Classes

Chapter 2
Descriptive Statistics
1Larson/Farber 4th ed.

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

Section 2.1
Frequency Distributions
and Their Graphs

Section 2.1 Objectives
• Construct frequency distributions
• Construct frequency histograms, frequency
polygons, relative frequency histograms, and ogives
Larson/Farber 4th ed. 4

Frequency Distribution
Frequency Distribution
• A table that shows
classes or 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
Upper class
limits
Class width
6 – 1 = 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.

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.

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.

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

Solution: Constructing a Frequency
Distribution
1. Number of classes = 7 (given)
2. Find the class width
max min 86 7
11.29
#classes 7
Round up to 12
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

Distribution
Lower
limit
Upper
limit
7Class
width = 12
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.
19
31
43
55
67
79

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
Upper
limit
7
19
31
43
55
67
79
Class
width = 12
30
42
54
66
78
90
18

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.
Larson/Farber 4th ed.
13
Class Tally Frequency, f
7 – 18 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

Determining the Midpoint
Midpoint of a class
(Lower class limit) (Upper class limit)
2
Class Midpoint Frequency, f
7 – 18 6
19 – 30 10
31 – 42 13
7 18
12.5
2
19 30
24.5
2
31 42
36.5
2
Class width = 12

Determining the Relative Frequency
Relative Frequency of a class
• Portion or percentage of the data that falls in a
particular class.
n
f
sizeSample
frequencyclass
frequencyrelative
Class Frequency, f Relative Frequency
7 – 18 6
19 – 30 10
31 – 42 13
6
0.12
50
10
0.20
50
13
0.26
50
•

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
19 – 30 10
31 – 42 13
+
+
6
16
29

Expanded Frequency Distribution
Class Frequency, f Midpoint
Relative
frequency
Cumulative
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 50
Σf = 50 1
n
f

Graphs of Frequency Distributions
Frequency Histogram
• A bar graph that represents the frequency distribution.
• The horizontal scale is quantitative and measures the
data values.
• The vertical scale measures the frequencies of the
classes.
• Consecutive bars must touch.
data valuesfrequency

Class Boundaries
Class boundaries
• The numbers that separate classes without forming
gaps between them.
Class
Class
Boundaries
Frequency,
f
7 – 18 6
19 – 30 10
31 – 42 13
• 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.
• First class lower boundary = 7 – 0.5 = 6.5
• First class upper boundary = 18 + 0.5 = 18.5
6.5 – 18.5

Class Boundaries
Class
Class
boundaries
Frequency,
f
7 – 18 6.5 – 18.5 6
19 – 30 18.5 – 30.5 10
31 – 42 30.5 – 42.5 13
43 – 54 42.5 – 54.5 8
55 – 66 54.5 – 66.5 5
67 – 78 66.5 – 78.5 6
79 – 90 78.5 – 90.5 2

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

Solution: Frequency Histogram
(using Midpoints)

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.

Frequency Polygon
• A line graph that emphasizes the continuous change
in frequencies.
data values
frequency

Example: Frequency Polygon
Construct a frequency polygon for the Internet usage
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

Solution: Frequency Polygon
0
2
4
6
8
10
12
14
0.5 12.5 24.5 36.5 48.5 60.5 72.5 84.5 96.5
Frequency
Time online (in minutes)
Internet Usage
You can see that the frequency of subscribers increases up to
36.5 minutes and then decreases.
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.

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 valuesrelative
frequency

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

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.

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
cumulative
frequency

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.

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).

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

Solution: Ogive
0
10
20
30
40
50
60
Cumulativefrequency
Time online (in minutes)
InternetUsage
6.5 18.5 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.

Textbook Exercises. Pages 49 - 53
# 14
a. Class Width: Difference between two consecutive lower / upper class
limits. So the class width in this case is 10 – 0 = 10
b. Class Midpoints: To find the class midpoints we add the lower and upper
limits of the first class and divide the sum by 2. We then simply add the
class width to the midpoint of the first class to obtain the second one and
so on. So for this problem the midpoint of the first class is (0 + 9) 2 =
4.5, midpoint of the second class is 4.5 + 10 (class width) = 14.5 and
thus the rest of the midpoints are 24.5, 34.5, 44.5, 54.5 and 64.5
c. Class Boundaries: To obtain class boundaries we first find the distance
between the first upper class limit and the second lower class limit. 10-
9=1. Half this distance is 0.5.
So the first lower class boundary is 0 - 0.5 = -0.5 and
the first upper class boundary is 9 + 0.5 = 9.5
To obtain the remaining of lower and upper class boundaries simply add
class width to the previous lower class boundary and the previous upper
class boundary

