 🤔

The ages of the last 9 children seen by a local children's dentist was recorded:

4, 5, 7, 8, 3, 6, 2, 9, 2

Find the mean age of the children.

✅ Solution:

Step 1 — Add the numbers: 4 + 5 + 7 + 8 + 3 + 6 + 2 + 9 + 2 = 46
Step 2 — Divide by how many numbers there are which is 9: 46 ÷ 9$\approx$5.1

Thus, the mean is 5.1 (or 5.11, if you want a more approximate answer).

Using the mathematical formula for mean, we have

\[\large\bar{x} = \frac{\sum x_i}{9} = \frac{x_1 + x_2 + x_3 + \dots + x_9}{9}=\frac{4 + 5 + 7 + 8 + 3 + 6 + 2 +9 + 2}{9}=\frac{46}{9}\approx5.1\nonumber\]

Example #6.8.3 🤔

Find the mean of the data set for a random sample of 10 GPA's from a particular English class:

4.0, 3.65, 2.15, 3.34, 2.87, 3.46, 2.75, 2.94, 2.66, 3.06

✅ Solution:

$\large\bar{x} = \frac{\sum x_i}{10} = \frac{x_1 + x_2 + x_3 + \dots + x_{10}}{10} = \frac{4.0+3.65+2.15+3.34+2.87+3.46+2.75+2.94+2.66+3.06}{10}=\frac{30.88}{10}=3.088\approx3.09\nonumber $

So, the mean is 3.09.

Definition: Median

The median is a type of average. it represents the "middle" value in a data set when the numbers are arranged in order from smallest to largest.

How to find the median:

Step 1 — Arrange the data in ascending order
Step 2 — Determine if the number of values in the data set is odd or even.
Step 3 —
- If the number of data points is odd, the median is the middle value. (Example: In the data set, {3, 5, 9}, the median is 5).
- If the number of data points is even, the median is the mean of the two middle values. (Example: In the data set, {1, 3, 4, 8}, the median is $\frac{3+4}{2}=\frac{7}{2}=3.5$.

The median is especially useful because it is less affected by extreme low and/or high values (outliers) than the mean (average). Let's say that the data set was actually {1, 3, 4, 88} instead of the above data set {1, 3, 4, 8}. The median would still be the same value of 3.5, but the mean would change from 4 to 24. In statistics, the median is considered resistant, if it is not strongly influenced by extreme values (outliers) or skewed data. A resistant statistic gives a reliable summary of the data even when there are unusually large or small values. A non-resistant statistic changes a lot when outliers are present.

Example #6.8.4 🤔

Find the median of the data set: 4, 8, 6, 2, 10

✅ Solution:

Step 1 — Arrange the data in ascending order: 2, 4, 6, 8, 10.
Step 2 — Determine if the number of values in the data set is odd or even. Since there are 5 values, the number of values in the set is odd.
Step 3 — An odd number of values means there is only one middle value which is the value of 6.

So, the median is 6.

Example #6.8.5 🤔

Ratings of a popular show exclusive to Paramount Plus are given in the following data set:

7, 10, 9, 9, 3, 5, 6, 9, 2, 5

Find the median of the data set above

✅ Solution:

Step 1 — Arrange the data in ascending order: 2, 3, 5, 5, 6, 7, 9, 9, 9, 10.
Step 2 — Determine if the number of values in the data set is odd or even. Since there are 10 values, the number of values in the set is even.
Step 3 — An even number of values means there are two middle values, which are 6 and 7. Now, find the mean of those two middle values 6 and 7 which is $\frac{6+7}{2}=\frac{13}{2}=6.5$.

So, the median is 6.5.

Example #6.8.6 🤔

A random sample of 15 monthly rent costs from Costa Mesa, CA were taken from apartments.com:

$2,240	$3,290	$3,150	$3,075	$2,819
$2,925	$2,725	$2,100	$2,200	$3,210
$2,300	$3,199	$2,352	$3,982	$2,259

Find the median of the rent costs.

✅ Solution:

Step 1 — Arrange the data in ascending order: 2100, 2200, 2259, 2240, 2300, 2352, 2725, 2819, 2925, 3075, 3150, 3199, 3210, 3290, 3982.
Step 2 — Determine if the number of values in the data set is odd or even. Since there are 15 values, the number of values in the set is odd.
Step 3 — An odd number of values means there is only one middle value which is the value of 2819.

So, the median is $2,819.

