Beginning Algebra
Tutorial 34: Central Tendencies
Learning Objectives
After completing this tutorial, you should be able to:
 Find the mean of a list of values.
 Find the median of a list of values.
 Find the mode of a list of values.
 Find the range of a list of values.
 Find the standard deviation of a list of values.
 Use a frequency distribution to find the mean.

Introduction
In this tutorial we will be looking at basic concepts
of central tendencies.
We will go over how to find the mean, median and mode of a list of
values
as well as the range and standard deviation. I think you are ready to get
started
on these central tendencies.

Tutorial
The mean of a list of values is the average of those
values.
You can find the mean by adding up all the values and
then dividing
that sum by the number of values that you have.
There is only one mean to a list of values.
The mean may or may not be a number that is in the
original list of
values. 
The median of a list of values is the middle value
compared to the
other values.
This does not necessarily mean it is the middle number
in the original
list. You need to make sure that your values are in numeric order
from smallest to largest before you find the median.
There is only one median to a list of values.
The median may or may not be a number that is in the
original list of
values. 
The mode of a list of values is the value that
occurs the most often.
You can have more than one mode, if more than one value
occurs the same
amount of times and that is the highest occurrence. 
Example
1: A student received the following grades on
quizzes
in a history course: 80, 88, 75, 93, 79, 95, 75, and 96. Find the mean, median and mode of the quizzes. 
The mean is the average of the
scores.
So we need to sum up all of the quizzes and then divide
by 8, since
there are 8 quizzes: 

*(sum of quiz)/(# of quizzes)
*Add numerator
*Divide by 8



The median is the middle value.
We need to list the numbers in numeric order:
75, 75, 79, 80, 88, 93, 95, 96
If we pick 80 for our median we have 3 values below it
and 4 above it.
If we pick 88 for our median then we have 4 values below it and 3 above
it. So neither of those values are the median. This
does
not mean we don’t have a median.
Note how there is an even number of values
listed. If that
is the case, we need to draw a line down the middle of the list and
take
the mean of the two numbers next to that line:
75, 75, 79, 80  88, 93,
95, 96
The mean of 80 and 88 is 

*Find number exactly in the
middle of 80 and
88

84 is the median. It is the value that is
right smack
dab in the middle of this list of values. 
The mode is the value that
occurs the most
often.
It helps to list the numbers in order to find the mode.
75, 75,
79, 80, 88, 93, 95, 96
Note how 75 occurs two times, which is the value that
occurs the most.
75 is the mode. 
Example
2: The number of points a kicker made during the
first five games of the season are given in the table:
Game

1

2

3

4

5

Points

3

9

3

6

6

Find the mean, median and mode of the points. 
The mean is the average of the
points.
So we need to sum up all of the points and then divide
by 5, since there
are 5 games: 

*(sum of points)/(# of points)
*Add numerator
*Divide by 5



The median is the middle value.
We need to list the numbers in numeric order:
3, 3, 6, 6, 9
This time we have an odd number of values. Our
median is going
to be 6 (the first 6 listed). That number has two values above it
and two below it, so it is the middle value.
6 is the median. It is the value that is
right smack dab
in the middle of this list of values. 
The mode is the value that
occurs the most
often.
It helps to list the numbers in order to find the mode.
3, 3, 6, 6,
9
Note how both 3 and 6 occur two times, which is the
most.
Both 3 and 6 are the mode. 
Example
3: If your scores of the first four exams are
98,
100, 90 and 97, what do you need to make on the next exam for your
overall
mean to be at least 90? 
This time we are given the mean and we need to find one
of our values.
Keep in mind that this is still a mean problem. We
will still
use the idea that we need to sum up the exams and then divide it by 5
to
get the mean. We can let our unknown exam be x. 

*(sum of tests)/(# of tests) =
mean
*Solve for x (missing
test)
*Inverse of div. by 5 is mult.
by 5
*Inverse of add 385 is sub. 385

You would need to make 65 on the next exam to have a
mean of 90. 
Example
4: A student has received scores of 88, 82, and 84 on
3 quizzes. If the tests count twice as much as the quizzes, what
is the lowest score the student can get on the next test to achieve an
average score of at least 80? 
This is similar to example 3, except that the test
score counts twice
instead of one time. So when we set this up we need to make sure
that we notate that properly. 