Textbook Exercises Pages 49-53
# 24
a. The class with the greatest relative frequency is the third class having the
midpoint of 19.5. And the class with the least relative frequency is the last
class having the midpoint of 21.5
b. The greatest relative frequency is approximately 40% and the least relative
frequency is approximately 2%
c. The relative frequency of the second class is approximately 33%

Textbook Exercises. Pages 49-53
classes Class
Boundaries
Midpoints frequency Relative
frequency
Cumulative
frequency
30 – 113 29.5 – 113.5 71.5 5 5/29 = 0.172 5
114 – 197 113.5 – 197.5 155.5 7 7/29 = 0.241 12
198 – 281 197.5 – 281.5 239.5 8 8/29 = 0.276 20
282 – 365 281.5 – 365.5 323.5 2 2/29 = 0.069 22
366 – 449 365.5 – 449.5 407.5 3 3/29 = 0.103 25
450 – 533 449.5 – 533.5 491.5 4 4/29 = 0.138 29
Total = 29 Total = 1
# 28 Book Spending
We will create the extended frequency distribution table

Section 2.1 Summary
• Constructed frequency distributions
• Constructed frequency histograms, frequency
polygons, relative frequency histograms and ogives

Section 2.2
More Graphs and Displays

• 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

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
26
2 1 5 5 6 7 8
3 0 6 6
4 5

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

Solution: Constructing a Stem-and-Leaf
Plot
• 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.
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

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.

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

Example: Constructing a Dot Plot
Use a dot plot organize the text messaging data.
• 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.
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

Solution: Constructing a Dot Plot
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.
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

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.

Example: Constructing a Pie Chart
The numbers of motor vehicle occupants killed in
crashes in 2005 are shown in the table. Use a pie chart
to organize the data. (Source: U.S. Department of
Transportation, National Highway Traffic Safety
Administration)
Vehicle type Killed
Cars 18,440
Trucks 13,778
Motorcycles 4,553
Other 823

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 4,553
Other 823
37,594
18440
0.49
37594
13778
0.37
37594
4553
0.12
37594
823
0.02
37594

• 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

Vehicle type Frequency, f
Relative
frequency Central angle
Cars 18,440 0.49
Trucks 13,778 0.37
Motorcycles 4,553 0.12
Other 823 0.02
360º(0.49)≈176º
360º(0.37)≈133º
360º(0.12)≈43º
360º(0.02)≈7º

Vehicle type
Relative
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.

Graphing Qualitative Data Sets
Pareto Chart
• A vertical bar graph in which the height of each bar
represents frequency or relative frequency.
• The bars are positioned in order of decreasing height,
with the tallest bar positioned at the left.
Categories
Frequency

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)

Solution: Constructing a Pareto Chart
Cause $ (million)
Admin. error 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.

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.
x
y

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)

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.

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.

Graphing Paired Data Sets
Time Series
• Data set is composed of quantitative entries taken at
regular intervals over a period of time.
 e.g., The amount of precipitation measured each
day for one month.
• Use a time series chart to graph.
time
Quantitative
data

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)

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.

Solution: Constructing a Time Series
Chart
The graph shows that the number of subscribers has been
increasing since 1995, with greater increases recently.

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

Section 2.3
Measures of Central Tendency

• 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

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

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.
• Population mean:
• Sample mean:
x
N
x
x
n

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

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
3695
527.9
7
x
x
n
The mean price of the flights is about $527.90.

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.

Example: Finding the Median
Find the median of the flight prices.
872 432 397 427 388 782 397

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.

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

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.
The median price of the flights is $412.
397 427
Median 412
2

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).

Example: Finding the Mode
Find the mode of the flight prices.
872 432 397 427 388 782 397

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.

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 Frequency, f
Democrat 34
Republican 56
Other 21
Did not respond 9

Solution: Finding the Mode
Political Party Frequency, f
Democrat 34
Republican 56
Other 21
Did not respond 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.

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).

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

Solution: Comparing the Mean, Median,
and Mode
Mean: 20 20 ... 24 65
23.8 years
20
x
x
n
Median:
21 22
21.5 years
2
20 years (the entry occurring with the
greatest frequency)
Ages in a class
20 20 20 20 20 20 21
21 21 21 22 22 22 23
23 23 23 24 24 65
Mode:

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.

