Transcript Ch. 2
Chapter 2
Descriptive Statistics
1
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 2
Section 2.1
Frequency Distributions and Their Graphs
3
Section 2.1 Objectives
• • Construct frequency distributions Construct frequency histograms, frequency polygons, relative frequency histograms, and ogives 4
Frequency Distribution
• •
Frequency Distribution
A table that shows
classes
or
intervals
of Class width 6 – 1 = 5 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.
Lower class limits Class 1 – 5 6 – 10 11 – 15 16 – 20 21 – 25 26 – 30 Frequency,
f
5 8 6 8 5 4 Upper class limits 5
Constructing a Frequency Distribution
1.
Decide on the number of classes. Usually between 5 and 20; otherwise, it may be difficult to detect any patterns. 2.
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.
6
Constructing a Frequency Distribution
3.
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.
7
Constructing a Frequency Distribution
4.
5.
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. 8
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 9
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 1.
2.
Number of classes = 7 (given) Find the class width #classes 7 11.29
Round up to 12 10
Solution: Constructing a Frequency Distribution
3.
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.
Class width = 12 Lower limit 7 19 31 43 55 67 79 Upper limit 11
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 7 19 31 43 55 67 79 Upper limit 18 30 42 54 66 78 90 Class width = 12 12
4.
5.
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 19 – 30 31 – 42 43 – 54 55 – 66 67 – 78 79 – 90 IIII I IIII IIII IIII IIII III IIII III IIII IIII I II 6 10 13 8 5 6 2 Σ
f
= 50 13
Determining the Midpoint
Midpoint of a class
(Lower class limit) (Upper class limit) 2 Class 7 – 18 19 – 30 31 – 42 Midpoint 12.5
2 24.5
2 36.5
2 Frequency,
f
6 Class width = 12 10 13 14
Determining the Relative Frequency
•
Relative Frequency of a class
Portion or percentage of the data that falls in a particular class. • relative frequency class frequency Sample size
f n
Class 7 – 18 19 – 30 31 – 42 Frequency,
f
6 10 13 Relative Frequency 6 50 0.12
10 50 0.20
13 50 0.26
15
Determining the Cumulative Frequency
•
Cumulative frequency of a class
The sum of the frequency for that class and all previous classes.
Class 7 – 18 19 – 30 31 – 42 Frequency,
f
6 + 10 + 13 Cumulative frequency 6 16 29 16
Expanded Frequency Distribution
Class 7 – 18 19 – 30 31 – 42 43 – 54 55 – 66 67 – 78 79 – 90 Frequency,
f
6 10 13 8 5 6 2 Σ
f
= 50 Midpoint 12.5
24.5
36.5
48.5
60.5
72.5
84.5
Relative frequency 0.12
0.20
0.26
0.16
0.10
0.12
0.04
f n
1 Cumulative frequency 6 16 29 37 42 48 50 17
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 18
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 7 – 18 19 – 30 31 – 42 Class Boundaries 6.5 – 18.5
Frequency,
f
6 10 13 • • First class lower boundary = 7 – 0.5 = 6.5
First class upper boundary = 18 + 0.5 = 18.5
19
Class 7 – 18 19 – 30 31 – 42 43 – 54 55 – 66 67 – 78 79 – 90
Class Boundaries
Class boundaries 6.5 – 18.5
18.5 – 30.5
30.5 – 42.5
42.5 – 54.5
54.5 – 66.5
66.5 – 78.5
78.5 – 90.5
Frequency,
f
6 10 13 8 5 6 2 20
Example: Frequency Histogram
Construct a frequency histogram for the Internet usage frequency distribution. Class 7 – 18 19 – 30 31 – 42 43 – 54 55 – 66 67 – 78 79 – 90 Class boundaries 6.5 – 18.5
18.5 – 30.5
30.5 – 42.5
42.5 – 54.5
54.5 – 66.5
66.5 – 78.5
78.5 – 90.5
Midpoint 12.5
24.5
36.5
48.5
60.5
72.5
84.5
Frequency,
f
6 10 13 8 5 6 2 21
Solution: Frequency Histogram (using Midpoints)
22
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.
