1. Descriptive Statistics
Chapter 2
Descriptive Statistics
Larson/Farber 4th ed.

2. 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
Larson/Farber 4th ed.

3. Frequency Distributions and Their Graphs
Section 2.1
Frequency Distributions
and Their Graphs
Larson/Farber 4th ed.

4. Section 2.1 Objectives Construct frequency distributions
Construct frequency histograms, frequency polygons, relative frequency histograms, and ogives
Larson/Farber 4th ed.

5. 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.
Class
Frequency, f
1 – 5 5
6 – 10 8
11 – 15 6
16 – 20
21 – 25
26 – 30 4
Class width 6 – 1 = 5
Lower class
limits
Upper class
limits
Larson/Farber 4th ed.

6. Constructing a Frequency Distribution
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.

7. 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.

8. 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.

9. 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.
Larson/Farber 4th ed.

10. Solution: Constructing a Frequency Distribution
Number of classes = 7 (given)
Find the class width
Round up to 12
Larson/Farber 4th ed.

11. Solution: Constructing a Frequency Distribution
Use 7 (minimum value) as first lower limit. Add the class width of 12 to get the lower limit of the next class.
= 19
Find the remaining lower limits.
Lower limit
Upper limit 7
Class width = 12 19 31 43 55 67 79
Larson/Farber 4th ed.

12. 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 = 30 Find the remaining upper limits.
Lower limit
Upper limit 7 19 31 43 55 67 79
Class width = 12 18 30 42 54 66 78 90
Larson/Farber 4th ed.

13. 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.
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
79 – 90 II 2
Σf = 50
Larson/Farber 4th ed.

14. Determining the Midpoint
Midpoint of a class
Class
Midpoint
Frequency, f
7 – 18 6
19 – 30 10
31 – 42 13
Class width = 12
Larson/Farber 4th ed.

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

16. Determining the Cumulative Frequency
Cumulative frequency of a class
The sum of the frequency for that class and all previous classes.
Class
Frequency, f
Cumulative frequency
7 – 18 6
19 – 30 10
31 – 42 13 6 + 16 + 29
Larson/Farber 4th ed.

17. Expanded Frequency Distribution
Class
Frequency, f
Midpoint
Relative
frequency
Cumulative frequency
7 – 18 6
12.5
0.12
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
72.5 48
79 – 90 2
84.5
0.04 50
Σf = 50
Larson/Farber 4th ed.

18. 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.
data values
frequency
Larson/Farber 4th ed.

19. Class Boundaries Class boundaries
The numbers that separate classes without forming gaps between them.
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.
Class
Boundaries
Frequency, f
7 – 18 6
19 – 30 10
31 – 42 13
6.5 – 18.5
First class lower boundary = 7 – 0.5 = 6.5
First class upper boundary = = 18.5
Larson/Farber 4th ed.

20. Class Boundaries Class Class boundaries Frequency, f 7 – 18 6.5 – 18.5
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
79 – 90
78.5 – 90.5 2
Larson/Farber 4th ed.

21. Example: Frequency Histogram
Construct a frequency histogram for the Internet usage frequency distribution.
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
79 – 90
78.5 – 90.5
84.5 2
Larson/Farber 4th ed.

22. Solution: Frequency Histogram (using Midpoints)
Larson/Farber 4th ed.

23. Solution: Frequency Histogram (using class boundaries)
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.

24. Graphs of Frequency Distributions
Frequency Polygon
A line graph that emphasizes the continuous change in frequencies.
data values
frequency
Larson/Farber 4th ed.

25. Example: Frequency Polygon
Construct a frequency polygon for the Internet usage frequency distribution.
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
79 – 90
84.5 2
Larson/Farber 4th ed.

26. Solution: Frequency Polygon
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.
You can see that the frequency of subscribers increases up to 36.5 minutes and then decreases.
Larson/Farber 4th ed.

27. 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.
data values
relative frequency
Larson/Farber 4th ed.

28. Example: Relative Frequency Histogram
Construct a relative frequency histogram for the Internet usage frequency distribution.
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
79 – 90
78.5 – 90.5 2
0.04
Larson/Farber 4th ed.

29. Solution: Relative Frequency Histogram
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.

30. 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.
data values
cumulative frequency
Larson/Farber 4th ed.

31. 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.

32. 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.

33. Example: Ogive
Construct an ogive for the Internet usage frequency distribution.
Class
Class boundaries
Frequency, f
Cumulative frequency
7 – 18
6.5 – 18.5 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 48
79 – 90
78.5 – 90.5 2 50
Larson/Farber 4th ed.

34. Solution: Ogive
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.

35. Section 2.1 Summary Constructed frequency distributions
Constructed frequency histograms, frequency polygons, relative frequency histograms and ogives
Larson/Farber 4th ed.

