1. 2 Chapter Descriptive Statistics © 2012 Pearson Education, Inc.
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2. 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
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3. Course Level Expectations
CLE Use technologies appropriately to develop understanding of abstract mathematical ideas, to facilitate problem solving, and to produce accurate and reliable models.
CLE Understand histograms, parallel box plots, and scatterplots, and use them to compare display data.
Larson/Farber 5th ed.

4. Checks for Understanding
Use properties of point estimators, including biased/unbiased, and variability.
Understand the meaning of confidence level, of confidence intervals, and the properties of confidence intervals.
Larson/Farber 5th ed.

5. Frequency Distributions and Their Graphs
Section 2.1
Frequency Distributions
and Their Graphs
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6. Section 2.1 Objectives Construct frequency distributions
Construct frequency histograms, frequency polygons, relative frequency histograms, and ogives
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7. Frequency Distribution
Why can’t we do ‘5-1’?
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
21–25
26–30 4
Class width 6 – 1 = 5
Lower class
limits
Upper class
limits
This table might represent the ages of people in a room for example
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8. Constructing a Frequency Distribution
Decide on the number of classes.
Usually between 5 and 20; otherwise, it may be difficult to detect any patterns.
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.
For example, you have a range of ages from 10 to 74
You decide you want 8 classes
The range of data is = 64
64/8 = 8.
Therefore, you have 8 classes with a width of 8
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9. Constructing a Frequency Distribution
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.
From the previous example, the minimum data entry is 10
The next lower limit would be 18
The first upper limit would be 17
The next would be 17+8 = 25
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10. Constructing a Frequency Distribution
Make a tally mark for each data entry in the row of the appropriate class.
Count the tally marks to find the total frequency f for each class.
For our example to continue, we would have to have a number for how many people were each age
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11. Example: Constructing a Frequency Distribution
The following sample data set lists the prices (in dollars) of 30 portable global positioning system (GPS) navigators. Construct a frequency distribution that has seven classes
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12. Solution: Constructing a Frequency Distribution
Number of classes = 7 (given)
Find the class width
Round up to 56
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13. Solution: Constructing a Frequency Distribution
Use 59 (minimum value) as first lower limit. Add the class width of 56 to get the lower limit of the next class.
= 115
Find the remaining lower limits.
Lower limit
Upper limit 59
115
171
227
283
339
395
Class width = 56
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14. Solution: Constructing a Frequency Distribution
The upper limit of the first class is 114 (one less than the lower limit of the second class). Add the class width of 56 to get the upper limit of the next class = 170 Find the remaining upper limits.
Lower limit
Upper limit 59
114
115
170
171
226
227
282
283
338
339
394
395
450
Class width = 56
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15. Solution: Constructing a Frequency Distribution
Make a tally mark for each data entry in the row of the appropriate class.
Count the tally marks to find the total frequency f for each class.
Class
Tally
Frequency, f
59–114
IIII 5
115–170
IIII III 8
171–226
IIII I 6
227–282
283–338 II 2
339–394 I 1
395–450
III 3
Note: you should sum the frequencies. This is the total sample size, and will be used later. Here, the sample size is 30
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16. Determining the Midpoint
Midpoint of a class
Class
Midpoint
Frequency, f
59–114 5
115–170 8
171–226 6
Class width = 56
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17. Determining the Relative Frequency
Relative Frequency of a class
Portion or percentage of the data that falls in a particular class.
Frequency is how often something occurs
Relative frequency is how often something occurs relative to the total occurrences (sample size)
Class
Frequency, f
Relative Frequency
59–114 5
115–170 8
171–226 6
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18. Determining the Cumulative Frequency
Cumulative frequency of a class
The sum of the frequencies for that class and all previous classes.
Class
Frequency, f
Cumulative frequency
