1. 1 of 96 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 Chapter 2 Descriptive Statistics

2. 2 of 96 Section 2.1 Objectives Construct frequency distributions Construct frequency histograms, relative frequency histograms, frequency polygons, and ogive Section 2.1 Frequency Distributions and Their Graphs

3. 3 of 96 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. 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

4. 4 of 96 Constructing a Frequency Distribution 1.Decide on the number of classes.  Usually between 5 and 20; otherwise, it may be difficult to detect any patterns. 2.Find the class width.  Determine the range, max – min, of the data.  Divide the range by the number of classes, k.  Round up to the next convenient/whole number. NOTE: If (max – min)/k = whole #, then let the class width = whole # + 1

5. 5 of 96 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.

6. 6 of 96 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.

7. 7 of 96 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

8. 8 of 96 Solution: Constructing a Frequency Distribution 1.Number of classes = 7 (given) 2.Find the class width Round up to 12 (Always round up to the next whole number) 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

9. 9 of 96 Solution: Constructing a Frequency Distribution 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

10. 10 of 96 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. Lower limit Upper limit 7 19 31 43 55 67 79 Class width = 12 30 42 54 66 78 90 18

11. 11 of 96 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. ClassTallyFrequency, f 7 – 18IIII I6 19 – 30IIII 10 31 – 42IIII IIII III13 43 – 54IIII III8 55 – 66IIII5 67 – 78IIII I6 79 – 90II2 Σf = 50

12. 12 of 96 Determining the Midpoint Midpoint of a class ClassMidpointFrequency, f 7 – 186 19 – 3010 31 – 4213 Class width = 12

13. 13 of 96 Determining the Relative Frequency Relative Frequency of a class Portion or percentage of the data that falls in a particular class. ClassFrequency, fRelative Frequency 7 – 186 19 – 3010 31 – 4213

14. 14 of 96 Determining the Cumulative Frequency Cumulative frequency of a class The sum of the frequency for that class and all previous classes. ClassFrequency, fCumulative frequency 7 – 186 19 – 3010 31 – 4213 + + 6 16 29

15. 15 of 96 Expanded Frequency Distribution 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

16. 16 of 96 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

17. 17 of 96 Class Boundaries Class boundaries The numbers that separate classes without forming gaps between them. 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

18. 18 of 96 Class Boundaries 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

19. 19 of 96 Example: Frequency Histogram Construct a frequency histogram for the Internet usage frequency distribution. 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

20. 20 of 96 Solution: Frequency Histogram (using class boundaries) 6.5 18.5 30.5 42.5 54.5 66.5 78.5 90.5 You can see that more than half of the subscribers spent between 19 and 54 minutes on the Internet during their most recent session.

21. 21 of 96 Graphs of Relative 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

22. 22 of 96 Example: Relative Frequency Histogram Construct a relative frequency histogram for the Internet usage frequency distribution. Class Class boundaries Frequency, f Relative frequency 7 – 18 6.5 – 18.560.12 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

23. 23 of 96 Solution: Relative Frequency Histogram 6.5 18.5 30.5 42.5 54.5 66.5 78.5 90.5 From this graph you can see that 20% of Internet subscribers spent between 18.5 minutes and 30.5 minutes online.

24. 24 of 96 Graphs of Frequency Distributions Frequency Polygon A line graph that emphasizes the continuous change in frequencies. data values frequency

25. 25 of 96 Example: Frequency Polygon Construct a frequency polygon for the Internet usage frequency distribution. ClassMidpointFrequency, f 7 – 1812.56 19 – 3024.510 31 – 4236.513 43 – 5448.58 55 – 6660.55 67 – 7872.56 79 – 9084.52

26. 26 of 96 Solution: Frequency Polygon 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. 27 of 96 Solution: Frequency Histogram and Frequency Polygon (using Midpoints) 0.596.5

28. 28 of 96 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

29. 29 of 96 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. 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).

30. 30 of 96 Example: Ogive Construct an ogive for the Internet usage frequency distribution. 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

31. 31 of 96 Solution: Ogive 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.

