#### Transcript Chapter 2

```Chapter
2
Descriptive Statistics
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Chapter Outline
• 2.1 Frequency Distributions and Their Graphs
• 2.2 More Graphs and Displays
• 2.3 Measures of Central Tendency
• 2.4 Measures of Variation
• 2.5 Measures of Position
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Course Level Expectations
• CLE 3136.1.7 Use technologies appropriately to
develop understanding of abstract mathematical
ideas, to facilitate problem solving, and to produce
accurate and reliable models.
• CLE 3136.2.1 Understand histograms, parallel box
plots, and scatterplots, and use them to compare
display data.
Larson/Farber 5th ed.
3
Checks for Understanding
• 3136.5.1 Use properties of point estimators, including
biased/unbiased, and variability.
• 3136.5.2 Understand the meaning of confidence
level, of confidence intervals, and the properties of
confidence intervals.
Larson/Farber 5th ed.
4
Section 2.1
Frequency Distributions
and Their Graphs
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Section 2.1 Objectives
• Construct frequency distributions
• Construct frequency histograms, frequency polygons,
relative frequency histograms, and ogives
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Frequency Distribution
Why can’t we do ‘5-1’?
Frequency Distribution
Class Frequency, f
Class width
• A table that shows
1–5
5
classes or intervals of 6 – 1 = 5
6–10
8
data with a count of the
11–15
6
number of entries in each
16–20
8
class.
21–25
5
• The frequency, f, of a
class is the number of
26–30
4
data entries in the class. Lower class
Upper class
This table might represent the ages of
people in a room for example
limits
limits
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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.
•For example, you have a range of
ages from 10 to 74
•You decide you want 8 classes
•The range of data is 74-10 = 64
•64/8 = 8.
•Therefore, you have 8 classes with
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a width of 8
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.
•From the previous example, the
minimum data entry is 10
•The next lower limit would be 18
•The first upper limit would be 17
•The next would be 17+8 = 25
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Constructing a Frequency Distribution
4. Make a tally mark for each data entry in the row of
the appropriate class.
5. Count the tally marks to find the total frequency f
for each class.
•For our example to continue, we
would have to have a number for
how many people were each age
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Example: Constructing a Frequency
Distribution
The following sample data set lists the prices (in
dollars) of 30 portable global positioning system (GPS)
navigators. Construct a frequency distribution that has
seven classes.
90 130 400 200 350 70 325 250 150 250
275 270 150 130 59 200 160 450 300 130
220 100 200 400 200 250 95 180 170 150
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Solution: Constructing a Frequency
Distribution
90 130 400 200 350 70 325 250 150 250
275 270 150 130 59 200 160 450 300 130
220 100 200 400 200 250 95 180 170 150
1. Number of classes = 7 (given)
2. Find the class width
max  min 450  59 391


 55.86
#classes
7
7
Round up to 56
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Solution: Constructing a Frequency
Distribution
3. Use 59 (minimum value)
the class width of 56 to
get the lower limit of the
next class.
59 + 56 = 115
Find the remaining
lower limits.
Lower
limit
Class
width = 56
Upper
limit
59
115
171
227
283
339
395
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Solution: Constructing a Frequency
Distribution
The upper limit of the
first class is 114 (one less
than the lower limit of the
second class).
Add the class width of 56
to get the upper limit of
the next class.
114 + 56 = 170
Find the remaining upper
limits.
Lower
limit
Upper
limit
59
115
171
227
283
339
114
170
226
282
338
394
395
450
Class
width = 56
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Solution: Constructing a Frequency
Distribution
4. Make a tally mark for each data entry in the row of
the appropriate class.
5. Count the tally marks to find the total frequency f
for each class.
Note: you should
sum the frequencies.
This is the total
sample size, and
will be used later.
Here, the sample
size is 30
Class
Tally
Frequency, f
IIII
5
115–170
IIII III
8
171–226
IIII I
6
227–282
IIII
5
283–338
II
2
339–394
I
1
395–450
III
3
59–114
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Determining the Midpoint
Midpoint of a class
(Lower class limit)  (Upper class limit)
2
Class
59–114
Midpoint
59  114
 86.5
2
115–170
115  170
 142.5
2
171–226
171  226
 198.5
2
Frequency, f
5
Class width = 56
8
6
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Determining the Relative Frequency
Relative Frequency of a class
• Portion or percentage of the data that falls in a
particular class.
•Frequency is
how often
Class frequency f
something

• Relative frequency 
occurs
Sample size
n
Class
Frequency, f
59–114
5
115–170
8
171–226
6
Relative Frequency
5
 0.17
30
8
 0.27
30
6
 0.2
30
•Relative
frequency is
how often
something
occurs relative
to the total
occurrences
(sample size)
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Determining the Cumulative Frequency
Cumulative frequency of a class
• The sum of the frequencies for that class and all
previous classes.