36. More Graphs and Displays
Section 2.2
More Graphs and Displays
Larson/Farber 4th ed.

37. 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.

38. 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. 26 2 3 4 5
Data: 21, 25, 25, 26, 27, 28, , 36, 36, 45
Larson/Farber 4th ed.

39. 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.

40. Solution: Constructing a Stem-and-Leaf Plot
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 | and = 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.
Larson/Farber 4th ed.

41. Solution: Constructing a Stem-and-Leaf Plot
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.

42. Graphing Quantitative Data Sets
Dot plot
Each data entry is plotted, using a point, above a horizontal axis
Data: 21, 25, 25, 26, 27, 28, 30, 36, 36, 45 26
Larson/Farber 4th ed.

43. Example: Constructing a Dot Plot
Use a dot plot organize the text messaging data.
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.
Larson/Farber 4th ed.

44. Solution: Constructing a Dot Plot
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.
Larson/Farber 4th ed.

45. 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.

46. 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)
Vehicle type
Killed
Cars
18,440
Trucks
13,778
Motorcycles
4,553
Other
823
Larson/Farber 4th ed.

47. Solution: Constructing a Pie Chart
Find the relative frequency (percent) of each category.
Vehicle type
Frequency, f
Relative frequency
Cars
18,440
Trucks
13,778
Motorcycles
4,553
Other
823
37,594
Larson/Farber 4th ed.

48. 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.

49. Solution: Constructing a Pie Chart
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
360º(0.49)≈176º
360º(0.37)≈133º
360º(0.12)≈43º
360º(0.02)≈7º
Larson/Farber 4th ed.

50. Solution: Constructing a Pie Chart
Vehicle type
Relative frequency
Central angle
Cars
0.49
176º
Trucks
0.37
133º
Motorcycles
0.12
43º
Other
0.02 7º
From the pie chart, you can see that most fatalities in motor vehicle crashes were those involving the occupants of cars.
Larson/Farber 4th ed.

51. 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.
Frequency
Categories
Larson/Farber 4th ed.

52. 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.

53. Solution: Constructing a Pareto Chart
Cause
$ (million)
Admin. error
7.8
Employee theft
15.6
Shoplifting
14.7
Vendor fraud
2.9
From the graph, it is easy to see that the causes of inventory shrinkage that should be addressed first are employee theft and shoplifting.
Larson/Farber 4th ed.

54. Graphing 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. y x
Larson/Farber 4th ed.

55. 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.

56. Example: Interpreting a Scatter Plot
As the petal length increases, what tends to happen to the petal width?
Each point in the scatter plot represents the
petal length and petal width of one flower.
Larson/Farber 4th ed.

57. Solution: Interpreting a Scatter Plot
Interpretation
From the scatter plot, you can see that as the petal length increases, the petal width also tends to increase.
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.
Larson/Farber 4th ed.

58. 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.
time
Quantitative data
Larson/Farber 4th ed.

59. Example: Constructing a Time Series Chart
The table lists the number of cellular telephone subscribers (in millions) for the years 1995 through Construct a time series chart for the number of cellular subscribers. (Source: Cellular Telecommunication & Internet Association)
Larson/Farber 4th ed.

60. 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.

61. Solution: Constructing a Time Series Chart
The graph shows that the number of subscribers has been increasing since 1995, with greater increases recently.
Larson/Farber 4th ed.

62. 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.

63. Measures of Central Tendency
Section 2.3
Measures of Central Tendency
Larson/Farber 4th ed.

64. 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.

65. 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.

66. 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.

67. 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?
Larson/Farber 4th ed.

68. Solution: Finding a Sample Mean
The sum of the flight prices is
Σx = = 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 $
Larson/Farber 4th ed.

69. Measure of Central Tendency: 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.

70. 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
Larson/Farber 4th ed.

71. Solution: Finding the Median
First order the data.
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.

72. Example: Finding the Median
The flight priced at $432 is no longer available. What is the median price of the remaining flights?
Larson/Farber 4th ed.

73. Solution: Finding the Median
First order the data.
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.

74. Measure of Central Tendency: 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.

75. 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
Larson/Farber 4th ed.

76. Solution: Finding the Mode
Ordering the data helps to find the mode.
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.

77. 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?
Political Party
Frequency, f
Democrat 34
Republican 56
Other 21
Did not respond 9
Larson/Farber 4th ed.

78. Solution: Finding the Mode
Political Party
Frequency, f
Democrat 34
Republican 56
Other 21
Did not respond 9
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.
Larson/Farber 4th ed.

79. 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).
Larson/Farber 4th ed.

80. 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?
Ages in a class 20 21 22 23 24 65
Larson/Farber 4th ed.

