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Transcript
```Chapter 2
Descriptive Statistics
1
Chapter Outline
• 2.1 Frequency Distributions and Their Graphs
• 2.2 More Graphs and Displays
• 2.3 Measures of Central Tendency
• 2.4 Measures of Variation
• 2.5 Measures of Position
2
Section 2.1
Frequency Distributions
and Their Graphs
3
Section 2.1 Objectives
• Construct frequency distributions
• Construct frequency histograms, frequency polygons,
relative frequency histograms, and ogives
4
Frequency Distribution
Frequency Distribution
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
limits
5
Constructing a Frequency Distribution
1. Decide on the number of classes.
 Usually between 5 and 20; otherwise, it may be
difficult to detect any patterns.
2. Find the class width.
 Determine the range of the data.
 Divide the range by the number of classes.
 Round up to the next convenient number.
6
Constructing a Frequency Distribution
3. Find the class limits.
 You can use the minimum data entry as the lower
limit of the first class.
 Find the remaining lower limits (add the class
width to the lower limit of the preceding class).
 Find the upper limit of the first class. Remember
that classes cannot overlap.
 Find the remaining upper class limits.
7
Constructing a Frequency Distribution
4. 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.
8
Example: Constructing a Frequency
Distribution
The following sample data set lists the number of
minutes 50 Internet subscribers spent on the Internet
during their most recent session. Construct a frequency
distribution that has seven classes.
50 40 41 17 11 7 22 44 28 21 19 23 37 51 54 42 86
41 78 56 72 56 17 7 69 30 80 56 29 33 46 31 39 20
18 29 34 59 73 77 36 39 30 62 54 67 39 31 53 44
9
Solution: Constructing a Frequency
Distribution
50 40 41 17 11 7 22 44 28 21 19 23 37 51 54 42 86
41 78 56 72 56 17 7 69 30 80 56 29 33 46 31 39 20
18 29 34 59 73 77 36 39 30 62 54 67 39 31 53 44
1. Number of classes = 7 (given)
2. Find the class width
max  min 86  7

 11.29
#classes
7
Round up to 12
10
Solution: Constructing a Frequency
Distribution
3. Use 7 (minimum value)
as first lower limit. Add
the class width of 12 to
get the lower limit of the
next class.
7 + 12 = 19
Find the remaining
lower limits.
Class
width = 12
Lower
limit
7
Upper
limit
19
31
43
55
67
79
11
Solution: Constructing a Frequency
Distribution
The upper limit of the
first class is 18 (one less
than the lower limit of the
second class).
Add the class width of 12
to get the upper limit of
the next class.
18 + 12 = 30
Find the remaining upper
limits.
Lower
limit
7
19
31
43
55
67
79
Upper
limit
18
30
42
54
66
Class
width = 12
78
90
12
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
7 – 18
Tally
Frequency, f
IIII I
6
19 – 30
IIII IIII
10
31 – 42
IIII IIII III
13
43 – 54
IIII III
8
55 – 66
IIII
5
67 – 78
IIII I
6
79 – 90
II
2
Σf = 50
13
Determining the Midpoint
Midpoint of a class
(Lower class limit)  (Upper class limit)
2
Class
7 – 18
Midpoint
7  18
 12.5
2
19 – 30
19  30
 24.5
2
31 – 42
31  42
 36.5
2
Frequency, f
6
Class width = 12
10
13
14
Determining the Relative Frequency
Relative Frequency of a class
• Portion or percentage of the data that falls in a
particular class.
class frequency
f

• relative frequency 
Sample size
n
Class
Frequency, f
7 – 18
6
19 – 30
10
31 – 42
13
Relative Frequency
6
 0.12
50
10
 0.20
50
13
 0.26
50
15
Determining the Cumulative Frequency
Cumulative frequency of a class
• The sum of the frequency for that class and all
previous classes.
Class
Frequency, f
Cumulative frequency
7 – 18
6
6
19 – 30
+ 10
16
31 – 42
+ 13
29
16
Expanded Frequency Distribution
Class
Frequency, f
Midpoint
Relative
frequency
7 – 18
6
12.5
0.12
6
19 – 30
10
24.5
0.20
16
31 – 42
13
36.5
0.26
29
43 – 54
8
48.5
0.16
37
55 – 66
5
60.5
0.10
42
67 – 78
6
72.5
0.12
48
79 – 90
2
84.5
0.04
f
 1
n
50
Σf = 50
Cumulative
frequency
17
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
18
Class Boundaries
Class boundaries
• The numbers that separate classes without forming
gaps between them.
• The distance from the upper
limit of the first class to the
lower limit of the second
class is 19 – 18 = 1.
• Half this distance is 0.5.
Class
Class
Frequency,
Boundaries
f
7 – 18
6.5 – 18.5
6
19 – 30
10
31 – 42
13
• First class lower boundary = 7 – 0.5 = 6.5
• First class upper boundary = 18 + 0.5 = 18.5
19
Class Boundaries
Class
7 – 18
19 – 30
31 – 42
43 – 54
55 – 66
67 – 78
79 – 90
Class
boundaries
6.5 – 18.5
18.5 – 30.5
30.5 – 42.5
42.5 – 54.5
54.5 – 66.5
66.5 – 78.5
78.5 – 90.5
Frequency,
f
6
10
13
8
5
6
2
20
Example: Frequency Histogram
Construct a frequency histogram for the Internet usage
frequency distribution.
