Classes

3,284 views

Published on

Published in: Technology, Business
0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total views
3,284
On SlideShare
0
From Embeds
0
Number of Embeds
4
Actions
Shares
0
Downloads
66
Comments
0
Likes
0
Embeds 0
No embeds

No notes for slide
  • 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.
  • Classes

    1. 1. Chapter 2 Descriptive Statistics 1Larson/Farber 4th ed.
    2. 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 2Larson/Farber 4th ed.
    3. 3. Section 2.1 Frequency Distributions and Their Graphs 3Larson/Farber 4th ed.
    4. 4. Section 2.1 Objectives • Construct frequency distributions • Construct frequency histograms, frequency polygons, relative frequency histograms, and ogives Larson/Farber 4th ed. 4
    5. 5. Frequency Distribution Frequency Distribution • A table that shows classes or intervals of data with a count of the number of entries in each class. • The frequency, f, of a class is the number of data entries in the class. Larson/Farber 4th ed. 5 Class Frequency, f 1 – 5 5 6 – 10 8 11 – 15 6 16 – 20 8 21 – 25 5 26 – 30 4 Lower class limits Upper class limits Class width 6 – 1 = 5
    6. 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. 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. 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. 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. 10. Solution: Constructing a Frequency Distribution 1. Number of classes = 7 (given) 2. Find the class width Larson/Farber 4th ed. 10 max min 86 7 11.29 #classes 7 Round up to 12 50 40 41 17 11 7 22 44 28 21 19 23 37 51 54 42 86 41 78 56 72 56 17 7 69 30 80 56 29 33 46 31 39 20 18 29 34 59 73 77 36 39 30 62 54 67 39 31 53 44
    11. 11. Solution: Constructing a Frequency Distribution Larson/Farber 4th ed. 11 Lower limit Upper limit 7Class width = 12 3. Use 7 (minimum value) as first lower limit. Add the class width of 12 to get the lower limit of the next class. 7 + 12 = 19 Find the remaining lower limits. 19 31 43 55 67 79
    12. 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. 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 Class Tally Frequency, f 7 – 18 IIII I 6 19 – 30 IIII IIII 10 31 – 42 IIII IIII III 13 43 – 54 IIII III 8 55 – 66 IIII 5 67 – 78 IIII I 6 79 – 90 II 2 Σf = 50
    14. 14. Determining the Midpoint Midpoint of a class Larson/Farber 4th ed. 14 (Lower class limit) (Upper class limit) 2 Class Midpoint Frequency, f 7 – 18 6 19 – 30 10 31 – 42 13 7 18 12.5 2 19 30 24.5 2 31 42 36.5 2 Class width = 12
    15. 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 n f sizeSample frequencyclass frequencyrelative Class Frequency, f Relative Frequency 7 – 18 6 19 – 30 10 31 – 42 13 6 0.12 50 10 0.20 50 13 0.26 50 •
    16. 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 Class Frequency, f Cumulative frequency 7 – 18 6 19 – 30 10 31 – 42 13 + + 6 16 29
    17. 17. Expanded Frequency Distribution Larson/Farber 4th ed. 17 Class Frequency, f Midpoint Relative frequency Cumulative frequency 7 – 18 6 12.5 0.12 6 19 – 30 10 24.5 0.20 16 31 – 42 13 36.5 0.26 29 43 – 54 8 48.5 0.16 37 55 – 66 5 60.5 0.10 42 67 – 78 6 72.5 0.12 48 79 – 90 2 84.5 0.04 50 Σf = 50 1 n f
    18. 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 valuesfrequency
    19. 19. Class Boundaries Class boundaries • The numbers that separate classes without forming gaps between them. Larson/Farber 4th ed. 19 Class Class Boundaries Frequency, f 7 – 18 6 19 – 30 10 31 – 42 13 • The distance from the upper limit of the first class to the lower limit of the second class is 19 – 18 = 1. • Half this distance is 0.5. • First class lower boundary = 7 – 0.5 = 6.5 • First class upper boundary = 18 + 0.5 = 18.5 6.5 – 18.5
    20. 20. Class Boundaries Larson/Farber 4th ed. 20 Class Class boundaries Frequency, f 7 – 18 6.5 – 18.5 6 19 – 30 18.5 – 30.5 10 31 – 42 30.5 – 42.5 13 43 – 54 42.5 – 54.5 8 55 – 66 54.5 – 66.5 5 67 – 78 66.5 – 78.5 6 79 – 90 78.5 – 90.5 2