Class
Frequency, f
Cumulative frequency
59–114
5
5
115–170
+ 8
13
171–226
+ 6
19
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Expanded Frequency Distribution
Remember what this is called
Class
Frequency, f
Midpoint
Relative
Frequency (f/n)
59–114
5
86.5
0.17
5
115–170
8
142.5
0.27
13
171–226
6
198.5
0.2
19
227–282
5
254.5
0.17
24
283–338
2
310.5
0.07
26
339–394
1
366.5
0.03
27
395–450
3
422.5
0.1
f
Σf = 30
 1
n
This is the uppercase Greek
Cumulative
frequency
30
letter “sigma” and means “to
sum”
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Graphs of Frequency Distributions
frequency
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
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Class Boundaries
Why do this?
Class boundaries
• The numbers that separate classes without forming
Because bars of a histogram must
gaps between them.
• The distance from the upper
limit of the first class to the
lower limit of the second
class is 115 – 114 = 1.
• Half this distance is 0.5.
touch, bars begin and end at class
Class
Class
boundaries
Frequency,
f
59–114
58.5–114.5
5
115–170
8
171–226
6
• First class lower boundary = 59 – 0.5 = 58.5
• First class upper boundary = 114 + 0.5 = 114.5
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Class Boundaries
Class
59–114
115–170
171–226
227–282
283–338
339–394
395–450
Class
boundaries
58.5–114.5
114.5–170.5
170.5–226.5
226.5–282.5
282.5–338.5
338.5–394.5
394.5–450.5
Frequency,
f
5
8
6
5
2
1
3
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Example: Frequency Histogram
Construct a frequency histogram for the Global
Positioning system (GPS) navigators.
Class
Class
boundaries
59–114
58.5–114.5
86.5
5
115–170
114.5–170.5
142.5
8
171–226
170.5–226.5
198.5
6
227–282
226.5–282.5
254.5
5
283–338
282.5–338.5
310.5
2
339–394
338.5–394.5
366.5
1
395–450
394.5–450.5
422.5
3
Frequency,
Midpoint
f
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Solution: Frequency Histogram
(using Midpoints)
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Solution: Frequency Histogram
(using class boundaries)
You can see that more than half of the GPS navigators are
priced below \$226.50.
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Graphs of Frequency Distributions
frequency
Frequency Polygon
• A line graph that emphasizes the continuous change
in frequencies.
data values
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Example: Frequency Polygon
Construct a frequency polygon for the GPS navigators
frequency distribution.
You will
use these
midpoints
to create
your
frequency
polygon
Class
Midpoint
Frequency, f
59–114
86.5
5
115–170
142.5
8
171–226
198.5
6
227–282
254.5
5
283–338
310.5
2
339–394
366.5
1
395–450
422.5
3
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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.
This
And
This
You can see that the frequency of GPS navigators increases
up to \$142.50 and then decreases.
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Graphs of Frequency Distributions
relative
frequency
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
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Example: Relative Frequency Histogram
Construct a relative frequency histogram for the GPS
navigators frequency distribution.
Class
Class
boundaries
Frequency,
f
Relative
frequency
59–114
58.5–114.5
5
0.17
115–170
114.5–170.5
8
0.27
171–226
170.5–226.5
6
0.2
227–282
226.5–282.5
5
0.17
283–338
282.5–338.5
2
0.07
339–394
338.5–394.5
1
0.03
395–450
394.5–450.5
3
0.1
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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 27% of GPS navigators are
priced between \$114.50 and \$170.50.
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Graphs of Frequency Distributions
cumulative
frequency
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
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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. Connect the points in order from left to right.
5. 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 GPS navigators frequency
distribution.
Class
Class
boundaries
Frequency,
f
Cumulative
frequency
59–114
58.5–114.5
5
5
115–170
114.5–170.5
8
13
171–226
170.5–226.5
6
19
227–282
226.5–282.5
5
24
283–338
282.5–338.5
2
26
339–394
338.5–394.5
1
27
395–450
394.5–450.5
3
30
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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 25 GPS navigators cost
\$300 or less. The greatest increase occurs between \$114.50 and
\$170.50.
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Section 2.1 Summary
• Constructed frequency distributions
• Constructed frequency histograms, frequency
polygons, relative frequency histograms and ogives
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Assignment
• Page 47-51 1-6 Orally
• 7-43 Odd
Larson/Farber 5th ed.
