Descriptive Statistics

1
Chapter 2

• Descriptive Statistics

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

3
Section 2.1

• Frequency Distributions
• and Their Graphs

4
Section 2.1 Objectives

• Construct frequency distributions
• Construct frequency histograms, frequency
  polygons, relative frequency histograms, and
  ogives

5
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.

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

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.

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.

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.
• 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

10
Solution Constructing a Frequency Distribution
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

• Number of classes 7 (given)
• Find the class width

Round up to 12
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.
• 7 12 19
• Find the remaining lower limits.

Lower limit Upper limit
7






Class width 12
19
31
43
55
67
79
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.
• 18 12 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
13
Solution Constructing a Frequency Distribution

1. Make a tally mark for each data entry in the row
   of the appropriate class.
2. 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 IIII I 6
79 90 II 2
Sf 50
14
Determining the Midpoint

• Midpoint of a class

Class Midpoint Frequency, f
7 18 6
19 30 10
31 42 13
Class width 12
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
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
17
Expanded Frequency Distribution
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
Sf 50
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.

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 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 0.5 18.5

20
Class Boundaries
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
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 6
79 90 78.5 90.5 84.5 2
22
Solution Frequency Histogram (using Midpoints)
23
Solution Frequency Histogram (using class
boundaries)
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.
24
Graphs of Frequency Distributions

• Frequency Polygon
• A line graph that emphasizes the continuous
  change in frequencies.

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 6
79 90 84.5 2
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.
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.

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 6 0.12
79 90 78.5 90.5 2 0.04
29
Solution Relative Frequency Histogram
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.
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.

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.

32
Constructing an Ogive

1. Connect the points in order from left to right.
2. 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).

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 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
34
Solution Ogive
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.
35
Section 2.1 Summary

• Constructed frequency distributions
• Constructed frequency histograms, frequency
  polygons, relative frequency histograms and ogives

36
Graphical Analysis Use the frequency polygon to
identify the class with the greatest, and the
class with the least, frequency.
37
Graphical Analysis Use the relative frequency
histogram to (a) identify the class with the
greatest, and the class with the least, relative
frequency. (b) approximate the greatest and
least relative frequencies. (c) approximate the
relative frequency of the second class.
38
Section 2.2

• More Graphs and Displays

39
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

40
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.

Data 21, 25, 25, 26, 27, 28, 30, 36,
36, 45
41
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.

• 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

42
Solution Constructing a Stem-and-Leaf Plot

• 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

• 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.

43
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.
44
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
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
45
Example Constructing a Dot Plot

• Use a dot plot organize the text messaging data.

• 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

• 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.

46
Solution Constructing a Dot Plot

• 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

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.
47
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.

48
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
49
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
50
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

51
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
52
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.
53
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
54
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)

55
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.
56
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57
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 aspoints in a
  coordinate plane.
• Used to show the relationship between two
  quantitative variables.

y
x
58
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)

59
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.
60
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.
61
Interpretation Stay Away From Meat!
62
Interpretation Stay Away From Guns!
63
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.

64
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)

65
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.

66
Solution Constructing a Time Series Chart
The graph shows that the number of subscribers
has been increasing since 1995, with greater
increases recently.
67
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

68
Section 2.3

• Measures of Central Tendency

69
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

70
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

71
Measure of Central Tendency Mean

• Mean (average)
• The sum of all the data entries divided by the
  number of entries.
• Sigma notation Sx add all of the data
  entries (x) in the data set.
• Population mean
• Sample mean

72
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

73
Solution Finding a Sample Mean

• 872 432 397 427 388 782 397

• The sum of the flight prices is
• Sx 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.
74
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.

75
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

76
Solution Finding the Median

• 872 432 397 427 388 782 397

• 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.
77
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

78
Solution Finding the Median

• 872 397 427 388 782 397

• 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.
79
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).

80
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

81
Solution Finding the Mode

• 872 432 397 427 388 782 397

• 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.
82
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
83
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.
84
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).

85
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 Ages in a class Ages in a class Ages in a class Ages in a class Ages in a class Ages in a class
20 20 20 20 20 20 21
21 21 21 22 22 22 23
23 23 23 24 24 65
86
Solution Comparing the Mean, Median, and Mode
Ages in a class Ages in a class Ages in a class Ages in a class Ages in a class Ages in a class Ages in a class
20 20 20 20 20 20 21
21 21 21 22 22 22 23
23 23 23 24 24 65
Mean
Median
Mode
20 years (the entry occurring with thegreatest
frequency)
87
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.

88
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.
89
Weighted Mean

• Weighted Mean
• The mean of a data set whose entries have varying
  weights.
• where w is the weight of
  each entry x

90
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?

91
Solution Finding a Weighted Mean
Source Score, x Weight, w xw
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
Sw 1 S(xw) 88.6
Your weighted mean for the course is 88.6. You
did not get an A.
92
Mean of Grouped Data

• Mean of a Frequency Distribution
• Approximated by
• where x and f are the midpoints and frequencies
  of a class, respectively

93
Finding the Mean of a Frequency Distribution

• In Words In Symbols

1. Find the midpoint of each class.
2. Find the sum of the products of the midpoints and
   the frequencies.
3. Find the sum of the frequencies.
4. Find the mean of the frequency distribution.

