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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

Larson/Farber 4th ed.

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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

Larson/Farber 4th ed.

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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.

Larson/Farber 4th ed.

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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.

Larson/Farber 4th ed.

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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

Larson/Farber 4th ed.

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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.

Larson/Farber 4th ed.

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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.

Larson/Farber 4th ed.

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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

Larson/Farber 4th ed.

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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

Larson/Farber 4th ed.

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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.

Larson/Farber 4th ed.

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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

Larson/Farber 4th ed.

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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

Larson/Farber 4th ed.

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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.

Larson/Farber 4th ed.

Graphing Quantitative Data Sets

Dot plot

  • Each data entry is plotted, using a point, above a horizontal axis

Larson/Farber 4th ed.

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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.

Larson/Farber 4th ed.

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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

Larson/Farber 4th ed.

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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.

Larson/Farber 4th ed.

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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

Larson/Farber 4th ed.

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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.

Larson/Farber 4th ed.

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Categories

Frequency

Larson/Farber 4th ed.

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

Larson/Farber 4th ed.

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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.

Larson/Farber 4th ed.

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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?

Larson/Farber 4th ed.

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Each point in the scatter plot represents the

petal length and petal width of one flower.

Larson/Farber 4th ed.

Solution: Interpreting a Scatter Plot

Larson/Farber 4th ed.

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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.

Larson/Farber 4th ed.

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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

Larson/Farber 4th ed.

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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

Larson/Farber 4th ed.

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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

Larson/Farber 4th ed.

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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

Larson/Farber 4th ed.

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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:

Larson/Farber 4th ed.

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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

Larson/Farber 4th ed.

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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).

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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

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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?

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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.

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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?

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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.

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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.

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The Shape of Distributions

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Skewed Left Distribution (negatively skewed)

  • The “tail” of the graph elongates more to the left.
  • The mean is to the left of the median.

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The Shape of Distributions

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Skewed Right Distribution (positively skewed)

  • The “tail” of the graph elongates more to the right.
  • The mean is to the right of the median.

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Section 2.3 Summary

  • Determined the mean, median, and mode of a population and of a sample
  • Determined the weighted mean of a data set and the mean of a frequency distribution
  • Described the shape of a distribution as symmetric, uniform, or skewed and compared the mean and median for each

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Larson/Farber 4th ed.

Section 2.4

Measures of Variation

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Section 2.4 Objectives

  • Determine the range of a data set
  • Determine the variance and standard deviation of a population and of a sample
  • Use the Empirical Rule and Chebychev’s Theorem to interpret standard deviation
  • Approximate the sample standard deviation for grouped data

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Larson/Farber 4th ed.

Range

Range

  • The difference between the maximum and minimum data entries in the set.
  • The data must be quantitative.
  • Range = (Max. data entry) – (Min. data entry)

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Larson/Farber 4th ed.

Example: Finding the Range

A corporation hired 10 graduates. The starting salaries for each graduate are shown. Find the range of the starting salaries.

Starting salaries (1000s of dollars)

41 38 39 45 47 41 44 41 37 42

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Larson/Farber 4th ed.

Solution: Finding the Range

  • Ordering the data helps to find the least and greatest salaries.

37 38 39 41 41 41 42 44 45 47

  • Range = (Max. salary) – (Min. salary)

= 47 – 37 = 10

The range of starting salaries is 10 or $10,000.

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minimum

maximum

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Deviation, Variance, and Standard Deviation

Deviation

  • The difference between the data entry, x, and the mean of the data set.
  • Population data set:
  • Deviation of x = x – μ
  • Sample data set:
  • Deviation of x = x – x

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Larson/Farber 4th ed.

Example: Finding the Deviation

A corporation hired 10 graduates. The starting salaries for each graduate are shown. Find the deviation of the starting salaries.

Starting salaries (1000s of dollars)

41 38 39 45 47 41 44 41 37 42

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Solution:

  • First determine the mean starting salary.

Larson/Farber 4th ed.

Solution: Finding the Deviation

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  • Determine the deviation for each data entry.