*(sum of scores)/(# of scores)
= mean
*Need 2 x's
since
tests count twice
*Solve for x (missing
test)
*Inverse of div. by 5 is mult.
by 5
*Inverse of add 254 is sub. 254

The student would have to score a 73 on the next
test to have a
mean of 80. 
One way to measure dispersion (variability) among
numerical values
is to find the range of those numbers. The range of a set of
numerical
data points is the difference between the largest value and the
smallest
value. In other words you take the greatest measurement minus
the least measurement. 
Another way to measure dispersion of a data set is to
find the standard
deviation of its values. The standard deviation is a relative
measure of the dispersion of a set of data.
Note that the range only involves two values in its
calculation  the
highest and the lowest. However, the standard deviation involves
every value of its data set. 
The steps to finding the standard deviation are as
follows:
Step 1: Find the mean of the values
of the data set.
Step 2: Find the difference between
the mean and each
separate value of the data set.
Step 3: Square each difference found
in step 2.
Step 4: Add up all of the squared
values found in step
3.
Step 5: Divide the sum found in
step 4 by the
number of data values in the set.
Step 6: Find the nonnegative square
root of the quotient
found in step 5. 
Example
5: Find the range and the standard deviation of the
following
sample: 3, 10, 8, 20, 4, 4, 3, 8, 8, 8, 12. 
I don't know about you, but I find it easier to work
with a group of
numbers like this when they are in chronological order. Let's put
them in order from lowest to highest: 3, 3, 4, 4, 8, 8, 8, 8, 10,
12, 20.
Let's find the range.
What
do you think it is?
Looking at the difference between the largest value,
which is 20 and
the smallest value, which is 3, it looks like the range is 17.
Now lets tackle the standard
deviation. 
Step 1: Find the mean of the values
of the data set. 
So we need to sum up all of the values and then divide
by 11, since
there are 11 numbers: 

*(sum of values)/(# of values)
*Add numerator
*Divide by 11

Step 2: Find the difference between
the mean and each
separate value of the data set,
AND
Step 3: Square each difference found in
step 2,
AND
Step 4: Add up all of the squared values
found in
step 3. 
x

x  8


3

5

25

3

5

25

4

4

16

4

4

16

8

0

0

8

0

0

8

0

0

8

0

0

10

2

4

12

4

16

20

12

144


SUM:

246

Step 5: Divide the sum found
in step 4 by the
number of data values in the set
AND
Step 6: Find the nonnegative square root
of the quotient
found in step 5. 

*Square root of [(sum of diff.
squared)/(#
of values)]

The standard deviation is approximately 4.729. 
Sometimes there are a lot of values in a data set and
some of them
are repeated. In that case, it may be easier to group those
values
using a frequency distribution. This is a chart that lists
each
unique value and then next to the number indicates the frequency, or
number
of times, that value occurs in the data set.
For example, if you had the list of test scores for a
class:
75, 80, 90, 80, 75, 75, 50, 65, 65, 50, 100, 90, 100, 90, 75, 40, 60,
60
Writing these values (x)
in a frequency
(f) distribution chart you would have: 
x

f

40

1

50

2

60

2

65

2

75

4

80

2

90

3

100

2

Total

18

Example
6: Find the mean of the test scores 75, 80, 90, 80,
75,
75, 50, 65, 65, 50, 100, 90, 100, 90, 75, 40, 60, 60, using a frequency
distribution. 
As shown above, the frequency distribution for
this set of numbers
is 
x

f

40

1

50

2

60

2

65

2

75

4

80

2

90

3

100

2

Total

18

As requested, I’m going to use the frequency
distribution to set up
my mean formula. Instead repeating numbers in my sum, I’m going
to
indicate a repetition by taking that value times the number of times it
occurs in the list. For example, 75 occurs 4 times. Instead
of writing it out 4 times in my sum, I will find 75(4) which is the
equivalent. 

*(sum of scores)/(# of scores)
*calculate numerator
*divide

Practice Problems
These are practice problems to help bring you to the
next level.
It will allow you to check and see if you have an understanding of
these
types of problems. Math works just like
anything
else, if you want to get good at it, then you need to practice
it.
Even the best athletes and musicians had help along the way and lots of
practice, practice, practice, to get good at their sport or instrument.
In fact there is no such thing as too much practice.
To get the most out of these, you should work the
problem out on
your own and then check your answer by clicking on the link for the
answer/discussion
for that problem. At the link you will find the answer
as well as any steps that went into finding that answer. 
Practice
Problem 1a: Find the mean, median, and mode.
1a. The number of cd’s sold by Dave’s Discs
for the last
6 days are given in the table.
Day

Day 1

Day 2

Day 3

Day 4

Day 5

Day 6

CD's

15

10

10

10

18

15

Find the mean, median, and mode.
(answer/discussion
to 1a)

Practice
Problem 2a: Find the test score.
2a. A student received scores of 92, 83, and 71
on three quizzes.
If tests count twice as much as quizzes, what is the lowest score that
the student can get on the next test to achieve a mean of at least 80?
(answer/discussion
to 2a)

Practice
Problem 3a: Find the range and standard deviation
of the list of
scores that were made by a football team during a season.
Practice
Problem 4a: Find the mean of the frequency
distribution.
Need Extra Help on these Topics?
Last revised on August 7, 2011 by Kim Seward.
All contents copyright (C) 2001  2011, WTAMU and Kim Seward. All rights reserved.