Class
Class
boundaries
Midpoint
Frequency,
f
7 – 18
6.5 – 18.5
12.5
6
19 – 30
18.5 – 30.5
24.5
10
31 – 42
30.5 – 42.5
36.5
13
43 – 54
42.5 – 54.5
48.5
8
55 – 66
54.5 – 66.5
60.5
5
67 – 78
66.5 – 78.5
72.5
6
79 – 90
78.5 – 90.5
84.5
2
21
Solution: Frequency Histogram
(using Midpoints)
22
Solution: Frequency Histogram
(using class boundaries)
6.5
18.5
30.5
42.5
54.5
66.5
78.5
90.5
You can see that more than half of the subscribers spent
between 19 and 54 minutes on the Internet during their most
recent session.
23
Graphs of Frequency Distributions
frequency
Frequency Polygon
• A line graph that emphasizes the continuous change
in frequencies.
data values
24
Example: Frequency Polygon
Construct a frequency polygon for the Internet usage
frequency distribution.
Class
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
25
Solution: Frequency Polygon
Internet Usage
Frequency
The graph should
begin and end on the
horizontal axis, so
extend the left side to
one class width before
the first class
midpoint and extend
the right side to one
class width after the
last class midpoint.
14
12
10
8
6
4
2
0
0.5
12.5
24.5
36.5
48.5
60.5
72.5
84.5
96.5
Time online (in minutes)
You can see that the frequency of subscribers increases up to
36.5 minutes and then decreases.
26
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
27
Example: Relative Frequency Histogram
Construct a relative frequency histogram for the Internet
usage frequency distribution.
Class
Class
boundaries
Frequency,
f
Relative
frequency
7 – 18
6.5 – 18.5
6
0.12
19 – 30
18.5 – 30.5
10
0.20
31 – 42
30.5 – 42.5
13
0.26
43 – 54
42.5 – 54.5
8
0.16
55 – 66
54.5 – 66.5
5
0.10
67 – 78
66.5 – 78.5
6
0.12
79 – 90
78.5 – 90.5
2
0.04
28
Solution: Relative Frequency Histogram
6.5
18.5
30.5
42.5
54.5
66.5
78.5
90.5
From this graph you can see that 20% of Internet subscribers
spent between 18.5 minutes and 30.5 minutes online.
29
Graphs of Frequency Distributions
cumulative
frequency
Cumulative Frequency Graph or Ogive
• A line graph that displays the cumulative frequency
of each class at its upper class boundary.
• The upper boundaries are marked on the horizontal
axis.
• The cumulative frequencies are marked on the
vertical axis.
data values
30
Constructing an Ogive
1. Construct a frequency distribution that includes
cumulative frequencies as one of the columns.
2. Specify the horizontal and vertical scales.
 The horizontal scale consists of the upper class
boundaries.
 The vertical scale measures cumulative
frequencies.
3. Plot points that represent the upper class boundaries
and their corresponding cumulative frequencies.
31
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).
32
Example: Ogive
Construct an ogive for the Internet usage frequency
distribution.
Class
Class
boundaries
Frequency,
f
Cumulative
frequency
7 – 18
6.5 – 18.5
6
6
19 – 30
18.5 – 30.5
10
16
31 – 42
30.5 – 42.5
13
29
43 – 54
42.5 – 54.5
8
37
55 – 66
54.5 – 66.5
5
42
67 – 78
66.5 – 78.5
6
48
79 – 90
78.5 – 90.5
2
50
33
Solution: Ogive
Internet Usage
Cumulative frequency
60
50
40
30
20
10
0
6.5
18.5
30.5
42.5
54.5
66.5
78.5
90.5
Time online (in minutes)
From the ogive, you can see that about 40 subscribers spent 60
minutes or less online during their last session. The greatest
increase in usage occurs between 30.5 minutes and 42.5 minutes.
34
Section 2.1 Summary
• Constructed frequency distributions
• Constructed frequency histograms, frequency
polygons, relative frequency histograms and ogives
35
Graphical Analysis: Use the frequency polygon to identify the
class with the greatest, and the class with the least, frequency.
36
Graphical Analysis: Use the relative frequency histogram to
(a) identify the class with the greatest, and the class with the
least, relative frequency.
(b) approximate the greatest and least relative frequencies.
(c) approximate the relative frequency of the second class.
37
Section 2.2
More Graphs and Displays
38
Section 2.2 Objectives
• Graph quantitative data using stem-and-leaf plots and
dot plots
• Graph qualitative data using pie charts and Pareto
charts
• Graph paired data sets using scatter plots and time
series charts
39
Graphing Quantitative Data Sets
Stem-and-leaf plot
• Each number is separated into a stem and a leaf.
• Similar to a histogram.