Definition: Mode

The mode is a type of average. It represents the value that occurs most frequently in a data set.

A data set can have:
- No mode → if no value repeats.
- One mode (unimodal) → if exactly one value occurs most often.
- Two modes (bimodal) → if exactly two values tie for highest frequency.
- More than two modes (multimodal) → if several values occur equally often.

Example #6.8.7 🤔

Find the mode of the following data set: 3, 7, 8, 4, 9, 6, 2, 8

✅ Solution:

Arranging the data in ascending order, we have 2, 3, 4, 6, 7, 8, 8, 9.
In this data set, there are two 8's and all other values have frequency one.

So, the mode is 8.

Example #6.8.8 🤔

Find the mode of the following data set: 5, 6, 8, 4, 8, 6, 5, 4

✅ Solution:

Arranging the data in ascending order, we have 4, 4, 5, 5, 6, 6, 8, 8.
In this data set, the values 4, 5, 6, and 8, all have frequency two and all other values have frequency one.

This data set is multimodal. The modes are 4, 5, 6, and 8.

Example #6.8.9 🤔

The following total days in each month of a non-leap year Gregorian calendar is the following:

31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31

Find the mode of days in the set.

✅ Solution:

Arranging the data in ascending order, we have 28, 30, 30, 30, 30, 31, 31, 31, 31, 31, 31, 31.
In this data set, the value 31 occurs (7) times, while 30 occurs (4) times, and 28 only once (1).

So, the mode is 31.

Example #6.8.10 🤔

he following data set is the total number of home runs each season by the Los Angeles Angels from the 2009 season to the 2024 season:

186, 155, 155, 145, 164, 172, 165, 156, 197, 175, 220, 85, 190, 190, 185, 183

Find the mode for the data set above.

✅ Solution:

Arranging the data in ascending order, we have 85, 145, 155, 155, 156, 164, 165, 172, 175, 183, 185, 186, 190, 190, 197, 220.
In this particular data set, there are two 155's and two 190's. All other values have frequency one.

This data set is bimodal. The modes are 155 and 190.

Example #6.8.11 🤔

According to science.nasa.gov, the radius of each of the 8 planets in our solar system is given below in the table:

Mercury	Neptune	Earth	Mars	Jupiter	Saturn	Uranus	Venus
1,516	3,760	3,959	2,106	43,441	36,184	15,759	15,299

Find the mode for the data set above.

✅ Solution:

You could go ahead and arrange the data in ascending order, however, a simple glance clearly shows that all of the values are different from one another. In order to qualify for a mode, you have to have at least one repetition of any value or values.

Thus, there is no mode.

Definition: Midrange

The midrange is a type of average. It is found by taking the mean of the highest and lowest values in a data set.

How to find the midrange:

Step 1 — Identify the lowest and the highest values in the data set. (Arrange the data in order, if necessary).
Step 2 — Add up the lowest and the highest values.
Step 3 — Divide the total by 2.

Or, equivalently, the formula

\[\text{Midrange} = \frac{\text{lowest value + highest value}}{2}\nonumber\]

Weaknesses of Midrange

The midrange statistic has many weaknesses. Here is a few of them:

Extremely sensitive to outliers - The midrange depends only on the smallest and largest values. If either extreme is unusually high or low, the midrange can shift dramatically, even if most of the data are clustered elsewhere.
Ignores most of the data - Unlike the mean or median, the midrange does not use information from the majority of observations, making it a poor representation of where the data are actually centered.
Often not a typical value - The midrange may fall in a region where no data values exist, especially in skewed distributions.

Despite its limitations, the midrange is still taught and occasionally used for several reasons. The midrange is very easy to compute and understand, making it useful as an introductory concept when first discussing measures of center. It is also useful in controlled situations. When data are uniformly distributed and free of outliers (such as bounded physical measurements), the midrange can provide a quick approximation of the center. Ot can also emphasize the role of extremes. Studying the midrange helps students see how extreme values affect measures of center, reinforcing why the mean and median are often preferred.

Example #6.8.12 🤔

Find the midrange of the following data set: 5, 8, 19, 10, 7, 11

✅ Solution:

Step 1 — Identify the lowest and the highest values in the data set: Notice that the data is unordered. The lowest value is 5 and the highest value is 19 (not 11).
Step 2 — Add up the lowest and the highest values: 5 + 19 = 24
Step 3 — Divide the total by 2: 24 ÷ 2 = 12

So, the midrange is 12.