38
Chapter 2 Quiz 1(20 points)
• Using the information from Question number 34 on
page 50, created an expanded Frequency Distribution
Chart.
• It should look like this:
Class
Larson/Farber 5th ed.
Frequency, f
Midpoint
Relative
Frequency (f/n)
Cumulative
frequency
39
Class
Frequency, f
Midpoint
Relative
Frequency (f/n)
2456-2542
7
2499
.28
7
2543-2629
3
2586
.12
10
2630-2716
2
2673
.08
12
2717-2803
4
2760
.16
16
2804-2888
9
2847
.36
25
Larson/Farber 5th ed.
Cumulative
frequency
40
Section 2.2
More Graphs and Displays
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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
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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, 27, 28,
30, 36, 36, 45
2
3
1 5 5 6 7 8
0 6 6
4
5
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Example: Constructing a Stem-and-Leaf
Plot
The following are the numbers of text messages sent
last week by the cellular phone users on one floor of a
college dormitory. Display the data in a stem-and-leaf
plot.
155 159
118 118
139 139
129 112
144
108
122
126
129
122
78
148
105 145 126 116 130 114 122 112 112 142 126
121 109 140 126 119 113 117 118 109 109 119
133 126 123 145 121 134 124 119 132 133 124
147
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Solution: Constructing a Stem-and-Leaf
Plot
155 159
118 118
139 139
129 112
144
108
122
126
129
122
78
148
105 145 126 116 130 114 122 112 112 142 126
121 109 140 126 119 113 117 118 109 109 119
133 126 123 145 121 134 124 119 132 133 124
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.
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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.
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Graphing Quantitative Data Sets
We did a version of this
on the 1st day of school
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
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Example: Constructing a Dot Plot
Use a dot plot organize the text messaging data.
155 159
118 118
139 139
129 112
144
108
122
126
129
122
78
148
105 145 126 116 130 114 122 112 112 142 126
121 109 140 126 119 113 117 118 109 109 119
133 126 123 145 121 134 124 119 132 133 124
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.
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Solution: Constructing a Dot Plot
155 159
118 118
139 139
129 112
144
108
122
126
129
122
78
148
105 145 126 116 130 114 122 112 112 142 126
121 109 140 126 119 113 117 118 109 109 119
133 126 123 145 121 134 124 119 132 133 124
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. This is commonly called an “outlier”
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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.
• Therefore you will need
Relative Frequency
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Example: Constructing a Pie Chart
The numbers of earned degrees conferred (in thousands)
in 2007 are shown in the table. Use a pie chart to
organize the data. (Source: U.S. National Center for
Educational Statistics)
Type of degree
Associate’s
Bachelor’s
Master’s
First professional
Doctoral
Number
(thousands)
728
1525
604
90
60
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Solution: Constructing a Pie Chart
• Find the relative frequency (percent) of each category.
Type of degree
Frequency, f
Associate’s
728
Bachelor’s
1525
Master’s
604
First professional
Doctoral
728
 0.24
3007
1525
 0.51
3007
604
 0.20
3007
90
90
 0.03
3007
60
60
 0.02
3007
Σf = 3007
Relative frequency
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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 associate’s
degrees is
360º(0.24) ≈ 86º
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Solution: Constructing a Pie Chart
Type of degree
Relative
Frequency, f frequency
Central angle
Associate’s
728
0.24
360º(0.24)≈86º
Bachelor’s
1525
0.51
360º(0.51)≈184º
604
0.20
360º(0.20)≈72º
First professional
90
0.03
360º(0.03)≈11º
Doctoral
60
0.02
360º(0.02)≈7º
Master’s
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Solution: Constructing a Pie Chart
Relative
frequency
Central
angle
Associate’s
0.24
86º
Bachelor’s
0.51
184º
Master’s
0.20
72º
First professional
0.03
11º
Doctoral
0.02
7º
Type of degree
From the pie chart, you can see that over one half of the
degrees conferred in 2007 were bachelor’s degrees.
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Graphing Qualitative Data Sets
Frequency
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
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Example: Constructing a Pareto Chart
In a recent year, the retail industry lost \$36.5 billion 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 (\$5.4 billion), employee theft
(\$15.9 billion), shoplifting (\$12.7 billion), and vendor
fraud (\$1.4 billion). Use a Pareto chart to organize this
data. (Source: National Retail Federation and Center for
Retailing Education, University of Florida)
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Chapter 2 Quiz 2(20 points)
• Using the information from Question number 32 on
page 50, created an expanded Frequency Distribution
Chart.
• It should look like this:
Class
Larson/Farber 5th ed.