23
Graphs of Frequency Distributions
•
Frequency Polygon
A line graph that emphasizes the continuous change in frequencies.
data values 24
Example: Frequency Polygon
Construct a frequency polygon for the Internet usage frequency distribution. Class 7 – 18 19 – 30 31 – 42 43 – 54 55 – 66 67 – 78 79 – 90 Midpoint 12.5
24.5
36.5
48.5
60.5
72.5
84.5
Frequency,
f
6 10 13 8 5 6 2 25
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.
14 12 10 8 6 4 2 0
Internet Usage
0.5
12.5
24.5
36.5
48.5
60.5
Time online (in minutes)
72.5
84.5
96.5
You can see that the frequency of subscribers increases up to 36.5 minutes and then decreases.
26
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 not frequencies.
relative frequencies
, data values 27
Example: Relative Frequency Histogram
Construct a relative frequency histogram for the Internet usage frequency distribution. Class 7 – 18 19 – 30 31 – 42 43 – 54 55 – 66 67 – 78 79 – 90 Class boundaries 6.5 – 18.5
18.5 – 30.5
30.5 – 42.5
42.5 – 54.5
54.5 – 66.5
66.5 – 78.5
78.5 – 90.5
Frequency,
f
6 10 13 8 5 6 2 Relative frequency 0.12
0.20
0.26
0.16
0.10
0.12
0.04
28
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.
29
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 30
Constructing an Ogive
1.
Construct a frequency distribution that includes cumulative frequencies as one of the columns.
2.
Specify the horizontal and vertical scales.
The horizontal scale consists of the upper class boundaries.
The vertical scale measures cumulative frequencies.
3.
Plot points that represent the upper class boundaries and their corresponding cumulative frequencies.
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Constructing an Ogive
4.
5.
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).
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Example: Ogive
Construct an ogive for the Internet usage frequency distribution. Class 7 – 18 19 – 30 31 – 42 43 – 54 55 – 66 67 – 78 79 – 90 Class boundaries 6.5 – 18.5
18.5 – 30.5
30.5 – 42.5
42.5 – 54.5
54.5 – 66.5
66.5 – 78.5
78.5 – 90.5
Frequency,
f
6 10 13 8 5 6 2 Cumulative frequency 6 16 29 37 42 48 50 33
Solution: Ogive Internet Usage
60 50 40 30 20 10 0 6.5 18.5 30.5 42.5 54.5 66.5 78.5 90.5
Time online (in minutes)
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.
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Section 2.1 Summary
• • Constructed frequency distributions Constructed frequency histograms, frequency polygons, relative frequency histograms and ogives 35
Graphical Analysis
: U
se the frequency polygon to identify the class with the greatest, and the class with the least, frequency.
36
Graphical Analysis
: U
se 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.
37
Section 2.2
More Graphs and Displays
38
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 39
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
Data: 21, 25, 25,
26
30, 36, 36, 45 , 27, 28, 2 3 4 1 5 5 6 7 8 0 6 6 5 40
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.
155 159 144 129 105 145 126 116 130 114 122 112 112 142 126 118 118 108 122 121 109 140 126 119 113 117 118 109 109 119 139 139 122 78 133 126 123 145 121 134 124 119 132 133 124 129 112 126 148 147 41
Solution: Constructing a Stem-and-Leaf Plot
155 159 144 129 105 145 126 116 130 114 122 112 112 142 126 118 118 108 122 121 109 140 126 119 113 117 118 109 109 119 139 139 122 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.
42
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.
43
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 44
Example: Constructing a Dot Plot
Use a dot plot organize the text messaging data.
155 159 144 129 105 145 126 116 130 114 122 112 112 142 126 118 118 108 122 121 109 140 126 119 113 117 118 109 109 119 139 139 122 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.
45
Solution: Constructing a Dot Plot
155 159 144 129 105 145 126 116 130 114 122 112 112 142 126 118 118 108 122 121 109 140 126 119 113 117 118 109 109 119 139 139 122 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.
46
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.
47
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 Trucks Motorcycles Other 18,440 13,778 4,553 823 48
Solution: Constructing a Pie Chart
• Find the relative frequency (percent) of each category.