Σx = 415

Σ(x – μ) = 0

Salary ($1000s), x Deviation: x – μ
41 41 – 41.5 = –0.5
38 38 – 41.5 = –3.5
39 39 – 41.5 = –2.5
45 45 – 41.5 = 3.5
47 47 – 41.5 = 5.5
41 41 – 41.5 = –0.5
44 44 – 41.5 = 2.5
41 41 – 41.5 = –0.5
37 37 – 41.5 = –4.5
42 42 – 41.5 = 0.5

Larson/Farber 4th ed.

Deviation, Variance, and Standard Deviation

Population Variance

Population Standard Deviation

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Sum of squares, SSx

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Finding the Population Variance & Standard Deviation

In Words In Symbols

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Find the mean of the population data set.

Find deviation of each entry.

Square each deviation.

Add to get the sum of squares.

x – μ

(x – μ)2

SSx = Σ(x – μ)2

Larson/Farber 4th ed.

Finding the Population Variance & Standard Deviation

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  • Divide by N to get the population variance.

Find the square root to get the population standard deviation.

In Words In Symbols

Larson/Farber 4th ed.

Example: Finding the Population Standard Deviation

A corporation hired 10 graduates. The starting salaries for each graduate are shown. Find the population variance and standard deviation of the starting salaries.

Starting salaries (1000s of dollars)

41 38 39 45 47 41 44 41 37 42

Recall μ = 41.5.

Larson/Farber 4th ed.

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Larson/Farber 4th ed.

Solution: Finding the Population Standard Deviation

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  • Determine SSx
  • N = 10

Σ(x – μ) = 0

SSx = 88.5

Salary, x Deviation: x – μ Squares: (x – μ)2
41 41 – 41.5 = –0.5 (–0.5)2 = 0.25
38 38 – 41.5 = –3.5 (–3.5)2 = 12.25
39 39 – 41.5 = –2.5 (–2.5)2 = 6.25
45 45 – 41.5 = 3.5 (3.5)2 = 12.25
47 47 – 41.5 = 5.5 (5.5)2 = 30.25
41 41 – 41.5 = –0.5 (–0.5)2 = 0.25
44 44 – 41.5 = 2.5 (2.5)2 = 6.25
41 41 – 41.5 = –0.5 (–0.5)2 = 0.25
37 37 – 41.5 = –4.5 (–4.5)2 = 20.25
42 42 – 41.5 = 0.5 (0.5)2 = 0.25

Larson/Farber 4th ed.

Solution: Finding the Population Standard Deviation

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Population Variance

Population Standard Deviation

The population standard deviation is about 3.0, or $3000.

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Deviation, Variance, and Standard Deviation

Sample Variance

Sample Standard Deviation

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Larson/Farber 4th ed.

Finding the Sample Variance & Standard Deviation

In Words In Symbols

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Find the mean of the sample data set.

Find deviation of each entry.

Square each deviation.

Add to get the sum of squares.

Larson/Farber 4th ed.

Finding the Sample Variance & Standard Deviation

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  • Divide by n – 1 to get the sample variance.

Find the square root to get the sample standard deviation.

In Words In Symbols

Larson/Farber 4th ed.

Example: Finding the Sample Standard Deviation

The starting salaries are for the Chicago branches of a corporation. The corporation has several other branches, and you plan to use the starting salaries of the Chicago branches to estimate the starting salaries for the larger population. Find the sample standard deviation of the starting salaries.

Starting salaries (1000s of dollars)

41 38 39 45 47 41 44 41 37 42

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Larson/Farber 4th ed.

Solution: Finding the Sample Standard Deviation

Larson/Farber 4th ed.

