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

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 2 Larson/Farber 4th ed.

3. Section 2.1 Frequency Distributions and Their Graphs 3 Larson/Farber 4th ed.

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

5. 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. Larson/Farber 4th ed. 5 ClassFrequency, f 1 – 55 6 – 108 11 – 156 16 – 208 21 – 255 26 – 304 Lower class limits Upper class limits Class width 6 – 1 = 5

6. Constructing a Frequency Distribution Larson/Farber 4th ed. 6 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.

7. Constructing a Frequency Distribution 3.Find the class limits.  You can use the minimum data entry as the lower limit of the first class.  Find the remaining lower limits (add the class width to the lower limit of the preceding class).  Find the upper limit of the first class. Remember that classes cannot overlap.  Find the remaining upper class limits. Larson/Farber 4th ed. 7

8. 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. Larson/Farber 4th ed. 8

9. Example: Constructing a Frequency Distribution The following sample data set lists the number of minutes 50 Internet subscribers spent on the Internet during their most recent session. Construct a frequency distribution that has seven classes. 50 40 41 17 11 7 22 44 28 21 19 23 37 51 54 42 86 41 78 56 72 56 17 7 69 30 80 56 29 33 46 31 39 20 18 29 34 59 73 77 36 39 30 62 54 67 39 31 53 44 Larson/Farber 4th ed. 9

10. Solution: Constructing a Frequency Distribution 1.Number of classes = 7 (given) 2.Find the class width Larson/Farber 4th ed. 10 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

11. Solution: Constructing a Frequency Distribution Larson/Farber 4th ed. 11 Lower limit Upper limit 7 Class 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

12. Solution: Constructing a Frequency Distribution The upper limit of the first class is 18 (one less than the lower limit of the second class). Add the class width of 12 to get the upper limit of the next class. 18 + 12 = 30 Find the remaining upper limits. Larson/Farber 4th ed. 12 Lower limit Upper limit 7 19 31 43 55 67 79 Class width = 12 30 42 54 66 78 90 18

13. 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. Larson/Farber 4th ed. 13 ClassTallyFrequency, f 7 – 18 IIII I 6 19 – 30 IIII 10 31 – 42 IIII IIII III 13 43 – 54 IIII III 8 55 – 66 IIII 5 67 – 78 IIII I 6 79 – 90 II 2 Σf = 50

14. Determining the Midpoint Midpoint of a class Larson/Farber 4th ed. 14 ClassMidpointFrequency, f 7 – 186 19 – 3010 31 – 4213 Class width = 12

15. Determining the Relative Frequency Relative Frequency of a class Portion or percentage of the data that falls in a particular class. Larson/Farber 4th ed. 15 ClassFrequency, fRelative Frequency 7 – 186 19 – 3010 31 – 4213

16. Determining the Cumulative Frequency Cumulative frequency of a class The sum of the frequency for that class and all previous classes. Larson/Farber 4th ed. 16 ClassFrequency, fCumulative frequency 7 – 186 19 – 3010 31 – 4213 + + 6 16 29

17. Expanded Frequency Distribution Larson/Farber 4th ed. 17 ClassFrequency, fMidpoint Relative frequency Cumulative frequency 7 – 18612.50.126 19 – 301024.50.2016 31 – 421336.50.2629 43 – 54848.50.1637 55 – 66560.50.1042 67 – 78672.50.1248 79 – 90284.50.0450 Σf = 50

18. 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. Larson/Farber 4th ed. 18 data values frequency

19. Class Boundaries Class boundaries The numbers that separate classes without forming gaps between them. Larson/Farber 4th ed. 19 Class Boundaries Frequency, f 7 – 186 19 – 3010 31 – 4213 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

20. Class Boundaries Larson/Farber 4th ed. 20 Class Class boundaries Frequency, f 7 – 18 6.5 – 18.56 19 – 3018.5 – 30.510 31 – 4230.5 – 42.513 43 – 5442.5 – 54.58 55 – 6654.5 – 66.55 67 – 7866.5 – 78.56 79 – 9078.5 – 90.52

21. Example: Frequency Histogram Construct a frequency histogram for the Internet usage frequency distribution. Larson/Farber 4th ed. 21 Class Class boundariesMidpoint Frequency, f 7 – 18 6.5 – 18.512.56 19 – 3018.5 – 30.524.510 31 – 4230.5 – 42.536.513 43 – 5442.5 – 54.548.58 55 – 6654.5 – 66.560.55 67 – 7866.5 – 78.572.56 79 – 9078.5 – 90.584.52

22. Solution: Frequency Histogram (using Midpoints) Larson/Farber 4th ed. 22

23. Solution: Frequency Histogram (using class boundaries) Larson/Farber 4th ed. 23 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.

