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.

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.

1

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.

2

155 159 144 129 105 145 126 116 130 114 122 112 112 142 126156 118 108 122 121 109 140 126 119 113 117 118 109 109 119139 139 122 78 133 126 123 145 121 134 124 119 132 133 124129 112 126 148 147

Solution: Constructing a Stem-and-Leaf Plot

3

• 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 126156 118 108 122 121 109 140 126 119 113 117 118 109 109 119139 139 122 78 133 126 123 145 121 134 124 119 132 133 124129 112 126 148 147

Solution: Constructing a Stem-and-Leaf Plot

4

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

5

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.

6

• 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 126156 118 108 122 121 109 140 126 119 113 117 118 109 109 119139 139 122 78 133 126 123 145 121 134 124 119 132 133 124129 112 126 148 147

Solution: Constructing a Dot Plot

7

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 126156 118 108 122 121 109 140 126 119 113 117 118 109 109 119139 139 122 78 133 126 123 145 121 134 124 119 132 133 124129 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.

8

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)

9

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.

10

Vehicle type Frequency, f

Relative frequency

Cars 18,440

Trucks 13,778

Motorcycles 4,553

Other 823

37,594

184400.49

37594

137780.37

37594

45530.12

37594

8230.02

37594

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

11

Solution: Constructing a Pie Chart

12

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

Solution: Constructing a Pie Chart

13

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

Bar graph

• 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.

14

Categories

Fre

quen

cy

Example: Constructing a Bar Graph

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)

15

Solution: Constructing a Bar Graph

16

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.

Frequency Distributions

and Their Graphs

17

Section Objectives

• Construct frequency distributions

• Construct frequency histograms, frequency polygons, relative frequency histograms, and ogives

18

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.

19

Class Frequency, f

1 – 5 5

6 – 10 8

11 – 15 6

16 – 20 8

21 – 25 5

26 – 30 4

Lower classlimits

Upper classlimits

Class width 6 – 1 = 5

Constructing a Frequency Distribution

20

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.

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.

21

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.

22

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

23

Solution: Constructing a Frequency Distribution

1. Number of classes = 7 (given)

2. Find the class width

24

max min 86 711.29

#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

Solution: Constructing a Frequency Distribution

25

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

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.

26

Lower limit

Upper limit

7

19

31

43

55

67

79

Class width = 1230

42

54

66

78

90

18

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.

27

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 2Σf = 50

Determining the Midpoint

Midpoint of a class

28

(Lower class limit) (Upper class limit)

2

Class Midpoint Frequency, f

7 – 18 6

19 – 30 10

31 – 42 13

7 1812.5

2

19 3024.5

2

31 4236.5

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.

29

n

f

sizeSample

frequencyclassfrequencyrelative

Class Frequency, f Relative Frequency

7 – 18 6

19 – 30 10

31 – 42 13

60.12

50

100.20

50

130.26

50

•

Determining the Cumulative Frequency

Cumulative frequency of a class

• The sum of the frequency for that class and all previous classes.

30

Class Frequency, f Cumulative frequency

7 – 18 6

19 – 30 10

31 – 42 13

+

+

6

16

29

Expanded Frequency Distribution

31

Class Frequency, f MidpointRelative

frequencyCumulative 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 1n

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.

32

data valuesfr

eque

ncy

Class Boundaries

Class boundaries

• The numbers that separate classes without forming gaps between them.

33

ClassClass

BoundariesFrequency,

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

34

ClassClass

boundariesFrequency,

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.

35

ClassClass

boundaries MidpointFrequency,

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)

36

Solution: Frequency Histogram (using class boundaries)

37

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.

Graphs of Frequency Distributions

Frequency Polygon

• A line graph that emphasizes the continuous change in frequencies.

38

data values

freq

uenc

y

Example: Frequency Polygon

Construct a frequency polygon for the Internet usage frequency distribution.

39

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

Solution: Frequency Polygon

02468101214

0.5 12.5 24.5 36.5 48.5 60.5 72.5 84.5 96.5

Freq

uenc

y

Time online (in minutes)

Internet Usage

40

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.

Graphs of Frequency Distributions

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.

41

data valuesre

lativ

e fr

eque

ncy

Example: Relative Frequency Histogram

Construct a relative frequency histogram for the Internet usage frequency distribution.

42

ClassClass

boundariesFrequency,

fRelative

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

43

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.

Graphs of Frequency Distributions

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.

44

data valuescu

mul

ativ

e fr

eque

ncy

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.

45

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).

46

Example: Ogive

Construct an ogive for the Internet usage frequency distribution.

47

ClassClass

boundariesFrequency,

fCumulative 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

Cu

mu

lati

ve fr

equ

ency

Time online (in minutes)

Internet Usage

48

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.

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.

Larson/Farber 4th ed. 49

x

y

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)

50

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.

51

Solution: Constructing a Time Series Chart

52

The graph shows that the number of subscribers has been increasing since 1995, with greater increases recently.