32. 32 of 96 Section 2.2 Objectives Graph qualitative data using pie charts Graph time series charts Section 2.2 More Graphs and Displays

33. 33 of 96 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.

34. 34 of 96 Example: Constructing a Pie Chart The numbers of motor vehicle occupants killed in crashes in 2005 are shown in the table. Use a pie chart to organize the data. (Source: U.S. Department of Transportation, National Highway Traffic Safety Administration) Vehicle typeKilled Cars18,440 Trucks13,778 Motorcycles4,553 Other823

35. 35 of 96 Solution: Constructing a Pie Chart Find the relative frequency (percent) of each category. Vehicle typeFrequency, fRelative frequency Cars18,440 Trucks13,778 Motorcycles4,553 Other823 37,594

36. 36 of 96 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

37. 37 of 96 Solution: Constructing a Pie Chart 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

38. 38 of 96 Solution: Constructing a Pie Chart 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.

39. 39 of 96 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)

40. 40 of 96 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.

41. 41 of 96 Solution: Constructing a Time Series Chart The graph shows that the number of subscribers has been increasing since 1995, with greater increases recently.

42. 42 of 96 Section 2.3 Objectives Determine the mean, median, and mode of a population and of a sample Determine the mean of grouped data Describe the shape of a distribution as symmetric, uniform, or skewed and compare the mean and median for each Section 2.3 Measures of Central Tendency

43. 43 of 96 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

44. 44 of 96 Measure of Central Tendency: Mean Mean (arithmetic average) The sum of all the data entries divided by the number of entries. Sigma notation: Σ x = add all of the data entries (x) in the data set. Population mean: (N = Population size) Sample mean: (n = Sample size)

45. 45 of 96 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

46. 46 of 96 Solution: Finding a Sample Mean 872 432 397 427 388 782 397 The sum of the flight prices is Σx = 872 + 432 + 397 + 427 + 388 + 782 + 397 = 3695 To find the mean price, divide the sum of the prices by the number of prices in the sample The mean price of the flights is about $ 527.90.

47. 47 of 96 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 average of the two middle data entries.

48. 48 of 96 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

49. 49 of 96 Solution: Finding the Median 872 432 397 427 388 782 397 First order the data. 388 397 397 427 432 782 872 There are seven entries (an odd number), the median is the middle, or fourth, data entry. The median price of the flights is $ 427.

50. 50 of 96 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

51. 51 of 96 Solution: Finding the Median 872 397 427 388 782 397 First order the data. 388 397 397 427 782 872 There are six entries (an even number), the median is the mean of the two middle entries. The median price of the flights is $ 412.

52. 52 of 96 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).

53. 53 of 96 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

54. 54 of 96 Solution: Finding the Mode 872 432 397 427 388 782 397 Ordering the data helps to find the mode. 388 397 397 427 432 782 872 The entry of 397 occurs twice, whereas the other data entries occur only once. The mode of the flight prices is $ 397.

55. 55 of 96 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 PartyFrequency, f Democrat34 Republican56 Other21 Did not respond9

56. 56 of 96 Solution: Finding the Mode 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.

57. 57 of 96 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), i.e, extreme values.

58. 58 of 96 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

59. 59 of 96 Solution: Comparing the Mean, Median, and Mode Mean: Median: 20 years (the entry occurring with the greatest frequency) Ages in a class 20 21 22 23 24 65 Mode:

60. 60 of 96 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.

61. 61 of 96 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.

62. 62 of 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. n is the sample size.

63. 63 of 96 Finding the Mean of a Frequency Distribution In Words In Symbols 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.