24. Graphs of Frequency Distributions Frequency Polygon A line graph that emphasizes the continuous change in frequencies. Larson/Farber 4th ed. 24 data values frequency

25. Example: Frequency Polygon Construct a frequency polygon for the Internet usage frequency distribution. Larson/Farber 4th ed. 25 ClassMidpointFrequency, f 7 – 1812.56 19 – 3024.510 31 – 4236.513 43 – 5448.58 55 – 6660.55 67 – 7872.56 79 – 9084.52

26. Solution: Frequency Polygon Larson/Farber 4th ed. 26 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.

27. 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. Larson/Farber 4th ed. 27 data values relative frequency

28. Example: Relative Frequency Histogram Construct a relative frequency histogram for the Internet usage frequency distribution. Larson/Farber 4th ed. 28 Class Class boundaries Frequency, f Relative frequency 7 – 18 6.5 – 18.560.12 19 – 3018.5 – 30.5100.20 31 – 4230.5 – 42.5130.26 43 – 5442.5 – 54.580.16 55 – 6654.5 – 66.550.10 67 – 7866.5 – 78.560.12 79 – 9078.5 – 90.520.04

29. Solution: Relative Frequency Histogram Larson/Farber 4th ed. 29 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.

30. 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. Larson/Farber 4th ed. 30 data values cumulative frequency

31. 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. Larson/Farber 4th ed. 31

32. 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). Larson/Farber 4th ed. 32

33. Example: Ogive Construct an ogive for the Internet usage frequency distribution. Larson/Farber 4th ed. 33 Class Class boundaries Frequency, f Cumulative frequency 7 – 18 6.5 – 18.566 19 – 3018.5 – 30.51016 31 – 4230.5 – 42.51329 43 – 5442.5 – 54.5837 55 – 6654.5 – 66.5542 67 – 7866.5 – 78.5648 79 – 9078.5 – 90.5250

34. Solution: Ogive Larson/Farber 4th ed. 34 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.

35. Section 2.1 Summary Constructed frequency distributions Constructed frequency histograms, frequency polygons, relative frequency histograms and ogives Larson/Farber 4th ed. 35

36. Section 2.2 More Graphs and Displays Larson/Farber 4th ed. 36

37. Section 2.2 Objectives Graph quantitative data using stem-and-leaf plots and dot plots Graph qualitative data using pie charts and Pareto charts Graph paired data sets using scatter plots and time series charts Larson/Farber 4th ed. 37

38. 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. Larson/Farber 4th ed. 38 Data: 21, 25, 25, 26, 27, 28, 30, 36, 36, 45 26 21 5 5 6 7 8 30 6 6 45

39. 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. Larson/Farber 4th ed. 39 155159 144 129 105 145 126 116 130 114 122 112 112 142 126 118118 108 122 121 109 140 126 119 113 117 118 109 109 119 139139 122 78 133 126 123 145 121 134 124 119 132 133 124 129 112 126 148 147

40. Solution: Constructing a Stem-and-Leaf Plot Larson/Farber 4th ed. 40 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. 155159 144 129 105 145 126 116 130 114 122 112 112 142 126 118118 108 122 121 109 140 126 119 113 117 118 109 109 119 139139 122 78 133 126 123 145 121 134 124 119 132 133 124 129 112 126 148 147

41. Solution: Constructing a Stem-and-Leaf Plot Larson/Farber 4th ed. 41 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.