59–114 5
115–170 8
171–226 6 5 + 13 + 19
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19. Expanded Frequency Distribution
Remember what this is called
Class
Frequency, f
Midpoint
Relative
Frequency (f/n)
Cumulative frequency
59–114 5
86.5
0.17
115–170 8
142.5
0.27 13
171–226 6
198.5
0.2 19
227–282
254.5 24
283–338 2
310.5
0.07 26
339–394 1
366.5
0.03 27
395–450 3
422.5
0.1 30
Σf = 30
This is the uppercase Greek letter “sigma” and means “to sum”
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20. 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 values
frequency
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21. Class Boundaries Why do this? Class boundaries
The numbers that separate classes without forming gaps between them.
Why do this?
Because bars of a histogram must touch, bars begin and end at class boundaries instead of class limits
The distance from the upper limit of the first class to the lower limit of the second class is 115 – 114 = 1.
Half this distance is 0.5.
Class
boundaries
Frequency, f
59–114 5
115–170 8
171–226 6
58.5–114.5
First class lower boundary = 59 – 0.5 = 58.5
First class upper boundary = = 114.5
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22. Class Boundaries Class Class boundaries Frequency, f 59–114 58.5–114.5
115–170
114.5–170.5 8
171–226
170.5–226.5 6
227–282
226.5–282.5
283–338
282.5–338.5 2
339–394
338.5–394.5 1
395–450
394.5–450.5 3
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23. Example: Frequency Histogram
Construct a frequency histogram for the Global Positioning system (GPS) navigators.
Class
Class boundaries
Midpoint
Frequency, f
59–114
58.5–114.5
86.5 5
115–170
114.5–170.5
142.5 8
171–226
170.5–226.5
198.5 6
227–282
226.5–282.5
254.5
283–338
282.5–338.5
310.5 2
339–394
338.5–394.5
366.5 1
395–450
394.5–450.5
422.5 3
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24. Solution: Frequency Histogram (using Midpoints)
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25. Solution: Frequency Histogram (using class boundaries)
You can see that more than half of the GPS navigators are priced below $
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26. Graphs of Frequency Distributions
Frequency Polygon
A line graph that emphasizes the continuous change in frequencies.
data values
frequency
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27. Example: Frequency Polygon
Construct a frequency polygon for the GPS navigators frequency distribution.
You will use these midpoints to create your frequency polygon
Class
Midpoint
Frequency, f
59–114
86.5 5
115–170
142.5 8
171–226
198.5 6
227–282
254.5
283–338
310.5 2
339–394
366.5 1
395–450
422.5 3
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28. Solution: Frequency Polygon
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.
Add This
And This
You can see that the frequency of GPS navigators increases up to $ and then decreases.
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29. 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.
data values
relative frequency
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30. Example: Relative Frequency Histogram
Construct a relative frequency histogram for the GPS navigators frequency distribution.
Class
Class boundaries
Frequency, f
Relative
frequency
59–114
58.5–114.5 5
0.17
115–170
114.5–170.5 8
0.27
171–226
170.5–226.5 6
0.2
227–282
226.5–282.5
283–338
282.5–338.5 2
0.07
339–394
338.5–394.5 1
0.03
395–450
394.5–450.5 3
0.1
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31. Solution: Relative Frequency Histogram
From this graph you can see that 27% of GPS navigators are priced between $ and $
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32. 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.
data values
cumulative frequency
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33. 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.
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34. Constructing an Ogive 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).
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35. Example: Ogive
Construct an ogive for the GPS navigators frequency distribution.
Class
Class boundaries
Frequency, f
Cumulative frequency
59–114
58.5–114.5 5
115–170
114.5–170.5 8 13
171–226
170.5–226.5 6 19
227–282
226.5–282.5 24
283–338
282.5–338.5 2 26
339–394
338.5–394.5 1 27
395–450
394.5–450.5 3 30
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36. Solution: Ogive
From the ogive, you can see that about 25 GPS navigators cost $300 or less. The greatest increase occurs between $ and $
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37. Section 2.1 Summary Constructed frequency distributions
Constructed frequency histograms, frequency polygons, relative frequency histograms and ogives
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38. Assignment
Page Orally
7-43 Odd
Larson/Farber 5th ed.