• Still contains original data values.
26
Data: 21, 25, 25, 26, 27, 28,
30, 36, 36, 45
2
3
1 5 5 6 7 8
0 6 6
4
5
40
Example: Constructing a Stem-and-Leaf
Plot
The following are the numbers of text messages sent
last month by the cellular phone users on one floor of a
college dormitory. Display the data in a stem-and-leaf
plot.
155 159 144 129 105 145 126 116 130 114 122 112 112 142 126
118 118 108 122 121 109 140 126 119 113 117 118 109 109 119
139 139 122 78 133 126 123 145 121 134 124 119 132 133 124
129 112 126 148 147
41
Solution: Constructing a Stem-and-Leaf
Plot
155 159 144 129 105 145 126 116 130 114 122 112 112 142 126
118 118 108 122 121 109 140 126 119 113 117 118 109 109 119
139 139 122 78 133 126 123 145 121 134 124 119 132 133 124
129 112 126 148 147
• The data entries go from a low of 78 to a high of 159.
• Use the rightmost digit as the leaf.
 For instance,
78 = 7 | 8
and 159 = 15 | 9
• List the stems, 7 to 15, to the left of a vertical line.
• For each data entry, list a leaf to the right of its stem.
42
Solution: Constructing a Stem-and-Leaf
Plot
Include a key to identify
the values of the data.
From the display, you can conclude that more than 50% of the
cellular phone users sent between 110 and 130 text messages.
43
Graphing Quantitative Data Sets
Dot plot
• Each data entry is plotted, using a point, above a
horizontal axis
Data: 21, 25, 25, 26, 27, 28, 30, 36, 36, 45
26
20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
44
Example: Constructing a Dot Plot
Use a dot plot organize the text messaging data.
155 159 144 129 105 145 126 116 130 114 122 112 112 142 126
118 118 108 122 121 109 140 126 119 113 117 118 109 109 119
139 139 122 78 133 126 123 145 121 134 124 119 132 133 124
129 112 126 148 147
• So that each data entry is included in the dot plot, the
horizontal axis should include numbers between 70 and
160.
• To represent a data entry, plot a point above the entry's
position on the axis.
• If an entry is repeated, plot another point above the
previous point.
45
Solution: Constructing a Dot Plot
155 159 144 129 105 145 126 116 130 114 122 112 112 142 126
118 118 108 122 121 109 140 126 119 113 117 118 109 109 119
139 139 122 78 133 126 123 145 121 134 124 119 132 133 124
129 112 126 148 147
From the dot plot, you can see that most values cluster
between 105 and 148 and the value that occurs the
most is 126. You can also see that 78 is an unusual data
value.
46
Graphing Qualitative Data Sets
Pie Chart
• A circle is divided into sectors that represent
categories.
• The area of each sector is proportional to the
frequency of each category.
47
Example: Constructing a Pie Chart
The numbers of motor vehicle occupants killed in
crashes in 2005 are shown in the table. Use a pie chart
to organize the data. (Source: U.S. Department of
Transportation, National Highway Traffic Safety
Vehicle type Killed
Cars
18,440
Trucks
13,778
Motorcycles
4,553
Other
823
48
Solution: Constructing a Pie Chart
• Find the relative frequency (percent) of each category.
Vehicle type Frequency, f
Cars
18,440
Trucks
13,778
Motorcycles
Other
4,553
823
Relative frequency
18440
 0.49
37594
13778
 0.37
37594
4553
 0.12
37594
823
 0.02
37594
37,594
49
Solution: Constructing a Pie Chart
• Construct the pie chart using the central angle that
corresponds to each category.
 To find the central angle, multiply 360º by the
category's relative frequency.
 For example, the central angle for cars is
360(0.49) ≈ 176º
50
Solution: Constructing a Pie Chart
Relative
Vehicle type Frequency, f frequency
Central angle
Cars
18,440
0.49
360º(0.49)≈176º
Trucks
13,778
0.37
360º(0.37)≈133º
4,553
0.12
360º(0.12)≈43º
823
0.02
360º(0.02)≈7º
Motorcycles
Other
51
Solution: Constructing a Pie Chart
Relative
Vehicle type frequency
Central
angle
Cars
0.49
176º
Trucks
0.37
133º
Motorcycles
0.12
43º
Other
0.02
7º
From the pie chart, you can see that most fatalities in motor
vehicle crashes were those involving the occupants of cars.
52
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
53
Example: Constructing a Pareto Chart
In a recent year, the retail industry lost \$41.0 million in
inventory shrinkage. Inventory shrinkage is the loss of
inventory through breakage, pilferage, shoplifting, and
so on. The causes of the inventory shrinkage are
administrative error (\$7.8 million), employee theft
(\$15.6 million), shoplifting (\$14.7 million), and vendor
fraud (\$2.9 million). Use a Pareto chart to organize this
data. (Source: National Retail Federation and Center for
Retailing Education, University of Florida)
54
Solution: Constructing a Pareto Chart
Cause
\$ (million)
7.8
Employee
theft
15.6
Shoplifting
14.7
Vendor fraud
2.9
From the graph, it is easy to see that the causes of inventory
shrinkage that should be addressed first are employee theft and
shoplifting.
