Lesson 2.9
Introduction
Course Objectives
This lesson will address the following course outcomes:
· 6. Demonstrate an understanding of the connection between the distribution of data and various mathematical summaries of data (measures of central tendency and of variation).
Specific Objectives
Students will understand
· that numerical data can be summarized using measures of central tendency.
· how each statistic—mean, median, and mode—provides a different snapshot of the data.
· that conclusions derived from statistical summaries are subject to error.
Students will be able to
· calculate the mean, median, and mode for numerical data.
· create a data set that meets certain criteria for measures of central tendency.
Measures of Center
People often talk about “averages,” and you probably have an idea of what is meant by that. Now, you will look at more formal mathematical ways of defining averages. In mathematics, you call an average, a measure of center because an average is a way of measuring or quantifying the center of a set of data. There are different measures of center because there are different ways to define the center.
Think about a long line of people waiting to buy tickets for a concert. (Figure A shows a line about 100-feet long and each dot represents a person in the line.) In some sections of the line people are grouped together very closely, while in other sections of the line people are spread out. How would you describe where the center of the line is?
Would you define the center of the line by finding the point at which half the people in the line are on one side and half are on the other (see Figure B)? Is the center based on the length of the line even though there would be more people on one side of the center than on the other (see Figure C)? Would you place the center among the largest groups of people (see Figure D)? The answer would depend on what you needed the center for. When working with data, you need different measures for different purposes.
Measures of Center
Mean (Arithmetic Average)
Find the average of numeric values by finding the sum of the values and dividing the sum by the number of values. The mean is what most people call the “average.”
Example
Find the mean of 18, 23, 45, 18, 36 Find the sum of the numbers: 18 + 23 + 45 + 18 +36 = 140 Divide the sum by 5 because there are 5 numbers: 140 ÷ 5 = 28 The mean is 28.
Example
Find the mean of 1.5, 1.2, 3.7, 5.3, 7.1, 2.9 Find the sum of the numbers: 1.5 + 1.2 + 3.7 + 5.3 + 7.1 + 2.9 = 21.7 Divide the sum by 6 because there are 6 numbers: 21.7 ÷ 6 = 3.6166666666
Since the original values were only accurate to one decimal place, reporting the mean as 3.61666666 would be misleading, as it would imply we knew the original values with a higher level of accuracy. To avoid this, we round the mean to one more decimal place than the original data.
The mean is 3.62.
Median
Find the median of numeric values by arranging the data in order of size. If there is an odd number of values, the median is the middle number. If there is an even number of values, the median is the mean of the two middle numbers.
Example (data set with odd number of values)
Find the median of 18, 23, 45, 18, 36. Write the numbers in order: 18, 18, 23, 36, 45 There is an odd number of values, so the median is the number in the middle. The median is 23.
Example (data set with even number of values)
Find the median of 18, 23, 45, 18, 12, 50. Write the numbers in order: 12, 18, 18, 23, 45, 50 There is an even number of values, so there is no one middle number. Find the median by finding the mean of the two middle numbers: 18 + 23 = 41 41 ÷ 2 = 20.5 The median is 20.5.
Mode
Find the mode by finding the number(s) that occur(s) most frequently. There may be more than one mode.
Example
Find the mode of 18, 23, 45, 18, 36. The number 18 occurs twice, more than any other number, so the mode is 18.
For another explanation, watch this video:
· Mean, Median, and Mode [+]
Note on terminology
The terms mean, median, and mode are well defined in mathematics and each gives a measure of center of a set of numbers. In everyday usage, the word “average” usually refers to the mean. But be aware that “average” is not clearly defined and someone might use it to refer to any measure of center.
#1 Points possible: 10. Total attempts: 5
Consider the data set.
| 5 | 5 | 4 | 4 | 7 | 8 | 3 | 1 | 9 | 6 | 7 |
Find the mean: Find the median:
#2 Points possible: 10. Total attempts: 5
Consider the data set
| 7 | 7 | 2 | 6 | 9 | 9 | 1 | 5 | 3 | 5 |
Find the mean: Find the median:
Credit Cards
Problem Situation: Summarizing Data About Credit Cards
A revolving line of credit is an agreement between a consumer and lender that allows the consumer to obtain credit for an undetermined amount of time. The debt is repaid periodically and can be borrowed again once it is repaid. The use of a credit card is an example of a revolving line of credit.