94
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 6
79 90 84.5 2
95
Solution Find the Mean of a Frequency
Distribution
Class Midpoint, x Frequency, f (xf)
7 18 12.5 6 12.56 75.0
19 30 24.5 10 24.510 245.0
31 42 36.5 13 36.513 474.5
43 54 48.5 8 48.58 388.0
55 66 60.5 5 60.55 302.5
67 78 72.5 6 72.56 435.0
79 90 84.5 2 84.52 169.0
n 50 S(xf) 2089.0
96
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.

97
The Shape of Distributions

• Uniform Distribution (rectangular)
• All entries or classes in the distribution have
  equal or approximately equal frequencies.
• Symmetric.

98
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.

99
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.

100
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

101
Section 2.4

• Measures of Variation

102
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 Chebychevs Theorem to
  interpret standard deviation
• Approximate the sample standard deviation for
  grouped data

103
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)

104
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

105
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.

106
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

107
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

• Solution
• First determine the mean starting salary.

108
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
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
Sx 415
S(x µ) 0
109
Deviation, Variance, and Standard Deviation

• Population Variance
• Population Standard Deviation

Sum of squares, SSx
110
Finding the Population Variance Standard
Deviation

• In Words In Symbols

1. Find the mean of the population data set.
2. Find deviation of each entry.
3. Square each deviation.
4. Add to get the sum of squares.

x µ
(x µ)2
SSx S(x µ)2
111
Finding the Population Variance Standard
Deviation
In Words In Symbols

1. Divide by N to get the population variance.
2. Find the square root to get the population
   standard deviation.

112
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.

113
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
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
S(x µ) 0
SSx 88.5
114
Solution Finding the Population Standard
Deviation

• Population Variance
• Population Standard Deviation

The population standard deviation is about 3.0,
or 3000.
115
Deviation, Variance, and Standard Deviation

• Sample Variance
• Sample Standard Deviation

116
Finding the Sample Variance Standard Deviation

• In Words In Symbols

1. Find the mean of the sample data set.
2. Find deviation of each entry.
3. Square each deviation.
4. Add to get the sum of squares.

117
Finding the Sample Variance Standard Deviation
In Words In Symbols

1. Divide by n 1 to get the sample variance.
2. Find the square root to get the sample standard
   deviation.

118
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

119
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
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
S(x µ) 0
SSx 88.5
120
Solution Finding the Sample Standard Deviation

• Sample Variance
• Sample Standard Deviation

The sample standard deviation is about 3.1, or
3100.
121
Example Using Technology to Find the Standard
Deviation

• Sample office rental rates (in dollars per square
  foot per year) for Miamis 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 Office Rental Rates 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
122
Solution Using Technology to Find the Standard
Deviation
Sample Mean Sample Standard Deviation
123
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.

124
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.

125
Interpreting Standard Deviation Empirical Rule
(68 95 99.7 Rule)
126
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.

127
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 13.5 47.5 of women are between 64 and
69.42 inches tall.
128
Chebychevs Theorem

• The portion of any data set lying within k
  standard deviations (k gt 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.
129
Example Using Chebychevs Theorem

• The age distribution for Florida is shown in the
  histogram. Apply Chebychevs Theorem to the data
  using k 2. What can you conclude?

130
Solution Using Chebychevs Theorem

• k 2 µ 2s 39.2 2(24.8) -10.4 (use 0
  since age cant be negative)
• µ 2s 39.2 2(24.8) 88.8

At least 75 of the population of Florida is
between 0 and 88.8 years old.
131
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 Sf (the number of entries in the data
set)
132
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 Number of Children in 50 Households Number of Children in 50 Households Number of Children in 50 Households 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
133
Solution Finding the Standard Deviation for
Grouped Data

• First construct a frequency distribution.
• Find the mean of the frequency distribution.

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
The sample mean is about 1.8 children.
Sf 50
S(xf ) 91
134
Solution Finding the Standard Deviation for
Grouped Data

• Determine the sum of squares.

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
135
Solution Finding the Standard Deviation for
Grouped Data

• Find the sample standard deviation.

The standard deviation is about 1.7 children.
136
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 Chebychevs Theorem
  to interpret standard deviation
• Approximated the sample standard deviation for
  grouped data

137
Section 2.5

• Measures of Position

138
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)

139
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.