Frequency, f
Midpoint
Relative
Frequency (f/n)
Cumulative
frequency
58
Class
Frequency, f
Midpoint
Relative
Frequency (f/n)
32-35
3
36.5
.125~.13
3
36-39
9
40.5
.375~.38
12
40-43
8
44.5
.33
20
44-47
3
48.5
.125~.13
23
48-51
1
51
.04
24
Larson/Farber 5th ed.
Cumulative
frequency
59
Solution: Constructing a Pareto Chart
Cause
\$ (billion)
5.4
Employee
theft
15.9
Shoplifting
12.7
Vendor fraud
Millions of dollars
Causes of Inventory Shrinkage
1.4
20
15
10
5
0
Employee
Theft
Cause
Vendor
fraud
From the graph, it is easy to see that the causes of inventory
shrinkage that should be addressed first are employee theft and
shoplifting.
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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
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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)
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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.
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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.
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Graphing Paired Data Sets
Quantitative
data
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
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Example: Constructing a Time Series
Chart
The table lists the number of cellular
telephone subscribers (in millions)
for the years 1998 through 2008.
Construct a time series chart for the
number of cellular subscribers.
(Source: Cellular Telecommunication &
Internet Association)
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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.
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Solution: Constructing a Time Series
Chart
The graph shows that the number of subscribers has been
increasing since 1998, with greater increases recently.
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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
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Assignment
• Page 60 1-4 orally
• Page 60-63 5-33 odd
Larson/Farber 5th ed.
70
Chapter 2 Quiz 3
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
Lower
limit
59
115
Upper
limit
114
170
What is the class width in this chart??
What are the class boundaries for the first class in this chart?
What is the midpoint of the first class in this chart?
What do you call a chart which has classes, frequency, midpoints, relative frequency and
cumulative frequency?
What does this ∑ symbol mean?
What is on the “y” axis on a frequency histogram?
What is on the “x” axis on a frequency histogram?
On the “x” axis of a frequency histogram, you may use two methods to indicate different classes.
What are they?
Does the first bar of a frequency histogram touch the “Y” axis?
True/False: A stem and leaf plot is a qualitative graph of data.
True/False: A pie chart is a qualitative graph of data.
True/False: A Pareto Graph has all bars touching each other, and is ordered greatest to least.
A ___________ uses a number line with marks above it to represent data values.
Each entry in one data set corresponds to one entry in a second data set on a _____________
Data set is composed of quantitative entries taken at regular intervals over a period of time in a
_________ _______ _________.
Larson/Farber 5th ed.
71
Section 2.3
Measures of Central Tendency
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Section 2.3 Objectives
• Determine the mean, median, and mode of a
population and of a sample
• Determine the weighted mean of a data set and the
mean of a frequency distribution
• Describe the shape of a distribution as symmetric,
uniform, or skewed and compare the mean and
median for each
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Measures of Central Tendency
Measure of central tendency
• A value that represents a typical, or central, entry of a
data set.
• Most common measures of central tendency:
 Mean
 Median
 Mode
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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.
x


• Population mean:
N
• Sample mean: x  x
n
Pronounced “X Bar”
always used for sample
means
Lowercase Greek
letter “mu” always
used for the mean of
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the population
Example: Finding a Sample Mean
The prices (in dollars) for a sample of round-trip flights
from Chicago, Illinois to Cancun, Mexico are listed.
What is the mean price of the flights?
872 432 397 427 388 782 397
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Solution: Finding a Sample Mean
872 432 397 427 388 782 397
• The sum of the flight prices is
Σx = 872 + 432 + 397 + 427 + 388 + 782 + 397 = 3695
• To find the mean price, divide the sum of the prices
by the number of prices in the sample
x 3695
x

 527.9
n
7
The mean price of the flights is about \$527.90.
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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.
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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
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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.
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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
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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.
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Measure of Central Tendency: Mode
Mode
• The data entry that occurs with the greatest frequency.
• A data set can have one mode, more than one mode,
or no mode.
• If no entry is repeated the data set has no mode.
• If two entries occur with the same greatest frequency,
each entry is a mode (bimodal).
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Example: Finding the Mode
The prices (in dollars) for a sample of roundtrip flights
from Chicago, Illinois to Cancun, Mexico are listed.
Find the mode of the flight prices.
872 432 397 427 388 782 397
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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.
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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
Frequency, f
34
Republican
Other
56
21
Did not respond
9
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Solution: Finding the Mode
Political Party
Democrat
Frequency, f
34
Republican
Other
Did not respond
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.
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Comparing the Mean, Median, and Mode
• All three measures describe a typical entry of a data
set.
• Advantage of using the mean:
 The mean is a reliable measure because it takes
into account every entry of a data set.