    21. 21. Example: Frequency Histogram Construct a frequency histogram for the Internet usage frequency distribution. Larson/Farber 4th ed. 21 Class Class boundaries Midpoint Frequency, f 7 – 18 6.5 – 18.5 12.5 6 19 – 30 18.5 – 30.5 24.5 10 31 – 42 30.5 – 42.5 36.5 13 43 – 54 42.5 – 54.5 48.5 8 55 – 66 54.5 – 66.5 60.5 5 67 – 78 66.5 – 78.5 72.5 6 79 – 90 78.5 – 90.5 84.5 2
    22. 22. Solution: Frequency Histogram (using Midpoints) Larson/Farber 4th ed. 22
    23. 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. 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. 25. Example: Frequency Polygon Construct a frequency polygon for the Internet usage frequency distribution. Larson/Farber 4th ed. 25 Class Midpoint Frequency, f 7 – 18 12.5 6 19 – 30 24.5 10 31 – 42 36.5 13 43 – 54 48.5 8 55 – 66 60.5 5 67 – 78 72.5 6 79 – 90 84.5 2
    26. 26. Solution: Frequency Polygon 0 2 4 6 8 10 12 14 0.5 12.5 24.5 36.5 48.5 60.5 72.5 84.5 96.5 Frequency Time online (in minutes) Internet Usage 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. 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 valuesrelative frequency
    28. 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 – 18.5 6 0.12 19 – 30 18.5 – 30.5 10 0.20 31 – 42 30.5 – 42.5 13 0.26 43 – 54 42.5 – 54.5 8 0.16 55 – 66 54.5 – 66.5 5 0.10 67 – 78 66.5 – 78.5 6 0.12 79 – 90 78.5 – 90.5 2 0.04
    29. 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. 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. 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. 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. 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 – 18.5 6 6 19 – 30 18.5 – 30.5 10 16 31 – 42 30.5 – 42.5 13 29 43 – 54 42.5 – 54.5 8 37 55 – 66 54.5 – 66.5 5 42 67 – 78 66.5 – 78.5 6 48 79 – 90 78.5 – 90.5 2 50
    34. 34. Solution: Ogive 0 10 20 30 40 50 60 Cumulativefrequency Time online (in minutes) InternetUsage 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. 35. Textbook Exercises. Pages 49 - 53 Larson/Farber 4th ed. 35 # 14 a. Class Width: Difference between two consecutive lower / upper class limits. So the class width in this case is 10 – 0 = 10 b. Class Midpoints: To find the class midpoints we add the lower and upper limits of the first class and divide the sum by 2. We then simply add the class width to the midpoint of the first class to obtain the second one and so on. So for this problem the midpoint of the first class is (0 + 9) 2 = 4.5, midpoint of the second class is 4.5 + 10 (class width) = 14.5 and thus the rest of the midpoints are 24.5, 34.5, 44.5, 54.5 and 64.5 c. Class Boundaries: To obtain class boundaries we first find the distance between the first upper class limit and the second lower class limit. 10- 9=1. Half this distance is 0.5. So the first lower class boundary is 0 - 0.5 = -0.5 and the first upper class boundary is 9 + 0.5 = 9.5 To obtain the remaining of lower and upper class boundaries simply add class width to the previous lower class boundary and the previous upper class boundary
    36. 36. Textbook Exercises Pages 49-53 Larson/Farber 4th ed. 36 # 24 a. The class with the greatest relative frequency is the third class having the midpoint of 19.5. And the class with the least relative frequency is the last class having the midpoint of 21.5 b. The greatest relative frequency is approximately 40% and the least relative frequency is approximately 2% c. The relative frequency of the second class is approximately 33%
    37. 37. Textbook Exercises. Pages 49-53 Larson/Farber 4th ed. 37 classes Class Boundaries Midpoints frequency Relative frequency Cumulative frequency 30 – 113 29.5 – 113.5 71.5 5 5/29 = 0.172 5 114 – 197 113.5 – 197.5 155.5 7 7/29 = 0.241 12 198 – 281 197.5 – 281.5 239.5 8 8/29 = 0.276 20 282 – 365 281.5 – 365.5 323.5 2 2/29 = 0.069 22 366 – 449 365.5 – 449.5 407.5 3 3/29 = 0.103 25 450 – 533 449.5 – 533.5 491.5 4 4/29 = 0.138 29 Total = 29 Total = 1 # 28 Book Spending We will create the extended frequency distribution table
    38. 38. Section 2.1 Summary • Constructed frequency distributions • Constructed frequency histograms, frequency polygons, relative frequency histograms and ogives Larson/Farber 4th ed. 38