39. Chapter 2 Quiz 1(20 points)
Using the information from Question number 34 on page 50, created an expanded Frequency Distribution Chart.
It should look like this:
Class
Frequency, f
Midpoint
Relative
Frequency (f/n)
Cumulative frequency
Larson/Farber 5th ed.

40. Answers to Quiz 1 Class Frequency, f Midpoint Relative Frequency (f/n)
Cumulative frequency 7
2499
.28 3
2586
.12 10 2
2673
.08 12 4
2760
.16 16 9
2847
.36 25
Larson/Farber 5th ed.

41. More Graphs and Displays
Section 2.2
More Graphs and Displays
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42. Section 2.2 Objectives
Graph quantitative data using stem-and-leaf plots and dot plots
Graph qualitative data using pie charts and Pareto charts
Graph paired data sets using scatter plots and time series charts
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43. Graphing Quantitative Data Sets
Stem-and-leaf plot
Each number is separated into a stem and a leaf.
Similar to a histogram.
Still contains original data values. 26 2 3 4 5
Data: 21, 25, 25, 26, 27, 28, , 36, 36, 45
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44. Example: Constructing a Stem-and-Leaf Plot
The following are the numbers of text messages sent last week by the cellular phone users on one floor of a college dormitory. Display the data in a stem-and-leaf plot.
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45. 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.
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46. 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.
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47. Graphing Quantitative Data Sets
We did a version of this on the 1st day of school
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
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48. 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.
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49. 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. This is commonly called an “outlier”
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50. 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.
Therefore you will need
Relative Frequency
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51. Example: Constructing a Pie Chart
The numbers of earned degrees conferred (in thousands) in 2007 are shown in the table. Use a pie chart to organize the data. (Source: U.S. National Center for Educational Statistics)
Type of degree
Number
(thousands)
Associate’s
728
Bachelor’s
1525
Master’s
604
First professional 90
Doctoral 60
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52. Solution: Constructing a Pie Chart
Find the relative frequency (percent) of each category.
Type of degree
Frequency, f
Relative frequency
Associate’s
728
Bachelor’s
1525
Master’s
604
First professional 90
Doctoral 60
Σf = 3007
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53. 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 associate’s degrees is
360º(0.24) ≈ 86º
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54. Solution: Constructing a Pie Chart
Type of degree
Frequency, f
Relative frequency
Central angle
Associate’s
728
0.24
Bachelor’s
1525
0.51
Master’s
604
0.20
First professional 90
0.03
Doctoral 60
0.02
360º(0.24)≈86º
360º(0.51)≈184º
360º(0.20)≈72º
360º(0.03)≈11º
360º(0.02)≈7º
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55. Solution: Constructing a Pie Chart
Type of degree
Relative frequency
Central angle
Associate’s
0.24
86º
Bachelor’s
0.51
184º
Master’s
0.20
72º
First professional
0.03
11º
Doctoral
0.02 7º
From the pie chart, you can see that over one half of the degrees conferred in 2007 were bachelor’s degrees.
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56. 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.
Frequency
Categories
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57. Example: Constructing a Pareto Chart
In a recent year, the retail industry lost $36.5 billion 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 ($5.4 billion), employee theft ($15.9 billion), shoplifting ($12.7 billion), and vendor fraud ($1.4 billion). Use a Pareto chart to organize this data. (Source: National Retail Federation and Center for Retailing Education, University of Florida)
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58. Chapter 2 Quiz 2(20 points)
Using the information from Question number 32 on page 50, created an expanded Frequency Distribution Chart.
It should look like this:
Class
Frequency, f
Midpoint
Relative
Frequency (f/n)
Cumulative frequency
Larson/Farber 5th ed.

59. Answers to Quiz 2 Class Frequency, f Midpoint Relative Frequency (f/n)
Cumulative frequency
32-35 3
36.5
.125~.13
36-39 9
40.5
.375~.38 12
40-43 8
44.5
.33 20
44-47
48.5 23
48-51 1 51
.04 24
Larson/Farber 5th ed.

60. Solution: Constructing a Pareto Chart
Cause
$ (billion)
Admin. error
5.4
Employee theft
15.9
Shoplifting
12.7
Vendor fraud
1.4
From the graph, it is easy to see that the causes of inventory shrinkage that should be addressed first are employee theft and shoplifting.
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61. Graphing Paired Data Sets
Each entry in one data set corresponds to one entry in a second data set.
Graph using a scatter plot.
The ordered pairs are graphed as points in a coordinate plane.
Used to show the relationship between two quantitative variables. y x
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62. 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)
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63. 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.
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64. 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.
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.
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65. 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
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66. Example: Constructing a Time Series Chart
The table lists the number of cellular telephone subscribers (in millions) for the years 1998 through Construct a time series chart for the number of cellular subscribers. (Source: Cellular Telecommunication & Internet Association)
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67. 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.
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68. Solution: Constructing a Time Series Chart
The graph shows that the number of subscribers has been increasing since 1998, with greater increases recently.
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69. 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
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70. Assignment Page 60 1-4 orally Page 60-63 5-33 odd
Larson/Farber 5th ed.