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

53

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

54

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:

55

x

N

xx

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

56

Solution: Finding a Sample Mean

872 432 397 427 388 782 397

57

• 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

3695527.9

7

xx

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.

58

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

59

Solution: Finding the Median

872 432 397 427 388 782 397

60

• 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

61

Solution: Finding the Median

872 397 427 388 782 397

62

• 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 427Median 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).

63

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

64

Solution: Finding the Mode

872 432 397 427 388 782 397

65

• 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?

66

Political Party Frequency, f

Democrat 34

Republican 56

Other 21

Did not respond 9

Solution: Finding the Mode

Larson/Farber 4th ed. 67

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).

68

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?

69

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

70

Mean: 20 20 ... 24 6523.8 years

20

xx

n

Median:21 22

21.5 years2

20 years (the entry occurring with thegreatest 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:

Solution: Comparing the Mean, Median, and Mode

71

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.

Solution: Comparing the Mean, Median, and Mode

72

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 because it is not as affected by outliers.

Weighted Mean

Weighted Mean

• The mean of a data set whose entries have varying weights.

• where w is the weight of each entry x

73

( )x wx

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?

74

Solution: Finding a Weighted Mean

75

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

( ) 88.688.6

1

x wx

w

Your weighted mean for the course is 88.6. You did not get an A.

Mean of Grouped Data

Mean of a Frequency Distribution

• Approximated by

where x and f are the midpoints and frequencies of a class, respectively

76

( )x fx n f

n

Finding the Mean of a Frequency Distribution

In Words In Symbols

77

( )x fx

n

(lower limit)+(upper limit)

2x

( )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 frequency distribution.

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.

78

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

Solution: Find the Mean of a Frequency Distribution

79

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

( ) 208941.8 minutes

50

x fx

n

The Shape of Distributions

80

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.

The Shape of Distributions

81

Uniform Distribution (rectangular)

• All entries or classes in the distribution have equal or approximately equal frequencies.

• Symmetric.

The Shape of Distributions

82

Skewed Left Distribution (negatively skewed)

• The “tail” of the graph elongates more to the left.

• The mean is to the left of the median mean always follows the tail.

The Shape of Distributions

83

Skewed Right Distribution (positively skewed)

• The “tail” of the graph elongates more to the right.

• The mean is to the right of the median.

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)

84

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

85

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.

86

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

87

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

88

Solution:• First determine the mean starting salary.

41541.5

10

x

N

Solution: Finding the Deviation

89

• 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, Variance, and Standard Deviation

Population Variance

•

Population Standard Deviation

•

90

22 ( )x

N

Sum of squares, SSx

22 ( )x

N

Finding the Population Variance & Standard Deviation

In Words In Symbols

91

1. Find the mean of the 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

92

5. Divide by N to get the population variance.

6. Find the square root to get the population standard deviation.

22 ( )x

N

2( )x

N

In Words In Symbols

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.

93

Solution: Finding the Population Standard Deviation

94

• 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

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Population Variance

•

Population Standard Deviation

•

22 ( ) 88.5

8.910

x

N

2 8.85 3.0

The population standard deviation is about 3.0, or $3000.

Deviation, Variance, and Standard Deviation

Sample Variance

•

Sample Standard Deviation

•

96

22 ( )

1

x xs

n

22 ( )

1

x xs s

n

Finding the Sample Variance & Standard Deviation

In Words In Symbols

97

1. Find the mean of the sample data set.

2. Find deviation of each entry.

3. Square each deviation.

4. Add to get the sum of squares.

xx

n

2( )xSS x x

2( )x x

x x

Finding the Sample Variance & Standard Deviation

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5. Divide by n – 1 to get the sample variance.

6. Find the square root to get the sample standard deviation.

In Words In Symbols2

2 ( )

1

x xs

n

2( )

1

x xs

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.

Starting salaries (1000s of dollars)

41 38 39 45 47 41 44 41 37 42

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Solution: Finding the Sample Standard Deviation

100

• 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

101

Sample Variance

•

Sample Standard Deviation

•

22 ( ) 88.5

9.81 10 1

x xs

n

2 88.53.1

9s 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.)

102

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

103

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.

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Quartiles

• Fractiles are numbers that partition (divide) an ordered data set into equal parts.

• Quartiles approximately divide an ordered data set into four equal parts. First quartile, Q1: About one quarter of the data fall on

or below Q1.

Second quartile, Q2: About one half of the data fall on or below Q2 (median).

Third quartile, Q3: About three quarters of the data fall on or below Q3.

105

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

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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

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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

• Is resistant to outliers…WHY???

108

Example: Finding the Interquartile Range

Find the interquartile range of the test scores.

Recall Q1 = 10, Q2 = 15, and Q3 = 18

109

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

110

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.

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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

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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

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Example: Interpreting Percentiles

The ogive represents the cumulative frequency distribution for SAT test scores of college-bound students in a recent year. What test score represents the 72nd percentile? How should you interpret this? (Source: College Board Online)

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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.

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