U.S. consumers own more than 600 million credit cards. As of 2015 the average credit card debt per household with a credit card was $15,863. In total, American consumers owe $901 billion in credit card debt. Worldwide the number of credit card transactions at merchants was over 135 billion in 2011.
Surveys indicate that the percentage of college freshmen with a credit card was 21% in 2012, while 60% of college seniors had a credit card. One-third of the college students reported having a zero balance on their credit card. The average balance carried across all students’ cards was $500. However, the median balance was $136.
At www.creditcards.com it explains that card issuers divide customers into two groups:
· “transactors” who use their cards for purchases and pay off the balances each month. Transactors pay off the balance before any interest charges are applied.
· “revolvers” who carry balances on their cards, paying interest charges month to month.
The number of people who carry credit card debt, the “revolvers”, has been steadily decreasing in the U.S. since 2009. By 2014 only one-third of adults surveyed said their household carries credit card debt. Fifteen percent of adults carry $2,500 or more in credit card debt each month.
In the first part of this lesson, you will use the information about credit cards given above to learn about some ways to summarize quantitative information.
#3 Points possible: 10. Total attempts: 5
The population of the United States is slightly more than 300 million people. There are about 100 million households in the United States. Use the information above to find:
a) What is the average number of credit cards per person?
cards per person
b) What is the average number of credit cards per household?
cards per household
#4 Points possible: 5. Total attempts: 5
In the U.S. the average number of credit cards held by cardholders is 3.7. Why is this number different than the average number of credit cards per person you found in the last question?
· The last question answer was when the credit cards are averaged over all the people in the country, including children and people who have no cards. 3.7 is the number of cards for person when averaged over only the people with cards.
· The last question answer was based on inaccurate, rounded values
· People lie about how many cards they have
· Some people have lots of cards
#5 Points possible: 5. Total attempts: 5
Consider the statement, “Average credit card debt per household with a credit card is $15,863.”
What does this statement tell us? (Assume the “average” they’re using is the mean)
· Half the households with credit cards owe less than $15,863, and half owe more
· That most households with credit cards owe $15,863
· That if we added up all the credit card debt and spread it evenly among the households with credit cards, each would owe $15,863
· That every household with a credit card has this much debt
· That if we added up all the credit card debt and spread it evenly among all households in the U.S., each would owe $15,863
#6 Points possible: 5. Total attempts: 5
If about 45% of households with credit cards carry no debt, what does that indicate about the amount of debt of some of the other 55% of households?
· The others must have debt that averages to 15,863
· Since a lot of households have no debt, the others must have debt much higher than $15,863 so the mean will come out to 15,863
· Since a lot of households have no debt, the others must have debt much lower than $15,863 so the mean will come out to 15,863
#7 Points possible: 24. Total attempts: 5
The introduction states that the average balance carried across all college students’ cards was $500. Imagine you ask four groups of six college students what their credit card debt is. The amount of dollars of debt for each student in each group is shown in the table, values listed in order of size.
Find the mean and median debt of each group of college students and record it in the table. Make sure the values you find are reasonable given the values for that group.
| Group A | Group B | Group C | Group D | |
| 0 | 500 | 410 | 0 | |
| 100 | 500 | 460 | 0 | |
| 110 | 500 | 480 | 0 | |
| 170 | 500 | 490 | 0 | |
| 1000 | 500 | 550 | 0 | |
| 1620 | 500 | 610 | 3000 | |
| Mean | ||||
| Median |
#8 Points possible: 10. Total attempts: 5
Observe the medians and the data values of each of the groups. For each statement below, indicate if it is true for: None of the groups, Some but not all of the groups, or All of the groups. i) Half of the data values are less than the median.
ii) Half of the data values are either less than or equal to the median.
iii) Half of the data values are greater than the median.
iv) Half of the data values are either greater than or equal to the median.
v) Half of the data values equal the median.
#9 Points possible: 5. Total attempts: 5
Recall that college students’ credit cards carry a mean balance of $500 while having a median balance of $136. What does this indicate about the distribution of credit card debt among various students? Does one of the Groups in the table have a distribution similar to this?