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.

Textbook Exercises. Pages 75-78
# 22 Power Failures
Calculate mean, median and mode of the given data, of possible.
Otherwise explain why not.
We will use the graphing calculator to find the above mentioned central tendencies.
Step 1: Press Stat Key on your graphing Calculator and you will see the following
image.
Step2: Press Enter to see six lists, namely L1, L2, L3, L4, L5, and L6, to enter the
data. In L1 enter the data values given in the problem.

# 22 cont.
Step 3: After entering the data values press the STAT key again and scroll over
to CALC menu.
Step 4: Press ENTER after selecting item 1 and then on following screen enter
L1 using second function key and # 1 key.
Step 5: After entering L1 press ENTER to see the results. Keep scrolling down
to see all the values.
Mean = 61.15
Median = 55

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

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?

Solution: Finding a Weighted Mean
Source Score, x Weight, w x∙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
( ) 88.6
88.6
1
x w
x
w
Your weighted mean for the course is 88.6. You did not
get an A.

Textbook Exercises. Pages 75 - 78
# 44 Account Balance
Using your graphing Calculator follow the steps shown below to calculate the
weighted mean of the data given in the problem.
Step 1: As shown earlier press STAT key, under EDIT menu select item 1 and press
ENTER to see the following screen. This time I am storing the data in L2 and L3.
Make sure to enter the balance amount first in L2 and then number of days in L3.
Reversing the order will change the results dramatically.
Step 2: After entering the data, press STAT
Key again, scroll to Calc menu and choose
Item 1. Enter L2 comma L3 and press the Enter key to
see the results. Mean = 982.19 dollars.

Mean of Grouped Data
Mean of a Frequency Distribution
• Approximated by
where x and f are the midpoints and frequencies of a
class, respectively
( )x f
x n f
n
Note: In section 2.4 the procedure to find the mean of the grouped
data using graphing calculator is discussed. Refer to slide 139

Finding the Mean of a Frequency
Distribution
In Words In Symbols
( )x f
x
n
(lower limit)+(upper limit)
2
x
( )x f
n f
1. Find the midpoint of each
class.
2. Find the sum of the
products of the midpoints
and the frequencies.
3. Find the sum of the
frequencies.
4. Find the mean of the

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.
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

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
( ) 2089
41.8 minutes
50
x f
x
n

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.

Uniform Distribution (rectangular)
• All entries or classes in the distribution have equal
or approximately equal frequencies.
• Symmetric.

Skewed Left Distribution (negatively skewed)
• The “tail” of the graph elongates more to the left.
• The mean is to the left of the median.

Skewed Right Distribution (positively skewed)
• The “tail” of the graph elongates more to the right.
• The mean is to the right of the median.

Section 2.3 Summary
• Determined the mean, median, and mode of a
• 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

Section 2.4
Measures of Variation

• Determine the range of a data set
• Determine the variance and standard deviation of a
• Use the Empirical Rule and Chebychev’s Theorem to
interpret standard deviation
• Approximate the sample standard deviation for
grouped data

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)

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

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
• Range = (Max. salary) – (Min. salary)
= 47 – 37 = 10
The range of starting salaries is 10 or $10,000.
minimum maximum

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

Example: Finding the Deviation
for each graduate are shown. Find the deviation of the
starting salaries.
41 38 39 45 47 41 44 41 37 42
Solution:
• First determine the mean starting salary.
415
41.5
10
x
N

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

Deviation
Population Variance
•
Population Standard Deviation
•
2
2 ( )x
N
Sum of squares, SSx
2
2 ( )x
N

Finding the Population Variance &
Standard Deviation
In Words In Symbols
population data set.
2. Find deviation of each
entry.
3. Square each deviation.
4. Add to get the sum of
squares.
x
N
x – μ
(x – μ)2
SSx = Σ(x – μ)2

Finding the Population Variance &
Standard Deviation
5. Divide by N to get the
population variance.
6. Find the square root to get
the population standard
deviation.
2
2 ( )x
N
2
( )x
N
In Words In Symbols

Example: Finding the Population
Standard Deviation
for each graduate are shown. Find the population
variance and standard deviation of the starting salaries.
41 38 39 45 47 41 44 41 37 42
Recall μ = 41.5.

Solution: Finding the Population
Standard Deviation
• Determine SSx
• N = 10
Salary, x 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

Solution: Finding the Population
Standard Deviation
Population Variance
•
Population Standard Deviation
•
2
2 ( ) 88.5
8.9
10
x
N
2
8.85 3.0
The population standard deviation is about 3.0, or $3000.