    39. 39. Section 2.2 More Graphs and Displays Larson/Farber 4th ed. 39
    40. 40. 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. 40
    41. 41. 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. 41 Data: 21, 25, 25, 26, 27, 28, 30, 36, 36, 45 26 2 1 5 5 6 7 8 3 0 6 6 4 5
    42. 42. 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. 42 155 159 144 129 105 145 126 116 130 114 122 112 112 142 126 118 118 108 122 121 109 140 126 119 113 117 118 109 109 119 139 139 122 78 133 126 123 145 121 134 124 119 132 133 124 129 112 126 148 147
    43. 43. Solution: Constructing a Stem-and-Leaf Plot Larson/Farber 4th ed. 43 • The data entries go from a low of 78 to a high of 159. • Use the rightmost digit as the leaf.  For instance, 78 = 7 | 8 and 159 = 15 | 9 • List the stems, 7 to 15, to the left of a vertical line. • For each data entry, list a leaf to the right of its stem. 155 159 144 129 105 145 126 116 130 114 122 112 112 142 126 118 118 108 122 121 109 140 126 119 113 117 118 109 109 119 139 139 122 78 133 126 123 145 121 134 124 119 132 133 124 129 112 126 148 147
    44. 44. Solution: Constructing a Stem-and-Leaf Plot Larson/Farber 4th ed. 44 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.
    45. 45. Graphing Quantitative Data Sets Dot plot • Each data entry is plotted, using a point, above a horizontal axis Larson/Farber 4th ed. 45 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
    46. 46. Example: Constructing a Dot Plot Use a dot plot organize the text messaging data. Larson/Farber 4th ed. 46 • So that each data entry is included in the dot plot, the horizontal axis should include numbers between 70 and 160. • To represent a data entry, plot a point above the entry's position on the axis. • If an entry is repeated, plot another point above the previous point. 155 159 144 129 105 145 126 116 130 114 122 112 112 142 126 118 118 108 122 121 109 140 126 119 113 117 118 109 109 119 139 139 122 78 133 126 123 145 121 134 124 119 132 133 124 129 112 126 148 147
    47. 47. Solution: Constructing a Dot Plot Larson/Farber 4th ed. 47 From the dot plot, you can see that most values cluster between 105 and 148 and the value that occurs the most is 126. You can also see that 78 is an unusual data value. 155 159 144 129 105 145 126 116 130 114 122 112 112 142 126 118 118 108 122 121 109 140 126 119 113 117 118 109 109 119 139 139 122 78 133 126 123 145 121 134 124 119 132 133 124 129 112 126 148 147
    48. 48. 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. 48
    49. 49. 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. 49 Vehicle type Killed Cars 18,440 Trucks 13,778 Motorcycles 4,553 Other 823
    50. 50. Solution: Constructing a Pie Chart • Find the relative frequency (percent) of each category. Larson/Farber 4th ed. 50 Vehicle type Frequency, f Relative frequency Cars 18,440 Trucks 13,778 Motorcycles 4,553 Other 823 37,594 18440 0.49 37594 13778 0.37 37594 4553 0.12 37594 823 0.02 37594
    51. 51. 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. 51
    52. 52. Solution: Constructing a Pie Chart Larson/Farber 4th ed. 52 Vehicle type Frequency, f Relative frequency Central angle Cars 18,440 0.49 Trucks 13,778 0.37 Motorcycles 4,553 0.12 Other 823 0.02 360º(0.49)≈176º 360º(0.37)≈133º 360º(0.12)≈43º 360º(0.02)≈7º
    53. 53. Solution: Constructing a Pie Chart Larson/Farber 4th ed. 53 Vehicle type Relative frequency Central angle Cars 0.49 176º Trucks 0.37 133º Motorcycles 0.12 43º Other 0.02 7º From the pie chart, you can see that most fatalities in motor vehicle crashes were those involving the occupants of cars.
    54. 54. 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. 54 Categories Frequency
    55. 55. 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. 55
    56. 56. Solution: Constructing a Pareto Chart Larson/Farber 4th ed. 56 Cause $ (million) Admin. error 7.8 Employee theft 15.6 Shoplifting 14.7 Vendor fraud 2.9 From the graph, it is easy to see that the causes of inventory shrinkage that should be addressed first are employee theft and shoplifting.