· Group A
· Group B
· Group C
· Group D
#10 Points possible: 5. Total attempts: 5
A survey in 2012 indicated that college freshmen carry a mean credit card debt of $611 but the median of their credit card debt is $47. Create a data set of five freshmen students so that the data set has a mean and median that is the same as that of the surveyed college freshmen.
, , , ,
Weighted Mean
Problem Situation 2: Weighted Means
Sometimes you want to find the mean average of some numbers, but the numbers are not “equally important” or in other words they don’t have the same “weight” or “frequency”. In that case we find the “weighted average” of the numbers.
Example: mean credit card debt:
Suppose researchers at a small two-year college surveyed students who have credit cards and report the following:
Mean credit card debt of $350 for 114 freshmen surveyed
Mean credit card debt of $285 for 220 sophomores surveyed
If we want to know overall for all the surveyed students what their mean credit card debt was, we cannot simply average $350 and $285 since these numbers are not equally “weighted”. The $350 was an average for only 114 people while the $285 was an average for a larger group of 220 people. The way to find the overall average credit card debt for all students surveyed is to multiply each of the values by its “weight” or “frequency” or “importance” before adding them together, and then divide by the total “weight” or “frequency”.
Overall mean = 350(114)+285(220)114+220=102600334=307.19350(114)+285(220)114+220=102600334=307.19
Conclude: The mean credit card debt of all the students surveyed was about $307
Example: Weighted grade averages in a course:
Suppose a History course has a midterm, a final exam, and a 20 page report, and the syllabus states they have these weights in determining the final course grade:
midterm – 20%; final exam – 45%; report – 35%.
Kim’s grades in the course were these:
Midterm – 82; final exam – 89; report – 93.
Kim’s final grade in the course is found by the weighted average:
82(.20)+89(.45)+93(.35).20+.45+.35=891.00=8982(.20)+89(.45)+93(.35).20+.45+.35=891.00=89 Kim’s final course grade is 89.
Note: If the grades were not weighted, Kim’s average would be (82 + 89 + 93)/3 = 88
#11 Points possible: 5. Total attempts: 5
Researchers at Acme groceries studied how long customers had to stand in the check-out line. One day 35 customers spent on average 7.7 minutes each in the check-out line. The next day 24 customers spent an average of 6.5 minutes each. What is the overall average time customers spent in line? Round your answer to two decimal places.
minutes
#12 Points possible: 5. Total attempts: 5
Your college GPA is a weighted mean. The grade earned in each class needs to be weighted by the number of credits for the class. Compute the GPA for a student who has earned the following credits:
· English 101 (5 credits) with a grade of 3.0,
· Math 96 (7 credits) with a grade of 3.2,
· Public Speaking (5 Credits) with a grade or 3.6,
· Chemistry 100 (5 credits) with a grade of 2.7, and
· College Success (3 credits) with a grade of 3.8.
Round the GPA to two decimal places.
GPA:
HW 2.9
#1 Points possible: 5. Total attempts: 5
Which of the following was one of the main mathematical ideas of the lesson?
· U.S. college students carry far too much credit card debt.
· The mean is calculated by adding all the numbers and dividing by the number of data points.
· The mean, median, and mode all give important information about a data set, but they do not give a complete picture of the data set.
· Any of the three measures of central tendency (mean, median, and mode) are good representations of data. It does not matter which one you use.
#2 Points possible: 10. Total attempts: 5
Consider the data set
| 2 | 6 | 1 | 2 | 1 | 4 |
Find the average (mean): Find the median:
#3 Points possible: 10. Total attempts: 5
Consider the data set
| 8 | 9 | 4 | 1 | 6 | 8 | 6 | 3 | 3 | 2 | 2 |
Find the average (mean): Find the median:
#4 Points possible: 24. Total attempts: 5
Use the following data set to answer the questions.
13 15 20 20 20 20 20 20 20 23 27 31
a. What is the mean?
b. What is the mode?
c. What is the median?
d. What fraction of the numbers in the data set are less than the median?
e. What fraction of the numbers in the data set are greater than the median?
f. Which of the following statements are correct?
· The median is the middle of a data set. Half of the data points are always less than the median, and half are always greater than the median.
· The median is the middle of a data set. Half of the data points are either less than or equal to the median.
· The median is the middle of a data set. At least half of the data points are always equal to the median.
· The median is not the middle of a data set. You cannot predict the distribution of the numbers in relationship to the median.