71. Chapter 2 Quiz 3 Lower limit Upper limit 59 114 115 170
What is the class width in this chart??
What are the class boundaries for the first class in this chart?
What is the midpoint of the first class in this chart?
What do you call a chart which has classes, frequency, midpoints, relative frequency and cumulative frequency?
What does this ∑ symbol mean?
What is on the “y” axis on a frequency histogram?
What is on the “x” axis on a frequency histogram?
On the “x” axis of a frequency histogram, you may use two methods to indicate different classes. What are they?
Does the first bar of a frequency histogram touch the “Y” axis?
True/False: A stem and leaf plot is a qualitative graph of data.
True/False: A pie chart is a qualitative graph of data.
True/False: A Pareto Graph has all bars touching each other, and is ordered greatest to least.
A ___________ uses a number line with marks above it to represent data values.
Each entry in one data set corresponds to one entry in a second data set on a _____________
Data set is composed of quantitative entries taken at regular intervals over a period of time in a _________ _______ _________.
Larson/Farber 5th ed.

72. Measures of Central Tendency
Section 2.3
Measures of Central Tendency
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73. Section 2.3 Objectives
Determine the mean, median, and mode of a population and of a sample
Determine the weighted mean of a data set and the mean of a frequency distribution
Describe the shape of a distribution as symmetric, uniform, or skewed and compare the mean and median for each
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74. 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
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75. 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:
Lowercase Greek letter “mu” always used for the mean of the population
Pronounced “X Bar” always used for sample means
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76. Example: Finding a Sample Mean
The prices (in dollars) for a sample of round-trip flights from Chicago, Illinois to Cancun, Mexico are listed. What is the mean price of the flights?
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77. 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 $
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78. 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.
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79. 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
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80. 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.
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81. Example: Finding the Median
The flight priced at $432 is no longer available. What is the median price of the remaining flights?
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82. Solution: Finding the Median
First order the data.
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.
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83. Measure of Central Tendency: Mode
The data entry that occurs with the greatest frequency.
A data set can have one mode, more than one mode, or no mode.
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).
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84. 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
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85. 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.
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86. 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
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87. 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.
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88. 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).
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89. 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
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90. 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)
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91. 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.
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92. 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.
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93. Weighted Mean Weighted Mean
The mean of a data set whose entries have varying weights.
where w is the weight of each entry x
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94. 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?
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95. 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.
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96. Mean of Grouped Data Mean of a Frequency Distribution Approximated by
where x and f are the midpoints and frequencies of a class, respectively
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97. 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.
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98. 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
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99. 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) =
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100. 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.
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101. The Shape of Distributions
Uniform Distribution (rectangular)
All entries or classes in the distribution have equal or approximately equal frequencies.
Symmetric.
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102. 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.
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103. 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.
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104. Section 2.3 Summary
Determined the mean, median, and mode of a population and of a sample
Determined the weighted mean of a data set and the mean of a frequency distribution
Described the shape of a distribution as symmetric, uniform, or skewed and compared the mean and median for each
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105. Assignment Page 72 1-4 orally Page 72-77 5-57 odd
Larson/Farber 5th ed.

106. Chapter 2 Quiz 4 (5 points)
T/F –In a uniform distribution, the mean, median and mode are all the same
T/F -If two entries occur with the same least frequency, each entry is a mode (bimodal).
T/F –An advantage of using the mean as a measure of central tendency is that it is not affected by outliers.
In a symmetrical distribution, the mean, median and mode are all the same
In a “Skewed Left” distribution, the median is to the ______ of the mean
In a “Skewed Right” distribution, the mean is to the _____ of the median
Quiz continues on next slide
Larson/Farber 5th ed.

