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```MM207-Statistics
Unit 2 Seminar-Descriptive Statistics
Dr Bridgette Stevens
AIM:BStevensKaplan
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
Larson/Farber 4th ed.
2
Section 2.1
Frequency Distributions
and Their Graphs
Larson/Farber 4th ed.
3
Frequency Distribution
Frequency Distribution
• A table that shows classes or Class width
6–1=5
intervals of data with a
count of the number of
entries in each class.
• The frequency, f, of a class
is the number of data entries
in the class.
Class
Frequency, f
1–5
5
6 – 10
8
11 – 15
6
16 – 20
8
21 – 25
5
26 – 30
4
Lower class
limits
Larson/Farber 4th ed.
Upper class
limits
4
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.
Larson/Farber 4th ed.
5
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.
Larson/Farber 4th ed.
6
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
Larson/Farber 4th ed.
7
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
Larson/Farber 4th ed.
8
Solution: Constructing a Frequency
Distribution
3. Use 7 (minimum value)
the class width of 12 to
get the lower limit of the
next class.
7 + 12 = 19
Find the remaining
lower limits.
Larson/Farber 4th ed.
Lower Upper
limit
limit
Class
width = 12
7
19
31
43
55
67
79
9
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.
Larson/Farber 4th ed.
Lower
limit
Upper
limit
7
19
31
43
55
18
30
42
54
66
67
79
78
90
Class
width = 12
10
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
Larson/Farber 4th ed.
11
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
Larson/Farber 4th ed.
Frequency, f
6
Class width = 12
10
13
12
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
Larson/Farber 4th ed.
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
13
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
Larson/Farber 4th ed.
14
Example: Frequency Histogram
Construct a frequency histogram for the Internet usage frequency
distribution.
Class
Class
boundaries
Midpoint
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
Larson/Farber 4th ed.
Frequency, f
15
Solution: Frequency Histogram
(using Midpoints)
Larson/Farber 4th ed.
16
Graphs of Frequency Distributions
frequency
Frequency Polygon
• A line graph that emphasizes the continuous change in
frequencies.
data
values
Larson/Farber 4th ed.
17
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.
Larson/Farber 4th ed.
18
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
Larson/Farber 4th ed.
19
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
Larson/Farber 4th ed.
20
Solution: Relative Frequency
Histogram
6.5
18.5
30.5
42.5
54.5
66.5
78.5
From this graph you can see that 20% of Internet
subscribers spent between 18.5 minutes and 30.5
minutes online.
21
90.5
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
Larson/Farber 4th ed.
22
Solution: Ogive
Internet Usage
Cumulative frequency
60
50
40
30
20
10
0
6.5
18.5
Time online (in minutes)
30.5 42.5
54.5
66.5
78.5
90.5
From the ogive, you can see that about 40 subscribers spent
60 minutes or less online during their last session. The
greatest increase in usage occurs between 30.5 minutes and
42.5 minutes.
23
Section 2.2
More Graphs and Displays
Larson/Farber 4th ed.
24
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
Larson/Farber 4th ed.
25
Graphing Quantitative Data Sets
Stem-and-leaf plot
• Each number is separated into a stem and a leaf.
• Similar to a histogram.
• Still contains original data values.
Data: 21, 25, 25, 26, 27, 28,
30, 36, 36, 45
Larson/Farber 4th ed.
26
2
1 5 5 6 7 8
3
0 6 6
4
5
26
Example: Constructing a Stem-andLeaf 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
118
139
129
159
118
139
112
144 129 105 145 126 116 130 114 122 112 112 142 126
108 122 121 109 140 126 119 113 117 118 109 109 119
122 78 133 126 123 145 121 134 124 119 132 133 124
126 148 147
Larson/Farber 4th ed.
27
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.
28
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
29
Solution: Constructing a Dot Plot
155
118
139
129
159
118
139
112
144 129 105 145 126 116 130 114 122 112 112 142 126
108 122 121 109 140 126 119 113 117 118 109 109 119
122 78 133 126 123 145 121 134 124 119 132 133 124
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.
30
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.
Larson/Farber 4th ed.
31
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
Larson/Farber 4th ed.
Vehicle type
Killed
Cars
18,440
Trucks
13,778
Motorcycles
4,553
Other
823
32
Solution: Constructing a Pie Chart
• Find the relative frequency (percent) of each category.
Vehicle type
Frequency, f Relative frequency
Cars
18,440
Trucks
13,778
Motorcycles
Other
4,553
823
18440
 0.49
37594
13778
 0.37
37594
4553
 0.12
37594
823
 0.02
37594
37,594
33
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.
34
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
Larson/Farber 4th ed.
35
Constructing a Pareto Chart
Cause
error
\$
(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.
36
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
points in a coordinate plane.
• Used to show the relationship
between two quantitative variables.
y
x
37
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.
38
Graphing Paired Data Sets
Time Series
• Data set is composed of quantitative entries taken at regular
intervals over a period of time.
• Use a time series chart to graph.
Quantitative
data
• e.g., The amount of precipitation measured each day for one
month.
time
39
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)
Larson/Farber 4th ed.
40
Solution: Constructing a Time Series Chart
The graph shows that the number of subscribers has
been increasing since 1995, with greater increases
recently.
41
Section 2.3
Measures of Central Tendency
Larson/Farber 4th ed.
42
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
Larson/Farber 4th ed.
43
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).
Larson/Farber 4th ed.
44
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).
Larson/Farber 4th ed.
45
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
Larson/Farber 4th ed.
20
20
20
20
20
20
21
21
21
21
22
22
22
23
23
23
23
24
24
65
46
Solution: Comparing the Mean,
Median, and Mode
Ages in a class
Mean:
Median:
Mode:
20
20
20
20
20
20
21
21
21
21
22
22
22
23
23
23
23
24
24
65
x 20  20  ...  24  65
x

