Section 2.1 frequency distributions and their graphs answers

Lecture 2

Descriptive Statistics

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Larson/Farber 4th ed.

Larson/Farber 4th ed.

Chapter Outline

2.1 Frequency Distributions and Their Graphs
2.2 More Graphs and Displays
2.3 Measures of Central Tendency
2.4 Measures of Variation
2.5 Measures of Position
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Larson/Farber 4th ed.

Larson/Farber 4th ed.

Section 2.1

Frequency Distributions

and Their Graphs

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Larson/Farber 4th ed.

Larson/Farber 4th ed.

Section 2.1 Objectives

Construct frequency distributions
Construct frequency histograms, frequency polygons, relative frequency histograms, and ogives
Larson/Farber 4th ed.

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Larson/Farber 4th ed.

Frequency Distribution

Frequency Distribution

A table that shows classes or intervals of data with a count of the number of entries in each class.
The frequency, f, of a class is the number of data entries in the class.
Larson/Farber 4th ed.

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Class Frequency, f
1 – 5 5
6 – 10 8
11 – 15 6
16 – 20 8
21 – 25 5
26 – 30 4
Lower class

limits

Upper class

limits

Class width 6 – 1 = 5

Larson/Farber 4th ed.

Constructing a Frequency Distribution

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Decide on the number of classes.

Usually between 5 and 20; otherwise, it may be difficult to detect any patterns.
Find the class width.

Determine the range of the data.
Divide the range by the number of classes.
Round up to the next convenient number.
Larson/Farber 4th ed.

Constructing a Frequency Distribution

Find the class limits.

You can use the minimum data entry as the lower limit of the first class.
Find the remaining lower limits (add the class width to the lower limit of the preceding class).
Find the upper limit of the first class. Remember that classes cannot overlap.
Find the remaining upper class limits.
Larson/Farber 4th ed.

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Larson/Farber 4th ed.

Constructing a Frequency Distribution

Make a tally mark for each data entry in the row of the appropriate class.

Count the tally marks to find the total frequency f for each class.

Larson/Farber 4th ed.

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Larson/Farber 4th ed.

Example: Constructing a Frequency Distribution

The following sample data set lists the number of minutes 50 Internet subscribers spent on the Internet during their most recent session. Construct a frequency distribution that has seven classes.

50 40 41 17 11 7 22 44 28 21 19 23 37 51 54 42 86

41 78 56 72 56 17 7 69 30 80 56 29 33 46 31 39 20

18 29 34 59 73 77 36 39 30 62 54 67 39 31 53 44

Larson/Farber 4th ed.

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Larson/Farber 4th ed.

Solution: Constructing a Frequency Distribution

Number of classes = 7 (given)

Find the class width

Larson/Farber 4th ed.

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Round up to 12

50 40 41 17 11 7 22 44 28 21 19 23 37 51 54 42 86

41 78 56 72 56 17 7 69 30 80 56 29 33 46 31 39 20

18 29 34 59 73 77 36 39 30 62 54 67 39 31 53 44

Larson/Farber 4th ed.

Solution: Constructing a Frequency Distribution

Larson/Farber 4th ed.

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Class width = 12

Use 7 (minimum value) as first lower limit. Add the class width of 12 to get the lower limit of the next class.

7 + 12 = 19

Find the remaining lower limits.

19

31

43

55

67

79

Lower limit Upper limit
7
Larson/Farber 4th ed.

Solution: Constructing a Frequency Distribution

The upper limit of the first class is 18 (one less than the lower limit of the second class).

Add the class width of 12 to get the upper limit of the next class.

18 + 12 = 30

Find the remaining upper limits.

Larson/Farber 4th ed.

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Class width = 12

30

42

54

66

78

90

18

Lower limit Upper limit
7
19
31
43
55
67
79
Larson/Farber 4th ed.

Solution: Constructing a Frequency Distribution

Make a tally mark for each data entry in the row of the appropriate class.

Count the tally marks to find the total frequency f for each class.