107. Chapter 2 Quiz 4 Continued (5 points)
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 8
19 – 30
24.5 11
31 – 42
36.5 16
43 – 54
48.5 10
55 – 66
60.5
67 – 78
72.5 5
79 – 90
84.5 2
Quiz continues on next slide
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108. Chapter 2 Quiz 4 Continued (5 points)
Use the scores earned by the student below, calculate his weighted grade. Did he earn an A? (assume the teacher rounds grades to whole numbers).
Source
Score, x
Weight, w
Test Mean 90
0.50
Midterm
0.15
Final Exam 98
0.20
Computer Lab
0.10
Homework
100
0.05
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109. Section 2.4 Measures of Variation
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110. Section 2.4 Objectives Determine the range of a data set
Determine the variance and standard deviation of a population and of a sample
Use the Empirical Rule and Chebychev’s Theorem to interpret standard deviation
Approximate the sample standard deviation for grouped data
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111. 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)
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112. 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)
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113. 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
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114. 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
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115. 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.
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116. Solution: Finding the Deviation
Salary ($1000s), x
Deviation ($1000s)
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.
Notice that the sum of the deviations is zero. This makes it hard to use the information without manipulation
Σx = 415
Σ(x – μ) = 0
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117. Deviation, Variance, and Standard Deviation
By squaring the deviations, we can get usable information which relates to central tendency
Population Variance
Population Standard Deviation
Sum of squares, SSx
Greek lower case sigma –Sigma squared is the standard symbol for Pop variance
Sigma is the standard symbol for Pop Standard Deviation
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118. Finding the Population Variance & Standard Deviation
In Words In Symbols
Find the mean of the population data set.
Find the deviation of each entry.
Square each deviation.
Add to get the sum of squares.
x – μ
(x – μ)2
SSx = Σ(x – μ)2
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119. Finding the Population Variance & Standard Deviation
In Words In Symbols
Divide by N to get the population variance.
Find the square root of the variance to get the population standard deviation.
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120. 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.
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121. 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 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
Σ(x – μ) = 0
SSx = 88.5
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122. Solution: Finding the Population Standard Deviation
Population Variance
Population Standard Deviation
We don’t use this value much
The population standard deviation is about 3.0, or $3000.
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123. Deviation, Variance, and Standard Deviation
Sample Variance
Sample Standard Deviation
Everything is basically the “same” except that we use n-1 instead of N. This is considered a more conservative value
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124. Finding the Sample Variance & Standard Deviation
In Words In Symbols
Find the mean of the sample data set.
Find the deviation of each entry.
Square each deviation.
Add to get the sum of squares.
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125. Finding the Sample Variance & Standard Deviation
In Words In Symbols
Divide by n – 1 to get the sample variance.
Find the square root of the variance to get the sample standard deviation.
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126. 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)
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127. 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 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
Σ(x – μ) = 0
SSx = 88.5
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128. Solution: Finding the Sample Standard Deviation
Sample Variance
Sample Standard Deviation
These were the same data values as before
Notice that the sample standard deviation is slightly larger. In simple terms, we aren’t quite as accurate
The sample standard deviation is about 3.1, or $3100.
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129. 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
29.00
40.50
24.50
33.00
38.00
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130. Using the Calculator Select “Stat” Select Edit Clear L1 (list 1)
Enter the data into L1
Select “Stat” and highlight “Calc” to display Calc menu
Select 1: “1-Var Stats” and press “2nd L1” Enter
Larson/Farber 5th ed.

131. Solution: Using Technology to Find the Standard Deviation
Sample Mean
Sample Standard Deviation
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132. Chapter 2 Quiz 5 Do all parts of problem 61 on page 78 Show all work
4 points for each part (16 points total)
Larson/Farber 5th ed.

133. 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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134. 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.
This may also be referred to as a “normal distribution”
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135. 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%
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136. 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 64.3 inches, with a sample standard deviation of 2.62 inches. Estimate the percent of the women whose heights are between inches and 64.3 inches.
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137. Solution: Using the Empirical Rule
Because the distribution is bell-shaped, you can use the Empirical Rule.
34% % = 47.5% of women are between and 64.3 inches tall.
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138. The empirical rule works on bell (normal) distributions
The empirical rule works on bell (normal) distributions. This works on ALL distributions
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.
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139. 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?
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140. 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 = (24.8) = 88.8
At least 75% of the population of Florida is between 0 and 88.8 years old.
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141. Standard Deviation for Grouped Data
Sample standard deviation for a frequency distribution
When a frequency distribution has classes, estimate the sample mean and the sample standard deviation by using the midpoint of each class.
where n = Σf (the number of entries in the data set)
This is the equation for Sample standard deviation:
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142. 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 2 5 6 4
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143. 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
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144. 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
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145. Solution: Finding the Standard Deviation for Grouped Data
Find the sample standard deviation.
The standard deviation is about 1.7 children.
What can I do with this information? Remember that the sample mean is 1.8 children. If we can assume that the distribution is normal, then about 68 percent of the sample lies within .1 children and 3.5 children
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146. Section 2.4 Summary Determined the range of a data set
Determined the variance and standard deviation of a population and of a sample
Used the Empirical Rule and Chebychev’s Theorem to interpret standard deviation
Approximated the sample standard deviation for grouped data
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146 of 149

147. Assignment Page 90 1-10 orally Page 90-97 11-47 odd
Larson/Farber 5th ed.

148. Chapter 2 Quiz 6 (10 points)
Using the data from question 53 on page 76:
Calculate the sample standard deviation (3 points)
What is the range of beds which would include 68% of the sample? (3 points)
What is the range of beds which would include 95% of the beds? (3 points)
Sx = 45.35, or about 46 beds
The mean is , or about 182 beds. Therefore, 68% = = 136 beds, = 228 beds
95% = 182 – 46 – 46 = 90 beds and = 274 beds
Larson/Farber 5th ed.