    57. 57. 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. 57 x y
    58. 58. 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. 58
    59. 59. Example: Interpreting a Scatter Plot As the petal length increases, what tends to happen to the petal width? Larson/Farber 4th ed. 59 Each point in the scatter plot represents the petal length and petal width of one flower.
    60. 60. Solution: Interpreting a Scatter Plot Larson/Farber 4th ed. 60 Interpretation From the scatter plot, you can see that as the petal length increases, the petal width also tends to increase.
    61. 61. 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. 61 time Quantitative data
    62. 62. 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. 62
    63. 63. 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. 63
    64. 64. Solution: Constructing a Time Series Chart Larson/Farber 4th ed. 64 The graph shows that the number of subscribers has been increasing since 1995, with greater increases recently.
    65. 65. 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. 65
    66. 66. Section 2.3 Measures of Central Tendency Larson/Farber 4th ed. 66
    67. 67. 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. 67
    68. 68. 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. 68
    69. 69. 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. 69 x N x x n
    70. 70. 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. 70
    71. 71. Solution: Finding a Sample Mean 872 432 397 427 388 782 397 Larson/Farber 4th ed. 71 • 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 3695 527.9 7 x x n The mean price of the flights is about $527.90.
    72. 72. 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. 72
    73. 73. 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. 73
    74. 74. Solution: Finding the Median 872 432 397 427 388 782 397 Larson/Farber 4th ed. 74 • 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.
    75. 75. 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. 75
    76. 76. Solution: Finding the Median 872 397 427 388 782 397 Larson/Farber 4th ed. 76 • First order the data. 388 397 397 427 782 872 • There are six entries (an even number), the median is the mean of the two middle entries. The median price of the flights is $412. 397 427 Median 412 2
    77. 77. 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. 77
    78. 78. 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. 78
    79. 79. Solution: Finding the Mode 872 432 397 427 388 782 397 Larson/Farber 4th ed. 79 • 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.
    80. 80. 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. 80 Political Party Frequency, f Democrat 34 Republican 56 Other 21 Did not respond 9
    81. 81. Solution: Finding the Mode Larson/Farber 4th ed. 81 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.
    82. 82. 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. 82
    83. 83. 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. 83 Ages in a class 20 20 20 20 20 20 21 21 21 21 22 22 22 23 23 23 23 24 24 65
    84. 84. Solution: Comparing the Mean, Median, and Mode Larson/Farber 4th ed. 84 Mean: 20 20 ... 24 65 23.8 years 20 x x n Median: 21 22 21.5 years 2 20 years (the entry occurring with the greatest frequency) Ages in a class 20 20 20 20 20 20 21 21 21 21 22 22 22 23 23 23 23 24 24 65 Mode:
    85. 85. Solution: Comparing the Mean, Median, and Mode Larson/Farber 4th ed. 85 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.
    86. 86. Solution: Comparing the Mean, Median, and Mode Larson/Farber 4th ed. 86 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.
    87. 87. Textbook Exercises. Pages 75-78 Larson/Farber 4th ed. 87 # 22 Power Failures Calculate mean, median and mode of the given data, of possible. Otherwise explain why not. We will use the graphing calculator to find the above mentioned central tendencies. Step 1: Press Stat Key on your graphing Calculator and you will see the following image. Step2: Press Enter to see six lists, namely L1, L2, L3, L4, L5, and L6, to enter the data. In L1 enter the data values given in the problem.
    88. 88. Larson/Farber 4th ed. 88 # 22 cont. Step 3: After entering the data values press the STAT key again and scroll over to CALC menu. Step 4: Press ENTER after selecting item 1 and then on following screen enter L1 using second function key and # 1 key. Step 5: After entering L1 press ENTER to see the results. Keep scrolling down to see all the values. Mean = 61.15 Median = 55
    89. 89. Weighted Mean 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. 89 ( )x w x w
    90. 90. 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. 90
    91. 91. Solution: Finding a Weighted Mean Larson/Farber 4th ed. 91 Source Score, x Weight, w x∙w Test Mean 86 0.50 86(0.50)= 43.0 Midterm 96 0.15 96(0.15) = 14.4 Final Exam 82 0.20 82(0.20) = 16.4 Computer Lab 98 0.10 98(0.10) = 9.8 Homework 100 0.05 100(0.05) = 5.0 Σw = 1 Σ(x∙w) = 88.6 ( ) 88.6 88.6 1 x w x w Your weighted mean for the course is 88.6. You did not get an A.