Vehicle type Frequency,
f
Cars Trucks Motorcycles Other 18,440 13,778 4,553 823 37,594 Relative frequency 18440 37594 0.49
13778 37594 0.37
4553 37594 0.12
823 37594 0.02
49
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 50
Solution: Constructing a Pie Chart
Vehicle type Frequency,
f
Cars 18,440 Relative frequency 0.49
Central angle 360º(0.49)≈176º Trucks 13,778 0.37
360º(0.37)≈133º Motorcycles 4,553 0.12
360º(0.12)≈43º Other 823 0.02
360º(0.02)≈7º 51
Solution: Constructing a Pie Chart
Vehicle type Cars Trucks Motorcycles Other Relative frequency 0.49
0.37
0.12
0.02
Central angle 176º 133º 43º 7º From the pie chart, you can see that most fatalities in motor vehicle crashes were those involving the occupants of cars.
52
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.
Categories 53
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)
54
Solution: Constructing a Pareto Chart
Cause Admin. error Employee theft Shoplifting Vendor fraud $ (million) 7.8
15.6
14.7
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.
55
.
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
y
points in a coordinate plane.
Used to show the relationship between two quantitative variables.
x
57
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)
58
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.
59
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.
60
Interpretation: Stay Away From Meat!
61
Interpretation: Stay Away From Guns!
62
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 63
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)
64
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.
65
Solution: Constructing a Time Series Chart
The graph shows that the number of subscribers has been increasing since 1995, with greater increases recently.
66
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 67
Section 2.3
Measures of Central Tendency
68
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 69
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 70
Measure of Central Tendency: Mean
• • • •
Mean
(average) The sum of all the data entries divided by the number of entries.
Sigma notation
in the data set.
: Σ
Population mean
:
x
= add all of the data entries (
x
)
x N
Sample mean
:
x
x n
71
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 72
Solution: Finding a Sample Mean
872 432 397 427 388 782 397 • The sum of the flight prices is Σ
x
= 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
x
3695
x
527.9
n
7 The mean price of the flights is about $527.90.
73
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
entry.
: median is the middle data
even number of entries
: median is the mean of the two middle data entries.
74
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 75
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.
76
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 77
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.
397 427 Median 412 2 The median price of the flights is $412.
78
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
).
79
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 80
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.
81
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
Democrat Republican Other Did not respond
Frequency, f
34 56 21 9 82
Solution: Finding the Mode Political Party
Democrat Republican Other Did not respond
Frequency, f
34 56 21 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.
83
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).
84
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?
20 21 23 20 21 23
Ages in a class
20 20 20 21 23 22 24 22 24 20 22 65 21 23 85
Solution: Comparing the Mean, Median, and Mode
20 21 23 20 21 23
Ages in a class
20 20 20 21 23 22 24 22 24 20 22 65 21 23 Mean: Median:
x
x n
20 21.5 years 2 Mode: greatest frequency) 23.8 years 20 years (the entry occurring with the 86
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. 87
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.
88
Weighted Mean
•
Weighted Mean
The mean of a data set whose entries have varying weights.
•
x
)
w
where
w
is the weight of each entry
x
89
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?
90
Solution: Finding a Weighted Mean Source
Test Mean Midterm Final Exam Computer Lab Homework
Score, x
86 96 82 98 100
Weight, w
0.50
0.15
0.20
0.10
0.05
Σ
w =
1
x∙w
86(0.50)= 43.0
96(0.15) = 14.4
82(0.20) = 16.4
98(0.10) = 9.8
100(0.05) = 5.0
Σ(
x∙w
)
=
88.6
x
(
w
) 88.6
1 88.6
Your weighted mean for the course is 88.6. You did not get an A.
91
Mean of Grouped Data
•
Mean of a Frequency Distribution
Approximated by
x
x f
)
n
f n
where
x
and
f
are the midpoints and frequencies of a class, respectively 92
Finding the Mean of a Frequency Distribution
In Words
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.
