Download MT4Ch2 - Spartanburg County School District 5

WARM UP
• Which of the following samples would best represent the associated populations?
A. lifetimes of 300 randomly chosen 100-watt bulbs, lifetimes of all 100-watt bulbs.
B. number of yearly lightning storms in Florida, number of yearly lightning storms in U. S.
C. body weights of 20 randomly picked NFL kickers, body weights of all NFL players
D. salaries of 750 randomly selected Boston doctors, salaries of all Boston employees
•
A.
B.
C.
D.
You are conducting a survey at the college you attend to determine the number of students
who own a personal computer. The college consists of four campuses, one each on the north,
south, east, and west sides of the city. Since you attend the south campus, you decide to ask
200 of the 5,000 students from that campus. Which sampling technique did you use?
cluster sampling
systematic sampling
stratified sampling
convenience sampling
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1
Dear Teacher
Write a letter to Mr Rider on a blank
sheet of paper telling him what you
know about organizing and
displaying data.
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2
Descriptive
Statistics
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3
Frequency Distributions
and Their Graphs
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4
Learning Goals
• How to construct a frequency distribution including
limits, midpoints, relative frequencies, cumulative
frequencies, and boundaries
• How to construct frequency histograms, frequency
polygons, relative frequency histograms, and ogives
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5
Frequency Distribution
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
limits
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limits
6
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.
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7
WARM UP
Lifetimes of Boat Batteries Find the class width of the frequency distribution.
Class
limits
Frequency
24 - 37
4
38 - 51
14
52 - 65
7
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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.
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9
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.
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10
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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12
Solution: Constructing a Frequency
Distribution
3. Use 59 (minimum value)
as first lower limit. Add
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
59
width = 56 115
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Upper
limit
171
227
283
339
395
13
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
59
115
Upper
limit
114
170
171
227
283
339
226
282
338
394
395
450
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Class
width = 56
14
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.
Class
59 – 114
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
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15
You Try It
The number of World Series appearances by each Major
League Baseball Team is listed below. Construct a
frequency distribution using 5 classes.
40 19 14 20 12 18 9 7 11 9 7 6 5 10 7
5 4 2 2 3 1 1 2 2 1 1 1 1 0 0
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16
SC Unemployment Rates
Using the Unemployment Rate by County Table, create
a frequency distribution with 5 classes for November
2014.
Note: Round the class width calculation up to the
next tenth.
After creating the frequency distribution, color code the
counties on the South Carolina map by their class.
(You will need a separate color for each class.)
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17
WARM UP
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18
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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19
Example: Midpoint
Find the midpoint for each class in the following frequency
distribution.
Unemployment Rates for SC Counties
November 2014
LCL
UCL
Frequency Midpoint
5.2
6.4
16
6.5
7.7
16
7.8
9
7
9.1
10.3
3
10.4
11.6
4
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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 115 – 114 = 1.
• Half this distance is 0.5.
Class
Class
Frequency,
Boundaries
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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21
Class Boundaries
Class
59 – 114
115 – 170
171 – 226
227 – 282
283 – 338
339 – 394
395 – 450
Class
Frequency,
boundaries
f
58.5 – 114.5
5
114.5 – 170.5
8
170.5 – 226.5
6
226.5 – 282.5
5
282.5 – 338.5
2
338.5 – 394.5
1
394.5 – 450.5
3
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22
You Try It
Construct a frequency distribution using 5 classes for
the given data. Find the midpoint and the boundaries
for each class.
Super Bowl Appearances By Team
88 8 7 6 5 5 5 5 5 4 4 4 3 3 2
2 2 2 2 1 1 1 1 11 1 1 0 0 0 0
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23
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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24
Example: Frequency Histogram
Construct a frequency histogram for the global
positioning system (GPS) navigators.
Class
Class
boundaries
Frequency,
Midpoint
f
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
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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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27
Graphs of Frequency Distributions
frequency
Frequency Polygon
• A line graph that emphasizes the continuous change
in frequencies.
data values
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28
Example: Frequency Polygon
Construct a frequency polygon for the GPS navigators
frequency distribution.
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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29
Solution: Frequency Polygon
The graph should
begin and end on the
horizontal axis, so
extend the left side to
one class width before
the first class
midpoint and extend
the right side to one
class width after the
last class midpoint.
You can see that the frequency of GPS navigators increases
up to $142.50 and then decreases.
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30
WARM UP
Find the class width then find the midpoint and class
boundaries for the first class.
LCL
UCL
Frequency
7
10
13
11
14
17
15
18
5
19
22
7
23
26
2
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32
Determining the Relative Frequency
Relative Frequency of a class
• Portion or percentage of the data that falls in a
particular class.
class frequency
f