149. Section 2.5 Measures of Position
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150. Section 2.5 Objectives Determine the quartiles of a data set
Determine the interquartile range of a data set
Create a box-and-whisker plot
Interpret other fractiles such as percentiles
Determine and interpret the standard score (z-score)
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151. 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.
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152. Example: Finding Quartiles
The number of nuclear power plants in the top 15 nuclear power-producing countries in the world are listed. Find the first, second, and third quartiles of the data set
Solution:
Q2 divides the data set into two halves.
Lower half
Upper half Q2
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153. 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 countries have 10 or fewer nuclear power plants; about one half have 18 or fewer; and about three fourths have 31 or fewer.
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154. Interquartile Range Interquartile Range (IQR)
The difference between the third and first quartiles.
IQR = Q3 – Q1
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155. Example: Finding the Interquartile Range
Find the interquartile range of the data set Recall Q1 = 10, Q2 = 18, and Q3 = 31
Solution:
IQR = Q3 – Q1 = 31 – 10 = 21
The number of power plants in the middle portion of the data set vary by at most 21.
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156. 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
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157. 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
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158. Example: Drawing a Box-and-Whisker Plot
Draw a box-and-whisker plot that represents the data set Min = 6, Q1 = 10, Q2 = 18, Q3 = 31, Max = 104,
Solution:
About half the data values are between 10 and 31. By looking at the length of the right whisker, you can conclude 104 is a possible outlier.
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159. Percentiles and Other Fractiles
Summary
Symbols
Quartiles
Divide a data set into 4 equal parts
Q1, Q2, Q3
Deciles
Divide a data set into 10 equal parts
D1, D2, D3,…, D9
Percentiles
Divide a data set into 100 equal parts
P1, P2, P3,…, P99
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160. 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 62nd percentile? How should you interpret this? (Source: College Board)
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161. Solution: Interpreting Percentiles
The 62nd percentile corresponds to a test score of This means that 62% of the students had an SAT score of 1600 or less.
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162. The Standard Score Standard Score (z-score)
Represents the number of standard deviations a given value x falls from the mean μ.
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163. Example: Comparing z-Scores from Different Data Sets
In 2009, Heath Ledger won the Oscar for Best Supporting Actor at age 29 for his role in the movie The Dark Knight. Penelope Cruz won the Oscar for Best Supporting Actress at age 34 for her role in Vicky Cristina Barcelona. The mean age of all Best Supporting Actor winners is 49.5, with a standard deviation of The mean age of all Best Supporting Actress winners is 39.9, with a standard deviation of Find the z-scores that correspond to the ages of Ledger and Cruz. Then compare your results.
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164. Solution: Comparing z-Scores from Different Data Sets
Heath Ledger
1.49 standard deviations below the mean
Penelope Cruz
0.42 standard deviations below the mean
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165. Solution: Comparing z-Scores from Different Data Sets
About 95 percent of the data should fall between and 2.
Both z-scores fall between –2 and 2, so neither score would be considered unusual. Compared with other Best Supporting Actor winners, Heath Ledger was relatively younger, whereas the age of Penelope Cruz was only slightly lower than the average age of other Best Supporting Actress winners.
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166. 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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167. Assignment
Page odd, 31,33, 37-50
Larson/Farber 5th ed.

168. Group Project (30 points)
Gather a sample of both men’s and women’s shoe sizes
Create an Expanded Frequency chart using 5 classes
Create a Frequency Histogram (using midpoints for x)
Calculate the mean, median, and mode of your data
Calculate the sample standard deviation of your data
If your sample IS normally distributed, then:
Using your mean and sample deviation, and the Empirical Rule:
Between what two shoe sizes would we find 68% of the shoe sizes?
Between what two shoe sizes would we find 95% of the shoe sizes?
If your sample is NOT normally distributed, then use Chebychev’s theorem to find:
Between what two shoe sizes would we find 75% of the shoe sizes?
Between what two shoe sizes would we find 88.9% of the shoe sizes?
5 pts
-Or
Larson/Farber 5th ed.