55
.
Graphing Paired Data Sets
Paired Data Sets
• Each entry in one data set corresponds to one entry in
a second data set.
• Graph using a scatter plot.
 The ordered pairs are graphed as y
points in a coordinate plane.
 Used to show the relationship
between two quantitative variables.
x
57
Example: Interpreting a Scatter Plot
The British statistician Ronald Fisher introduced a
famous data set called Fisher's Iris data set. This data set
describes various physical characteristics, such as petal
length and petal width (in millimeters), for three species
of iris. The petal lengths form the first data set and the
petal widths form the second data set. (Source: Fisher, R.
A., 1936)
58
Example: Interpreting a Scatter Plot
As the petal length increases, what tends to happen to
the petal width?
Each point in the
scatter plot
represents the
petal length and
petal width of one
flower.
59
Solution: Interpreting a Scatter Plot
Interpretation
From the scatter plot, you can see that as the petal
length increases, the petal width also tends to
increase.
60
Interpretation: Stay Away From Meat!
61
Interpretation: Stay Away From Guns!
62
Graphing Paired Data Sets
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
63
Example: Constructing a Time Series
Chart
The table lists the number of cellular
telephone subscribers (in millions)
for the years 1995 through 2005.
Construct a time series chart for the
number of cellular subscribers.
(Source: Cellular Telecommunication &
Internet Association)
64
Solution: Constructing a Time Series
Chart
• Let the horizontal axis represent
the years.
• Let the vertical axis represent the
number of subscribers (in
millions).
• Plot the paired data and connect
them with line segments.
65
Solution: Constructing a Time Series
Chart
The graph shows that the number of subscribers has been
increasing since 1995, with greater increases recently.
66
Section 2.2 Summary
• Graphed quantitative data using stem-and-leaf plots
and dot plots
• Graphed qualitative data using pie charts and Pareto
charts
• Graphed paired data sets using scatter plots and time
series charts
67
Section 2.3
Measures of Central Tendency
68
Section 2.3 Objectives
• Determine the mean, median, and mode of a
population and of a sample
• Determine the weighted mean of a data set and the
mean of a frequency distribution
• Describe the shape of a distribution as symmetric,
uniform, or skewed and compare the mean and
median for each
69
Measures of Central Tendency
Measure of central tendency
• A value that represents a typical, or central, entry of a
data set.
• Most common measures of central tendency:
 Mean
 Median
 Mode
70
Measure of Central Tendency: Mean
Mean (average)
• The sum of all the data entries divided by the number
of entries.
• Sigma notation: Σx = add all of the data entries (x)
in the data set.
x
• Population mean:  
N
• Sample mean:
x
x
n
71
Example: Finding a Sample Mean
The prices (in dollars) for a sample of roundtrip flights
from Chicago, Illinois to Cancun, Mexico are listed.
What is the mean price of the flights?
872 432 397 427 388 782 397
72
Solution: Finding a Sample Mean
872 432 397 427 388 782 397
• The sum of the flight prices is
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.
73
Measure of Central Tendency: Median
Median
• The value that lies in the middle of the data when the
data set is ordered.
• Measures the center of an ordered data set by dividing
it into two equal parts.
• If the data set has an
 odd number of entries: median is the middle data
entry.
 even number of entries: median is the mean of
the two middle data entries.
74
Example: Finding the Median
The prices (in dollars) for a sample of roundtrip flights
from Chicago, Illinois to Cancun, Mexico are listed.
Find the median of the flight prices.
872 432 397 427 388 782 397
75
Solution: Finding the Median
872 432 397 427 388 782 397
• First order the data.
388 397 397 427 432 782 872
• There are seven entries (an odd number), the median
is the middle, or fourth, data entry.
The median price of the flights is \$427.
76
Example: Finding the Median
The flight priced at \$432 is no longer available. What is
the median price of the remaining flights?
872 397 427 388 782 397
77
Solution: Finding the Median
872 397 427 388 782 397
• First order the data.
388 397 397 427 782 872
• There are six entries (an even number), the median is
the mean of the two middle entries.
397  427
Median 
 412
2
The median price of the flights is \$412.
78
Measure of Central Tendency: Mode
Mode
• The data entry that occurs with the greatest frequency.
• If no entry is repeated the data set has no mode.
• If two entries occur with the same greatest frequency,
each entry is a mode (bimodal).
79
Example: Finding the Mode
The prices (in dollars) for a sample of roundtrip flights
from Chicago, Illinois to Cancun, Mexico are listed.
Find the mode of the flight prices.
872 432 397 427 388 782 397
80
Solution: Finding the Mode
872 432 397 427 388 782 397
• Ordering the data helps to find the mode.
388 397 397 427 432 782 872
• The entry of 397 occurs twice, whereas the other
data entries occur only once.
The mode of the flight prices is \$397.
81
Example: Finding the Mode
At a political debate a sample of audience members was
asked to name the political party to which they belong.