• relative frequency 
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
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33
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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34
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
86.5
0.17
115 – 170
114.5 – 170.5
142.5
0.27
171 – 226
170.5 – 226.5
198.5
0.2
227 – 282
226.5 – 282.5
254.5
0.17
283 – 338
282.5 – 338.5
310.5
0.07
339 – 394
338.5 – 394.5
366.5
0.03
395 – 450
394.5 – 450.5
422.5
0.1
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35
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 GPS navigators are
priced between $114.50 and $170.50.
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36
AIR TIME
Find some material to
read quietly.
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37
WARM UP
1. Identify the class width and the class boundaries in
the frequency distribution.
2. Find the relative frequency for each class.
LCL
UCL
Frequency
Relative
Frequency
50
59
10
60
69
30
70
79
32
80
89
18
90
99
10
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38
 and X Cards
After each statement made, raise your card.
A check means you agree with the statement.
An X means you disagree with the statement.
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39
Determining the Cumulative Frequency
Cumulative frequency of a class
• The sum of the frequency for that class and all
previous classes.
Class
Frequency, f
Cumulative frequency
59 – 114
5
5
115 – 170
+ 8
13
171 – 226
+ 6
19
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40
Expanded Frequency Distribution
Class
Frequency, f
Midpoint
Relative
frequency
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
 1
n
30
Σf = 30
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Cumulative
frequency
41
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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42
WARM UP
Find the cumulative frequency for each class in the
distribution.
SC Unemployment Rates By County, Nov. 2014
LCL UCL
Frequency
Cumulative
Frequency
5.2
6.4
16
6.5
7.7
16
7.8
9
7
9.1
10
3
10
12
4
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43
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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44
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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45
Example: Ogive
Construct an ogive for the GPS navigators frequency
distribution.
Midpoint
Class
Class
boundaries
Cumulative
frequency
59 – 114
58.5 – 114.5
86.5
5
115 – 170
114.5 – 170.5
142.5
13
171 – 226
170.5 – 226.5
198.5
19
227 – 282
226.5 – 282.5
254.5
24
283 – 338
282.5 – 338.5
310.5
26
339 – 394
338.5 – 394.5
366.5
27
395 – 450
394.5 – 450.5
422.5
30
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46
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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47
Example: Ogive
Create a Cumulative Frequency Distribution and an
Ogive for the following distribution.
LCL
UCL
Frequency
7
10
13
11
14
17
15
18
5
19
22
7
23
26
2
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49
ASSIGNMENT
Use the Federal Report Card Data Sheet to create a
cumulative frequency distribution and an ogive for the
Total Column. Use 5 classes to create the frequency
distribution. Round the class width up to the next tenth.
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50
Section 2.1 Summary
• Constructed frequency distributions
• Constructed frequency histograms, frequency
polygons, relative frequency histograms and ogives
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51
WARM UP
Construct an ogive for the frequency
table.
LCL
UCL
Frequency
8
11
10
12
15
12
16
19
3
20
23
0
24
27
2
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Cumulative
Frequency
52
THINK-PAIR-WRITE
With a partner, brainstorm about other types of graphs
that are used to display data. Each student should write
a response with a list of the types of graphs that was
discussed with a partner.
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53
More Graphs and Displays
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54
Learning Goals
• How to graph and interpret quantitative data using
stem-and-leaf plots and dot plots
• How to graph and interpret qualitative data using pie
charts and Pareto charts
• How to graph and interpret paired data sets using
scatter plots and time series charts
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55
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
.
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2
3
1 5 5 6 7 8
0 6 6
4
5
56
Stem-and-Leaf Plot
In a stem-and-leaf plot, each number is separated into a
stem (usually the entry’s leftmost digits) and a leaf
(usually the rightmost digit). This is an example of
exploratory data analysis.
Example: The following data represents the ages of 30
students in a college statistics class. Display the data in a
stem-and-leaf plot.
Ages of Students
18
20
21
27
29
20
19
30
32
19
34
19
24
29
18
37
38
22
30
39
32
44
33
46
54
49
18
51
21
21
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57
Stem-and-Leaf Plot
Ages of Students
Key: 1|8 = 18
1 888999
2 0011124799
3 002234789
4 469
5 14
Most of the values lie
between 20 and 39.
This graph allows us to see
the shape of the data as well
as the actual values.
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58
Stem-and-Leaf Plot
Stem-and-leaf displays are quick ways to make a picture for small sets of
data, like these numbers posted by NFL quarterback Brett Favre during his