42. Graphing Quantitative Data Sets Dot plot Each data entry is plotted, using a point, above a horizontal axis Larson/Farber 4th ed. 42 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

43. Example: Constructing a Dot Plot Use a dot plot organize the text messaging data. Larson/Farber 4th ed. 43 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. 155159 144 129 105 145 126 116 130 114 122 112 112 142 126 118118 108 122 121 109 140 126 119 113 117 118 109 109 119 139139 122 78 133 126 123 145 121 134 124 119 132 133 124 129 112 126 148 147

44. Solution: Constructing a Dot Plot Larson/Farber 4th ed. 44 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. 155159 144 129 105 145 126 116 130 114 122 112 112 142 126 118118 108 122 121 109 140 126 119 113 117 118 109 109 119 139139 122 78 133 126 123 145 121 134 124 119 132 133 124 129 112 126 148 147

45. Graphing Qualitative Data Sets Pie Chart A circle is divided into sectors that represent categories. The area of each sector is proportional to the frequency of each category. Larson/Farber 4th ed. 45

46. 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) Larson/Farber 4th ed. 46 Vehicle typeKilled Cars18,440 Trucks13,778 Motorcycles4,553 Other823

47. Solution: Constructing a Pie Chart Find the relative frequency (percent) of each category. Larson/Farber 4th ed. 47 Vehicle typeFrequency, fRelative frequency Cars18,440 Trucks13,778 Motorcycles4,553 Other823 37,594

48. 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 Larson/Farber 4th ed. 48

49. Solution: Constructing a Pie Chart Larson/Farber 4th ed. 49 Vehicle typeFrequency, f Relative frequencyCentral angle Cars18,4400.49 Trucks13,7780.37 Motorcycles4,5530.12 Other8230.02 360º(0.49)≈176º 360º(0.37)≈133º 360º(0.12)≈43º 360º(0.02)≈7º

50. Solution: Constructing a Pie Chart Larson/Farber 4th ed. 50 Vehicle type Relative frequency Central angle Cars0.49176º Trucks0.37133º Motorcycles0.1243º Other0.027º From the pie chart, you can see that most fatalities in motor vehicle crashes were those involving the occupants of cars.

51. 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. Larson/Farber 4th ed. 51 Categories Frequency

52. 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) Larson/Farber 4th ed. 52

53. Solution: Constructing a Pareto Chart Larson/Farber 4th ed. 53 Cause$ (million) Admin. error7.8 Employee theft 15.6 Shoplifting14.7 Vendor fraud2.9 From the graph, it is easy to see that the causes of inventory shrinkage that should be addressed first are employee theft and shoplifting.

54. 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. 54 x y

55. 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) Larson/Farber 4th ed. 55

56. Example: Interpreting a Scatter Plot As the petal length increases, what tends to happen to the petal width? Larson/Farber 4th ed. 56 Each point in the scatter plot represents the petal length and petal width of one flower.

57. Solution: Interpreting a Scatter Plot Larson/Farber 4th ed. 57 Interpretation From the scatter plot, you can see that as the petal length increases, the petal width also tends to increase.

58. 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. Larson/Farber 4th ed. 58 time Quantitative data

59. Example: Constructing a Time Series Chart The table lists the number of cellular telephone subscribers (in millions) for the years 1995 through 2005. Construct a time series chart for the number of cellular subscribers. (Source: Cellular Telecommunication & Internet Association) Larson/Farber 4th ed. 59

60. 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. Larson/Farber 4th ed. 60

61. Solution: Constructing a Time Series Chart Larson/Farber 4th ed. 61 The graph shows that the number of subscribers has been increasing since 1995, with greater increases recently.

62. 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 Larson/Farber 4th ed. 62

63. Section 2.3 Measures of Central Tendency Larson/Farber 4th ed. 63

64. 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 Larson/Farber 4th ed. 64

65. Measures of Central Tendency Measure of central tendency A value that represents a typical, or central, entry of a data set. Most common measures of central tendency:  Mean  Median  Mode Larson/Farber 4th ed. 65

66. 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: Larson/Farber 4th ed. 66

67. 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 Larson/Farber 4th ed. 67

68. Solution: Finding a Sample Mean 872 432 397 427 388 782 397 Larson/Farber 4th ed. 68 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 The mean price of the flights is about $527.90.

