1. Descriptive Statistics
Chapter 2
Descriptive Statistics

2. 2.1 Frequency Distributions and Their Graphs
In the previous chapter we learned that we can categorize the data in a set by type and level of measurement. Now we will learn different ways to organize and describe a data set.
Important characteristics to look for when organizing and describing a data set are its center, variability (or spread), and its shape. Center and shape will be covered in section 2.3
In this section you will learn how to organize data sets by grouping data into intervals called classes and forming a frequency distribution as well as constructing graphs.

3. Frequency Distributions
A frequency distribution is a table that shows classes or intervals of data entries 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.
A class is a grouping of data.
Each class has a lower class limit, which is the least number that can belong to the class, and an upper class limit, which is the greatest number that can belong to the class.
The class width is the distance between lower (or upper) limits of consecutive classes.
The difference between the maximum and minimum data entries is the range.

4. Constructing a Frequency Distribution
Decide on the number of classes and it should be between 5 and 20 in order to detect patterns in the data set.
Find the class width by determining the range of the data and dividing the range by the number of classes and rounding up to the next whole number.
Find the class limits. The minimum data entry is the lower limit of the first class. To find the remaining lower limits, add the class width to the lower limit of the preceding class. Similarly for the upper limit.
Tally the number of data entries in each class, then count the number of tallies to find the frequency f of each class.
The sum of all the frequencies f of each class should total to the number of entries.

5. Example 1

6. Example 2 a. State the number of classes.
b. Find the minimum and maximum values and the class width.
c. Find the class limits.
d. Tally the data entries.
e. Write the frequency f for each class.

7. Additional features of Frequency Distributions
The midpoint (class mark) of a class is the sum of the lower and upper limits of the class divided by two.
The relative frequency of a class is the portion or percentage of the data that falls in that class.
The cumulative frequency of a class is the sum of the frequency for that class and all previous classes.

8. Example 3

9. Example 4
a. Use the formulas to find each midpoint, relative frequency, and cumulative frequency.
b. Organize your results in a frequency distribution.
c. Identify patterns that emerge from the data.

10. Graphs of Frequency Distributions
Sometimes it is easier to identify patterns of a data set by looking at the graph of the frequency distribution. The first graph is a frequency histogram.
A frequency histogram is a bar graph that represents the frequency distribution of a data set and satisfies the following properties
The horizontal scale is quantitative and measure the data values.
The vertical scale measures the frequencies of the classes.
Consecutive bars must touch.

11. Frequency Histograms
Since consecutive bars of a histogram must touch, bars must begin and end at class boundaries instead of class limits.
Class boundaries are the numbers that separate classes without forming gaps between them.
data values
frequency

12. How to find the class boundaries
The class boundaries can be found by subtracting the lower limit of the second class from the upper limit of the first class.
Then you take that number and divide by 2.
Then you subtract and add that number from the lower and upper limits respectively.

13. Constructing a Frequency Histogram
Find the class boundaries.
The horizontal scale will be either the class midpoints or the class boundaries.
The vertical scale will be the frequency values for the classes.
Construct the histogram.

14. Example 5
Using the frequency distribution from example 1 construct a frequency histogram labeled with class midpoints and class boundaries.

15. Example 6
Using the frequency distribution from example 2 construct a frequency histogram labeled with class midpoints and class boundaries.

16. Frequency Polygon
A frequency polygon is a line graph that emphasizes the continuous change in frequencies.
To construct a frequency polygon use the same horizontal and vertical scales that were used in the histogram labeled with class midpoints.
Then plot points that represent the midpoint and frequency of each class and connect the points in order from left to right.
The graph should begin and end on the horizontal
axis, so extend the first and last class midpoints by
the class width.
data values
frequency

17. Example 7
Using the frequency distribution from example 1 construct a frequency polygon.

18. Example 8
Using the frequency distribution from example 2 construct a frequency polygon.

19. Relative Frequency Histogram
A relative frequency histogram has the same shape and the same horizontal scale as the corresponding frequency histogram but instead of having the frequencies as the vertical scale it has the relative frequencies.
data values
relative frequency

20. Example 9
Using the frequency distribution from example 1 construct a relative frequency histogram.

21. Example 10
Using the frequency distribution from example 2 construct a relative frequency histogram.

22. Cumulative Frequency Graph or Ogive
A cumulative frequency graph (or ogive) is 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

23. Constructing an Ogive
Construct a frequency distribution that includes cumulative frequencies as one of the columns.
Specify the horizontal and vertical scales.
The horizontal scale consists of the upper class boundaries.
The vertical scale measures cumulative frequencies.
Plot points that represent the upper class boundaries and their corresponding cumulative frequencies.
Connect the points in order from left to right.
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).