Deviation
Sample Variance
•
Sample Standard Deviation
•
2
2 ( )
1
x x
s
n
2
2 ( )
1
x x
s s
n

Finding the Sample Variance & Standard
Deviation
In Words In Symbols
sample data set.
2. Find deviation of each
entry.
3. Square each deviation.
4. Add to get the sum of
squares.
x
x
n
2
( )xSS x x
2
( )x x
x x

Finding the Sample Variance & Standard
Deviation
5. Divide by n – 1 to get the
sample variance.
6. Find the square root to get
the sample standard
deviation.
In Words In Symbols
2
2 ( )
1
x x
s
n
2
( )
1
x x
s
n

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.
41 38 39 45 47 41 44 41 37 42

Solution: Finding the Sample Standard
Deviation
• Determine SSx
• n = 10
Salary, x 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

Solution: Finding the Sample Standard
Deviation
Sample Variance
•
Sample Standard Deviation
•
2
2 ( ) 88.5
9.8
1 10 1
x x
s
n
2 88.5
3.1
9
s s
The sample standard deviation is about 3.1, or $3100.

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

Solution: Using Technology to Find the
Standard Deviation
Sample Mean
Sample Standard
Deviation

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.

Textbook Exercises. Page 94.
Exercise # 22 Annual Salaries
Public Teachers:
 Range: 5.1 thousand dollars
Variance: 2.96 square thousand dollars (Why ?)
 Standard Deviation: 1.72 thousand dollars
Private Teachers:
Range: 4.2 thousand dollars
Variance: 1.99 square thousand dollars (Why ?)
 Standard Deviation: 1.41 thousand dollars
From the values of sample standard deviation it is clear that the annual
salaries of Public teachers varies more than that of Private teachers.

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.

Interpreting Standard Deviation:
Empirical Rule (68 – 95 – 99.7 Rule)
3x s x s 2x s 3x sx s x2x s
68% within 1
standard deviation
34% 34%
99.7% within 3 standard deviations
2.35% 2.35%
95% within 2 standard deviations
13.5% 13.5%

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.

Solution: Using the Empirical Rule
3x s x s 2x s 3x sx s x2x s
55.87 58.58 61.29 64 66.71 69.42 72.13
34%
13.5%
• Because the distribution is bell-shaped, you can use
the Empirical Rule.
34% + 13.5% = 47.5% of women are between 64 and
69.42 inches tall.

Textbook Exercises. Page 96
Problem # 30
Since the data is bell shaped, according to the Empirical Rule 95% of the data
values falls within two standard deviation from the mean.
So, 95% of the data values lies between 2400 – 2(450) and 2400 + 2(450)
That is, between $1500 and $3300.
Problem # 32
a. Since 95% of the data values lies between $1500 and $3300, the
number of farms whose land and building values per acre are between
$1500 and $3300 are 0.95 * 40 = 38 farms.
b. On adding 20 more farms we can expect 95% of them to be between
$1500 and $3300 per acre. That is 0.95 * 20 = 19 farms.

Chebychev’s Theorem
• The portion of any data set lying within k standard
deviations (k > 1) of the mean is at least:
2
1
1
k
• k = 2: In any data set, at least 2
1 3
1 or 75%
2 4
of the data lie within 2 standard deviations of the
mean.
• k = 3: In any data set, at least 2
1 8
1 or 88.9%
3 9
of the data lie within 3 standard deviations of the
mean.

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?

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.

Standard Deviation for Grouped Data
Sample standard deviation for a frequency distribution
•
• When a frequency distribution has classes, estimate the
sample mean and standard deviation by using the
midpoint of each class.
2
( )
1
x x f
s
n
where n= Σf (the number of
entries in the data set)

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

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
Solution: Finding the Standard Deviation
for Grouped Data
• First construct a frequency distribution.
• Find the mean of the frequency
distribution.
Σf = 50 Σ(xf )= 91
91
1.8
50
xf
x
n
The sample mean is about 1.8
children.

for Grouped Data
• Determine the sum of squares.
x f
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
( )x x 2
( )x x f
2
( ) 145.40x x f

for Grouped Data
• Find the sample standard deviation.
x x 2
( )x x 2
( )x x f2
( ) 145.40
1.7
1 50 1
x x f
s
n
The standard deviation is about 1.7 children.