69. 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. Larson/Farber 4th ed. 69

70. 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 Larson/Farber 4th ed. 70

71. Solution: Finding the Median 872 432 397 427 388 782 397 Larson/Farber 4th ed. 71 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.

72. 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 Larson/Farber 4th ed. 72

73. Solution: Finding the Median 872 397 427 388 782 397 Larson/Farber 4th ed. 73 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.

74. Measure of Central Tendency: Mode Mode The data entry that occurs with the greatest frequency. If no entry is repeated the data set has no mode. If two entries occur with the same greatest frequency, each entry is a mode (bimodal). Larson/Farber 4th ed. 74

75. 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 Larson/Farber 4th ed. 75

76. Solution: Finding the Mode 872 432 397 427 388 782 397 Larson/Farber 4th ed. 76 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.

77. 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? Larson/Farber 4th ed. 77 Political PartyFrequency, f Democrat34 Republican56 Other21 Did not respond9

78. Solution: Finding the Mode Larson/Farber 4th ed. 78 Political PartyFrequency, f Democrat34 Republican56 Other21 Did not respond9 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.

79. Comparing the Mean, Median, and Mode All three measures describe a typical entry of a data set. Advantage of using the mean:  The mean is a reliable measure because it takes into account every entry of a data set. Disadvantage of using the mean:  Greatly affected by outliers (a data entry that is far removed from the other entries in the data set). Larson/Farber 4th ed. 79

80. 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? Larson/Farber 4th ed. 80 Ages in a class 20 21 22 23 24 65

81. Solution: Comparing the Mean, Median, and Mode Larson/Farber 4th ed. 81 Mean: Median: 20 years (the entry occurring with the greatest frequency) Ages in a class 20 21 22 23 24 65 Mode:

82. Solution: Comparing the Mean, Median, and Mode Larson/Farber 4th ed. 82 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.

83. Solution: Comparing the Mean, Median, and Mode Larson/Farber 4th ed. 83 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.

84. Weighted Mean The mean of a data set whose entries have varying weights. where w is the weight of each entry x Larson/Farber 4th ed. 84

85. 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? Larson/Farber 4th ed. 85

86. Solution: Finding a Weighted Mean Larson/Farber 4th ed. 86 SourceScore, xWeight, wx∙w Test Mean860.5086(0.50)= 43.0 Midterm960.1596(0.15) = 14.4 Final Exam820.2082(0.20) = 16.4 Computer Lab980.1098(0.10) = 9.8 Homework1000.05100(0.05) = 5.0 Σw = 1Σ(x∙w) = 88.6 Your weighted mean for the course is 88.6. You did not get an A.

87. Mean of Grouped Data Mean of a Frequency Distribution Approximated by where x and f are the midpoints and frequencies of a class, respectively Larson/Farber 4th ed. 87

88. Finding the Mean of a Frequency Distribution In Words In Symbols Larson/Farber 4th ed. 88 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.

89. Example: Find the Mean of a Frequency Distribution Use the frequency distribution to approximate the mean number of minutes that a sample of Internet subscribers spent online during their most recent session. Larson/Farber 4th ed. 89 ClassMidpointFrequency, f 7 – 1812.56 19 – 3024.510 31 – 4236.513 43 – 5448.58 55 – 6660.55 67 – 7872.56 79 – 9084.52

90. Solution: Find the Mean of a Frequency Distribution Larson/Farber 4th ed. 90 ClassMidpoint, xFrequency, f(x∙f) 7 – 1812.5612.5∙6 = 75.0 19 – 3024.51024.5∙10 = 245.0 31 – 4236.51336.5∙13 = 474.5 43 – 5448.5848.5∙8 = 388.0 55 – 6660.5560.5∙5 = 302.5 67 – 7872.5672.5∙6 = 435.0 79 – 9084.5284.5∙2 = 169.0 n = 50Σ(x∙f) = 2089.0

91. The Shape of Distributions Larson/Farber 4th ed. 91 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.

92. The Shape of Distributions Larson/Farber 4th ed. 92 Uniform Distribution (rectangular) All entries or classes in the distribution have equal or approximately equal frequencies. Symmetric.

93. The Shape of Distributions Larson/Farber 4th ed. 93 Skewed Left Distribution (negatively skewed) The “tail” of the graph elongates more to the left. The mean is to the left of the median.