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  • Determine SSx
  • n = 10

Σ(x – μ) = 0

SSx = 88.5

Salary, x Deviation: x – μ Squares: (x – μ)2
41 41 – 41.5 = –0.5 (–0.5)2 = 0.25
38 38 – 41.5 = –3.5 (–3.5)2 = 12.25
39 39 – 41.5 = –2.5 (–2.5)2 = 6.25
45 45 – 41.5 = 3.5 (3.5)2 = 12.25
47 47 – 41.5 = 5.5 (5.5)2 = 30.25
41 41 – 41.5 = –0.5 (–0.5)2 = 0.25
44 44 – 41.5 = 2.5 (2.5)2 = 6.25
41 41 – 41.5 = –0.5 (–0.5)2 = 0.25
37 37 – 41.5 = –4.5 (–4.5)2 = 20.25
42 42 – 41.5 = 0.5 (0.5)2 = 0.25

Larson/Farber 4th ed.

Solution: Finding the Sample Standard Deviation

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Sample Variance

Sample Standard Deviation

The sample standard deviation is about 3.1, or $3100.

Larson/Farber 4th ed.

Example: Using Technology to Find the Standard Deviation

Sample office rental rates (in dollars per square foot per year) for Miami’s central business district are shown in the table. Use a calculator or a computer to find the mean rental rate and the sample standard deviation. (Adapted from: Cushman & Wakefield Inc.)

Larson/Farber 4th ed.

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Office Rental Rates
35.00 33.50 37.00
23.75 26.50 31.25
36.50 40.00 32.00
39.25 37.50 34.75
37.75 37.25 36.75
27.00 35.75 26.00
37.00 29.00 40.50
24.50 33.00 38.00

Larson/Farber 4th ed.

Solution: Using Technology to Find the Standard Deviation

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Sample Mean

Sample Standard Deviation

Larson/Farber 4th ed.

Interpreting Standard Deviation

  • Standard deviation is a measure of the typical amount an entry deviates from the mean.
  • The more the entries are spread out, the greater the standard deviation.

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Larson/Farber 4th ed.

Interpreting Standard Deviation: Empirical Rule (68 – 95 – 99.7 Rule)

For data with a (symmetric) bell-shaped distribution, the standard deviation has the following characteristics:

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  • About 68% of the data lie within one standard deviation of the mean.
  • About 95% of the data lie within two standard deviations of the mean.
  • About 99.7% of the data lie within three standard deviations of the mean.

Larson/Farber 4th ed.

Interpreting Standard Deviation: Empirical Rule (68 – 95 – 99.7 Rule)

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68% within 1 standard deviation

34%

34%

99.7% within 3 standard deviations

2.35%

2.35%

95% within 2 standard deviations

13.5%

13.5%

Larson/Farber 4th ed.

Example: Using the Empirical Rule

In a survey conducted by the National Center for Health Statistics, the sample mean height of women in the United States (ages 20-29) was 64 inches, with a sample standard deviation of 2.71 inches. Estimate the percent of the women whose heights are between 64 inches and 69.42 inches.

Larson/Farber 4th ed.

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Larson/Farber 4th ed.

Solution: Using the Empirical Rule

Larson/Farber 4th ed.

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55.87

58.58

61.29

64

66.71

69.42

72.13

34%

13.5%

  • Because the distribution is bell-shaped, you can use the Empirical Rule.

34% + 13.5% = 47.5% of women are between 64 and 69.42 inches tall.

Larson/Farber 4th ed.

Chebychev’s Theorem

  • The portion of any data set lying within k standard deviations (k > 1) of the mean is at least:

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  • k = 2: In any data set, at least

of the data lie within 2 standard deviations of the mean.

  • k = 3: In any data set, at least

of the data lie within 3 standard deviations of the mean.

Larson/Farber 4th ed.

Example: Using Chebychev’s Theorem

The age distribution for Florida is shown in the histogram. Apply Chebychev’s Theorem to the data using k = 2. What can you conclude?

Larson/Farber 4th ed.

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Larson/Farber 4th ed.

Solution: Using Chebychev’s Theorem

k = 2: μ – 2σ = 39.2 – 2(24.8) = -10.4 (use 0 since age
can’t be negative)

μ + 2σ = 39.2 + 2(24.8) = 88.8

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At least 75% of the population of Florida is between 0 and 88.8 years old.