24. Example 11
Using the frequency distribution from example 1 construct an ogive.

25. Example 12
Using the frequency distribution from example 2 construct an ogive.

26. 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

27. Graphing Quantitative Data Sets
Stem-and-leaf plot
Each number is separated into a stem and a leaf.
Still contains original data values. 26 2 3 4 5
Data: 21, 25, 25, 26, 27, 28, , 36, 36, 45

28. 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.

29. 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 | and = 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.

30. Example: Constructing a Stem-and-Leaf Plot

31. 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

32. 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.

33. 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.

34. Example: Constructing a Dot Plot

35. 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.
Vehicle type
Killed
Cars
18,440
Trucks
13,778
Motorcycles
4,553
Other
823

36. How to construct a pie chart
First find the relative frequency, or percent, of each category.
Second find the central angle that corresponds to each category.
Find the central angle by multiplying 360 degrees by the category’s relative frequency.
Construct the pie chart.

37. Example: Constructing a Pie Chart
The numbers of motor vehicle occupants killed in crashes in are shown in the table. Use a pie chart to organize the data. What can you conclude? (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

38. 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

39. Solution: Constructing a Pie Chart
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

40. Example: Constructing a Pie Chart
The numbers of motor vehicle occupants killed in crashes in are shown in the table. Use a pie chart to organize the data. Compare the 1995 data with the 2005 data. (Source: U.S. Department of Transportation, National Highway Traffic Safety Administration)
Vehicle type
Killed
Cars
22,423
Trucks
10216
Motorcycles
2227
Other
425

41. 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.
Such positioning helps highlight important data and is used frequently in business.
Frequency
Categories

42. 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)

43. 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.

44. Example: Constructing a Pareto Chart
Every year, the Better Business Bureau (BBB) receives complaints from customers. In a recent year, the BBB received the following complaints.
7792 complaints about home furnishing stores.
5733 complaints about computer sales and service stores.
14668 complaints about auto dealers
9728 complaints about auto repair shops
4649 complaints about dry cleaning companies
Find the relative frequency for each data entry.
Position the bars in decreasing order according to relative frequency.
Interpret the results in the context of the data.

45. Graphing Paired Data Sets
Each entry in one data set corresponds to one entry in a second data set.
For instance, suppose a data set contains the costs of an item and a second data set contains sales amounts for the item at each cost. Because each cost corresponds to a sales amount, the data sets are paired.
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

46. 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)

47. 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.

48. Solution: Interpreting a Scatter Plot
A complete discussion of types of correlation occurs in chapter 9. You may want, however, to discuss positive correlation, negative correlation, and no correlation at this point.
Be sure that students do not confuse correlation with causation.
Interpretation
From the scatter plot, you can see that as the petal length increases, the petal width also tends to increase.

49. 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

50. Example: Constructing a Time Series Chart
The table lists the number of cellular telephone subscribers (in millions) for the years 1995 through Construct a time series chart for the number of cellular subscribers. (Source: Cellular Telecommunication & Internet Association)

51. 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.

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

53. 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

54. 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

55. 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:

56. 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?

57. Solution: Finding a Sample Mean
The sum of the flight prices is
Σx = = 3695
To find the mean price, divide the sum of the prices by the number of prices in the sample
The mean price of the flights is about $

58. Measure of Central Tendency: 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.

59. 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.

60. Solution: Finding the Median
First order the data.
There are seven entries (an odd number), the median is the middle, or fourth, data entry.
The median price of the flights is $427.

61. Measure of Central Tendency: 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).

62. 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

63. Solution: Finding the Mode
Ordering the data helps to find the mode.
The entry of 397 occurs twice, whereas the other data entries occur only once.
The mode of the flight prices is $397.

64. 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

65. 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.

66. 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).

67. 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 21 22 23 24 65

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

69. 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.

70. Solution: Comparing the Mean, Median, and Mode
Sometimes a graphical comparison can help you decide which measure of central tendency best represents a data set.
In this case, it appears that the median best describes the data set.