In Symbols x
(lower limit)+(upper limit) 2
x
x f
)
n
f x f
)
n
93
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 7 – 18 19 – 30 31 – 42 43 – 54 55 – 66 67 – 78 79 – 90 Midpoint 12.5
24.5
36.5
48.5
60.5
72.5
84.5
Frequency,
f
6 10 13 8 5 6 2 94
Solution: Find the Mean of a Frequency Distribution
Class 7 – 18 19 – 30 31 – 42 43 – 54 55 – 66 67 – 78 79 – 90 Midpoint,
x
12.5
24.5
36.5
48.5
60.5
72.5
84.5
Frequency,
f
6 10 13 8 5 6 2 n = 50 (
x∙f
) 12.5∙6 = 75.0
24.5∙10 = 245.0
36.5∙13 = 474.5
48.5∙8 = 388.0
60.5∙5 = 302.5
72.5∙6 = 435.0
84.5∙2 = 169.0
Σ(
x∙f
) = 2089.0
x
x f
)
n
2089 41.8 minutes 50 95
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.
96
The Shape of Distributions
• •
Uniform Distribution (rectangular)
All entries or classes in the distribution have equal or approximately equal frequencies.
Symmetric.
97
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.
98
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.
99
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 100
Section 2.4
Measures of Variation
101
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 102
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) 103
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 42 104
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 maximum minimum • Range = (Max. salary) – (Min. salary) = 47 – 37 = 10 The range of starting salaries is 10 or $10,000.
105
Deviation, Variance, and Standard Deviation
• • •
Deviation
The difference between the data entry, mean of the data set.
x
, and the Population data set: Deviation of
x
=
x
– μ Sample data set: Deviation of
x
=
x
–
x
106
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.
x N
415 10 41.5
107
Solution: Finding the Deviation
• Determine the deviation for each data entry.
Salary ($1000s), x
41 38 39 45 47 41 44 41 37 42 Σ
x
= 415
Deviation: x – μ
41 – 41.5 = –0.5
38 – 41.5 = –3.5
39 – 41.5 = –2.5
45 – 41.5 = 3.5
47 – 41.5 = 5.5
41 – 41.5 = –0.5
44 – 41.5 = 2.5
41 – 41.5 = –0.5
37 – 41.5 = –4.5
42 – 41.5 = 0.5
Σ(
x
– μ) = 0 108
Deviation, Variance, and Standard Deviation
•
Population Variance
2
x N
) 2 Sum of squares, SS
x
•
Population Standard Deviation
2
x N
) 2 109
Finding the Population Variance & Standard Deviation
In Words
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.
In Symbols
x N x
– μ (
x
– μ) 2 SS
x
= Σ(
x
– μ) 2 110
Finding the Population Variance & Standard Deviation
In Words
5.
Divide by
N
to get the
population variance
.
6.
Find the square root to get the
population standard deviation
.
2
In Symbols
x
) 2
N
x N
) 2 111
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.
112
Solution: Finding the Population Standard Deviation
• • Determine SS
x N
= 10
Salary, x
41 38 39 45 47 41 44 41 37 42
Deviation: x – μ
41 – 41.5 = –0.5
38 – 41.5 = –3.5
39 – 41.5 = –2.5
45 – 41.5 = 3.5
47 – 41.5 = 5.5
41 – 41.5 = –0.5
44 – 41.5 = 2.5
41 – 41.5 = –0.5
37 – 41.5 = –4.5
42 – 41.5 = 0.5
Σ(
x
– μ) = 0
Squares: (x – μ) 2
(–0.5) 2 = 0.25
(–3.5) 2 = 12.25
(–2.5) 2 = 6.25
(3.5) 2 = 12.25
(5.5) 2 = 30.25
(–0.5) 2 = 0.25
(2.5) 2 = 6.25
(–0.5) 2 = 0.25
(–4.5) 2 = 20.25
(0.5) 2 = 0.25
SS
x
= 88.5
113
Solution: Finding the Population Standard Deviation
•
Population Variance
2
x N
) 2 88.5
10 8.9
Population Standard Deviation
• 2 8.85
3.0
The population standard deviation is about 3.0, or $3000.
114
Deviation, Variance, and Standard Deviation
•
Sample Variance
s
2
x x
) 2
n
1 •
Sample Standard Deviation
s
s
2
x x
) 2
n
1 115
Finding the Sample Variance & Standard Deviation
In Words
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.
In Symbols x
x n x
x SS x
(
x
x
) 2
x x
) 2 116
Finding the Sample Variance & Standard Deviation
In Words
5.