 23.8 years
n
20
21  22
 21.5 years
2
20 years (the entry occurring with the
greatest frequency)
Larson/Farber 4th ed.
47
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.
Larson/Farber 4th ed.
48
Solution: Comparing the Mean,
Median, and Mode
measure of central tendency best represents a data set.
In this case, it appears that the median best describes the data set.
Larson/Farber 4th ed.
49
Weighted Mean
Weighted Mean
• The mean of a data set whose entries have varying
weights.
•
where w is the weight of each entry x
( x  w)
x
w
Larson/Farber 4th ed.
50
Example: Finding a Weighted Mean
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?
Larson/Farber 4th ed.
51
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.
Larson/Farber 4th ed.
52
Mean of Grouped Data
Mean of a Frequency Distribution
• Approximated by
( x  f )
x xand f are the midpoints
n  fand frequencies of a
where
n
class, respectively
Larson/Farber 4th ed.
53
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.
Larson/Farber 4th ed.
( x  f )
x
n
54
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.
Larson/Farber 4th ed.
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
55
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
Larson/Farber 4th ed.
56
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.
Larson/Farber 4th ed.
57
The Shape of Distributions
Uniform Distribution (rectangular)
• All entries or classes in the distribution have equal or approximately equal
frequencies.
• Symmetric.
Larson/Farber 4th ed.
58
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.
Larson/Farber 4th ed.
59
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.
Larson/Farber 4th ed.
60
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
Larson/Farber 4th ed.
61
Section 2.4
Measures of Variation
Larson/Farber 4th ed.
62
Deviation, Variance, and Standard Deviation
Population Variance
•
( x   )
 
N
2
2
Sum of squares, SSx
Population Standard Deviation
•
2

(
x


)
  2 
N
Larson/Farber 4th ed.
63
Example: Using Technology to Find
the Standard Deviation
Sample office rental rates (in dollars
per square foot per year) for
shown in the table. Use a calculator
or a computer to find the mean rental
rate and the sample standard
Wakefield Inc.)
Larson/Farber 4th ed.
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
64
Solution: Using Technology to Find
the Standard Deviation
Sample Mean
Sample
Standard
Deviation
Larson/Farber 4th ed.
65
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.
66
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.
67
Interpreting Standard Deviation: Empirical
Rule (68 – 95 – 99.7 Rule)
99.7% within 3 SD
95% within 2 SD
68% within 1 SD
34%
2.35%
x  3s
x  2s
Larson/Farber 4th ed.
34%
13.5%
x s
2.35%
13.5%
x
xs
x  2s
68
x  3s
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.
Larson/Farber 4th ed.
69
Solution: Using the Empirical Rule
• Because the distribution is bell-shaped, you can use the Empirical
Rule.
34%
13.5%
55.8
x  3s
7
58.5
x  2s
8
61.2
9x  s
64
x
66.7
xs
1
69.4
x  2s
2
34% + 13.5% = 47.5% of women are between 64 and 69.42
inches tall.
Larson/Farber 4th ed.
70
72.1
3x  3s
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
• k = 2: In any data set, at least
1 3
1  2  or 75%
2
4
of the data lie within 2 standard deviations of the mean.
• k = 3: In any data set, at least
1 8
1  2  or 88.9%
3
9
of the data lie within 3 standard deviations of the mean.
Larson/Farber 4th ed.
71
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
Larson/Farber 4th ed.
72
Drawing a Box-and-Whisker Plot
1.
2.
3.
4.
Find the five-number summary of the data set.
Construct a horizontal scale that spans the range of the data.
Plot the five numbers above the horizontal scale.
Draw a box above the horizontal scale from 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.
Whiske
r
Minimu
m entry
Q1
Larson/Farber 4th ed.
Box
Median,
Q2
Whiske
r
Maximu
Q m entry
3
73
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