record breaking career with the Green Bay Packers, New York Jets, and
Minnesota Vikings.
Here are the number of passes Favre completed each season:
363 343 356 343 372 346
308 341 314 338 341 347
304 325 359 363 318 302
Make a stem-leaf-display using the first two digits as the stem and the last digit as the
leaf. For example, 363 completions should look like this: 36 | 3
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59
Stem-and-Leaf Plot
Make a stem-and-leaf plot showing the number of touchdown passes Favre threw each
season.
33
22
28
18
20
30
32
27
32
20
22
31
35
39
38
33
19
18
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60
Graphing Quantitative Data Sets
Dot plot
In a 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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61
Example:
Use a dot plot to display the ages of the 30 students in the
statistics class.
Ages of Students
18
20
21
19
23
20
19
19
22
19
20
19
24
29
18
20
20
22
30
18
32
19
33
19
54
20
18
19
21
21
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62
Dot Plot
Ages of Students
1
5
1
8
2
1
2
4
2
7
3
0
3 3
3 6
3
9
4
2
4
5
4
8
5
1
5 5
4 7
From this graph, we can conclude that most of the
values lie between 18 and 32.
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63
Dot Plot
Using the rosters of the two Super Bowl teams, create a
dot plot of the ages for players from each team. You
will have two dot plots; one for the Patriots and one for
the Seahawks.
Turn the rosters back in because we will use them again
later.
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64
WARM UP
Create a stem-and-leaf plot for the data.
15 15 17 19 22 25 27 27 27 29 30 30
32 35 41
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65
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.
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66
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
Associate’s
Bachelor’s
Master’s
First professional
Doctoral
.
Frequency, f
Relative frequency
728
728
 0.24
3007
1525
1525
 0.51
3007
604
604
 0.20
3007
90
 0.03
3007
60
 0.02
3007
90
60
3007
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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 cars is
360(0.24) ≈ 86º
.
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Solution: Constructing a Pie Chart
Relative
Frequency, f frequency
Type of degree
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
.
Central angle
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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 most fatalities in motor
vehicle crashes were those involving the occupants of cars.
.
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71
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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72
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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73
Solution: Constructing a Pareto Chart
Cause
$ (billion)
Admin. error
5.4
Employee
theft
15.9
Shoplifting
12.7
Vendor fraud
1.4
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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Assignment
Create a question with 4 possible
responses that will give you qualitative
data.
Have classmates respond to the
question.
Create a Pie Chart and a Pareto Chart
for the responses to the question.
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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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77
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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79
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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80
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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81
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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82
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 and interpreted quantitative data using stemand-leaf plots and dot plots
• Graphed and interpreted qualitative data using pie
charts and Pareto charts
• Graphed and interpreted paired data sets using scatter
plots and time series charts
.
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84
AIR TIME
Find some material to
read quietly.
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Section 2.3
Measures of Central Tendency
.
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Learning Goals
• How to find the mean, median, and mode of a
population and of a sample
• How to find the weighted mean of a data set and the
mean of a frequency distribution
• How to describe the shape of a distribution as
symmetric, uniform, or skewed and how to 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
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89
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
.
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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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95
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.
• 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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99
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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100
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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102
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
20
20
20
20
20
21
21
21
21
22
22
22
23
23
23
23
24
24
65
Mean:
x 20  20  ...  24  65
x