Their responses are shown in the table. What is the
mode of the responses?
Political Party
Democrat
Frequency, f
34
Republican
Other
56
21
Did not respond
9
82
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.
83
Comparing the Mean, Median, and Mode
• All three measures describe a typical entry of a data
set.
• Advantage of using the mean:
 The mean is a reliable measure because it takes
into account every entry of a data set.
• Disadvantage of using the mean:
 Greatly affected by outliers (a data entry that is far
removed from the other entries in the data set).
84
Example: Comparing the Mean, Median,
and Mode
Find the mean, median, and mode of the sample ages of
a class shown. Which measure of central tendency best
describes a typical entry of this data set? Are there any
outliers?
Ages in a class
20
20
20
20
20
20
21
21
21
21
22
22
22
23
23
23
23
24
24
65
85
Solution: Comparing the Mean, Median,
and Mode
Ages in a class
20
20
20
20
20
20
21
21
21
21
22
22
22
23
23
23
23
24
24
65
Mean:
x 20  20  ...  24  65
x

 23.8 years
n
20
Median:
21  22
 21.5 years
2
Mode:
20 years (the entry occurring with the
greatest frequency)
86
Solution: Comparing the Mean, Median,
and Mode
Mean ≈ 23.8 years
Median = 21.5 years
Mode = 20 years
• The mean takes every entry into account, but is
influenced by the outlier of 65.
• The median also takes every entry into account, and
it is not affected by the outlier.
• In this case the mode exists, but it doesn't appear to
represent a typical entry.
87
Solution: Comparing the Mean, Median,
and Mode
Sometimes a graphical comparison can help you decide
which measure of central tendency best represents a
data set.
In this case, it appears that the median best describes
the data set.
88
Weighted Mean
Weighted Mean
• The mean of a data set whose entries have varying
weights.
( x  w)
• x 
where w is the weight of each entry x
w
89
Example: Finding a Weighted Mean
You are taking a class in which your grade is
determined from five sources: 50% from your test
mean, 15% from your midterm, 20% from your final
exam, 10% from your computer lab work, and 5% from
your homework. Your scores are 86 (test mean), 96
(midterm), 82 (final exam), 98 (computer lab), and 100
(homework). What is the weighted mean of your
scores? If the minimum average for an A is 90, did you
get an A?
90
Solution: Finding a Weighted Mean
Source
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.
91
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
92
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
93
Example: Find the Mean of a Frequency
Distribution
Use the frequency distribution to approximate the mean
number of minutes that a sample of Internet subscribers
spent online during their most recent session.
Class
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
94
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
95
The Shape of Distributions
Symmetric Distribution
• A vertical line can be drawn through the middle of
a graph of the distribution and the resulting halves
are approximately mirror images.
96
The Shape of Distributions
Uniform Distribution (rectangular)
• All entries or classes in the distribution have equal
or approximately equal frequencies.
• Symmetric.
97
The Shape of Distributions
Skewed Left Distribution (negatively skewed)
• The “tail” of the graph elongates more to the left.
• The mean is to the left of the median.
98
The Shape of Distributions
Skewed Right Distribution (positively skewed)
• The “tail” of the graph elongates more to the right.
• The mean is to the right of the median.
99
Section 2.3 Summary
• Determined the mean, median, and mode of a
population and of a sample
• Determined the weighted mean of a data set and the
mean of a frequency distribution
• Described the shape of a distribution as symmetric,
uniform, or skewed and compared the mean and
median for each
100
Section 2.4
Measures of Variation
101
Section 2.4 Objectives
• Determine the range of a data set
• Determine the variance and standard deviation of a
population and of a sample
• Use the Empirical Rule and Chebychev’s Theorem to
interpret standard deviation
• Approximate the sample standard deviation for
grouped data
102
Range
Range
• The difference between the maximum and minimum
data entries in the set.
• The data must be quantitative.
• Range = (Max. data entry) – (Min. data entry)
103
Example: Finding the Range
A corporation hired 10 graduates. The starting salaries
for each graduate are shown. Find the range of the
starting salaries.
Starting salaries (1000s of dollars)
41 38 39 45 47 41 44 41 37 42
104
Solution: Finding the Range
• Ordering the data helps to find the least and greatest
salaries.
37 38 39 41 41 41 42 44 45 47
minimum
maximum
• Range = (Max. salary) – (Min. salary)
= 47 – 37 = 10
The range of starting salaries is 10 or \$10,000.
105
Deviation, Variance, and Standard
Deviation
Deviation
• The difference between the data entry, x, and the
mean of the data set.
• Population data set:
 Deviation of x = x – μ
• Sample data set:
 Deviation of x = x – x
106
Example: Finding the Deviation
A corporation hired 10 graduates. The starting salaries
for each graduate are shown. Find the deviation of the
starting salaries.
Starting salaries (1000s of dollars)
41 38 39 45 47 41 44 41 37 42
Solution:
• First determine the mean starting salary.
x 415


 41.5
N
10
107
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
108
Deviation, Variance, and Standard
Deviation
Population Variance
( x   )
•  
N
2
2
Sum of squares, SSx
Population Standard Deviation
2

(
x


)
2
•    
N
109
Finding the Population Variance &
Standard Deviation
In Words
1. Find the mean of the
population data set.
