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Chapter 2 Descriptive Statistics 1 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 Section 2.1 Frequency Distributions and Their Graphs 3 Section 2.1 Objectives • Construct frequency distributions • Construct frequency histograms, frequency polygons, relative frequency histograms, and ogives 4 Frequency Distribution Frequency Distribution Class Frequency, f Class width • A table that shows 1–5 5 classes or intervals of 6 – 1 = 5 6 – 10 8 data with a count of the 11 – 15 6 number of entries in each 16 – 20 8 class. 21 – 25 5 • The frequency, f, of a class is the number of 26 – 30 4 data entries in the class. Lower class Upper class limits limits 5 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 of the data. Divide the range by the number of classes. Round up to the next convenient number. 6 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. 7 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. 8 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 9 Solution: Constructing a Frequency Distribution 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 1. Number of classes = 7 (given) 2. Find the class width max min 86 7 11.29 #classes 7 Round up to 12 10 Solution: Constructing a Frequency Distribution 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. Class width = 12 Lower limit 7 Upper limit 19 31 43 55 67 79 11 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 7 19 31 43 55 67 79 Upper limit 18 30 42 54 66 Class width = 12 78 90 12 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. Class 7 – 18 Tally Frequency, f 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 13 Determining the Midpoint Midpoint of a class (Lower class limit) (Upper class limit) 2 Class 7 – 18 Midpoint 7 18 12.5 2 19 – 30 19 30 24.5 2 31 – 42 31 42 36.5 2 Frequency, f 6 Class width = 12 10 13 14 Determining the Relative Frequency Relative Frequency of a class • Portion or percentage of the data that falls in a particular class. class frequency f • relative frequency Sample size n Class Frequency, f 7 – 18 6 19 – 30 10 31 – 42 13 Relative Frequency 6 0.12 50 10 0.20 50 13 0.26 50 15 Determining the Cumulative Frequency Cumulative frequency of a class • The sum of the frequency for that class and all previous classes. Class Frequency, f Cumulative frequency 7 – 18 6 6 19 – 30 + 10 16 31 – 42 + 13 29 16 Expanded Frequency Distribution Class Frequency, f Midpoint Relative 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 f 1 n 50 Σf = 50 Cumulative frequency 17 Graphs of Frequency Distributions frequency 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 18 Class Boundaries Class boundaries • The numbers that separate classes without forming gaps between them. • 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. Class Class Frequency, Boundaries f 7 – 18 6.5 – 18.5 6 19 – 30 10 31 – 42 13 • First class lower boundary = 7 – 0.5 = 6.5 • First class upper boundary = 18 + 0.5 = 18.5 19 Class Boundaries Class 7 – 18 19 – 30 31 – 42 43 – 54 55 – 66 67 – 78 79 – 90 Class boundaries 6.5 – 18.5 18.5 – 30.5 30.5 – 42.5 42.5 – 54.5 54.5 – 66.5 66.5 – 78.5 78.5 – 90.5 Frequency, f 6 10 13 8 5 6 2 20 Example: Frequency Histogram Construct a frequency histogram for the Internet usage frequency distribution. 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 21 Solution: Frequency Histogram (using Midpoints) 22 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. 23 Graphs of Frequency Distributions frequency Frequency Polygon • A line graph that emphasizes the continuous change in frequencies. data values 24 Example: Frequency Polygon Construct a frequency polygon for the Internet usage frequency distribution. 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 25 Solution: Frequency Polygon Internet Usage Frequency 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. 14 12 10 8 6 4 2 0 0.5 12.5 24.5 36.5 48.5 60.5 72.5 84.5 96.5 Time online (in minutes) You can see that the frequency of subscribers increases up to 36.5 minutes and then decreases. 26 Graphs of Frequency Distributions relative frequency 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 27 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 – 18.5 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 28 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. 29 Graphs of Frequency Distributions cumulative frequency 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 30 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. 31 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). 32 Example: Ogive Construct an ogive for the Internet usage frequency distribution. 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 33 Solution: Ogive Internet Usage Cumulative frequency 60 50 40 30 20 10 0 6.5 18.5 30.5 42.5 54.5 66.5 78.5 90.5 Time online (in minutes) 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. 34 Section 2.1 Summary • Constructed frequency distributions • Constructed frequency histograms, frequency polygons, relative frequency histograms and ogives 35 Graphical Analysis: Use the frequency polygon to identify the class with the greatest, and the class with the least, frequency. 