    92. 92. Textbook Exercises. Pages 75 - 78 Larson/Farber 4th ed. 92 # 44 Account Balance Using your graphing Calculator follow the steps shown below to calculate the weighted mean of the data given in the problem. Step 1: As shown earlier press STAT key, under EDIT menu select item 1 and press ENTER to see the following screen. This time I am storing the data in L2 and L3. Make sure to enter the balance amount first in L2 and then number of days in L3. Reversing the order will change the results dramatically. Step 2: After entering the data, press STAT Key again, scroll to Calc menu and choose Item 1. Enter L2 comma L3 and press the Enter key to see the results. Mean = 982.19 dollars.
    93. 93. 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. 93 ( )x f x n f n Note: In section 2.4 the procedure to find the mean of the grouped data using graphing calculator is discussed. Refer to slide 139
    94. 94. Finding the Mean of a Frequency Distribution In Words In Symbols Larson/Farber 4th ed. 94 ( )x f x n (lower limit)+(upper limit) 2 x ( )x f n f 1. Find the midpoint of each class. 2. Find the sum of the products of the midpoints and the frequencies. 3. Find the sum of the frequencies. 4. Find the mean of the frequency distribution.
    95. 95. 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. 95 Class Midpoint Frequency, f 7 – 18 12.5 6 19 – 30 24.5 10 31 – 42 36.5 13 43 – 54 48.5 8 55 – 66 60.5 5 67 – 78 72.5 6 79 – 90 84.5 2
    96. 96. Solution: Find the Mean of a Frequency Distribution Larson/Farber 4th ed. 96 Class Midpoint, x Frequency, f (x∙f) 7 – 18 12.5 6 12.5∙6 = 75.0 19 – 30 24.5 10 24.5∙10 = 245.0 31 – 42 36.5 13 36.5∙13 = 474.5 43 – 54 48.5 8 48.5∙8 = 388.0 55 – 66 60.5 5 60.5∙5 = 302.5 67 – 78 72.5 6 72.5∙6 = 435.0 79 – 90 84.5 2 84.5∙2 = 169.0 n = 50 Σ(x∙f) = 2089.0 ( ) 2089 41.8 minutes 50 x f x n
    97. 97. The Shape of Distributions Larson/Farber 4th ed. 97 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.
    98. 98. The Shape of Distributions Larson/Farber 4th ed. 98 Uniform Distribution (rectangular) • All entries or classes in the distribution have equal or approximately equal frequencies. • Symmetric.
    99. 99. The Shape of Distributions Larson/Farber 4th ed. 99 Skewed Left Distribution (negatively skewed) • The “tail” of the graph elongates more to the left. • The mean is to the left of the median.
    100. 100. The Shape of Distributions Larson/Farber 4th ed. 100 Skewed Right Distribution (positively skewed) • The “tail” of the graph elongates more to the right. • The mean is to the right of the median.
    101. 101. 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. 101
    102. 102. Section 2.4 Measures of Variation Larson/Farber 4th ed. 102
    103. 103. 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. 103
    104. 104. 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) Larson/Farber 4th ed. 104
    105. 105. 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. 105
    106. 106. 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. 106 minimum maximum
    107. 107. 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. 107
    108. 108. 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. 108 Solution: • First determine the mean starting salary. 415 41.5 10 x N
    109. 109. Solution: Finding the Deviation Larson/Farber 4th ed. 109 • Determine the deviation for each data entry. Salary ($1000s), x Deviation: x – μ 41 41 – 41.5 = –0.5 38 38 – 41.5 = –3.5 39 39 – 41.5 = –2.5 45 45 – 41.5 = 3.5 47 47 – 41.5 = 5.5 41 41 – 41.5 = –0.5 44 44 – 41.5 = 2.5 41 41 – 41.5 = –0.5 37 37 – 41.5 = –4.5 42 42 – 41.5 = 0.5 Σx = 415 Σ(x – μ) = 0
    110. 110. Deviation, Variance, and Standard Deviation Population Variance • Population Standard Deviation • Larson/Farber 4th ed. 110 2 2 ( )x N Sum of squares, SSx 2 2 ( )x N