• Disadvantage of using the mean:
 Greatly affected by outliers (a data entry that is far
removed from the other entries in the data set).
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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
20
20
20
20
20
21
21
21
21
22
22
22
23
23
23
23
24
24
65
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Solution: Comparing the Mean, Median,
and Mode
Ages in a class
Mean:
Median:
Mode:
20
20
20
20
20
20
21
21
21
21
22
22
22
23
23
23
23
24
24
65
x 20  20  ...  24  65
x

 23.8 years
n
20
21  22
 21.5 years
2
20 years (the entry occurring with the
greatest frequency)
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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.
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Solution: Comparing the Mean, Median,
and Mode
which measure of central tendency best represents a
data set.
In this case, it appears that the median best describes
the data set.
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Weighted Mean
Weighted Mean
• The mean of a data set whose entries have varying
weights.
 ( x  w)
• x 
where w is the weight of each entry x
w
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Example: Finding a Weighted Mean
determined from five sources: 50% from your test
exam, 10% from your computer lab work, and 5% from
(midterm), 82 (final exam), 98 (computer lab), and 100
(homework). What is the weighted mean of your
scores? If the minimum average for an A is 90, did you
get an A?
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Solution: Finding a Weighted Mean
Source
x∙w
Score, x
Weight, 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
( x  w) 88.6
x

 88.6
w
1
Your weighted mean for the course is 88.6. You did not
get an A.
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Mean of Grouped Data
Mean of a Frequency Distribution
• Approximated by
( x  f )
x
n
n  f
where x and f are the midpoints and frequencies of a
class, respectively
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Finding the Mean of a Frequency
Distribution
In Words
1. Find the midpoint of each
class.
In Symbols
(lower limit)+(upper limit)
x
2
2. Find the sum of the
products of the midpoints
and the frequencies.
( x  f )
3. Find the sum of the
frequencies.
n  f
4. Find the mean of the
frequency distribution.
( x  f )
x
n
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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
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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
6
72.5∙6 = 435.0
79 – 90
84.5
2
84.5∙2 = 169.0
n = 50
Σ(x∙f) = 2089.0
( x  f ) 2089
x

 41.8 minutes
n
50
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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.
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The Shape of Distributions
Uniform Distribution (rectangular)
• All entries or classes in the distribution have equal
or approximately equal frequencies.
• Symmetric.
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The Shape of Distributions
Skewed Left Distribution (negatively skewed)
• The “tail” of the graph elongates more to the left.
• The mean is to the left of the median.
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The Shape of Distributions
Skewed Right Distribution (positively skewed)
• The “tail” of the graph elongates more to the right.
• The mean is to the right of the median.
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Section 2.3 Summary
• Determined the mean, median, and mode of a
population and of a sample
• Determined the weighted mean of a data set and the
mean of a frequency distribution
• Described the shape of a distribution as symmetric,
uniform, or skewed and compared the mean and
median for each
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Assignment
• Page 72 1-4 orally
• Page 72-77 5-57 odd
Larson/Farber 5th ed.
105
Chapter 2 Quiz 4 (5 points)
1. T/F –In a uniform distribution, the mean, median and mode
are all the same
2. T/F -If two entries occur with the same least frequency, each
entry is a mode (bimodal).
3. T/F –An advantage of using the mean as a measure of central
tendency is that it is not affected by outliers.
4. In a symmetrical distribution, the mean, median and mode
are all the same
5. In a “Skewed Left” distribution, the median is to the ______
of the mean
6. In a “Skewed Right” distribution, the mean is to the _____ of
the median
• Quiz continues on next slide
Larson/Farber 5th ed.
106
Chapter 2 Quiz 4 Continued (5 points)
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
8
19 – 30
24.5
11
31 – 42
36.5
16
43 – 54
48.5
10
55 – 66
60.5
8
67 – 78
72.5
5
79 – 90
84.5
2
Quiz
continues on
next slide
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Chapter 2 Quiz 4 Continued (5 points)
Use the scores earned by the student below, calculate his
weighted grade. Did he earn an A? (assume the teacher
Source
Score, x
Weight, w
Test Mean
90
0.50
Midterm
90
0.15
Final Exam
98
0.20
Computer Lab
98
0.10
Homework
100
0.05
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Section 2.4
Measures of Variation
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Section 2.4 Objectives
• Determine the range of a data set
• Determine the variance and standard deviation of a
population and of a sample
• Use the Empirical Rule and Chebychev’s Theorem to
interpret standard deviation
• Approximate the sample standard deviation for
grouped data
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Range
Range
• The difference between the maximum and minimum
data entries in the set.
• The data must be quantitative.