Larson/Farber 4th ed.

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Σf = 50

Class Tally Frequency, f
7 – 18 IIII I 6
19 – 30 IIII IIII 10
31 – 42 IIII IIII III 13
43 – 54 IIII III 8
55 – 66 IIII 5
67 – 78 IIII I 6
79 – 90 II 2
Larson/Farber 4th ed.

Determining the Midpoint

Midpoint of a class

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Class width = 12

Class Midpoint Frequency, f
7 – 18 6
19 – 30 10
31 – 42 13
Larson/Farber 4th ed.

Determining the Relative Frequency

Relative Frequency of a class

Portion or percentage of the data that falls in a particular class.
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Class Frequency, f Relative Frequency
7 – 18 6
19 – 30 10
31 – 42 13
Larson/Farber 4th ed.

Determining the Cumulative Frequency

Cumulative frequency of a class

The sum of the frequency for that class and all previous classes.
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+

+

6

16

29

Class Frequency, f Cumulative frequency
7 – 18 6
19 – 30 10
31 – 42 13
Larson/Farber 4th ed.

Expanded Frequency Distribution

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Σf = 50

Class Frequency, f Midpoint Relative frequency Cumulative frequency
7 – 18 6 12.5 0.12 6
19 – 30 10 24.5 0.20 16
31 – 42 13 36.5 0.26 29
43 – 54 8 48.5 0.16 37
55 – 66 5 60.5 0.10 42
67 – 78 6 72.5 0.12 48
79 – 90 2 84.5 0.04 50
Larson/Farber 4th ed.

Graphs of Frequency Distributions

Frequency Histogram

A bar graph that represents the frequency distribution.
The horizontal scale is quantitative and measures the data values.
The vertical scale measures the frequencies of the classes.
Consecutive bars must touch.
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data values

frequency

Larson/Farber 4th ed.

Class Boundaries

Class boundaries

The numbers that separate classes without forming gaps between them.
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The distance from the upper limit of the first class to the lower limit of the second class is 19 – 18 = 1.
Half this distance is 0.5.
First class lower boundary = 7 – 0.5 = 6.5
First class upper boundary = 18 + 0.5 = 18.5
6.5 – 18.5

Class Class Boundaries Frequency, f
7 – 18 6
19 – 30 10
31 – 42 13
Larson/Farber 4th ed.

Class Boundaries

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Class Class boundaries Frequency, f
7 – 18 6.5 – 18.5 6
19 – 30 18.5 – 30.5 10
31 – 42 30.5 – 42.5 13
43 – 54 42.5 – 54.5 8
55 – 66 54.5 – 66.5 5
67 – 78 66.5 – 78.5 6
79 – 90 78.5 – 90.5 2
Larson/Farber 4th ed.

Example: Frequency Histogram

Construct a frequency histogram for the Internet usage frequency distribution.

Larson/Farber 4th ed.

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Class Class boundaries Midpoint Frequency, f
7 – 18 6.5 – 18.5 12.5 6
19 – 30 18.5 – 30.5 24.5 10
31 – 42 30.5 – 42.5 36.5 13
43 – 54 42.5 – 54.5 48.5 8
55 – 66 54.5 – 66.5 60.5 5
67 – 78 66.5 – 78.5 72.5 6
79 – 90 78.5 – 90.5 84.5 2
Larson/Farber 4th ed.

Solution: Frequency Histogram
(using Midpoints)

Larson/Farber 4th ed.

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Larson/Farber 4th ed.

Chart1
6
10
13
8
5
6
2
Time Online (minutes)
Time online (in minutes)
Frequency
Internet Usage
Sheet1
Column1 Time Online (minutes)
6
10
13
8
5
6
2
Solution: Frequency Histogram
(using class boundaries)

Larson/Farber 4th ed.

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6.5 18.5 30.5 42.5 54.5 66.5 78.5 90.5

You can see that more than half of the subscribers spent between 19 and 54 minutes on the Internet during their most recent session.

Larson/Farber 4th ed.