94. The Shape of Distributions Larson/Farber 4th ed. 94 Skewed Right Distribution (positively skewed) The “tail” of the graph elongates more to the right. The mean is to the right of the median.

95. Section 2.3 Summary Determined the mean, median, and mode of a population and of a sample Determined the weighted mean of a data set and the mean of a frequency distribution Described the shape of a distribution as symmetric, uniform, or skewed and compared the mean and median for each Larson/Farber 4th ed. 95

96. Section 2.4 Measures of Variation Larson/Farber 4th ed. 96

97. 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 Larson/Farber 4th ed. 97

98. 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) Larson/Farber 4th ed. 98

99. 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 Larson/Farber 4th ed. 99

100. 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. Larson/Farber 4th ed. 100 minimum maximum

101. 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 Larson/Farber 4th ed. 101

102. 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 Larson/Farber 4th ed. 102 Solution: First determine the mean starting salary.

103. Solution: Finding the Deviation Larson/Farber 4th ed. 103 Determine the deviation for each data entry. Salary ($1000s), xDeviation: x – μ 4141 – 41.5 = –0.5 3838 – 41.5 = –3.5 3939 – 41.5 = –2.5 4545 – 41.5 = 3.5 4747 – 41.5 = 5.5 4141 – 41.5 = –0.5 4444 – 41.5 = 2.5 4141 – 41.5 = –0.5 3737 – 41.5 = –4.5 4242 – 41.5 = 0.5 Σx = 415 Σ(x – μ) = 0

104. Deviation, Variance, and Standard Deviation Population Variance Population Standard Deviation Larson/Farber 4th ed. 104 Sum of squares, SS x

105. Finding the Population Variance & Standard Deviation In Words In Symbols Larson/Farber 4th ed. 105 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 – μ (x – μ) 2 SS x = Σ(x – μ) 2

106. Finding the Population Variance & Standard Deviation Larson/Farber 4th ed. 106 5.Divide by N to get the population variance. 6.Find the square root to get the population standard deviation. In Words In Symbols

107. 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. Larson/Farber 4th ed. 107

108. Solution: Finding the Population Standard Deviation Larson/Farber 4th ed. 108 Determine SS x N = 10 Salary, xDeviation: x – μSquares: (x – μ) 2 4141 – 41.5 = –0.5(–0.5) 2 = 0.25 3838 – 41.5 = –3.5(–3.5) 2 = 12.25 3939 – 41.5 = –2.5(–2.5) 2 = 6.25 4545 – 41.5 = 3.5(3.5) 2 = 12.25 4747 – 41.5 = 5.5(5.5) 2 = 30.25 4141 – 41.5 = –0.5(–0.5) 2 = 0.25 4444 – 41.5 = 2.5(2.5) 2 = 6.25 4141 – 41.5 = –0.5(–0.5) 2 = 0.25 3737 – 41.5 = –4.5(–4.5) 2 = 20.25 4242 – 41.5 = 0.5(0.5) 2 = 0.25 Σ(x – μ) = 0 SS x = 88.5

109. Solution: Finding the Population Standard Deviation Larson/Farber 4th ed. 109 Population Variance Population Standard Deviation The population standard deviation is about 3.0, or $3000.

110. Deviation, Variance, and Standard Deviation Sample Variance Sample Standard Deviation Larson/Farber 4th ed. 110

111. Finding the Sample Variance & Standard Deviation In Words In Symbols Larson/Farber 4th ed. 111 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.

112. Finding the Sample Variance & Standard Deviation Larson/Farber 4th ed. 112 5.Divide by n – 1 to get the sample variance. 6.Find the square root to get the sample standard deviation. In Words In Symbols

113. 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 Larson/Farber 4th ed. 113