#5 Points possible: 10. Total attempts: 5
Consider the statement “Worldwide, there are more than $2.5 trillion in credit card transactions annually.”
a. What is the daily average dollar amount of transactions? Round to the nearest hundred million dollars. $
b. How many dollars in credit card transactions are made on any particular day?
· $6,849,315
· $6,849,315,068
· $1,460,000,000
· $1,460,000
· It is impossible to know.
#6 Points possible: 5. Total attempts: 5
Students at Dover Community College (DCC) have a mean credit card debt of $3,600 with a median of $1,500. Students at Ralton Community College (RCC) have a mean credit card debt of $3,000 with a median of $2,800. Which statements about the two groups are true based on this information? There may be more than one correct answer.
· About three-fourths of DCC students have debt less than $3,600.
· The total debt of RCC students is less than the total debt of DCC students.
· No more than half of RCC students have debt less than $2,800.
· The total debt of RCC students is larger than the total debt of DCC students.
· At least one DCC student has debt in excess of $3,600.
· The largest debt of the RCC students is less than the largest debt of the DCC students.
#7 Points possible: 12. Total attempts: 5
Decide whether the following statements must be true or might be false, based on the information provided. Be prepared to explain your reasoning.
a. The median of 25 numbers is 13. Twelve of the numbers must be greater than 13.
b. The average of 11 numbers is 130. None of the 11 numbers are more than 260.
c. The average of 25 numbers is 100, and the median of those 25 numbers is also 100. The mode of the 25 numbers must be 100.
d. The mean of 45 numbers is 70. If you pick any group of 10 numbers from the 45, the mean will be 70.
e. The average of 42 numbers is 20. The sum of all 42 numbers is 840.
f. The average of 49 numbers is 100. If a 50th number is added and the average remains at 100, the 50th number must have been 100.
#8 Points possible: 9. Total attempts: 5
Rio Blanca City Hall publishes the following statistics on household incomes of the town’s citizens. The mode is given as a range.
Mean: $257,000 Median: $65,000 Mode: $20,000–$30,000
Which measure would be the most useful for each of the following situations?
a. State officials want to estimate the total amount of income of the citizens of Rio Blanca.
· mean
· median
· mode
b. The school district wants to know the income level of the largest number of students.
· mean
· median
· mode
c. A businesswoman is thinking about opening an expensive restaurant in the town. She wants to know how many people in town could afford to eat at her restaurant.
· mean
· median
· mode
#9 Points possible: 5. Total attempts: 5
A course has five exams, and passing the course requires a 75% average on the exams. Maria scored 60%, 72%, 80%, and 70% on the first four exams. What is the minimum score on the fifth exam that will let Maria pass the class? %
#10 Points possible: 12. Total attempts: 5
Use Figures 1 and 2 for the following questions.
a. Select the figure(s) with 40% shaded.
· Figure 1
· Figure 2
· Both Figure 1 and Figure 2
· Neither Figure 1 or Figure 2
b. Which of the following statements are correct? There may be more than one correct answer.
· The shaded area of Figure 1 is larger than the shaded area of Figure 2 because Figure 1 is larger than Figure 2.
· The shaded area of Figure 1 is the same as the shaded area of Figure 2 because they are both 40% of the square.
· The shaded area of Figure 1 is the same proportion of the figure as the shaded area of Figure 2 because they are both 40% of the square.
c. These figures illustrate what important concept?
· Percentages cannot be used for comparisons unless the reference values are equal.
· Percentages compare measures relative to the size of the reference values, but do not give information about absolute measures.
· Percentages are a ratio out of 100, so they can always be compared directly. In other words, 60% of one value is equal to 60% of another value.
#11 Points possible: 5. Total attempts: 5
In a psychology class, 58 students have a mean score of 81.4 on a test. Then 21 more students take the test and their mean score is 73.7. What is the mean score of all of these students together? Round to one decimal place. mean of the scores of all the students =
#12 Points possible: 5. Total attempts: 5
To compute a student’s Grade Point Average (GPA) for a term, the student’s grades for each course are weighted by the number of credits for the course. Suppose a student had these grades: 4.0 in a 5 credit Math course 2.6 in a 2 credit Music course 2.6 in a 4 credit Chemistry course 3.1 in a 6 credit Journalism course What is the student’s GPA for that term? Round to two decimal places. Student’s GPA =