Chart1
6
10
13
8
5
6
2
Time Online (minutes)
Time online (in minutes)
Frequency
Internet Usage
Sheet1
Column1 Time Online (minutes)
12.5 6
24.5 10
36.5 13
48.5 8
60.5 5
72.5 6
84.5 2
Graphs of Frequency Distributions

Frequency Polygon

A line graph that emphasizes the continuous change in frequencies.
Larson/Farber 4th ed.

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data values

frequency

Larson/Farber 4th ed.

Example: Frequency Polygon

Construct a frequency polygon for the Internet usage frequency distribution.

Larson/Farber 4th ed.

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Class Midpoint Frequency, f
7 – 18 12.5 6
19 – 30 24.5 10
31 – 42 36.5 13
43 – 54 48.5 8
55 – 66 60.5 5
67 – 78 72.5 6
79 – 90 84.5 2
Larson/Farber 4th ed.

Solution: Frequency Polygon

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You can see that the frequency of subscribers increases up to 36.5 minutes and then decreases.

The graph should begin and end on the horizontal axis, so extend the left side to one class width before the first class midpoint and extend the right side to one class width after the last class midpoint.

Larson/Farber 4th ed.

Chart1
0.5
12.5
24.5
36.5
48.5
60.5
72.5
84.5
96.5
Time Online (minutes)
Time online (in minutes)
Frequency
Internet Usage
0
6
10
13
8
5
6
2
0
Sheet1
Column1 Time Online (minutes)
0.5 0
12.5 6
24.5 10
36.5 13
48.5 8
60.5 5
72.5 6
84.5 2
96.5 0
Graphs of Frequency Distributions

Relative Frequency Histogram

Has the same shape and the same horizontal scale as the corresponding frequency histogram.
The vertical scale measures the relative frequencies, not frequencies.
Larson/Farber 4th ed.

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data values

relative frequency

Larson/Farber 4th ed.

Example: Relative Frequency Histogram

Construct a relative frequency histogram for the Internet usage frequency distribution.

Larson/Farber 4th ed.

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Class Class boundaries Frequency, f Relative frequency
7 – 18 6.5 – 18.5 6 0.12
19 – 30 18.5 – 30.5 10 0.20
31 – 42 30.5 – 42.5 13 0.26
43 – 54 42.5 – 54.5 8 0.16
55 – 66 54.5 – 66.5 5 0.10
67 – 78 66.5 – 78.5 6 0.12
79 – 90 78.5 – 90.5 2 0.04
Larson/Farber 4th ed.

Solution: Relative Frequency Histogram

Larson/Farber 4th ed.

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6.5 18.5 30.5 42.5 54.5 66.5 78.5 90.5

From this graph you can see that 20% of Internet subscribers spent between 18.5 minutes and 30.5 minutes online.

Larson/Farber 4th ed.

Chart1
0.12
0.2
0.26
0.16
0.1
0.12
0.04
Time Online (minutes)
Time online (in minutes)
Relative Frequency
Internet Usage
Sheet1
Column1 Time Online (minutes)
12.5 0.12
24.5 0.2
36.5 0.26
48.5 0.16
60.5 0.1
72.5 0.12
84.5 0.04
Graphs of Frequency Distributions

Cumulative Frequency Graph or Ogive

A line graph that displays the cumulative frequency of each class at its upper class boundary.
The upper boundaries are marked on the horizontal axis.
The cumulative frequencies are marked on the vertical axis.
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data values

cumulative frequency

Larson/Farber 4th ed.

Constructing an Ogive

Construct a frequency distribution that includes cumulative frequencies as one of the columns.

Specify the horizontal and vertical scales.

The horizontal scale consists of the upper class boundaries.
The vertical scale measures cumulative frequencies.
Plot points that represent the upper class boundaries and their corresponding cumulative frequencies.

Larson/Farber 4th ed.

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Larson/Farber 4th ed.

Constructing an Ogive

Connect the points in order from left to right.