71. Weighted Mean Weighted Mean
The mean of a data set whose entries have varying weights.
where w is the weight of each entry x

72. 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?

73. 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
Your weighted mean for the course is You did not get an A.

74. Mean of Grouped Data Mean of a Frequency Distribution Approximated by
where x and f are the midpoints and frequencies of a class, respectively

75. Finding the Mean of a Frequency Distribution
In Words In Symbols
Find the midpoint of each class.
Find the sum of the products of the midpoints and the frequencies.
Find the sum of the frequencies.
Find the mean of the frequency distribution.

76. Example: Find the Mean of a Frequency Distribution
Use the frequency distribution to approximate the mean number of minutes that a sample of Internet subscribers spent online during their most recent session.
Class
Midpoint
Frequency, f
7 – 18
12.5 6
19 – 30
24.5 10
31 – 42
36.5 13
43 – 54
48.5 8
55 – 66
60.5 5
67 – 78
72.5
79 – 90
84.5 2

77. 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
72.5∙6 = 435.0
79 – 90
84.5 2
84.5∙2 = 169.0
n = 50
Σ(x∙f) =

78. 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.

79. The Shape of Distributions
Uniform Distribution (rectangular)
All entries or classes in the distribution have equal or approximately equal frequencies.
Symmetric.

80. 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.

81. 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.

82. The Shape of Distributions
When a distribution is symmetric and unimodal, the mean median and mode are equal.
If a distribution is skewed left, the mean is less than the median and the median is usually less than the mode.
If a distribution is skewed right, the mean is greater than the median and the median is usually greater than the mode.
In some cases, the shape cannot be classified as symmetric, uniform, or skewed.

83. Section 2.4 Measures of Variation
Determine the range of a data set
Determine the variance and standard deviation of a population and of a sample
Use the Empirical Rule and Chebychev’s Theorem to interpret standard deviation
Approximate the sample standard deviation for grouped data

84. 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)

85. 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)

86. Solution: Finding the Range
Ordering the data helps to find the least and greatest salaries.
Range = (Max. salary) – (Min. salary)
= 47 – 37 = 10
The range of starting salaries is 10 or $10,000.
minimum
maximum

87. Deviation, Variance, and Standard Deviation
As a measure of variation, the range has the advantage of being easy to compute.
Its disadvantage, however, is that it uses only two entries from the data set.
Two measures of variation that use all the entries in a data set are the variance and the standard deviation.
However, before you learn about these measures of variation, you need to know what is meant by the deviation of an entry in a data set.

88. Deviation, Variance, and Standard 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

89. 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)
Solution:
First determine the mean starting salary.

90. Solution: Finding the Deviation
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 44
44 – 41.5 = 2.5 37
37 – 41.5 = –4.5 42
42 – 41.5 = 0.5
Determine the deviation for each data entry.
Σx = 415
Σ(x – μ) = 0

91. Deviation, Variance, and Standard Deviation
In Example 2, notice that the sum of the deviations is zero. Because this is true for any data set, it doesn’t make sense to find the average of the deviations.
To overcome this problem, you can square each deviation. When you add the squares of the deviations, you compute a quantity called the sum of squares, denoted SSx
In a population data set, the mean of the squares of the deviations is called the population variance.

92. Deviation, Variance, and Standard Deviation
Population Variance
Population Standard Deviation
Sum of squares, SSx

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

94. Finding the Population Variance & Standard Deviation
In Words In Symbols
Divide by N to get the population variance.
Find the square root to get the population standard deviation.

95. 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) Recall μ = 41.5.

96. Solution: Finding the Population Standard Deviation
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 44
44 – 41.5 = 2.5
(2.5)2 = 6.25 37
37 – 41.5 = –4.5
(–4.5)2 = 20.25 42
42 – 41.5 = 0.5
(0.5)2 = 0.25
Determine SSx
N = 10
Σ(x – μ) = 0
SSx = 88.5

97. Solution: Finding the Population Standard Deviation
Population Variance
Population Standard Deviation
The population standard deviation is about 3.0, or $3000.

98. Deviation, Variance, and Standard Deviation
Sample Variance
Sample Standard Deviation

99. Finding the Sample Variance & Standard Deviation
In Words In Symbols
Find the mean of the sample data set.
Find deviation of each entry.
Square each deviation.
Add to get the sum of squares.