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Standard Deviation for Grouped Data

Sample standard deviation for a frequency distribution

  • When a frequency distribution has classes, estimate the sample mean and standard deviation by using the midpoint of each class.

Larson/Farber 4th ed.

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where n= Σf (the number of entries in the data set)

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Example: Finding the Standard Deviation for Grouped Data

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You collect a random sample of the number of children per household in a region. Find the sample mean and the sample standard deviation of the data set.

Number of Children in 50 Households
1 3 1 1 1
1 2 2 1 0
1 1 0 0 0
1 5 0 3 6
3 0 3 1 1
1 1 6 0 1
3 6 6 1 2
2 3 0 1 1
4 1 1 2 2
0 3 0 2 4

Larson/Farber 4th ed.

Solution: Finding the Standard Deviation for Grouped Data

  • First construct a frequency distribution.

  • Find the mean of the frequency distribution.

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Σf = 50

Σ(xf )= 91

The sample mean is about 1.8 children.

x f xf
0 10 0(10) = 0
1 19 1(19) = 19
2 7 2(7) = 14
3 7 3(7) =21
4 2 4(2) = 8
5 1 5(1) = 5
6 4 6(4) = 24

Larson/Farber 4th ed.

Solution: Finding the Standard Deviation for Grouped Data

  • Determine the sum of squares.

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x f
0 10 0 – 1.8 = –1.8 (–1.8)2 = 3.24 3.24(10) = 32.40
1 19 1 – 1.8 = –0.8 (–0.8)2 = 0.64 0.64(19) = 12.16
2 7 2 – 1.8 = 0.2 (0.2)2 = 0.04 0.04(7) = 0.28
3 7 3 – 1.8 = 1.2 (1.2)2 = 1.44 1.44(7) = 10.08
4 2 4 – 1.8 = 2.2 (2.2)2 = 4.84 4.84(2) = 9.68
5 1 5 – 1.8 = 3.2 (3.2)2 = 10.24 10.24(1) = 10.24
6 4 6 – 1.8 = 4.2 (4.2)2 = 17.64 17.64(4) = 70.56

Larson/Farber 4th ed.

Solution: Finding the Standard Deviation for Grouped Data

  • Find the sample standard deviation.

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The standard deviation is about 1.7 children.

Larson/Farber 4th ed.

Section 2.4 Summary

  • Determined the range of a data set
  • Determined the variance and standard deviation of a population and of a sample
  • Used the Empirical Rule and Chebychev’s Theorem to interpret standard deviation
  • Approximated the sample standard deviation for grouped data

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Larson/Farber 4th ed.

Section 2.5

Measures of Position

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Larson/Farber 4th ed.

Section 2.5 Objectives

  • Determine the quartiles of a data set
  • Determine the interquartile range of a data set
  • Create a box-and-whisker plot
  • Interpret other fractiles such as percentiles
  • Determine and interpret the standard score (z-score)

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Larson/Farber 4th ed.

Quartiles

  • Fractiles are numbers that partition (divide) an ordered data set into equal parts.
  • Quartiles approximately divide an ordered data set into four equal parts.
  • First quartile, Q1: About one quarter of the data fall on or below Q1.
  • Second quartile, Q2: About one half of the data fall on or below Q2 (median).
  • Third quartile, Q3: About three quarters of the data fall on or below Q3.

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Larson/Farber 4th ed.

Example: Finding Quartiles

The test scores of 15 employees enrolled in a CPR training course are listed. Find the first, second, and third quartiles of the test scores.

13 9 18 15 14 21 7 10 11 20 5 18 37 16 17

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Solution:

  • Q2 divides the data set into two halves.

5 7 9 10 11 13 14 15 16 17 18 18 20 21 37

Q2

Lower half

Upper half

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Solution: Finding Quartiles

The first and third quartiles are the medians of the lower and upper halves of the data set.

5 7 9 10 11 13 14 15 16 17 18 18 20 21 37

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About one fourth of the employees scored 10 or less, about one half scored 15 or less; and about three fourths scored 18 or less.