Or, using the formula for midrange, we have

\[\text{Midrange} = \frac{5 + 19}{2}=\frac{24}{2}=12\nonumber\]

Example #6.8.13 🤔

Find the midrange of the data set: 12, 4, 6, 4, 7, 11, 8, 10, 2, 17, 9

✅ Solution:

Step 1 — Identify the lowest and the highest values in the data set: Notice that the data is unordered. The lowest value is 2 and the highest value is 17.
Step 2 — Add up the lowest and the highest values: 2 + 17 = 19
Step 3 — Divide the total by 2: 19 ÷ 2 = 9.5

So, the midrange is 9.5.

Or, using the formula for midrange, we have

\[\text{Midrange} = \frac{2 + 17}{2}=\frac{19}{2}=9.5\nonumber\]

Example #6.8.14 🤔

Temperature highs over the last 20 days were recorded below:

88, 86, 92, 81, 85, 88, 87, 90, 82, 95, 92, 84, 93, 86, 91, 92, 91, 89, 87, 83

Find the midrange of the data set above.

✅ Solution:

Step 1 — Identify the lowest and the highest values in the data set: Notice that the data is unordered. The lowest value is 81 and the highest value is 95.
Step 2 — Add up the lowest and the highest values: 81 + 95 = 176
Step 3 — Divide the total by 2: 176 ÷ 2 = 88

So, the midrange is 88.

Or, using the formula for midrange, we have

\[\text{Midrange} = \frac{81 + 95}{2}=\frac{176}{2}=88\nonumber\]

Section 6.8: Measures of Center [In-Class Exercises]

The time, in minutes, it took 9 students to complete their first quiz in a History class was:

5, 7, 5, 2, 9, 4, 9, 9, 10

Find the mean, median, mode, and midrange for the above data set of times.

Find the mean, median, mode, and midrange of the following data set: 6, 10, 5, 9, 7, 12, 5, 11, 8, 6.
Find the median and mode of the following data set: 5, 6, 7, 6, 7, 8, 8, 5.
Find the median and mode of the following data set: 13, 17, 19, 23, 24, 27, 11.
According to the California Department of Conservation, the table below shows the California's largest ever recorded earthquakes since 1800, ranked by date.

Magnitude	Date	Location	Damage
7.9	Jan. 9, 1857	Fort Tejon	Two killed; 220-mile surface scar
7.4	Mar. 26, 1872	Owens Valley	27 killed; three aftershocks of magnitude >6
7.8	April 18, 1906	San Francisco	Possibly 3,000 killed; 225,000 displaced
7.2	Jan. 22, 1923	Mendocino*	Damaged homes in several towns
7.1	Nov. 4, 1927	SW of Lompoc*	No major injuries; slight damage in 2 counties
6.4	March 10, 1933	SE of Long Beach	115 killed; led to new building codes for schools
7.0	May 18, 1940	El Centro	9 killed; $6 million in damage
7.3	July 21, 1952	Kern County	12 killed; 3 magnitude >6 aftershocks in 5 days
6.6	Feb. 9, 1971	San Fernando	65 killed; 2,000 injured; $505 million in damage
7.4	Nov. 8, 1980	W. of Eureka*	Injured 6; $2 million in damage
6.9	Oct. 17, 1989	Bay Area	63 killed; 3,753 hurt; up to $10 billion in damage
7.2	April 25, 1992	Cape Mendocino	356 injuries; $48.3 million in damage
7.3	June 28, 1992	Landers	One killed; 400 injured; $9.1 million in damage
6.7	Jan. 17, 1994	Northridge	57 killed; 9,000 hurt, up to $40 billion in damage
7.1	Oct. 16, 1999	Ludlow	Minimal damage due to remote location
7.1	July 5, 2019	Ridgecrest/Trona	Preceded by M6.4 quake; no fatalities

Find the mean, median, mode, and midrange of the magnitude ratings.

Answers

Mean = 6.7 , Median = 7, Mode = 9, Midrange = 6.
Mean = 7.9 , Median = 7.5, Mode = 5 & 6, Midrange = 8.5.
Median = 6.5, Mode = 5, 6, 7 & 8.
Median = 19, Mode = None.
Mean = 7.15 , Median = 7.15, Mode = 7., Midrange = 7.15.

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Example #6.8.12 🤔

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Example #6.8.14 🤔

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Section 6.8: Measures of Center [In-Class Exercises]