 23.8 years
n
20
Median:
21  22
 21.5 years
2
Mode:
.
20
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
Sometimes a graphical comparison can help you decide
which measure of central tendency best represents a
data set.
In this case, it appears that the median best describes
the data set.
.
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Weighted Mean
Weighted Mean
• The mean of a data set whose entries have varying
weights.
( x  w)
• x 
where w is the weight of each entry x
w
.
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Example: Finding a Weighted Mean
You are taking a class in which your grade is
determined from five sources: 50% from your test
mean, 15% from your midterm, 20% from your final
exam, 10% from your computer lab work, and 5% from
your homework. Your scores are 86 (test mean), 96
(midterm), 82 (final exam), 98 (computer lab), and 100
(homework). What is the weighted mean of your
scores? If the minimum average for an A is 90, did you
get an A?
.
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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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Example: College GPA
Jason wanted to find out what his GPA would be at the
end of his first semester of college. He estimated his
grades in the following classes.
Course
Credits
English 101
3
Biology I
4
Math 120
3
History 101
3
PE 125
2
College 101
1
Grade
A
B
A
C
A
D
A=4 points B=3 points C=2 points D=1 point F=0 points
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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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Solution: College GPA
Course Credits Grade Quality Points Grade Points
English 101
3
A
4
12
Biology I
4
B
3
12
Math 120
3
A
4
12
History 101
3
C
2
6
PE 125
2
A
4
8
College 101
1
D
1
1
Total
16
51
51/16 = 3.1875
Jason will have a 3.1875 GPA at the end of the semester with
these grades.
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Finding the Mean of a Frequency
Distribution
In Words
In Symbols
1. Find the midpoint of each
class.
(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.
.
Copyright © 2015, 2012, and 2009 Pearson Education, Inc.
( x  f )
x
n
113
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
• Found the mean, median, and mode of a population
and of a sample
• Found 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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120
Measures of Variation
.
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Learning Goals
• How to find the range of a data set
• How to find the variance and standard deviation of a
population and of a sample
• How to use the Empirical Rule and Chebychev’s
Theorem to interpret standard deviation
• How to approximate the sample standard deviation
for grouped data
• How to use the coefficient of variation to compare
variation in different data sets
.
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122
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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123
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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126
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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127
Solution: Finding the Deviation
• Determine the
deviation for each
data entry.
.
Salary ($1000s), x Deviation: x – μ
41
41 – 41.5 = –0.5
38
38 – 41.5 = –3.5
39
39 – 41.5 = –2.5
45
45 – 41.5 = 3.5
47
47 – 41.5 = 5.5
41
41 – 41.5 = –0.5
44
44 – 41.5 = 2.5
41
41 – 41.5 = –0.5
37
37 – 41.5 = –4.5
42
42 – 41.5 = 0.5
Σx = 415
Σ(x – μ) = 0
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Deviation, Variance, and Standard
Deviation
Population Variance
( x   )
•  
N
2
2
Sum of squares, SSx
Population Standard Deviation
2