Divide by
n
– 1 to get the
sample variance
.
6.
Find the square root to get the
sample standard deviation
.
s In Symbols s
2
x x
) 2
n
1
x x
) 2
n
1 117
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 42 118
Solution: Finding the Sample Standard Deviation
• • Determine SS
x n
= 10
Salary, x
41 38 39 45 47 41 44 41 37 42
Deviation: x – μ
41 – 41.5 = –0.5
38 – 41.5 = –3.5
39 – 41.5 = –2.5
45 – 41.5 = 3.5
47 – 41.5 = 5.5
41 – 41.5 = –0.5
44 – 41.5 = 2.5
41 – 41.5 = –0.5
37 – 41.5 = –4.5
42 – 41.5 = 0.5
Σ(
x
– μ) = 0
Squares: (x – μ) 2
(–0.5) 2 = 0.25
(–3.5) 2 = 12.25
(–2.5) 2 = 6.25
(3.5) 2 = 12.25
(5.5) 2 = 30.25
(–0.5) 2 = 0.25
(2.5) 2 = 6.25
(–0.5) 2 = 0.25
(–4.5) 2 = 20.25
(0.5) 2 = 0.25
SS
x
= 88.5
119
Solution: Finding the Sample Standard Deviation
•
Sample Variance
s
2
x x
) 2
n
1 88.5
9.8
Sample Standard Deviation
•
s
s
2 88.5
3.1
9 The sample standard deviation is about 3.1, or $3100.
120
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
23.75
33.50
26.50
37.00
31.25
36.50
39.25
37.75
27.00
37.00
24.50
40.00
37.50
37.25
35.75
29.00
33.00
32.00
34.75
36.75
26.00
40.50
38.00
121
Solution: Using Technology to Find the Standard Deviation
Sample Mean Sample Standard Deviation 122
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.
123
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.
124
Interpreting Standard Deviation: Empirical Rule (68 – 95 – 99.7 Rule)
99.7% within 3 standard deviations 95% within 2 standard deviations 68% within 1 standard deviation
x
3
s
2.35%
x
2
s
13.5%
x
s
34% 34%
x
2.35%
x
s
13.5%
x
2
s x
3
s
125
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.
126
Solution: Using the Empirical Rule
• Because the distribution is bell-shaped, you can use the Empirical Rule. 34% 13.5% 55.87
x
3
s
58.58
x
2
s
61.29
x
s
64
x
66.71
x
s
69.42
x
2
s
72.13
x
3
s
34% + 13.5% = 47.5% of women are between 64 and 69.42 inches tall.
127
Chebychev’s Theorem
• The portion of any data set lying within
k
standard deviations (
k
> 1) of the mean is at least: • 1 1
k
2
k
= 2: In any data set, at least 1 1 2 2 3 or 75% 4 of the data lie within 2 standard deviations of the mean.
•
k
= 3: In any data set, at least 1 1 3 2 8 or 88.9% 9 of the data lie within 3 standard deviations of the mean.
128
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?
129
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
At least 75% of the population of Florida is between 0 and 88.8 years old.
130
Standard Deviation for Grouped Data
•
Sample standard deviation for a frequency distribution
s
x x
) 2
n
1
f
where
n
= Σ
f
(the number of entries in the data set) • When a frequency distribution has classes, estimate the sample mean and standard deviation by using the midpoint of each class. 131
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.
3 1 3 2 4 0 1 1
Number of Children in 50 Households
3 2 1 2 1 1 1 0 1 1 1 5 0 0 0 3 0 6 0 1 6 3 1 3 3 6 6 0 1 0 1 0 1 1 2 2 1 1 2 1 2 4 132
Solution: Finding the Standard Deviation for Grouped Data
• • First construct a frequency distribution.
Find the mean of the frequency distribution.
x
xf
children.
n
91 1.8
50 The sample mean is about 1.8
x
0 1 2 3 4 5 6
f
10 19 7 7 2 1 4 Σ
f
= 50
xf
0(10) = 0 1(19) = 19 2(7) = 14 3(7) =21 4(2) = 8 5(1) = 5 6(4) = 24 Σ(
xf
)= 91 133
Solution: Finding the Standard Deviation for Grouped Data
• Determine the sum of squares.