36 Graphical Analysis: Use the relative frequency histogram to (a) identify the class with the greatest, and the class with the least, relative frequency. (b) approximate the greatest and least relative frequencies. (c) approximate the relative frequency of the second class. 37 Section 2.2 More Graphs and Displays 38 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 39 Graphing Quantitative Data Sets Stem-and-leaf plot • Each number is separated into a stem and a leaf. • Similar to a histogram. • Still contains original data values. 26 Data: 21, 25, 25, 26, 27, 28, 30, 36, 36, 45 2 3 1 5 5 6 7 8 0 6 6 4 5 40 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. 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 41 Solution: Constructing a Stem-and-Leaf Plot 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 • 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. 42 Solution: Constructing a Stem-and-Leaf Plot Include a key to identify the values of the data. From the display, you can conclude that more than 50% of the cellular phone users sent between 110 and 130 text messages. 43 Graphing Quantitative Data Sets Dot plot • Each data entry is plotted, using a point, above a horizontal axis Data: 21, 25, 25, 26, 27, 28, 30, 36, 36, 45 26 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 44 Example: Constructing a Dot Plot Use a dot plot organize the text messaging data. 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 • 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. 45 Solution: Constructing a Dot Plot 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 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. 46 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. 47 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 type Killed Cars 18,440 Trucks 13,778 Motorcycles 4,553 Other 823 48 Solution: Constructing a Pie Chart • Find the relative frequency (percent) of each category. Vehicle type Frequency, f Cars 18,440 Trucks 13,778 Motorcycles Other 4,553 823 Relative frequency 18440 0.49 37594 13778 0.37 37594 4553 0.12 37594 823 0.02 37594 37,594 49 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 50 Solution: Constructing a Pie Chart Relative Vehicle type Frequency, f frequency Central angle Cars 18,440 0.49 360º(0.49)≈176º Trucks 13,778 0.37 360º(0.37)≈133º 4,553 0.12 360º(0.12)≈43º 823 0.02 360º(0.02)≈7º Motorcycles Other 51 Solution: Constructing a Pie Chart Relative Vehicle type 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. 52 Graphing Qualitative Data Sets Frequency 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. Categories 53 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) 54 Solution: Constructing a Pareto Chart Cause $ (million) Admin. error 7.8 Employee theft 15.6 Shoplifting 14.7 Vendor fraud 2.9 From the graph, it is easy to see that the causes of inventory shrinkage that should be addressed first are employee theft and shoplifting. 55 . 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 y points in a coordinate plane. Used to show the relationship between two quantitative variables. x 57 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) 58 Example: Interpreting a Scatter Plot As the petal length increases, what tends to happen to the petal width? Each point in the scatter plot represents the petal length and petal width of one flower. 59 Solution: Interpreting a Scatter Plot Interpretation From the scatter plot, you can see that as the petal length increases, the petal width also tends to increase. 60 Interpretation: Stay Away From Meat! 61 Interpretation: Stay Away From Guns! 62 Graphing Paired Data Sets Quantitative data Time Series • Data set is composed of quantitative entries taken at regular intervals over a period of time. e.g., The amount of precipitation measured each day for one month. • Use a time series chart to graph. time 63 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) 64 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. 65 Solution: Constructing a Time Series Chart The graph shows that the number of subscribers has been increasing since 1995, with greater increases recently. 66 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 67 Section 2.3 Measures of Central Tendency 68 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 69 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 70 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. x • Population mean: N • Sample mean: x x n 71 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 72 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 x 3695 x 527.9 n 7 The mean price of the flights is about $527.90. 73 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. 74 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 75 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. 76 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 77 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. 397 427 Median 412 2 The median price of the flights is $412. 78 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). 