In Symbols
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
110
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
111
Example: Finding the Population
Standard Deviation
A corporation hired 10 graduates. The starting salaries
for each graduate are shown. Find the population
variance and standard deviation of the starting salaries.
Starting salaries (1000s of dollars)
41 38 39 45 47 41 44 41 37 42
Recall μ = 41.5.
112
Solution: Finding the Population
Standard Deviation
• Determine 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
113
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.
114
Deviation, Variance, and Standard
Deviation
Sample Variance
( x  x )
• s 
n 1
2
2
Sample Standard Deviation
•
2

(
x

x
)
s  s2 
n 1
115
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
116
Finding the Sample Variance & Standard
Deviation
In Words
5. Divide by n – 1 to get the
sample variance.
6. Find the square root to get
the sample standard
deviation.
In Symbols
2

(
x

x
)
s2 
n 1
( x  x ) 2
s
n 1
117
Example: Finding the Sample Standard
Deviation
The starting salaries are for the Chicago branches of a
corporation. The corporation has several other branches,
and you plan to use the starting salaries of the Chicago
branches to estimate the starting salaries for the larger
population. Find the sample standard deviation of the
starting salaries.
Starting salaries (1000s of dollars)
41 38 39 45 47 41 44 41 37 42
118
Solution: Finding the Sample Standard
Deviation
• Determine 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
119
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.
120
Example: Using Technology to Find the
Standard Deviation
Sample office rental rates (in
dollars per square foot per year)
for Miami’s central business
district are shown in the table.
Use a calculator or a computer
to find the mean rental rate and
the sample standard deviation.
(Adapted from: Cushman &
Wakefield Inc.)
Office Rental Rates
35.00
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
121
Solution: Using Technology to Find the
Standard Deviation
Sample Mean
Sample Standard
Deviation
122
Interpreting Standard Deviation
• Standard deviation is a measure of the typical amount
an entry deviates from the mean.
• The more the entries are spread out, the greater the
standard deviation.
123
Interpreting Standard Deviation:
Empirical Rule (68 – 95 – 99.7 Rule)
For data with a (symmetric) bell-shaped distribution, the
standard deviation has the following characteristics:
• About 68% of the data lie within one standard
deviation of the mean.
• About 95% of the data lie within two standard
deviations of the mean.
• About 99.7% of the data lie within three standard
deviations of the mean.
124
Interpreting Standard Deviation:
Empirical Rule (68 – 95 – 99.7 Rule)
99.7% within 3 standard deviations
95% within 2 standard deviations
68% within 1
standard deviation
34%
34%
2.35%
2.35%
13.5%
x  3s
x  2s
13.5%
x s
x
xs
x  2s
x  3s
125
Example: Using the Empirical Rule
In a survey conducted by the National Center for Health
Statistics, the sample mean height of women in the
United States (ages 20-29) was 64 inches, with a sample
standard deviation of 2.71 inches. Estimate the percent
of the women whose heights are between 64 inches and
69.42 inches.
126
Solution: Using the Empirical Rule
• Because the distribution is bell-shaped, you can use
the Empirical Rule.
34%
13.5%
55.87
x  3s
58.58
x  2s
61.29
x s
64
x
66.71
xs
69.42
x  2s
72.13
x  3s
34% + 13.5% = 47.5% of women are between 64 and
69.42 inches tall.
127
Chebychev’s Theorem
• The portion of any data set lying within k standard
deviations (k > 1) of the mean is at least:
1
1 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.
128
Example: Using Chebychev’s Theorem
The age distribution for Florida is shown in the
histogram. Apply Chebychev’s Theorem to the data
using k = 2. What can you conclude?
129
Solution: Using Chebychev’s Theorem
k = 2: μ – 2σ = 39.2 – 2(24.8) = -10.4 (use 0 since age
can’t be negative)
μ + 2σ = 39.2 + 2(24.8) = 88.8
At least 75% of the population of Florida is between 0
and 88.8 years old.
130
Standard Deviation for Grouped Data
Sample standard deviation for a frequency distribution
•
( 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.
131
Example: Finding the Standard Deviation
for Grouped Data
You collect a random sample of the
number of children per household in
a region. Find the sample mean and
the sample standard deviation of the
data set.
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
132
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
133
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
134
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.
135
Section 2.4 Summary
• Determined the range of a data set
• Determined the variance and standard deviation of a
population and of a sample
• Used the Empirical Rule and Chebychev’s Theorem
to interpret standard deviation
• Approximated the sample standard deviation for
grouped data
136
Section 2.5
Measures of Position
137
Section 2.5 Objectives
•
•
•
•
•
Determine the quartiles of a data set
Determine the interquartile range of a data set
Create a box-and-whisker plot
Interpret other fractiles such as percentiles
Determine and interpret the standard score (z-score)
138
Quartiles
• Fractiles are numbers that partition (divide) an
ordered data set into equal parts.