    111. 111. Finding the Population Variance & Standard Deviation In Words In Symbols Larson/Farber 4th ed. 111 1. Find the mean of the population data set. 2. Find deviation of each entry. 3. Square each deviation. 4. Add to get the sum of squares. x N x – μ (x – μ)2 SSx = Σ(x – μ)2
    112. 112. Finding the Population Variance & Standard Deviation Larson/Farber 4th ed. 112 5. Divide by N to get the population variance. 6. Find the square root to get the population standard deviation. 2 2 ( )x N 2 ( )x N In Words In Symbols
    113. 113. 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. 113
    114. 114. Solution: Finding the Population Standard Deviation Larson/Farber 4th ed. 114 • Determine SSx • N = 10 Salary, x Deviation: x – μ Squares: (x – μ)2 41 41 – 41.5 = –0.5 (–0.5)2 = 0.25 38 38 – 41.5 = –3.5 (–3.5)2 = 12.25 39 39 – 41.5 = –2.5 (–2.5)2 = 6.25 45 45 – 41.5 = 3.5 (3.5)2 = 12.25 47 47 – 41.5 = 5.5 (5.5)2 = 30.25 41 41 – 41.5 = –0.5 (–0.5)2 = 0.25 44 44 – 41.5 = 2.5 (2.5)2 = 6.25 41 41 – 41.5 = –0.5 (–0.5)2 = 0.25 37 37 – 41.5 = –4.5 (–4.5)2 = 20.25 42 42 – 41.5 = 0.5 (0.5)2 = 0.25 Σ(x – μ) = 0 SSx = 88.5
    115. 115. Solution: Finding the Population Standard Deviation Larson/Farber 4th ed. 115 Population Variance • Population Standard Deviation • 2 2 ( ) 88.5 8.9 10 x N 2 8.85 3.0 The population standard deviation is about 3.0, or $3000.
    116. 116. Deviation, Variance, and Standard Deviation Sample Variance • Sample Standard Deviation • Larson/Farber 4th ed. 116 2 2 ( ) 1 x x s n 2 2 ( ) 1 x x s s n
    117. 117. Finding the Sample Variance & Standard Deviation In Words In Symbols Larson/Farber 4th ed. 117 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. x x n 2 ( )xSS x x 2 ( )x x x x
    118. 118. Finding the Sample Variance & Standard Deviation Larson/Farber 4th ed. 118 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 2 2 ( ) 1 x x s n 2 ( ) 1 x x s n
    119. 119. 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. 119
    120. 120. Solution: Finding the Sample Standard Deviation Larson/Farber 4th ed. 120 • Determine SSx • n = 10 Salary, x Deviation: x – μ Squares: (x – μ)2 41 41 – 41.5 = –0.5 (–0.5)2 = 0.25 38 38 – 41.5 = –3.5 (–3.5)2 = 12.25 39 39 – 41.5 = –2.5 (–2.5)2 = 6.25 45 45 – 41.5 = 3.5 (3.5)2 = 12.25 47 47 – 41.5 = 5.5 (5.5)2 = 30.25 41 41 – 41.5 = –0.5 (–0.5)2 = 0.25 44 44 – 41.5 = 2.5 (2.5)2 = 6.25 41 41 – 41.5 = –0.5 (–0.5)2 = 0.25 37 37 – 41.5 = –4.5 (–4.5)2 = 20.25 42 42 – 41.5 = 0.5 (0.5)2 = 0.25 Σ(x – μ) = 0 SSx = 88.5
    121. 121. Solution: Finding the Sample Standard Deviation Larson/Farber 4th ed. 121 Sample Variance • Sample Standard Deviation • 2 2 ( ) 88.5 9.8 1 10 1 x x s n 2 88.5 3.1 9 s s The sample standard deviation is about 3.1, or $3100.
    122. 122. 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. 122 Office Rental Rates 35.00 33.50 37.00 23.75 26.50 31.25 36.50 40.00 32.00 39.25 37.50 34.75 37.75 37.25 36.75 27.00 35.75 26.00 37.00 29.00 40.50 24.50 33.00 38.00
    123. 123. Solution: Using Technology to Find the Standard Deviation Larson/Farber 4th ed. 123 Sample Mean Sample Standard Deviation
    124. 124. 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. 124
    125. 125. Textbook Exercises. Page 94. Larson/Farber 4th ed. 125 Exercise # 22 Annual Salaries Public Teachers:  Range: 5.1 thousand dollars Variance: 2.96 square thousand dollars (Why ?)  Standard Deviation: 1.72 thousand dollars Private Teachers: Range: 4.2 thousand dollars Variance: 1.99 square thousand dollars (Why ?)  Standard Deviation: 1.41 thousand dollars From the values of sample standard deviation it is clear that the annual salaries of Public teachers varies more than that of Private teachers.
    126. 126. 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. 126 • 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.