• Range = (Max. data entry) – (Min. data entry)
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Example: Finding the Range
A corporation hired 10 graduates. The starting salaries
for each graduate are shown. Find the range of the
starting salaries.
Starting salaries (1000s of dollars)
41 38 39 45 47 41 44 41 37 42
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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
minimum
maximum
• Range = (Max. salary) – (Min. salary)
= 47 – 37 = 10
The range of starting salaries is 10 or \$10,000.
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Deviation, Variance, and Standard
Deviation
Deviation
• The difference between the data entry, x, and the
mean of the data set.
• Population data set:
 Deviation of x = x – μ
• Sample data set:
 Deviation of x  x  x
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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 415


 41.5
N
10
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Solution: Finding the Deviation
• Determine the
deviation for each
data entry.
Notice that the sum
of the deviations is
zero. This makes it
hard to use the
information without
manipulation
Deviation (\$1000s)
Salary (\$1000s), x
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
Σx = 415
42 – 41.5 = 0.5
Σ(x – μ) = 0
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Deviation, Variance, and Standard
Deviation By squaring the deviations,
we can get usable
information which relates to
central tendency
Population Variance
( x   )
•  
N
2
2
Sum of squares, SSx
Population Standard Deviation
2

(
x


)
2
•    
N
Greek lower case
sigma –Sigma squared
is the standard symbol
for Pop variance
Sigma is the standard
symbol for Pop
Standard Deviation
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Finding the Population Variance &
Standard Deviation
In Words
1. Find the mean of the
population data set.
2. Find the deviation of each
entry.
In Symbols
x

N
x–μ
3. Square each deviation.
(x – μ)2
4. Add to get the sum of
squares.
SSx = Σ(x – μ)2
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Finding the Population Variance &
Standard Deviation
In Words
5. Divide by N to get the
population variance.
6. Find the square root of the
variance to get the
population standard
deviation.
In Symbols
2

(
x


)
2 
N
( x   ) 2

N
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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.
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Solution: Finding the Population
Standard Deviation
• Determine SSx
• N = 10
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
Σ(x – μ) = 0
SSx = 88.5
Salary, x
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Solution: Finding the Population
Standard Deviation
Population Variance
( x   )
88.5

 8.9
•  
N
10
2
2
We don’t use
this value much
Population Standard Deviation
•    2  8.85  3.0
The population standard deviation is about 3.0, or \$3000.
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Deviation, Variance, and Standard
Deviation
Sample Variance
( x  x )
• s 
n 1
2
2
Everything is basically the “same”
except that we use n-1 instead of N.
This is considered a more
conservative value
Sample Standard Deviation
•
2

(
x

x
)
s  s2 
n 1
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Finding the Sample Variance & Standard
Deviation
In Words
In Symbols
x
n
1. Find the mean of the
sample data set.
x
2. Find the deviation of each
entry.
xx
3. Square each deviation.
( x  x )2
4. Add to get the sum of
squares.
SS x  ( x  x ) 2
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Finding the Sample Variance & Standard
Deviation
In Words
5. Divide by n – 1 to get the
sample variance.
6. Find the square root of the
variance to get the sample
standard deviation.
In Symbols
2

(
x

x
)
s2 
n 1
( x  x ) 2
s
n 1
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Example: Finding the Sample Standard
Deviation
The starting salaries are for the Chicago branches of a
corporation. The corporation has several other branches,
and you plan to use the starting salaries of the Chicago
branches to estimate the starting salaries for the larger
population. Find the sample standard deviation of the
starting salaries.
Starting salaries (1000s of dollars)
41 38 39 45 47 41 44 41 37 42
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Solution: Finding the Sample Standard
Deviation
• Determine SSx
• n = 10
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
Σ(x – μ) = 0
SSx = 88.5
Salary, x
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Solution: Finding the Sample Standard
Deviation
Sample Variance
These were the same
data values as before
( x  x )
88.5

 9.8 Notice that the
• s 
n 1
10  1
2
2
Sample Standard Deviation
88.5
 3.1
• s s 
9
2
sample standard
deviation is slightly
larger. In simple
terms, we aren’t quite
as accurate
The sample standard deviation is about 3.1, or \$3100.
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Example: Using Technology to Find the
Standard Deviation
Sample office rental rates (in
dollars per square foot per year)
district are shown in the table.
Use a calculator or a computer
to find the mean rental rate and
the sample standard deviation.
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
37.00
29.00
40.50
24.50
33.00
38.00
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Using the Calculator
•
•
•
•
•
Select “Stat”
Select Edit
Clear L1 (list 1)
Enter the data into L1
Select “Stat” and highlight “Calc” to display Calc
• Select 1: “1-Var Stats” and press “2nd L1” Enter
Larson/Farber 5th ed.