The graph should start at the lower boundary of the first class (cumulative frequency is zero) and should end at the upper boundary of the last class (cumulative frequency is equal to the sample size).

Larson/Farber 4th ed.

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Larson/Farber 4th ed.

Example: Ogive

Construct an ogive for the Internet usage frequency distribution.

Larson/Farber 4th ed.

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Class Class boundaries Frequency, f Cumulative frequency
7 – 18 6.5 – 18.5 6 6
19 – 30 18.5 – 30.5 10 16
31 – 42 30.5 – 42.5 13 29
43 – 54 42.5 – 54.5 8 37
55 – 66 54.5 – 66.5 5 42
67 – 78 66.5 – 78.5 6 48
79 – 90 78.5 – 90.5 2 50
Larson/Farber 4th ed.

Solution: Ogive

Larson/Farber 4th ed.

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6.5 18.5 30.5 42.5 54.5 66.5 78.5 90.5

From the ogive, you can see that about 40 subscribers spent 60 minutes or less online during their last session. The greatest increase in usage occurs between 30.5 minutes and 42.5 minutes.

Larson/Farber 4th ed.

Chart1
0
6
16
29
37
42
48
50
Time Online (minutes)
Time online (in minutes)
Cumulative frequency
Internet Usage
Sheet1
Column1 Time Online (minutes)
0
12.5 6
24.5 16
36.5 29
48.5 37
60.5 42
72.5 48
84.5 50
Section 2.1 Summary

Constructed frequency distributions
Constructed frequency histograms, frequency polygons, relative frequency histograms and ogives
Larson/Farber 4th ed.

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Larson/Farber 4th ed.

Section 2.2

More Graphs and Displays

Larson/Farber 4th ed.

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Larson/Farber 4th ed.

Section 2.2 Objectives

Graph quantitative data using stem-and-leaf plots and dot plots
Graph qualitative data using pie charts and Pareto charts
Graph paired data sets using scatter plots and time series charts
Larson/Farber 4th ed.

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Larson/Farber 4th ed.

Graphing Quantitative Data Sets

Stem-and-leaf plot

Each number is separated into a stem and a leaf.
Similar to a histogram.
Still contains original data values.
Larson/Farber 4th ed.

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Data: 21, 25, 25, 26, 27, 28,
30, 36, 36, 45

26

2

1 5 5 6 7 8

3

0 6 6

4

5

Larson/Farber 4th ed.

Example: Constructing a Stem-and-Leaf Plot

The following are the numbers of text messages sent last month by the cellular phone users on one floor of a college dormitory. Display the data in a stem-and-leaf plot.

Larson/Farber 4th ed.

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159 144 129 105 145 126 116 130 114 122 112 112 142 126
118 108 122 121 109 140 126 119 113 117 118 109 109 119
139 122 78 133 126 123 145 121 134 124 119 132 133 124
129 112 126 148 147

Larson/Farber 4th ed.

Solution: Constructing a Stem-and-Leaf Plot

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The data entries go from a low of 78 to a high of 159.
Use the rightmost digit as the leaf.
For instance,
78 = 7 | 8 and 159 = 15 | 9

List the stems, 7 to 15, to the left of a vertical line.
For each data entry, list a leaf to the right of its stem.
159 144 129 105 145 126 116 130 114 122 112 112 142 126
118 108 122 121 109 140 126 119 113 117 118 109 109 119
139 122 78 133 126 123 145 121 134 124 119 132 133 124
129 112 126 148 147

Larson/Farber 4th ed.

Solution: Constructing a Stem-and-Leaf Plot

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Include a key to identify the values of the data.

From the display, you can conclude that more than 50% of the cellular phone users sent between 110 and 130 text messages.

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Graphing Quantitative Data Sets

Dot plot

Each data entry is plotted, using a point, above a horizontal axis
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Data: 21, 25, 25, 26, 27, 28, 30, 36, 36, 45

26

20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

Larson/Farber 4th ed.

Example: Constructing a Dot Plot

Use a dot plot organize the text messaging data.