81. Solution: Comparing the Mean, Median, and Mode
Ages in a class 20 21 22 23 24 65
Mean:
Median:
Mode:
20 years (the entry occurring with the greatest frequency)
Larson/Farber 4th ed.

82. Solution: Comparing the Mean, Median, and Mode
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.

83. Solution: Comparing the Mean, Median, and Mode
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.

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

85. 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.

86. Solution: Finding a Weighted Mean
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
Your weighted mean for the course is You did not get an A.
Larson/Farber 4th ed.

87. Mean of Grouped Data Mean of a Frequency Distribution Approximated by
where x and f are the midpoints and frequencies of a class, respectively
Larson/Farber 4th ed.

88. Finding the Mean of a Frequency Distribution
In Words In Symbols
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.

89. 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.
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
79 – 90
84.5 2
Larson/Farber 4th ed.

90. Solution: Find the Mean of a Frequency Distribution
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
72.5∙6 = 435.0
79 – 90
84.5 2
84.5∙2 = 169.0
n = 50
Σ(x∙f) =
Larson/Farber 4th ed.

91. The Shape of Distributions
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.
Larson/Farber 4th ed.

92. The Shape of Distributions
Uniform Distribution (rectangular)
All entries or classes in the distribution have equal or approximately equal frequencies.
Symmetric.
Larson/Farber 4th ed.

93. The Shape of Distributions
Skewed Left Distribution (negatively skewed)
The “tail” of the graph elongates more to the left.
The mean is to the left of the median.
Larson/Farber 4th ed.

94. The Shape of Distributions
Skewed Right Distribution (positively skewed)
The “tail” of the graph elongates more to the right.
The mean is to the right of the median.
Larson/Farber 4th ed.

95. 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
Larson/Farber 4th ed.

96. Section 2.4
Measures of Variation
Larson/Farber 4th ed.

97. 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
Larson/Farber 4th ed.

98. 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)
Larson/Farber 4th ed.

99. 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)
Larson/Farber 4th ed.

100. Solution: Finding the Range
Ordering the data helps to find the least and greatest salaries.
Range = (Max. salary) – (Min. salary)
= 47 – 37 = 10
The range of starting salaries is 10 or $10,000.
minimum
maximum
Larson/Farber 4th ed.

101. Deviation, Variance, and Standard 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
Larson/Farber 4th ed.

102. 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)
Solution:
First determine the mean starting salary.
Larson/Farber 4th ed.

103. Solution: Finding the Deviation
Determine the deviation for each data entry.
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 44
44 – 41.5 = 2.5 37
37 – 41.5 = –4.5 42
42 – 41.5 = 0.5
Σx = 415
Σ(x – μ) = 0
Larson/Farber 4th ed.

104. Deviation, Variance, and Standard Deviation
Population Variance
Population Standard Deviation
Sum of squares, SSx
Larson/Farber 4th ed.

105. Finding the Population Variance & Standard Deviation
In Words In Symbols
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.

106. Finding the Population Variance & Standard Deviation
In Words In Symbols
Divide by N to get the population variance.
Find the square root to get the population standard deviation.
Larson/Farber 4th ed.

107. 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) Recall μ = 41.5.
Larson/Farber 4th ed.

108. Solution: Finding the Population Standard Deviation
Determine SSx
N = 10
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 44
44 – 41.5 = 2.5
(2.5)2 = 6.25 37
37 – 41.5 = –4.5
(–4.5)2 = 20.25 42
42 – 41.5 = 0.5
(0.5)2 = 0.25
Σ(x – μ) = 0
SSx = 88.5
Larson/Farber 4th ed.

109. Solution: Finding the Population Standard Deviation
Population Variance
Population Standard Deviation
The population standard deviation is about 3.0, or $3000.
Larson/Farber 4th ed.

110. Deviation, Variance, and Standard Deviation
Sample Variance
Sample Standard Deviation
Larson/Farber 4th ed.

111. Finding the Sample Variance & Standard Deviation
In Words In Symbols
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.

112. Finding the Sample Variance & Standard Deviation
In Words In Symbols
Divide by n – 1 to get the sample variance.
Find the square root to get the sample standard deviation.
Larson/Farber 4th ed.

113. 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)
Larson/Farber 4th ed.

114. Solution: Finding the Sample Standard Deviation
Determine SSx
n = 10
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 44
44 – 41.5 = 2.5
(2.5)2 = 6.25 37
37 – 41.5 = –4.5
(–4.5)2 = 20.25 42
42 – 41.5 = 0.5
(0.5)2 = 0.25
Σ(x – μ) = 0
SSx = 88.5
Larson/Farber 4th ed.

115. Solution: Finding the Sample Standard Deviation
Sample Variance
Sample Standard Deviation
The sample standard deviation is about 3.1, or $3100.
Larson/Farber 4th ed.