130
Chapter 2 Quiz 5
• Do all parts of problem 61 on page 78
• Show all work
• 4 points for each part (16 points total)
Larson/Farber 5th ed.
132
Interpreting Standard Deviation
• Standard deviation is a measure of the typical amount
an entry deviates from the mean.
• The more the entries are spread out, the greater the
standard deviation.
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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.
This may also be
referred to as a “normal
distribution”
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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
34%
34%
2.35%
2.35%
13.5%
x  3s
x  2s
13.5%
x s
x
x s
x  2s
x  3s
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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 64.3 inches, with a
sample standard deviation of 2.62 inches. Estimate the
percent of the women whose heights are between 59.06
inches and 64.3 inches.
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Solution: Using the Empirical Rule
• Because the distribution is bell-shaped, you can use
the Empirical Rule.
34% + 13.5% = 47.5% of women are between 59.06
and 64.3 inches tall.
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Chebychev’s Theorem
The empirical rule
works on bell (normal)
distributions. This
works on ALL
distributions
• The portion of any data set lying within k standard
deviations (k > 1) of the mean is at least:
1
1 2
k
1 3
• k = 2: In any data set, at least 1  2  or 75%
2
4
of the data lie within 2 standard deviations of the
mean.
1 8
• k = 3: In any data set, at least 1  2  or 88.9%
3
9
of the data lie within 3 standard deviations of the
mean.
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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?
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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.
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Standard Deviation for Grouped Data
Sample standard deviation for a frequency distribution
•
( x  x ) 2 f
s
n 1
where n = Σf (the number of
entries in the data set)
• When a frequency distribution has classes, estimate the
sample mean and the sample standard deviation by
using the midpoint of each class.
This is the equation for Sample standard deviation:
( x  x )
s
n 1
2
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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
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
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Solution: Finding the Standard Deviation
for Grouped Data
• First construct a frequency distribution.
• Find the mean of the frequency
distribution.
xf 91
x

 1.8
n
50
The sample mean is about 1.8
children.
x
f
xf
0
10
0(10) = 0
1
19
1(19) = 19
2
7
2(7) = 14
3
7
3(7) =21
4
2
4(2) = 8
5
1
5(1) = 5
6
4
6(4) = 24
Σf = 50 Σ(xf )= 91
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Solution: Finding the Standard Deviation
for Grouped Data
• Determine the sum of squares.
x
f
xx
( x  x )2
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
( x  x )2 f
( x  x ) 2 f  145.40
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Solution: Finding the Standard Deviation
for Grouped Data
• Find the sample standard deviation.
x 2 x
( x  x )2
( x  x ) f
145.40
s

 1.7
n 1
50  1
( x  x )2 f
The standard deviation is about 1.7 children.
What can I do with this information? Remember that
the sample mean is 1.8 children. If we can assume that
the distribution is normal, then about 68 percent of the
sample lies within .1 children and 3.5 children
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Section 2.4 Summary
• Determined the range of a data set
• Determined the variance and standard deviation of a
population and of a sample
• Used the Empirical Rule and Chebychev’s Theorem
to interpret standard deviation
• Approximated the sample standard deviation for
grouped data
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Assignment
• Page 90 1-10 orally
• Page 90-97 11-47 odd
Larson/Farber 5th ed.
147
Chapter 2 Quiz 6 (10 points)
• Using the data from question 53 on page 76:
1. Calculate the sample standard deviation (3 points)
2. What is the range of beds which would include 68%
of the sample? (3 points)
3. What is the range of beds which would include 95%
of the beds? (3 points)
Larson/Farber 5th ed.
148
Section 2.5
Measures of Position
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Section 2.5 Objectives
•
•
•
•
•
Determine the quartiles of a data set
Determine the interquartile range of a data set
Create a box-and-whisker plot
Interpret other fractiles such as percentiles
Determine and interpret the standard score (z-score)
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Quartiles
• Fractiles are numbers that partition (divide) an
ordered data set into equal parts.
• Quartiles approximately divide an ordered data set
into four equal parts.
 First quartile, Q1: About one quarter of the data
fall on or below Q1.
 Second quartile, Q2: About one half of the data
fall on or below Q2 (median).
 Third quartile, Q3: About three quarters of the
data fall on or below Q3.
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Example: Finding Quartiles
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.
7 18 11 6 59 17 18 54 104 20 31 8 10 15 19
Solution:
• Q2 divides the data set into two halves.