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So that each data entry is included in the dot plot, the horizontal axis should include numbers between 70 and 160.
To represent a data entry, plot a point above the entry's position on the axis.
If an entry is repeated, plot another point above the previous point.
159 144 129 105 145 126 116 130 114 122 112 112 142 126
118 108 122 121 109 140 126 119 113 117 118 109 109 119
139 122 78 133 126 123 145 121 134 124 119 132 133 124
129 112 126 148 147

Larson/Farber 4th ed.

Solution: Constructing a Dot Plot

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From the dot plot, you can see that most values cluster between 105 and 148 and the value that occurs the most is 126. You can also see that 78 is an unusual data value.

159 144 129 105 145 126 116 130 114 122 112 112 142 126
118 108 122 121 109 140 126 119 113 117 118 109 109 119
139 122 78 133 126 123 145 121 134 124 119 132 133 124
129 112 126 148 147

Larson/Farber 4th ed.

Graphing Qualitative Data Sets

Pie Chart

A circle is divided into sectors that represent categories.
The area of each sector is proportional to the frequency of each category.
Larson/Farber 4th ed.

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Larson/Farber 4th ed.

Example: Constructing a Pie Chart

The numbers of motor vehicle occupants killed in crashes in 2005 are shown in the table. Use a pie chart to organize the data. (Source: U.S. Department of Transportation, National Highway Traffic Safety Administration)

Larson/Farber 4th ed.

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Vehicle type Killed
Cars 18,440
Trucks 13,778
Motorcycles 4,553
Other 823
Larson/Farber 4th ed.

Solution: Constructing a Pie Chart

Find the relative frequency (percent) of each category.
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37,594

Vehicle type Frequency, f Relative frequency
Cars 18,440
Trucks 13,778
Motorcycles 4,553
Other 823
Larson/Farber 4th ed.

Solution: Constructing a Pie Chart

Construct the pie chart using the central angle that corresponds to each category.
To find the central angle, multiply 360º by the category's relative frequency.
For example, the central angle for cars is
360(0.49) ≈ 176º

Larson/Farber 4th ed.

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Larson/Farber 4th ed.

Solution: Constructing a Pie Chart

Larson/Farber 4th ed.

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360º(0.49)≈176º

360º(0.37)≈133º

360º(0.12)≈43º

360º(0.02)≈7º

Vehicle type Frequency, f Relative frequency Central angle
Cars 18,440 0.49
Trucks 13,778 0.37
Motorcycles 4,553 0.12
Other 823 0.02
Larson/Farber 4th ed.

Solution: Constructing a Pie Chart

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From the pie chart, you can see that most fatalities in motor vehicle crashes were those involving the occupants of cars.

Vehicle type Relative frequency Central angle
Cars 0.49 176º
Trucks 0.37 133º
Motorcycles 0.12 43º
Other 0.02 7º
Larson/Farber 4th ed.

Graphing Qualitative Data Sets

Pareto Chart

A vertical bar graph in which the height of each bar represents frequency or relative frequency.
The bars are positioned in order of decreasing height, with the tallest bar positioned at the left.
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Categories

Frequency

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Chart1
10
7
6
2
Series 1
Sheet1
Series 1
10
7
6
2
To resize chart data range, drag lower right corner of range.
Example: Constructing a Pareto Chart

In a recent year, the retail industry lost $41.0 million in inventory shrinkage. Inventory shrinkage is the loss of inventory through breakage, pilferage, shoplifting, and so on. The causes of the inventory shrinkage are administrative error ($7.8 million), employee theft ($15.6 million), shoplifting ($14.7 million), and vendor fraud ($2.9 million). Use a Pareto chart to organize this data. (Source: National Retail Federation and Center for Retailing Education, University of Florida)

Larson/Farber 4th ed.

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Larson/Farber 4th ed.

Solution: Constructing a Pareto Chart

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From the graph, it is easy to see that the causes of inventory shrinkage that should be addressed first are employee theft and shoplifting.