4 5 6
x
0 1 2 3 2 1 4
f
10 19 7 7
x
x
0 – 1.8 = –1.8
1 – 1.8 = –0.8
2 – 1.8 = 0.2
3 – 1.8 = 1.2
4 – 1.8 = 2.2
5 – 1.8 = 3.2
6 – 1.8 = 4.2
(
x
x
) 2 (–1.8) 2 = 3.24
(–0.8) 2 = 0.64
(0.2) 2 = 0.04
(1.2) 2 = 1.44
(2.2) 2 = 4.84
(3.2) 2 = 10.24
(4.2) 2 = 17.64
(
x
x
) 2
f
3.24(10) = 32.40
0.64(19) = 12.16
0.04(7) = 0.28
1.44(7) = 10.08
4.84(2) = 9.68
10.24(1) = 10.24
x
17.64(4) = 70.56
x
) 2
f
145.40
134
Solution: Finding the Standard Deviation for Grouped Data
• Find the sample standard deviation.
s
x x x n
1 ) 2
x f
(
x
145.40
x
) 2 1.7
(
x
x
) 2
f
The standard deviation is about 1.7 children.
135
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 136
Section 2.5
Measures of Position
137
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) 138
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, Q 1
: About one quarter of the data fall on or below
Q
1 .
Second quartile, Q
fall on or below
Q
2
2
: About one half of the data (median).
Third quartile, Q 3
: About three quarters of the data fall on or below
Q
3 .
139
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:
•
Q
2 divides the data set into two halves. Lower half Upper half 5 7 9 10 11 13 14 15 16 17 18 18 20 21 37
Q
2
140
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 5 7 9 10 11 13 14 15 16 17 18 18 20 21 37
Q
1
Q
2
Q
3
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.
141
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
Rank
(
N
1)
Pth
(
N
100 1)
Qth Rank Rank
(
N
4 1)
Dth
10
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
.
Slide 2- 143
Interquartile Range
• •
Interquartile Range (IQR)
The difference between the third and first quartiles.
IQR =
Q
3 –
Q
1 144
Example: Finding the Interquartile Range
Find the interquartile range of the test scores.
Recall
Q
1 = 10,
Q
2 = 15, and
Q
3 = 18
Solution:
• IQR =
Q
3 –
Q
1 = 18 – 10 = 8 The test scores in the middle portion of the data set vary by at most 8 points.
145
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
Q
1 Median
Q
2 Third quartile
Q
3 Maximum entry 146
Drawing a Box-and-Whisker Plot
1.
2.
3.
4.
5.
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
Q
1 and draw a vertical line in the box at
Q
2 .
to
Q
3 Draw whiskers from the box to the minimum and maximum entries.
Box Whisker Whisker Minimum entry
Q
1 Median,
Q
2
Q
3 Maximum entry 147
Example: Drawing a Box-and-Whisker Plot
Draw a box-and-whisker plot that represents the 15 test scores. Recall Min = 5
Q
1 = 10
Q
2 = 15
Q
3 = 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.
148
Percentiles and Other Fractiles
Fractiles
Quartiles Deciles Percentiles
Summary
Divides data into 4 equal parts Divides data into 10 equal parts Divides data into 100 equal parts
Symbols
Q
1 ,
Q
2 ,
Q
3
D
1 ,
D
2 ,
D
3 ,…,
D
9
P
1
, P
2 ,
P
3 ,…,
P
99 149
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 72 nd percentile? How should you interpret this?
(Source: College Board Online)
150
Solution: Interpreting Percentiles
The 72 nd percentile corresponds to a test score of 1700. This means that 72% of the students had an SAT score of 1700 or less.
151
The Standard Score
• •
Standard Score (z-score)
Represents the number of standard deviations a given value
x
falls from the mean μ.
z
value - mean standard deviation
x
152
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.
153
Solution: Comparing z-Scores from Different Data Sets
• Forest Whitaker
z
x
8.8
0.15
0.15 standard deviations above the mean • Helen Mirren
z
x
11.5
2.17
2.17 standard deviations above the mean 154
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.
155
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) 156
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.
.