100. Finding the Sample Variance & Standard Deviation
In Words In Symbols
Divide by n – 1 to get the sample variance.
Find the square root to get the sample standard deviation.

101. 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)

102. Solution: Finding the Sample Standard Deviation
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 44
44 – 41.5 = 2.5
(2.5)2 = 6.25 37
37 – 41.5 = –4.5
(–4.5)2 = 20.25 42
42 – 41.5 = 0.5
(0.5)2 = 0.25
Determine SSx
n = 10
Σ(x – μ) = 0
SSx = 88.5

103. Solution: Finding the Sample Standard Deviation
Sample Variance
Sample Standard Deviation
The sample standard deviation is about 3.1, or $3100.

104. 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.

105. 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.

106. Interpreting Standard Deviation: Empirical Rule (68 – 95 – 99.7 Rule)
99.7% within 3 standard deviations
2.35%
95% within 2 standard deviations
13.5%
68% within 1 standard deviation
34%

107. 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 inches.

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

109. Chebychev’s Theorem
The Empirical Rule applies only to (symmetric) bell- shaped distributions. What if the distribution is not bell-shaped, or what if the shape of the distribution is not known? The following theorem gives an inequality statement that applies to all distributions. It is named after the Russian statistician Pafnuti Chebychev( ).

110. Chebychev’s Theorem
The portion of any data set lying within k standard deviations (k > 1) of the mean is at least:
k = 2: In any data set, at least
of the data lie within 2 standard deviations of the mean.
k = 3: In any data set, at least
of the data lie within 3 standard deviations of the mean.

111. 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?

112. Solution: Using Chebychev’s Theorem
k = 2: μ – 2σ = 39.2 – 2(24.8) = (use 0 since age can’t be negative) μ + 2σ = (24.8) = 88.8
At least 75% of the population of Florida is between 0 and 88.8 years old.

113. 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.
where n= Σf (the number of entries in the data set)

114. Example: Finding the Standard Deviation for Grouped Data
Number of Children in 50 Households 1 3 2 5 6 4
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.

115. Solution: Finding the Standard Deviation for Grouped Data
First construct a frequency distribution.
Find the mean of the frequency distribution. x f xf 10
0(10) = 0 1 19
1(19) = 19 2 7
2(7) = 14 3
3(7) =21 4
4(2) = 8 5
5(1) = 5 6
6(4) = 24
The sample mean is about 1.8 children.
Σf = 50
Σ(xf )= 91

116. Solution: Finding the Standard Deviation for Grouped Data
Determine the sum of squares. x f 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
3 – 1.8 = 1.2
(1.2)2 = 1.44
1.44(7) = 10.08 4
4 – 1.8 = 2.2
(2.2)2 = 4.84
4.84(2) = 9.68 5
5 – 1.8 = 3.2
(3.2)2 = 10.24
10.24(1) = 10.24 6
6 – 1.8 = 4.2
(4.2)2 = 17.64
17.64(4) = 70.56

117. Solution: Finding the Standard Deviation for Grouped Data
Find the sample standard deviation.
The standard deviation is about 1.7 children.

118. 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
Determine and interpret the standard score (z-score)

119. 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.

120. 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
Solution:
Q2 divides the data set into two halves.
Lower half
Upper half Q2

121. Solution: Finding Quartiles
The first and third quartiles are the medians of the lower and upper halves of the data set.
Lower half
Upper half Q1 Q2 Q3
About one fourth of the employees scored 10 or less, about one half scored 15 or less; and about three fourths scored 18 or less.

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

123. 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.

124. 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

125. Drawing a Box-and-Whisker Plot
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.
Draw whiskers from the box to the minimum and maximum entries.
Whisker
Maximum entry
Minimum entry
Box
Median, Q2 Q3 Q1

126. Example: Drawing a Box-and-Whisker Plot
Draw a box-and-whisker plot that represents the 15 test scores. Recall Min = 5 Q1 = 10 Q2 = 15 Q3 = 18 Max = 37
Solution: 5 10 15 18 37
About half the scores are between 10 and 18. By looking at the length of the right whisker, you can conclude 37 is a possible outlier.

127. Percentiles and Other 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

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

129. 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 Find the z-score that corresponds to the age for each actor or actress. Then compare your results.

130. Solution: Comparing z-Scores from Different Data Sets
Forest Whitaker
Helen Mirren
0.15 standard deviations above the mean
2.17 standard deviations above the mean

131. 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.