(
x


)
2
•    
N
.
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129
Finding the Population Variance &
Standard Deviation
In Words
In Symbols
1. Find the mean of the
population data set.
.
x

N
2. Find deviation of each
entry.
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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130
Finding the Population Variance &
Standard Deviation
In Words
.
In Symbols
5. Divide by N to get the
population variance.
2

(
x


)
2 
N
6. Find the square root to get
the population standard
deviation.
( x   ) 2

N
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131
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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132
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
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
Sample Standard Deviation
•
.
2

(
x

x
)
s  s2 
n 1
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135
Finding the Sample Variance & Standard
Deviation
In Words
.
In Symbols
x
n
1. Find the mean of the
sample data set.
x
2. Find 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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136
Finding the Sample Variance & Standard
Deviation
In Words
In Symbols
5. Divide by n – 1 to get the
sample variance.
6. Find the square root to get
the sample standard
deviation.
.
Copyright © 2015, 2012, and 2009 Pearson Education, Inc.
2

(
x

x
)
s2 
n 1
( x  x ) 2
s
n 1
137
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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138
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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139
Solution: Finding the Sample Standard
Deviation
Sample Variance
( x  x )
88.5

 9.8
• s 
n 1
10  1
2
2
Sample Standard Deviation
88.5
 3.1
• s s 
9
2
The sample standard deviation is about 3.1, or $3100.
.
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140
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.)
.
Copyright © 2015, 2012, and 2009 Pearson Education, 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
141
Solution: Using Technology to Find the
Standard Deviation
Sample Mean
Sample Standard
Deviation
.
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WARM UP
Find the population standard deviation and the sample
standard deviation for the data set.
Graduation Rates for Southeastern States, 2010-2011
86 83 80 75 75 75 70 72 71
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143
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.
.
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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
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x  2s
x  3s
146
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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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 standard deviation by using the
midpoint of each class.
.
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149
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.
.
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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
150
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.
.
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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
151
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.
.
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153
Coefficient of Variation
Coefficient of Variation (CV)
• Describes the standard deviation of a data set as a
percent of the mean.
• Population data set:
 CV   100%

• Sample data set:
s

CV  100%
x
.
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154
Example: Comparing Variation in
Different Data Sets
The table shows the population
heights (in inches) and weights (in
pounds) of the members of a
basketball team. Find the
coefficient of variation for the
heights and the weighs. Then
compare the results.
.
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Solution: Comparing Variation in
Different Data Sets
The mean height is   72.8 inches with a standard
deviation of   3.3 inches. The coefficient of variation
for the heights is
CVheight

 100%

3.3

100%
72.8
 4.5%
.
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Solution: Comparing Variation in
Different Data Sets
The mean weight is   187.8 pounds with a standard
deviation of   17.7 pounds. The coefficient of
variation for the weights is
CVw eight

 100%

17.7

100%
187.8
 9.4%
The weights (9.4%) are more variable than the heights
(4.5%).
.
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157
Section 2.4 Summary
• Found the range of a data set
• Found 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
• Used the coefficient of variation to compare variation
in different data sets
.
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158
Measures of Position
.
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159
Learning Goals
• How to find the first, second, and third quartiles of a
data set, how to find the interquartile range of a data
set, and how to represent a data set graphically using
a box-and whisker plot
• How to interpret other fractiles such as percentiles
and how to find percentiles for a specific data entry
• Determine and interpret the standard score (z-score)
.
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160
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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161
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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162
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 less,
about one half have 18 or less; and about three fourths
have 31 or less.
.
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163
WARM UP
Find the first quartile, second quartile and third quartile
for the data set.
Graduation Rates for Southeastern States, 2010-2011
86 83 80 75 75 75 70 72 71
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164
Interquartile Range
Interquartile Range (IQR)
• The difference between the third and first quartiles.
• IQR = Q3 – Q1
.
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165
Example: Finding the Interquartile Range
Find the interquartile range of the data set.
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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166
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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167
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
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Q3
Maximum
entry
168
Example: Drawing a Box-and-Whisker
Plot
Draw a box-and-whisker plot that represents the 15 data
set.
Min = 6, Q1 = 10, Q2 = 18, Q3 = 31, Max = 104,
Solution:
About half the scores 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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169
Percentiles and Other Fractiles
Fractiles
Quartiles
Summary
Divides data into 4 equal
parts
Symbols
Q1, Q2, Q3
Deciles
Divides data into 10 equal
parts
Divides data into 100 equal
parts
D1, D2, D3,…, D9
Percentiles
.
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P1, P2, P3,…, P99
170
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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171
Solution: Interpreting Percentiles
The 62nd percentile
corresponds to a test score
of 1600.
This means that 62% of the
students had an SAT score
of 1600 or less.
.
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172
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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173
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-score that corresponds to the ages of
Ledger and Cruz. Then compare your results.
.
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174
Solution: Comparing z-Scores from
Different Data Sets
• Heath Ledger
z
x

29  49.5

 1.49
13.8
1.49 standard
deviations above
the mean
• Penelope Cruz
z
.
x

34  39.9

 0.42
14.0
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0.42 standard
deviations below
the mean
175
Solution: Comparing z-Scores from
Different Data Sets
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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176
Assignment
Use the US Album Sales By Year Statistics to create a
box-and-whisker plot for the Physical Units Sold
column.
Also, find the mean and the population standard
deviation for the Physical Units Sold column.
Find the z-score for 362.6
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Section 2.5 Summary
• Found the first, second, and third quartiles of a data
set, how to find the interquartile range of a data set,
and represented a data set graphically using a box-and
whisker plot
• Interpreted other fractiles such as percentiles and
percentiles for a specific data entry
• Determined and interpreted the standard score
(z-score)
.
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178