Cause $ (million)
Admin. error 7.8
Employee theft 15.6
Shoplifting 14.7
Vendor fraud 2.9
Larson/Farber 4th ed.

Chart1
Employee Theft
Shoplifting
Admin. Error
Vendor fraud
Series 1
Cause
Millions of dollars
Causes of Inventory Shrinkage
15.6
14.7
7.8
2.9
Sheet1
Series 1
Employee Theft 15.6
Shoplifting 14.7
Admin. Error 7.8
Vendor fraud 2.9
To resize chart data range, drag lower right corner of range.
Graphing Paired Data Sets

Paired Data Sets

Each entry in one data set corresponds to one entry in a second data set.
Graph using a scatter plot.
The ordered pairs are graphed as
points in a coordinate plane.
Used to show the relationship
between two quantitative variables.
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x

y

Larson/Farber 4th ed.

Example: Interpreting a Scatter Plot

The British statistician Ronald Fisher introduced a famous data set called Fisher's Iris data set. This data set describes various physical characteristics, such as petal length and petal width (in millimeters), for three species of iris. The petal lengths form the first data set and the petal widths form the second data set. (Source: Fisher, R. A., 1936)

Larson/Farber 4th ed.

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Larson/Farber 4th ed.

Example: Interpreting a Scatter Plot

As the petal length increases, what tends to happen to the petal width?

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Each point in the scatter plot represents the

petal length and petal width of one flower.

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Solution: Interpreting a Scatter Plot

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Interpretation

From the scatter plot, you can see that as the petal length increases, the petal width also tends to increase.

Larson/Farber 4th ed.

A complete discussion of types of correlation occurs in chapter 9. You may want, however, to discuss positive correlation, negative correlation, and no correlation at this point.

Be sure that students do not confuse correlation with causation.

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Graphing Paired Data Sets

Time Series

Data set is composed of quantitative entries taken at regular intervals over a period of time.
e.g., The amount of precipitation measured each day for one month.
Use a time series chart to graph.
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time

Quantitative data

Larson/Farber 4th ed.

Example: Constructing a Time Series Chart

The table lists the number of cellular telephone subscribers (in millions) for the years 1995 through 2005. Construct a time series chart for the number of cellular subscribers. (Source: Cellular Telecommunication & Internet Association)

Larson/Farber 4th ed.

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Larson/Farber 4th ed.

Solution: Constructing a Time Series Chart

Let the horizontal axis represent the years.
Let the vertical axis represent the number of subscribers (in millions).
Plot the paired data and connect them with line segments.
Larson/Farber 4th ed.

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Larson/Farber 4th ed.

Solution: Constructing a Time Series Chart

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The graph shows that the number of subscribers has been increasing since 1995, with greater increases recently.

Larson/Farber 4th ed.

Section 2.2 Summary

Graphed quantitative data using stem-and-leaf plots and dot plots
Graphed qualitative data using pie charts and Pareto charts
Graphed paired data sets using scatter plots and time series charts
Larson/Farber 4th ed.

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Larson/Farber 4th ed.

Section 2.3

Measures of Central Tendency

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Larson/Farber 4th ed.

Section 2.3 Objectives

Determine the mean, median, and mode of a population and of a sample
Determine the weighted mean of a data set and the mean of a frequency distribution
Describe the shape of a distribution as symmetric, uniform, or skewed and compare the mean and median for each
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Larson/Farber 4th ed.

Measures of Central Tendency

Measure of central tendency

A value that represents a typical, or central, entry of a data set.
Most common measures of central tendency:
Mean
Median
Mode
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Larson/Farber 4th ed.

Measure of Central Tendency: Mean

Mean (average)

The sum of all the data entries divided by the number of entries.
Sigma notation: Σx = add all of the data entries (x) in the data set.
Population mean:
Sample mean:
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Larson/Farber 4th ed.

Example: Finding a Sample Mean

The prices (in dollars) for a sample of roundtrip flights from Chicago, Illinois to Cancun, Mexico are listed. What is the mean price of the flights?