Finding Standard Deviation of the Grouped Data
Example: The heights (in inches) of 23 male students in a physical education
class.
 First, determine the midpoints of each classes.
Enter the midpoints in one of the lists, say L1,
of the graphing calculator.
 Enter the frequency into the next list of the
graphing calculator.
 Using 1-Var stats from the CALC menu
obtain the results to determine the value of the mean and the standard deviation
of the given data.
Go to the next slide to view the results.
Height
(in inches)
Frequency
63 – 65 3
66 – 68 6
69 – 71 7
72 – 74 4
75 – 77 3

Finding Standard deviation of the Grouped Data
Mean = 69.74 inches
Sample Standard Deviation = 3.72 inches
The midpoints are entered in list L4 and the
frequency in L5. You can use any lists but
make sure to enter them in this order,
midpoints followed by frequency

Section 2.4 Summary
• Determined the range of a data set
• Determined the variance and standard deviation of a
• Used the Empirical Rule and Chebychev’s Theorem
to interpret standard deviation
• Approximated the sample standard deviation for
grouped data

Section 2.5
Measures of Position

• 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)

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.

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.
5 7 9 10 11 13 14 15 16 17 18 18 20 21 37
Q2
Lower half Upper half

Solution: Finding Quartiles
• The first and third quartiles are the medians of the
lower and upper halves of the data set.
5 7 9 10 11 13 14 15 16 17 18 18 20 21 37
Q2
Lower half Upper half
Q1 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.

Interquartile Range
Interquartile Range (IQR)
• The difference between the third and first quartiles.
• IQR = Q3 – Q1

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.

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

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.
Whisker Whisker
Maximum
entry
Minimum
entry
Box
Median, Q2 Q3Q1

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
5 10 15 18 37
Solution:
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.

Percentiles and Other Fractiles
Fractiles Summary Symbols
Quartiles Divides data into 4 equal
parts
Q1, Q2, Q3
Deciles Divides data into 10 equal
parts
D1, D2, D3,…, D9
Percentiles Divides data into 100 equal
parts
P1, P2, P3,…, P99
Note: Along with the mean the graphing calculator also gives you values of Q1, Q2
(median) and Q3.

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)

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.

The Standard Score
Standard Score (z-score)
• Represents the number of standard deviations a given
value x falls from the mean μ.
•
value - mean
standarddeviation
x
z

Example: Comparing z-Scores from
Different Data Sets
In 2007, Forest Whitaker won the Best Actor Oscar at
age 45 for his role in the movie The Last King of
Scotland. Helen Mirren won the Best Actress Oscar at
age 61 for her role in The Queen. The mean age of all
best actor winners is 43.7, with a standard deviation of
8.8. The mean age of all best actress winners is 36, with
a standard deviation of 11.5. Find the z-score that
corresponds to the age for each actor or actress. Then
compare your results.

Solution: Comparing z-Scores from
Different Data Sets
• Forest Whitaker
45 43.7
0.15
8.8
x
z
• Helen Mirren
61 36
2.17
11.5
x
z
0.15 standard
deviations above
the mean
2.17 standard
deviations above
the mean

Solution: Comparing z-Scores from
Different Data Sets
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.
z = 0.15 z = 2.17

Textbook Exercises. Page 112
Problem 36: Life Span of Fruit Flies
Mean life span = 33 days and Standard deviation = 4 days
a. z- score of life span of 34 days is
similarly the z-scores of life span of 30 days and 42 days are -0.75 and 2.25
respectively. Since the z-score for the life span of 42 days is 2.25 (beyond 2
standard deviations )it is an unusual life span.
b. z-scores for the life span of 29 days, 41 days and 25 days are -1, 2 and -2
respectively.
Since 29 days with z-score of -1 is 1 standard deviation below the mean, its
percentile using the Empirical rule is 16th percentile. (sum of values to the left of
1 standard deviation below the mean). Refer to the figure of Empirical Rule.
Since 41 days is 2 standard deviation above the mean, its percentile is 97.5th
percentile. (sum of values to the left of 2 standard deviations above the mean).
Since 25 days is 2 standard deviations below the mean, its percentile is 2.5th
percentile. (sum of values to the left of 2 standard deviations below the mean.
25.0
4
3334x
z

Section 2.5 Summary
• Determined the quartiles of a data set
• Determined the interquartile range of a data set
• Created a box-and-whisker plot
• Interpreted other fractiles such as percentiles
• Determined and interpreted the standard score
(z-score)

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