116. 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.)
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
29.00
40.50
24.50
33.00
38.00
Larson/Farber 4th ed.

117. Solution: Using Technology to Find the Standard Deviation
Sample Mean
Sample Standard Deviation
Larson/Farber 4th ed.

118. 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.
Larson/Farber 4th ed.

119. 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:
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.

120. Interpreting Standard Deviation: Empirical Rule (68 – 95 – 99.7 Rule)
99.7% within 3 standard deviations
2.35%
95% within 2 standard deviations
13.5%
68% within 1 standard deviation
34%
Larson/Farber 4th ed.

121. 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 inches.
Larson/Farber 4th ed.

122. Solution: Using the Empirical Rule
Because the distribution is bell-shaped, you can use the Empirical Rule.
34%
13.5%
55.87
58.58
61.29 64
66.71
69.42
72.13
34% % = 47.5% of women are between 64 and inches tall.
Larson/Farber 4th ed.

123. Chebychev’s Theorem
The portion of any data set lying within k standard deviations (k > 1) of the mean is at least:
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.

124. 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.

125. Solution: Using Chebychev’s Theorem
k = 2: μ – 2σ = 39.2 – 2(24.8) = (use 0 since age can’t be negative) μ + 2σ = (24.8) = 88.8
At least 75% of the population of Florida is between 0 and 88.8 years old.
Larson/Farber 4th ed.

126. 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.
where n= Σf (the number of entries in the data set)
Larson/Farber 4th ed.

127. Example: Finding the Standard Deviation for Grouped Data
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 2 5 6 4
Larson/Farber 4th ed.

128. Solution: Finding the Standard Deviation for Grouped Data
First construct a frequency distribution.
Find the mean of the frequency distribution. x f xf 10
0(10) = 0 1 19
1(19) = 19 2 7
2(7) = 14 3
3(7) =21 4
4(2) = 8 5
5(1) = 5 6
6(4) = 24
The sample mean is about 1.8 children.
Σf = 50
Σ(xf )= 91
Larson/Farber 4th ed.

129. Solution: Finding the Standard Deviation for Grouped Data
Determine the sum of squares. x f 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
3 – 1.8 = 1.2
(1.2)2 = 1.44
1.44(7) = 10.08 4
4 – 1.8 = 2.2
(2.2)2 = 4.84
4.84(2) = 9.68 5
5 – 1.8 = 3.2
(3.2)2 = 10.24
10.24(1) = 10.24 6
6 – 1.8 = 4.2
(4.2)2 = 17.64
17.64(4) = 70.56
Larson/Farber 4th ed.

130. Solution: Finding the Standard Deviation for Grouped Data
Find the sample standard deviation.
The standard deviation is about 1.7 children.
Larson/Farber 4th ed.

131. 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
Larson/Farber 4th ed.

132. Section 2.5
Measures of Position
Larson/Farber 4th ed.

133. 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)
Larson/Farber 4th ed.

134. 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.
Larson/Farber 4th ed.

135. 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
Solution:
Q2 divides the data set into two halves.
Lower half
Upper half Q2
Larson/Farber 4th ed.

136. Solution: Finding Quartiles
The first and third quartiles are the medians of the lower and upper halves of the data set.
Lower half
Upper half Q1 Q2 Q3
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.
Larson/Farber 4th ed.

137. Interquartile Range Interquartile Range (IQR)
The difference between the third and first quartiles.
IQR = Q3 – Q1
Larson/Farber 4th ed.

138. Example: Finding the Interquartile Range
Find the interquartile range of the test scores. Recall Q1 = 10, Q2 = 15, and Q3 = 18
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.

139. 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
Larson/Farber 4th ed.

140. 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.
Whisker
Maximum entry
Minimum entry
Box
Median, Q2 Q3 Q1
Larson/Farber 4th ed.

141. 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
Solution: 5 10 15 18 37
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.
Larson/Farber 4th ed.

142. Percentiles and Other 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.

143. 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)
Larson/Farber 4th ed.

144. Solution: Interpreting Percentiles
The 72nd percentile corresponds to a test score of This means that 72% of the students had an SAT score of 1700 or less.
Larson/Farber 4th ed.

145. The Standard Score Standard Score (z-score)
Represents the number of standard deviations a given value x falls from the mean μ.
Larson/Farber 4th ed.

146. 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 Find the z-score that corresponds to the age for each actor or actress. Then compare your results.
Larson/Farber 4th ed.

147. Solution: Comparing z-Scores from Different Data Sets
Forest Whitaker
0.15 standard deviations above the mean
Helen Mirren
2.17 standard deviations above the mean
Larson/Farber 4th ed.

148. Solution: Comparing z-Scores from Different Data Sets
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.
Larson/Farber 4th ed.

149. 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.