140
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

• Solution
• Q2 divides the data set into two halves.
• 5 7 9 10 11 13 14 15 16 17 18 18 20
  21 37

141
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

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.
142
The number of nuclear power plants in the top 15
nuclear power-producing countries in the world
are listed. Find the first, second, and third
quartiles of the data set. What can you conclude?
National Institute of Standards and Technology
 
143
The 2-quantile is called the median The
3-quantiles are called tertiles or terciles ?
T The 4-quantiles are called quartiles ? Q The
5-quantiles are called quintiles ? QU The
6-quantiles are called sextiles ? S The
10-quantiles are called deciles ? D The
12-quantiles are called duo-deciles ? Dd The
20-quantiles are called vigintiles ? V The
100-quantiles are called percentiles ? P The
1000-quantiles are called permilles ? Pr
144
Interquartile Range

• Interquartile Range (IQR)
• The difference between the third and first
  quartiles.
• IQR Q3 Q1

145
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.
146
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

147
Drawing a Box-and-Whisker Plot

1. Find the five-number summary of the data set.
2. Construct a horizontal scale that spans the range
   of the data.
3. Plot the five numbers above the horizontal scale.
4. Draw a box above the horizontal scale from Q1 to
   Q3 and draw a vertical line in the box at Q2.
5. Draw whiskers from the box to the minimum and
   maximum entries.

148
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
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.
149
Percentiles and Other Fractiles
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
150
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)

151
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.

152
The Standard Score

• Standard Score (z-score)
• Represents the number of standard deviations a
  given value x falls from the mean µ.

153
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.

154
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
155
Solution Comparing z-Scores from Different Data
Sets
z 0.15
z 2.17
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.
156
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)

157
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158
Over the same period, the lowest-income fifth saw
a decrease in real income of 7.4 percent.
Between 1979 and 2009, the top 5 percent of
American families saw their real incomes increase
72.7 percent, according to Census data.
159
(No Transcript)
160
Chapter 2 Descriptive Statistics
Elementary Statistics Picturing the World Fifth
Edition by Larson and Farber
161

• Find the class width
• 3
• 4
• 5
• 19

Class Frequency, f
1 5 21
6 10 16
11 15 28
16 20 13
162

• Find the class width
• 3
• 4
• 5
• 19

Class Frequency, f
1 5 21
6 10 16
11 15 28
16 20 13
163

• Estimate the frequency of the class with the
  greatest frequency.
• 28
• 21
• 58
• 53

164

• Estimate the frequency of the class with the
  greatest frequency.
• 28
• 21
• 58
• 53

165

• What is the maximum data entry?
• 96
• 38
• 9
• 41

Key 3 8 38
166

• What is the maximum data entry?
• 96
• 38
• 9
• 41

Key 3 8 38
167

• The heights (in inches) of a sample of basketball
  players are shown
• 76 79 81 78 82 78
• Find the mean.
• 78.5
• 79
• 474
• 78

168

• The heights (in inches) of a sample of basketball
  players are shown
• 76 79 81 78 82 78
• Find the mean.
• 78.5
• 79
• 474
• 78

 
169

• The heights (in inches) of a sample of basketball
  players are shown
• 76 79 81 78 82 78
• Find the median.
• 78.5
• 79
• 79.5
• 78

170

• The heights (in inches) of a sample of basketball
  players are shown
• 76 79 81 78 82 78
• Find the median.
• 78.5
• 79
• 79.5
• 78

171

• The heights (in inches) of a sample of basketball
  players are shown
• 76 79 81 78 82 78
• Find the mode.
• 78.5
• 79
• 79.5
• 78

172

• The heights (in inches) of a sample of basketball
  players are shown
• 76 79 81 78 82 78
• Find the mode.
• 78.5
• 79
• 79.5
• 78

173

• The heights (in inches) of a sample of basketball
  players are shown
• 76 79 81 78 82 78
• Find the standard deviation.
• 2.2
• 6
• 2
• 4.8

174

• The heights (in inches) of a sample of basketball
  players are shown
• 76 79 81 78 82 78
• Find the standard deviation.
• 2.2
• 6
• 2
• 4.8

 
 
175

• The mean annual automobile insurance premium is
  950, with a standard deviation of 175. The data
  set has a bell-shaped distribution. Estimate the
  percent of premiums that are between 600 and
  1300.
• 68
• 75
• 95
• 99.7

176

• The mean annual automobile insurance premium is
  950, with a standard deviation of 175. The data
  set has a bell-shaped distribution. Estimate the
  percent of premiums that are between 600 and
  1300.
• 68
• 75
• 95
• 99.7

177

• Use the box-and-whisker plot to identify the
  first quartile.
• 10
• 18
• 24
• 26

10
18
24
26
30

10 12 14 16 18 20
22 24 26 28 30
178

• Use the box-and-whisker plot to identify the
  first quartile.
• 10
• 18
• 24
• 26

10
18
24
26
30

10 12 14 16 18 20
22 24 26 28 30
179

• The mean annual automobile insurance premium is
  950, with a standard deviation of 175. Find the
  z-score that corresponds to a premium of 1250.
• 1.13
• 1.13
• 1.71
• 1.71

180

• The mean annual automobile insurance premium is
  950, with a standard deviation of 175. Find the
  z-score that corresponds to a premium of 1250.
• 1.13
• 1.13
• 1.71
• 1.71