114. Solution: Finding the Sample Standard Deviation Larson/Farber 4th ed. 114 Determine SS x n = 10 Salary, xDeviation: x – μSquares: (x – μ) 2 4141 – 41.5 = –0.5(–0.5) 2 = 0.25 3838 – 41.5 = –3.5(–3.5) 2 = 12.25 3939 – 41.5 = –2.5(–2.5) 2 = 6.25 4545 – 41.5 = 3.5(3.5) 2 = 12.25 4747 – 41.5 = 5.5(5.5) 2 = 30.25 4141 – 41.5 = –0.5(–0.5) 2 = 0.25 4444 – 41.5 = 2.5(2.5) 2 = 6.25 4141 – 41.5 = –0.5(–0.5) 2 = 0.25 3737 – 41.5 = –4.5(–4.5) 2 = 20.25 4242 – 41.5 = 0.5(0.5) 2 = 0.25 Σ(x – μ) = 0 SS x = 88.5

115. Solution: Finding the Sample Standard Deviation Larson/Farber 4th ed. 115 Sample Variance Sample Standard Deviation The sample standard deviation is about 3.1, or $3100.

116. 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.) Larson/Farber 4th ed. 116 Office Rental Rates 35.0033.5037.00 23.7526.5031.25 36.5040.0032.00 39.2537.5034.75 37.7537.2536.75 27.0035.7526.00 37.0029.0040.50 24.5033.0038.00

117. Solution: Using Technology to Find the Standard Deviation Larson/Farber 4th ed. 117 Sample Mean Sample Standard Deviation

118. 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. Larson/Farber 4th ed. 118

119. 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: Larson/Farber 4th ed. 119 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.

120. Interpreting Standard Deviation: Empirical Rule (68 – 95 – 99.7 Rule) Larson/Farber 4th ed. 120 68% within 1 standard deviation 34% 99.7% within 3 standard deviations 2.35% 95% within 2 standard deviations 13.5%

121. Example: Using the Empirical Rule In a survey conducted by the National Center for Health Statistics, the sample mean height of women in the United States (ages 20-29) was 64 inches, with a sample standard deviation of 2.71 inches. Estimate the percent of the women whose heights are between 64 inches and 69.42 inches. Larson/Farber 4th ed. 121

122. Solution: Using the Empirical Rule Larson/Farber 4th ed. 122 55.8758.5861.296466.7169.4272.13 34% 13.5% Because the distribution is bell-shaped, you can use the Empirical Rule. 34% + 13.5% = 47.5% of women are between 64 and 69.42 inches tall.

123. Chebychev’s Theorem The portion of any data set lying within k standard deviations (k > 1) of the mean is at least: Larson/Farber 4th ed. 123 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.

124. 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? Larson/Farber 4th ed. 124

125. Solution: Using Chebychev’s Theorem k = 2: μ – 2σ = 39.2 – 2(24.8) = -10.4 (use 0 since age can’t be negative) μ + 2σ = 39.2 + 2(24.8) = 88.8 Larson/Farber 4th ed. 125 At least 75% of the population of Florida is between 0 and 88.8 years old.

126. 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. Larson/Farber 4th ed. 126 where n= Σf (the number of entries in the data set)

127. Example: Finding the Standard Deviation for Grouped Data Larson/Farber 4th ed. 127 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 13111 12210 11000 15036 30311 11601 36612 23011 41122 03024

128. xfxf 0100(10) = 0 1191(19) = 19 272(7) = 14 373(7) =21 424(2) = 8 515(1) = 5 646(4) = 24 Solution: Finding the Standard Deviation for Grouped Data First construct a frequency distribution. Find the mean of the frequency distribution. Larson/Farber 4th ed. 128 Σf = 50 Σ(xf )= 91 The sample mean is about 1.8 children.

129. Solution: Finding the Standard Deviation for Grouped Data Determine the sum of squares. Larson/Farber 4th ed. 129 xf 0100 – 1.8 = –1.8(–1.8) 2 = 3.243.24(10) = 32.40 1191 – 1.8 = –0.8(–0.8) 2 = 0.640.64(19) = 12.16 272 – 1.8 = 0.2(0.2) 2 = 0.040.04(7) = 0.28 373 – 1.8 = 1.2(1.2) 2 = 1.441.44(7) = 10.08 424 – 1.8 = 2.2(2.2) 2 = 4.844.84(2) = 9.68 515 – 1.8 = 3.2(3.2) 2 = 10.2410.24(1) = 10.24 646 – 1.8 = 4.2(4.2) 2 = 17.6417.64(4) = 70.56

130. Solution: Finding the Standard Deviation for Grouped Data Find the sample standard deviation. Larson/Farber 4th ed. 130 The standard deviation is about 1.7 children.