Lower half
Upper half
6 7 8 10 11 15 17 18 18 19 20 31 54 59 104
Q2
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Solution: Finding Quartiles
• The first and third quartiles are the medians of the
lower and upper halves of the data set.
Lower half
Upper half
6 7 8 10 11 15 17 18 18 19 20 31 54 59 104
Q1
Q2
Q3
About one fourth of the countries have 10 or fewer
nuclear power plants; about one half have 18 or fewer;
and about three fourths have 31 or fewer.
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Interquartile Range
Interquartile Range (IQR)
• The difference between the third and first quartiles.
• IQR = Q3 – Q1
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Example: Finding the Interquartile Range
Find the interquartile range of the data set.
7 18 11 6 59 17 18 54 104 20 31 8 10 15 19
Recall Q1 = 10, Q2 = 18, and Q3 = 31
Solution:
• IQR = Q3 – Q1 = 31 – 10 = 21
The number of power plants in the middle portion of
the data set vary by at most 21.
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Box-and-Whisker Plot
Box-and-whisker plot
• Exploratory data analysis tool.
• Highlights important features of a data set.
• Requires (five-number summary):
 Minimum entry
 First quartile Q1
 Median Q2
 Third quartile Q3
 Maximum entry
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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.
Box
Whisker
Minimum
entry
Whisker
Q1
Median, Q2
Q3
Maximum
entry
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Example: Drawing a Box-and-Whisker
Plot
Draw a box-and-whisker plot that represents the data set.
7 18 11 6 59 17 18 54 104 20 31 8 10 15 19
Min = 6, Q1 = 10, Q2 = 18, Q3 = 31, Max = 104,
Solution:
About half the data values are between 10 and 31. By
looking at the length of the right whisker, you can
conclude 104 is a possible outlier.
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Percentiles and Other Fractiles
Fractiles
Quartiles
Deciles
Percentiles
Summary
Divide a data set into 4 equal
parts
Divide a data set into 10
equal parts
Divide a data set into 100
equal parts
Symbols
Q1, Q2, Q3
D1, D2, D3,…, D9
P1, P2, P3,…, P99
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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 62nd
percentile? How should you
interpret this? (Source: College
Board)
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Solution: Interpreting Percentiles
The 62nd percentile
corresponds to a test score
of 1600.
This means that 62% of the
of 1600 or less.
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The Standard Score
Standard Score (z-score)
• Represents the number of standard deviations a given
value x falls from the mean μ.
value  mean
x
• z

standard deviation

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Example: Comparing z-Scores from
Different Data Sets
In 2009, Heath Ledger won the Oscar for Best
Supporting Actor at age 29 for his role in the movie The
Dark Knight. Penelope Cruz won the Oscar for Best
Supporting Actress at age 34 for her role in Vicky
Cristina Barcelona. The mean age of all Best
Supporting Actor winners is 49.5, with a standard
deviation of 13.8. The mean age of all Best Supporting
Actress winners is 39.9, with a standard deviation of
14.0. Find the z-scores that correspond to the ages of
Ledger and Cruz. Then compare your results.
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Solution: Comparing z-Scores from
Different Data Sets
• Heath Ledger
z
x

29  49.5

 1.49
13.8
1.49 standard
deviations below
the mean
• Penelope Cruz
z
x

34  39.9

 0.42
14.0
0.42 standard
deviations below
the mean
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Solution: Comparing z-Scores from
Different Data Sets
percent of the
data should
fall between
-2 and 2.
Both z-scores fall between –2 and 2, so neither score
would be considered unusual. Compared with other
Best Supporting Actor winners, Heath Ledger was
relatively younger, whereas the age of Penelope Cruz
was only slightly lower than the average age of other
Best Supporting Actress winners.
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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)
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Assignment
• Page 107 7-13 odd, 31,33, 37-50
Larson/Farber 5th ed.
167
Group Project (30 points)
•
5 pts •
5 pts •
5 pts •
5 pts •
•
•
Gather a sample of both men’s and women’s shoe sizes
Create an Expanded Frequency chart using 5 classes
Create a Frequency Histogram (using midpoints for x)
Calculate the mean, median, and mode of your data
Calculate the sample standard deviation of your data
If your sample IS normally distributed, then:
Using your mean and sample deviation, and the Empirical Rule:
5 pts
 Between what two shoe sizes would we find 68% of the shoe sizes?
5 pts
 Between what two shoe sizes would we find 95% of the shoe sizes?
-Or • If your sample is NOT normally distributed, then use Chebychev’s theorem
to find:
 Between what two shoe sizes would we find 75% of the shoe sizes?
5 pts
 Between what two shoe sizes would we find 88.9% of the shoe sizes?
5 pts
Larson/Farber 5th ed.
168
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