64. 64 of 96 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. ClassMidpointFrequency, f 7 – 1812.56 19 – 3024.510 31 – 4236.513 43 – 5448.58 55 – 6660.55 67 – 7872.56 79 – 9084.52

65. 65 of 96 Solution: Find the Mean of a Frequency Distribution 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

66. 66 of 96 In a common system for finding a grade-point average (GPA), an A grade is assigned 4 points, with 3 points for a B, 2 for C, 1 for D, and 0 for F. Find the grade-point average by multiplying the number of units/credits for a course and the number assigned to each grade, and then adding these products. Finally, divide this sum by the total number of units/credits. This calculation of a grade- point average in an example of a weighted mean. Example: Grade-Point Average (GPA)

67. 67 of 96 Find the grade-point average (weighted mean) for the grades below. CourseGrade Points, xUnits (Credits), f Math4 (A)5 History3 (B)3 Health4 (A)2 Art2 (C)2 Example: Grade-Point Average

68. 68 of 96 Solution CourseGrade Pts, xUnits (Credits), f(Grade pts)*(units) Math4 (A)520 History3 (B)39 Health4 (A)28 Art2 (C)24 Grade-point average = Example: Grade Point Average

69. 69 of 96 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.

70. 70 of 96 The Shape of Distributions Uniform Distribution (rectangular) All entries or classes in the distribution have equal or approximately equal frequencies. Symmetric.

71. 71 of 96 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.

72. 72 of 96 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.

73. 73 of 96 Section 2.4 Objectives Determine the range of a data set Determine the variance and standard deviation of a population and of a sample Calculate the sample standard deviation for grouped data Section 2.4 Measures of Variation

74. 74 of 96 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)

75. 75 of 96 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 (thousand dollars) 41 38 39 45 47 41 44 41 37 42

76. 76 of 96 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 $ 10,000. minimum maximum

77. 77 of 96 Deviation, Variance, and Standard Deviation Deviation The difference between the data entry, x, and the mean of the data set. Population data set: (μ = mu = population mean)  Deviation of x = x – μ Sample data set:  Deviation of x = x – x

78. 78 of 96 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 (thousand dollars) 41 38 39 45 47 41 44 41 37 42 Solution: First determine the mean starting salary.

79. 79 of 96 Solution: Finding the Deviation 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

80. 80 of 96 Deviation, Variance, and Standard Deviation Population Variance Population Standard Deviation Sum of Squares, SS x

81. 81 of 96 Finding the Population Variance & Standard Deviation In Words In Symbols 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

82. 82 of 96 Finding the Population Variance & Standard Deviation 5.Divide by N to get the population variance. 6.Find the square root to get the population standard deviation. In Words In Symbols

83. 83 of 96 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.

84. 84 of 96 Solution: Finding the Population Standard Deviation 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

85. 85 of 96 Solution: Finding the Population Standard Deviation Population Variance Population Standard Deviation The population standard deviation is about $ 3,000.

86. 86 of 96 Deviation, Variance, and Standard Deviation Sample Variance Sample Standard Deviation

87. 87 of 96 Finding the Sample Variance & Standard Deviation In Words In Symbols 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.

88. 88 of 96 Finding the Sample Variance & Standard Deviation 5.Divide by n – 1 to get the sample variance. 6.Find the square root to get the sample standard deviation. In Words In Symbols

89. 89 of 96 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 (thousand dollars) 41 38 39 45 47 41 44 41 37 42

90. 90 of 96 Solution: Finding the Sample Standard Deviation 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

91. 91 of 96 Solution: Finding the Sample Standard Deviation Sample Variance Sample Standard Deviation The sample standard deviation is about 3.1, or $ 3,100.

92. 92 of 96 Standard Deviation for Grouped Data Sample standard deviation for a frequency distribution When a frequency distribution has classes, estimate the sample mean and standard deviation by using the midpoint of each class. where n= Σf (the number of entries in the data set)

93. 93 of 96 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 13111 12210 11000 15036 30311 11601 36612 23011 41122 03024

94. 94 of 96 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. Σf = 50 Σ(xf )= 91 The sample mean is about 1.8 children.

95. 95 of 96 Solution: Finding the Standard Deviation for Grouped Data Determine the sum of squares. 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

96. 96 of 96 Solution: Finding the Standard Deviation for Grouped Data Find the sample standard deviation. The standard deviation is about 1.7 children.