    127. 127. Interpreting Standard Deviation: Empirical Rule (68 – 95 – 99.7 Rule) Larson/Farber 4th ed. 127 3x s x s 2x s 3x sx s x2x s 68% within 1 standard deviation 34% 34% 99.7% within 3 standard deviations 2.35% 2.35% 95% within 2 standard deviations 13.5% 13.5%
    128. 128. 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. 128
    129. 129. Solution: Using the Empirical Rule Larson/Farber 4th ed. 129 3x s x s 2x s 3x sx s x2x s 55.87 58.58 61.29 64 66.71 69.42 72.13 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.
    130. 130. Textbook Exercises. Page 96 Larson/Farber 4th ed. 130 Problem # 30 Since the data is bell shaped, according to the Empirical Rule 95% of the data values falls within two standard deviation from the mean. So, 95% of the data values lies between 2400 – 2(450) and 2400 + 2(450) That is, between $1500 and $3300. Problem # 32 a. Since 95% of the data values lies between $1500 and $3300, the number of farms whose land and building values per acre are between $1500 and $3300 are 0.95 * 40 = 38 farms. b. On adding 20 more farms we can expect 95% of them to be between $1500 and $3300 per acre. That is 0.95 * 20 = 19 farms.
    131. 131. 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. 131 2 1 1 k • k = 2: In any data set, at least 2 1 3 1 or 75% 2 4 of the data lie within 2 standard deviations of the mean. • k = 3: In any data set, at least 2 1 8 1 or 88.9% 3 9 of the data lie within 3 standard deviations of the mean.
    132. 132. 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. 132
    133. 133. 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. 133 At least 75% of the population of Florida is between 0 and 88.8 years old.
    134. 134. 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. 134 2 ( ) 1 x x f s n where n= Σf (the number of entries in the data set)
    135. 135. Example: Finding the Standard Deviation for Grouped Data Larson/Farber 4th ed. 135 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 1 1 1 1 2 2 1 0 1 1 0 0 0 1 5 0 3 6 3 0 3 1 1 1 1 6 0 1 3 6 6 1 2 2 3 0 1 1 4 1 1 2 2 0 3 0 2 4
    136. 136. x f xf 0 10 0(10) = 0 1 19 1(19) = 19 2 7 2(7) = 14 3 7 3(7) =21 4 2 4(2) = 8 5 1 5(1) = 5 6 4 6(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. 136 Σf = 50 Σ(xf )= 91 91 1.8 50 xf x n The sample mean is about 1.8 children.
    137. 137. Solution: Finding the Standard Deviation for Grouped Data • Determine the sum of squares. Larson/Farber 4th ed. 137 x f 0 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 7 3 – 1.8 = 1.2 (1.2)2 = 1.44 1.44(7) = 10.08 4 2 4 – 1.8 = 2.2 (2.2)2 = 4.84 4.84(2) = 9.68 5 1 5 – 1.8 = 3.2 (3.2)2 = 10.24 10.24(1) = 10.24 6 4 6 – 1.8 = 4.2 (4.2)2 = 17.64 17.64(4) = 70.56 x x 2 ( )x x 2 ( )x x f 2 ( ) 145.40x x f
    138. 138. Solution: Finding the Standard Deviation for Grouped Data • Find the sample standard deviation. Larson/Farber 4th ed. 138 x x 2 ( )x x 2 ( )x x f2 ( ) 145.40 1.7 1 50 1 x x f s n The standard deviation is about 1.7 children.
    139. 139. Finding Standard Deviation of the Grouped Data Larson/Farber 4th ed. 139 Example: The heights (in inches) of 23 male students in a physical education class.  First, determine the midpoints of each classes. Enter the midpoints in one of the lists, say L1, of the graphing calculator.  Enter the frequency into the next list of the graphing calculator.  Using 1-Var stats from the CALC menu obtain the results to determine the value of the mean and the standard deviation of the given data. Go to the next slide to view the results. Height (in inches) Frequency 63 – 65 3 66 – 68 6 69 – 71 7 72 – 74 4 75 – 77 3
    140. 140. Finding Standard deviation of the Grouped Data Larson/Farber 4th ed. 140 Mean = 69.74 inches Sample Standard Deviation = 3.72 inches The midpoints are entered in list L4 and the frequency in L5. You can use any lists but make sure to enter them in this order, midpoints followed by frequency
    141. 141. 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. 141
    142. 142. Section 2.5 Measures of Position Larson/Farber 4th ed. 142
    143. 143. 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. 143
    144. 144. 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. Larson/Farber 4th ed. 144
    145. 145. 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. 145 Solution: • Q2 divides the data set into two halves. 5 7 9 10 11 13 14 15 16 17 18 18 20 21 37 Q2 Lower half Upper half
    146. 146. 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. 146 Q2 Lower half Upper half Q1 Q3 About one fourth of the employees scored 10 or less, about one half scored 15 or less; and about three fourths scored 18 or less.