872 432 397 427 388 782 397

Larson/Farber 4th ed.

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Larson/Farber 4th ed.

Solution: Finding a Sample Mean

872 432 397 427 388 782 397
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The sum of the flight prices is
Σx = 872 + 432 + 397 + 427 + 388 + 782 + 397 = 3695
To find the mean price, divide the sum of the prices by the number of prices in the sample
The mean price of the flights is about $527.90.

Larson/Farber 4th ed.

Measure of Central Tendency: Median

Median

The value that lies in the middle of the data when the data set is ordered.
Measures the center of an ordered data set by dividing it into two equal parts.
If the data set has an
odd number of entries: median is the middle data entry.
even number of entries: median is the mean of the two middle data entries.
Larson/Farber 4th ed.

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Larson/Farber 4th ed.

Example: Finding the Median

The prices (in dollars) for a sample of roundtrip flights from Chicago, Illinois to Cancun, Mexico are listed. Find the median of the flight prices.

872 432 397 427 388 782 397

Larson/Farber 4th ed.

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Larson/Farber 4th ed.

Solution: Finding the Median

872 432 397 427 388 782 397
Larson/Farber 4th ed.

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First order the data.

388 397 397 427 432 782 872
There are seven entries (an odd number), the median is the middle, or fourth, data entry.

The median price of the flights is $427.

Larson/Farber 4th ed.

Example: Finding the Median

The flight priced at $432 is no longer available. What is the median price of the remaining flights?

872 397 427 388 782 397

Larson/Farber 4th ed.

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Larson/Farber 4th ed.

Solution: Finding the Median

872 397 427 388 782 397
Larson/Farber 4th ed.

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First order the data.

388 397 397 427 782 872
There are six entries (an even number), the median is the mean of the two middle entries.

The median price of the flights is $412.

Larson/Farber 4th ed.

Measure of Central Tendency: Mode

Mode

The data entry that occurs with the greatest frequency.
If no entry is repeated the data set has no mode.
If two entries occur with the same greatest frequency, each entry is a mode (bimodal).
Larson/Farber 4th ed.

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Larson/Farber 4th ed.

Example: Finding the Mode

The prices (in dollars) for a sample of roundtrip flights from Chicago, Illinois to Cancun, Mexico are listed. Find the mode of the flight prices.

872 432 397 427 388 782 397

Larson/Farber 4th ed.

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Larson/Farber 4th ed.

Solution: Finding the Mode

872 432 397 427 388 782 397
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Ordering the data helps to find the mode.

388 397 397 427 432 782 872
The entry of 397 occurs twice, whereas the other data entries occur only once.

The mode of the flight prices is $397.

Larson/Farber 4th ed.

Example: Finding the Mode

At a political debate a sample of audience members was asked to name the political party to which they belong. Their responses are shown in the table. What is the mode of the responses?

Larson/Farber 4th ed.

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Political Party Frequency, f
Democrat 34
Republican 56
Other 21
Did not respond 9
Larson/Farber 4th ed.

Solution: Finding the Mode

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The mode is Republican (the response occurring with the greatest frequency). In this sample there were more Republicans than people of any other single affiliation.

Political Party Frequency, f
Democrat 34
Republican 56
Other 21
Did not respond 9
Larson/Farber 4th ed.

Comparing the Mean, Median, and Mode

All three measures describe a typical entry of a data set.
Advantage of using the mean:
The mean is a reliable measure because it takes into account every entry of a data set.
Disadvantage of using the mean:
Greatly affected by outliers (a data entry that is far removed from the other entries in the data set).
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Larson/Farber 4th ed.

Example: Comparing the Mean, Median, and Mode

Find the mean, median, and mode of the sample ages of a class shown. Which measure of central tendency best describes a typical entry of this data set? Are there any outliers?

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Ages in a class
20 20 20 20 20 20 21
21 21 21 22 22 22 23
23 23 23 24 24 65
Larson/Farber 4th ed.

