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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

 Mean

 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.

 Median

 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.

 Mode

 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 mean is 85.125.

 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 mean is 5.4.

 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.

 Measures of Dispersion

 Range

 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.

 Standard Deviation

 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.

 Frequency Distributions

 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.

 This combines two ideas covered in this tutorial, finding the mean and setting up 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.

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.

 3a.  7, 21, 21, 17, 17, 14, 7, 0 (answer/discussion to 3a)

Practice Problem 4a: Find the mean of the frequency distribution.

4a.

 x f 9 3 10 5 15 4 20 10 Total 22

Need Extra Help on these Topics?

 Go to Get Help Outside the Classroom found in Tutorial 1: How to Succeed in a Math Class for some more suggestions.

Last revised on August 7, 2011 by Kim Seward.