    147. 147. Interquartile Range Interquartile Range (IQR) • The difference between the third and first quartiles. • IQR = Q3 – Q1 Larson/Farber 4th ed. 147
    148. 148. Example: Finding the Interquartile Range Find the interquartile range of the test scores. Recall Q1 = 10, Q2 = 15, and Q3 = 18 Larson/Farber 4th ed. 148 Solution: • IQR = Q3 – Q1 = 18 – 10 = 8 The test scores in the middle portion of the data set vary by at most 8 points.
    149. 149. 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 Larson/Farber 4th ed. 149
    150. 150. Drawing a Box-and-Whisker Plot 1. Find the five-number summary of the data set. 2. Construct a horizontal scale that spans the range of the data. 3. Plot the five numbers above the horizontal scale. 4. Draw a box above the horizontal scale from Q1 to Q3 and draw a vertical line in the box at Q2. 5. Draw whiskers from the box to the minimum and maximum entries. Larson/Farber 4th ed. 150 Whisker Whisker Maximum entry Minimum entry Box Median, Q2 Q3Q1
    151. 151. Example: Drawing a Box-and-Whisker Plot Draw a box-and-whisker plot that represents the 15 test scores. Recall Min = 5 Q1 = 10 Q2 = 15 Q3 = 18 Max = 37 Larson/Farber 4th ed. 151 5 10 15 18 37 Solution: About half the scores are between 10 and 18. By looking at the length of the right whisker, you can conclude 37 is a possible outlier.
    152. 152. Percentiles and Other Fractiles Fractiles Summary Symbols Quartiles Divides data into 4 equal parts Q1, Q2, Q3 Deciles Divides data into 10 equal parts D1, D2, D3,…, D9 Percentiles Divides data into 100 equal parts P1, P2, P3,…, P99 Larson/Farber 4th ed. 152 Note: Along with the mean the graphing calculator also gives you values of Q1, Q2 (median) and Q3.
    153. 153. Example: Interpreting Percentiles The ogive represents the cumulative frequency distribution for SAT test scores of college-bound students in a recent year. What test score represents the 72nd percentile? How should you interpret this? (Source: College Board Online) Larson/Farber 4th ed. 153
    154. 154. Solution: Interpreting Percentiles The 72nd percentile corresponds to a test score of 1700. This means that 72% of the students had an SAT score of 1700 or less. Larson/Farber 4th ed. 154
    155. 155. 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. 155 value - mean standarddeviation x z
    156. 156. 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. 156
    157. 157. Solution: Comparing z-Scores from Different Data Sets Larson/Farber 4th ed. 157 • Forest Whitaker 45 43.7 0.15 8.8 x z • Helen Mirren 61 36 2.17 11.5 x z 0.15 standard deviations above the mean 2.17 standard deviations above the mean
    158. 158. Solution: Comparing z-Scores from Different Data Sets Larson/Farber 4th ed. 158 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 = 0.15 z = 2.17
    159. 159. Textbook Exercises. Page 112 Larson/Farber 4th ed. 159 Problem 36: Life Span of Fruit Flies Mean life span = 33 days and Standard deviation = 4 days a. z- score of life span of 34 days is similarly the z-scores of life span of 30 days and 42 days are -0.75 and 2.25 respectively. Since the z-score for the life span of 42 days is 2.25 (beyond 2 standard deviations )it is an unusual life span. b. z-scores for the life span of 29 days, 41 days and 25 days are -1, 2 and -2 respectively. Since 29 days with z-score of -1 is 1 standard deviation below the mean, its percentile using the Empirical rule is 16th percentile. (sum of values to the left of 1 standard deviation below the mean). Refer to the figure of Empirical Rule. Since 41 days is 2 standard deviation above the mean, its percentile is 97.5th percentile. (sum of values to the left of 2 standard deviations above the mean). Since 25 days is 2 standard deviations below the mean, its percentile is 2.5th percentile. (sum of values to the left of 2 standard deviations below the mean. 25.0 4 3334x z
    160. 160. 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. 160

    ×