Chapter 2: Descriptive Statistics
Elementary Statistics:
Picturing the World
Fifth Edition by Larson and Farber Slide 4- 160
.
Find the class width: Class
1 – 5 6 – 10 11 – 15 16 – 20
Frequency, f
21 16 28 13
A. 3 B. 4 C. 5 D. 19
Slide 2- 161
.
Find the class width: Class
1 – 5 6 – 10 11 – 15 16 – 20
Frequency, f
21 16 28 13
A. 3 B. 4 C. 5 D. 19
or
or
Slide 2- 162
.
Estimate the frequency of the class with the greatest frequency. Ages of Concert Attendees A. 28 B. 21 C. 58 D. 53
60 50 40 30 20 10 0 18 28 38
Age
48 58 Slide 2- 163
.
Estimate the frequency of the class with the greatest frequency. Ages of Concert Attendees A. 28 B. 21 C. 58 D. 53
60 50 40 30 20 10 0 18 28 38
Age
48 58 Slide 2- 164
.
What is the maximum data entry? A. 96 B. 38 C. 9 D. 41
Key: 3 | 8 = 38 3 8 9 4 0 2 7 5 1 1 4 8 6 3 3 3 8 9 9 7 0 0 1 1 2 4 7 8 8 8 8 9 8 2 2 3 4 7 7 8 9 9 9 1 1 4 5 6 Slide 2- 165
What is the maximum data entry? A. 96 B. 38 C. 9 D. 41
Key: 3 | 8 = 38 3 8 9 4 0 2 7 5 1 1 4 8 6 3 3 3 8 9 9 7 0 0 1 1 2 4 7 8 8 8 8 9 8 2 2 3 4 7 7 8 9 9 9 1 1 4 5 6 Slide 2- 166
.
The heights (in inches) of a sample of basketball players are shown: 76 79 81 78 82 78 Find the mean.
A. 78.5
B. 79 C. 474 D. 78
Slide 2- 167
.
The heights (in inches) of a sample of basketball players are shown: 76 79 81 78 82 78 Find the mean.
A. 78.5
B. 79 C. 474 D. 78
Slide 2- 168
.
The heights (in inches) of a sample of basketball players are shown: 76 79 81 78 82 78 Find the median.
A. 78.5
B. 79 C. 79.5
D. 78
Slide 2- 169
.
The heights (in inches) of a sample of basketball players are shown: 76 79 81 78 82 78 Find the median.
A. 78.5
B. 79 C. 79.5
D. 78
Slide 2- 170
.
The heights (in inches) of a sample of basketball players are shown: 76 79 81 78 82 78 Find the mode.
A. 78.5
B. 79 C. 79.5
D. 78
Slide 2- 171
.
The heights (in inches) of a sample of basketball players are shown: 76 79 81 78 82 78 Find the mode.
A. 78.5
B. 79 C. 79.5
D. 78
Slide 2- 172
.
The heights (in inches) of a sample of basketball players are shown: 76 79 81 78 82 78 Find the standard deviation.
A. 2.2
B. 6 C. 2 D. 4.8
Slide 2- 173
.
The heights (in inches) of a sample of basketball players are shown: 76 79 81 78 82 78 Find the standard deviation.
A. 2.2
B. 6 C. 2 D. 4.8
Slide 2- 174
.
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.
A. 68% B. 75% C. 95% D. 99.7%
Slide 2- 175
.
600 1300
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.
A. 68% B. 75% C. 95% D. 99.7%
Slide 2- 176
.
Use the box-and-whisker plot to identify the first quartile.
A. 10 B. 18 C. 24 D. 26
10 18 24 26 30 | | | | | | | | | | | 10 12 14 16 18 20 22 24 26 28 30 Slide 2- 177
.
Use the box-and-whisker plot to identify the first quartile.
A. 10 B. 18 C. 24 D. 26
10 18 24 26 30 | | | | | | | | | | | 10 12 14 16 18 20 22 24 26 28 30 Slide 2- 178
.
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. A. 1.13
B.
–1.13
C. 1.71
D.
–1.71
Slide 2- 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. A. 1.13
B.
–1.13
C. 1.71
D.
–1.71
z
x
175 1.71
Slide 2- 180