• Quartiles approximately divide an ordered data set
into four equal parts.
 First quartile, 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.
139
Example: Finding Quartiles
The test scores of 15 employees enrolled in a CPR
training course are listed. Find the first, second, and
third quartiles of the test scores.
13 9 18 15 14 21 7 10 11 20 5 18 37 16 17
Solution:
• Q2 divides the data set into two halves.
Lower half
Upper half
5 7 9 10 11 13 14 15 16 17 18 18 20 21 37
Q2
140
Solution: Finding Quartiles
• The first and third quartiles are the medians of the
lower and upper halves of the data set.
Lower half
Upper half
5 7 9 10 11 13 14 15 16 17 18 18 20 21 37
Q1
Q2
Q3
About one fourth of the employees scored 10 or less,
about one half scored 15 or less; and about three
fourths scored 18 or less.
141
The number of nuclear power plants in the top 15 nuclear
power-producing countries in the world are listed. Find the
first, second, and third quartiles of the data set. What can you
conclude?
National Institute of Standards and Technology
( N  1) Pth
Rank  n 
100
( N  1)Qth
Rank  n 
4
( N  1) Dth
Rank  n 
10
The 2-quantile is called the median
The 3-quantiles are called tertiles or terciles → T
The 4-quantiles are called quartiles → Q
The 5-quantiles are called quintiles → QU
The 6-quantiles are called sextiles → S
The 10-quantiles are called deciles → D
The 12-quantiles are called duo-deciles → Dd
The 20-quantiles are called vigintiles → V
The 100-quantiles are called percentiles → P
The 1000-quantiles are called permilles → Pr
.
Slide 2- 143
Interquartile Range
Interquartile Range (IQR)
• The difference between the third and first quartiles.
• IQR = Q3 – Q1
144
Example: Finding the Interquartile Range
Find the interquartile range of the test scores.
Recall Q1 = 10, Q2 = 15, and Q3 = 18
Solution:
• IQR = Q3 – Q1 = 18 – 10 = 8
The test scores in the middle portion of the data set
vary by at most 8 points.
145
Box-and-Whisker Plot
Box-and-whisker plot
• Exploratory data analysis tool.
• Highlights important features of a data set.
• Requires (five-number summary):
 Minimum entry
 First quartile Q1
 Median Q2
 Third quartile Q3
 Maximum entry
146
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
147
Example: Drawing a Box-and-Whisker
Plot
Draw a box-and-whisker plot that represents the 15 test
scores.
Recall Min = 5 Q1 = 10 Q2 = 15 Q3 = 18 Max = 37
Solution:
5
10
15
18
37
About half the scores are between 10 and 18. By looking
at the length of the right whisker, you can conclude 37 is
a possible outlier.
148
Percentiles and Other Fractiles
Fractiles
Quartiles
Deciles
Percentiles
Summary
Divides data into 4 equal
parts
Divides data into 10 equal
parts
Divides data into 100 equal
parts
Symbols
Q1, Q2, Q3
D1, D2, D3,…, D9
P1, P2, P3,…, P99
149
Example: Interpreting Percentiles
The ogive represents the
cumulative frequency
distribution for SAT test
scores of college-bound
students in a recent year. What
test score represents the 72nd
percentile? How should you
interpret this? (Source: College
Board Online)
150
Solution: Interpreting Percentiles
The 72nd percentile
corresponds to a test score
of 1700.
This means that 72% of the
students had an SAT score
of 1700 or less.
151
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

152
Example: Comparing z-Scores from
Different Data Sets
In 2007, Forest Whitaker won the Best Actor Oscar at
age 45 for his role in the movie The Last King of
Scotland. Helen Mirren won the Best Actress Oscar at
age 61 for her role in The Queen. The mean age of all
best actor winners is 43.7, with a standard deviation of
8.8. The mean age of all best actress winners is 36, with
a standard deviation of 11.5. Find the z-score that
corresponds to the age for each actor or actress. Then
153
Solution: Comparing z-Scores from
Different Data Sets
• Forest Whitaker
z
x

• Helen Mirren
z
x

45  43.7

 0.15
8.8
0.15 standard
deviations above
the mean
61  36

 2.17
11.5
2.17 standard
deviations above
the mean
154
Solution: Comparing z-Scores from
Different Data Sets
z = 0.15
z = 2.17
The z-score corresponding to the age of Helen Mirren
is more than two standard deviations from the mean,
so it is considered unusual. Compared to other Best
Actress winners, she is relatively older, whereas the
age of Forest Whitaker is only slightly higher than the
average age of other Best Actor winners.
155
Section 2.5 Summary
•
•
•
•
•
Determined the quartiles of a data set
Determined the interquartile range of a data set
Created a box-and-whisker plot
Interpreted other fractiles such as percentiles
Determined and interpreted the standard score
(z-score)
156
Over the same period, the
lowest-income fifth saw a
decrease in real income of 7.4
percent.
Between 1979 and 2009, the top 5 percent of American
families saw their real incomes increase 72.7 percent,
according to Census data.