Solution: Comparing the Mean, Median, and Mode

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Mean:

Median:

20 years (the entry occurring with the
greatest frequency)

Mode:

Ages in a class
20 20 20 20 20 20 21
21 21 21 22 22 22 23
23 23 23 24 24 65
Larson/Farber 4th ed.

Solution: Comparing the Mean, Median, and Mode

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Mean ≈ 23.8 years Median = 21.5 years Mode = 20 years

The mean takes every entry into account, but is influenced by the outlier of 65.
The median also takes every entry into account, and it is not affected by the outlier.
In this case the mode exists, but it doesn't appear to represent a typical entry.
Larson/Farber 4th ed.

Solution: Comparing the Mean, Median, and Mode

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Sometimes a graphical comparison can help you decide which measure of central tendency best represents a data set.

In this case, it appears that the median best describes the data set.

Larson/Farber 4th ed.

Weighted Mean

Weighted Mean

The mean of a data set whose entries have varying weights.
where w is the weight of each entry x
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Larson/Farber 4th ed.

Example: Finding a Weighted Mean

You are taking a class in which your grade is determined from five sources: 50% from your test mean, 15% from your midterm, 20% from your final exam, 10% from your computer lab work, and 5% from your homework. Your scores are 86 (test mean), 96 (midterm), 82 (final exam), 98 (computer lab), and 100 (homework). What is the weighted mean of your scores? If the minimum average for an A is 90, did you get an A?

Larson/Farber 4th ed.

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Larson/Farber 4th ed.

Solution: Finding a Weighted Mean

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Your weighted mean for the course is 88.6. You did not get an A.

Source Score, x Weight, w x∙w
Test Mean 86 0.50 86(0.50)= 43.0
Midterm 96 0.15 96(0.15) = 14.4
Final Exam 82 0.20 82(0.20) = 16.4
Computer Lab 98 0.10 98(0.10) = 9.8
Homework 100 0.05 100(0.05) = 5.0
Σw = 1 Σ(x∙w) = 88.6
Larson/Farber 4th ed.

Mean of Grouped Data

Mean of a Frequency Distribution

Approximated by
where x and f are the midpoints and frequencies of a class, respectively

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Larson/Farber 4th ed.

Finding the Mean of a Frequency Distribution

In Words In Symbols

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Find the midpoint of each class.

Find the sum of the products of the midpoints and the frequencies.

Find the sum of the frequencies.

Find the mean of the frequency distribution.

Larson/Farber 4th ed.

Example: Find the Mean of a Frequency Distribution

Use the frequency distribution to approximate the mean number of minutes that a sample of Internet subscribers spent online during their most recent session.

Larson/Farber 4th ed.

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Class Midpoint Frequency, f
7 – 18 12.5 6
19 – 30 24.5 10
31 – 42 36.5 13
43 – 54 48.5 8
55 – 66 60.5 5
67 – 78 72.5 6
79 – 90 84.5 2
Larson/Farber 4th ed.

Solution: Find the Mean of a Frequency Distribution

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Class Midpoint, x Frequency, f (x∙f)
7 – 18 12.5 6 12.5∙6 = 75.0
19 – 30 24.5 10 24.5∙10 = 245.0
31 – 42 36.5 13 36.5∙13 = 474.5
43 – 54 48.5 8 48.5∙8 = 388.0
55 – 66 60.5 5 60.5∙5 = 302.5
67 – 78 72.5 6 72.5∙6 = 435.0
79 – 90 84.5 2 84.5∙2 = 169.0
n = 50 Σ(x∙f) = 2089.0
Larson/Farber 4th ed.

The Shape of Distributions

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Symmetric Distribution

A vertical line can be drawn through the middle of a graph of the distribution and the resulting halves are approximately mirror images.
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The Shape of Distributions

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Uniform Distribution (rectangular)

All entries or classes in the distribution have equal or approximately equal frequencies.
Symmetric.
Larson/Farber 4th ed.

The Shape of Distributions