Q2

Lower half

Upper half

Q1

Q3

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Interquartile Range

Interquartile Range (IQR)

  • The difference between the third and first quartiles.
  • IQR = Q3 – Q1

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Example: Finding the Interquartile Range

Find the interquartile range of the test scores.

Recall Q1 = 10, Q2 = 15, and Q3 = 18

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Solution:

  • IQR = Q3 – Q1 = 18 – 10 = 8

The test scores in the middle portion of the data set vary by at most 8 points.

Larson/Farber 4th ed.

Box-and-Whisker Plot

Box-and-whisker plot

  • Exploratory data analysis tool.
  • Highlights important features of a data set.
  • Requires (five-number summary):
  • Minimum entry
  • First quartile Q1
  • Median Q2
  • Third quartile Q3
  • Maximum entry

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Larson/Farber 4th ed.

Drawing a Box-and-Whisker Plot

Find the five-number summary of the data set.

Construct a horizontal scale that spans the range of the data.

Plot the five numbers above the horizontal scale.

Draw a box above the horizontal scale from Q1 to Q3 and draw a vertical line in the box at Q2.

Draw whiskers from the box to the minimum and maximum entries.

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Whisker

Whisker

Maximum entry

Minimum entry

Box

Median, Q2

Q3

Q1

Larson/Farber 4th ed.

Example: Drawing a Box-and-Whisker Plot

Draw a box-and-whisker plot that represents the 15 test scores.

Recall Min = 5 Q1 = 10 Q2 = 15 Q3 = 18 Max = 37

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Solution:

About half the scores are between 10 and 18. By looking at the length of the right whisker, you can conclude 37 is a possible outlier.

5

10

15

18

37

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Percentiles and Other Fractiles

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Fractiles Summary Symbols
Quartiles Divides data into 4 equal parts Q1, Q2, Q3
Deciles Divides data into 10 equal parts D1, D2, D3,…, D9
Percentiles Divides data into 100 equal parts P1, P2, P3,…, P99

Larson/Farber 4th ed.

Example: Interpreting Percentiles

The ogive represents the cumulative frequency distribution for SAT test scores of college-bound students in a recent year. What test score represents the 72nd percentile? How should you interpret this? (Source: College Board Online)

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Larson/Farber 4th ed.

Solution: Interpreting Percentiles

The 72nd percentile corresponds to a test score of 1700.

This means that 72% of the students had an SAT score of 1700 or less.

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Larson/Farber 4th ed.

The Standard Score

Standard Score (z-score)

  • Represents the number of standard deviations a given value x falls from the mean μ.

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Example: Comparing z-Scores from Different Data Sets

In 2007, Forest Whitaker won the Best Actor Oscar at age 45 for his role in the movie The Last King of Scotland. Helen Mirren won the Best Actress Oscar at age 61 for her role in The Queen. The mean age of all best actor winners is 43.7, with a standard deviation of 8.8. The mean age of all best actress winners is 36, with a standard deviation of 11.5. Find the z-score that corresponds to the age for each actor or actress. Then compare your results.

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Larson/Farber 4th ed.

Solution: Comparing z-Scores from Different Data Sets

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  • Forest Whitaker
  • Helen Mirren

0.15 standard deviations above the mean

2.17 standard deviations above the mean

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Solution: Comparing z-Scores from Different Data Sets

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The z-score corresponding to the age of Helen Mirren is more than two standard deviations from the mean, so it is considered unusual. Compared to other Best Actress winners, she is relatively older, whereas the age of Forest Whitaker is only slightly higher than the average age of other Best Actor winners.

z = 0.15

z = 2.17

Larson/Farber 4th ed.

Section 2.5 Summary

  • Determined the quartiles of a data set
  • Determined the interquartile range of a data set
  • Created a box-and-whisker plot
  • Interpreted other fractiles such as percentiles
  • Determined and interpreted the standard score
    (z-score)

Larson/Farber 4th ed.

*

Larson/Farber 4th ed.

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