Chapter 2: Descriptive Statistics
Elementary Statistics:
Picturing the World
Fifth Edition
by Larson and Farber
.
Slide 4- 160
Find the class width:
A. 3
Class
1– 5
Frequency, f
21
6 – 10
11 – 15
16 – 20
16
28
13
B. 4
C. 5
D. 19
.
Slide 2- 161
Find the class width:
A. 3
Class
1– 5
Frequency, f
21
6 – 10
11 – 15
16 – 20
16
28
13
B. 4
C. 5
D. 19
(6  1)or (11  6)or (10  5)
.
Slide 2- 162
Estimate the frequency of the class with
the greatest frequency.
60
Ages of Concert Attendees
A. 28
B. 21
C. 58
D. 53
Frequency
50
40
30
20
10
0
18
28
38
48
58
Age
.
Slide 2- 163
Estimate the frequency of the class with
the greatest frequency.
60
Ages of Concert Attendees
A. 28
B. 21
C. 58
D. 53
Frequency
50
40
30
20
10
0
18
28
38
48
58
Age
.
Slide 2- 164
What is the maximum data entry?
Key: 3 | 8 = 38
3 8 9
A. 96
4 0 2 7
B. 38
5 1 1 4 8
C. 9
6 3 3 3 8 9 9
D. 41
7 0 0 1 1 2 4 7 8 8 8 8 9
8 2 2 3 4 7 7 8 9 9
9 1 1 4 5 6
.
Slide 2- 165
What is the maximum data entry?
Key: 3 | 8 = 38
3 8 9
A. 96
4 0 2 7
B. 38
5 1 1 4 8
C. 9
6 3 3 3 8 9 9
D. 41
7 0 0 1 1 2 4 7 8 8 8 8 9
8 2 2 3 4 7 7 8 9 9
9 1 1 4 5 6
Slide 2- 166
The heights (in inches) of a sample of
basketball players are shown:
76
79
81
78
82
78
Find the mean.
A. 78.5
B. 79
C. 474
D. 78
.
Slide 2- 167
The heights (in inches) of a sample of
basketball players are shown:
76
79
81
78
82
78
Find the mean.
A. 78.5
B. 79
C. 474
D. 78
.
Slide 2- 168
The heights (in inches) of a sample of
basketball players are shown:
76
79
81
78
82
78
Find the median.
A. 78.5
B. 79
C. 79.5
D. 78
.
Slide 2- 169
The heights (in inches) of a sample of
basketball players are shown:
76
79
81
78
82
78
Find the median.
A. 78.5
B. 79
C. 79.5
D. 78
.
Slide 2- 170
The heights (in inches) of a sample of
basketball players are shown:
76
79
81
78
82
78
Find the mode.
A. 78.5
B. 79
C. 79.5
D. 78
.
Slide 2- 171
The heights (in inches) of a sample of
basketball players are shown:
76
79
81
78
82
78
Find the mode.
A. 78.5
B. 79
C. 79.5
D. 78
.
Slide 2- 172
The heights (in inches) of a sample of
basketball players are shown:
76
79
81
78
82
78
Find the standard deviation.
A. 2.2
B. 6
C. 2
D. 4.8
.
Slide 2- 173
The heights (in inches) of a sample of
basketball players are shown:
76
79
81
78
82
78
Find the standard deviation.
A. 2.2
B. 6
C. 2
D. 4.8
.
Slide 2- 174
The mean annual automobile insurance
premium is \$950, with a standard deviation
of \$175. The data set has a bell-shaped
distribution. Estimate the percent of
premiums that are between \$600 and \$1300.
A. 68%
B. 75%
C. 95%
D. 99.7%
.
Slide 2- 175
600  950  2(175)  x  950  2(175)  1300
The mean annual automobile insurance
premium is \$950, with a standard deviation
of \$175. The data set has a bell-shaped
distribution. Estimate the percent of
premiums that are between \$600 and \$1300.
A. 68%
B. 75%
C. 95%
D. 99.7%
.
Slide 2- 176
Use the box-and-whisker plot to identify the
first quartile.
10
|
18
|
|
|
|
24 26
|
|
|
|
30
|
|
10 12 14 16 18 20 22 24 26 28 30
A. 10
B. 18
C. 24
D. 26
.
Slide 2- 177
Use the box-and-whisker plot to identify the
first quartile.
10
|
18
|
|
|
|
24 26
|
|
|
|
30
|
|
10 12 14 16 18 20 22 24 26 28 30
A. 10
B. 18
C. 24
D. 26
.
Slide 2- 178
The mean annual automobile insurance
premium is \$950, with a standard deviation
of \$175. Find the z-score that corresponds to
a premium of \$1250.
A. 1.13
B. –1.13
C. 1.71
D. –1.71
.
Slide 2- 179
The mean annual automobile insurance
premium is \$950, with a standard deviation
of \$175. Find the z-score that corresponds to
a premium of \$1250.
A. 1.13
B. –1.13
C. 1.71
z
x

1250  950

 1.71
175
D. –1.71
.
Slide 2- 180
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