131. 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 Larson/Farber 4th ed. 131

132. Section 2.5 Measures of Position Larson/Farber 4th ed. 132

133. 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) Larson/Farber 4th ed. 133

134. 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, Q 1 : About one quarter of the data fall on or below Q 1.  Second quartile, Q 2 : About one half of the data fall on or below Q 2 (median).  Third quartile, Q 3 : About three quarters of the data fall on or below Q 3. Larson/Farber 4th ed. 134

135. 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 Larson/Farber 4th ed. 135 Solution: Q 2 divides the data set into two halves. 5 7 9 10 11 13 14 15 16 17 18 18 20 21 37 Q2Q2 Lower half Upper half

136. 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 Larson/Farber 4th ed. 136 Q2Q2 Lower half Upper half Q1Q1 Q3Q3 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.

137. Interquartile Range Interquartile Range (IQR) The difference between the third and first quartiles. IQR = Q 3 – Q 1 Larson/Farber 4th ed. 137

138. Example: Finding the Interquartile Range Find the interquartile range of the test scores. Recall Q 1 = 10, Q 2 = 15, and Q 3 = 18 Larson/Farber 4th ed. 138 Solution: IQR = Q 3 – Q 1 = 18 – 10 = 8 The test scores in the middle portion of the data set vary by at most 8 points.

139. 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 Q 1  Median Q 2  Third quartile Q 3  Maximum entry Larson/Farber 4th ed. 139

140. 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 Q 1 to Q 3 and draw a vertical line in the box at Q 2. 5.Draw whiskers from the box to the minimum and maximum entries. Larson/Farber 4th ed. 140 Whisker Maximum entry Minimum entry Box Median, Q 2 Q3Q3 Q1Q1

141. Example: Drawing a Box-and-Whisker Plot Draw a box-and-whisker plot that represents the 15 test scores. Recall Min = 5 Q 1 = 10 Q 2 = 15 Q 3 = 18 Max = 37 Larson/Farber 4th ed. 141 510151837 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.

142. Percentiles and Other Fractiles FractilesSummarySymbols QuartilesDivides data into 4 equal parts Q 1, Q 2, Q 3 DecilesDivides data into 10 equal parts D 1, D 2, D 3,…, D 9 PercentilesDivides data into 100 equal parts P 1, P 2, P 3,…, P 99 Larson/Farber 4th ed. 142

143. 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 72 nd percentile? How should you interpret this? (Source: College Board Online) Larson/Farber 4th ed. 143

144. Solution: Interpreting Percentiles The 72 nd percentile corresponds to a test score of 1700. This means that 72% of the students had an SAT score of 1700 or less. Larson/Farber 4th ed. 144

145. The Standard Score Standard Score (z-score) Represents the number of standard deviations a given value x falls from the mean μ. Larson/Farber 4th ed. 145

146. Example: Comparing z-Scores from Different Data Sets In 2007, Forest Whitaker won the Best Actor Oscar at age 45 for his role in the movie The Last King of Scotland. Helen Mirren won the Best Actress Oscar at age 61 for her role in The Queen. The mean age of all best actor winners is 43.7, with a standard deviation of 8.8. The mean age of all best actress winners is 36, with a standard deviation of 11.5. Find the z-score that corresponds to the age for each actor or actress. Then compare your results. Larson/Farber 4th ed. 146

147. Solution: Comparing z-Scores from Different Data Sets Larson/Farber 4th ed. 147 Forest Whitaker Helen Mirren 0.15 standard deviations above the mean 2.17 standard deviations above the mean

148. Solution: Comparing z-Scores from Different Data Sets Larson/Farber 4th ed. 148 The z-score corresponding to the age of Helen Mirren is more than two standard deviations from the mean, so it is considered unusual. Compared to other Best Actress winners, she is relatively older, whereas the age of Forest Whitaker is only slightly higher than the average age of other Best Actor winners. z = 0.15z = 2.17

149. 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) Larson/Farber 4th ed. 149