79 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 80 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. 81 Example: Finding the Mode At a political debate a sample of audience members was asked to name the political party to which they belong. Their responses are shown in the table. What is the mode of the responses? Political Party Democrat Frequency, f 34 Republican Other 56 21 Did not respond 9 82 Solution: Finding the Mode Political Party Democrat Frequency, f 34 Republican Other Did not respond 56 21 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. 83 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). 84 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 20 20 20 20 20 21 21 21 21 22 22 22 23 23 23 23 24 24 65 85 Solution: Comparing the Mean, Median, and Mode Ages in a class 20 20 20 20 20 20 21 21 21 21 22 22 22 23 23 23 23 24 24 65 Mean: x 20 20 ... 24 65 x 23.8 years n 20 Median: 21 22 21.5 years 2 Mode: 20 years (the entry occurring with the greatest frequency) 86 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. 87 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. 88 Weighted Mean Weighted Mean • The mean of a data set whose entries have varying weights. ( x w) • x where w is the weight of each entry x w 89 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? 90 Solution: Finding a Weighted Mean Source x∙w Score, x Weight, 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 ( x w) 88.6 x 88.6 w 1 Your weighted mean for the course is 88.6. You did not get an A. 91 Mean of Grouped Data Mean of a Frequency Distribution • Approximated by ( x f ) x n n f where x and f are the midpoints and frequencies of a class, respectively 92 Finding the Mean of a Frequency Distribution In Words 1. Find the midpoint of each class. In Symbols (lower limit)+(upper limit) x 2 2. Find the sum of the products of the midpoints and the frequencies. ( x f ) 3. Find the sum of the frequencies. n f 4. Find the mean of the frequency distribution. ( x f ) x n 93 Example: Find the Mean of a Frequency Distribution Use the frequency distribution to approximate the mean number of minutes that a sample of Internet subscribers spent online during their most recent session. Class Midpoint Frequency, f 7 – 18 12.5 6 19 – 30 24.5 10 31 – 42 36.5 13 43 – 54 48.5 8 55 – 66 60.5 5 67 – 78 72.5 6 79 – 90 84.5 2 94 Solution: Find the Mean of a Frequency Distribution Class Midpoint, x Frequency, f (x∙f) 7 – 18 12.5 6 12.5∙6 = 75.0 19 – 30 24.5 10 24.5∙10 = 245.0 31 – 42 36.5 13 36.5∙13 = 474.5 43 – 54 48.5 8 48.5∙8 = 388.0 55 – 66 60.5 5 60.5∙5 = 302.5 67 – 78 72.5 6 72.5∙6 = 435.0 79 – 90 84.5 2 84.5∙2 = 169.0 n = 50 Σ(x∙f) = 2089.0 ( x f ) 2089 x 41.8 minutes n 50 95 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. 96 The Shape of Distributions Uniform Distribution (rectangular) • All entries or classes in the distribution have equal or approximately equal frequencies. • Symmetric. 97 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. 98 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. 99 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 100 Section 2.4 Measures of Variation 101 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 102 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) 103 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 42 104 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 minimum maximum • Range = (Max. salary) – (Min. salary) = 47 – 37 = 10 The range of starting salaries is 10 or $10,000. 105 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 106 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 Solution: • First determine the mean starting salary. x 415 41.5 N 10 107 Solution: Finding the Deviation • 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) = 0 108 Deviation, Variance, and Standard Deviation Population Variance ( x ) • N 2 2 Sum of squares, SSx Population Standard Deviation 2 ( x ) 2 • N 109 Finding the Population Variance & Standard Deviation In Words 1. Find the mean of the population data set. In Symbols x N 2. Find deviation of each entry. x–μ 3. Square each deviation. (x – μ)2 4. Add to get the sum of squares. SSx = Σ(x – μ)2 110 Finding the Population Variance & Standard Deviation In Words In Symbols 5. Divide by N to get the population variance. 2 ( x ) 2 N 6. Find the square root to get the population standard deviation. ( x ) 2 N 111 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. 112 Solution: Finding the Population Standard Deviation • Determine SSx • N = 10 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 Salary, x 113 Solution: Finding the Population Standard Deviation Population Variance ( x ) 88.5 8.9 • N 10 2 2 Population Standard Deviation • 2 8.85 3.0 The population standard deviation is about 3.0, or $3000. 114 Deviation, Variance, and Standard Deviation Sample Variance ( x x ) • s n 1 2 2 Sample Standard Deviation • 2 ( x x ) s s2 n 1 115 Finding the Sample Variance & Standard Deviation In Words In Symbols x n 1. Find the mean of the sample data set. x 2. Find deviation of each entry. xx 3. Square each deviation. ( x x )2 4. Add to get the sum of squares. SS x ( x x ) 2 116 Finding the Sample Variance & Standard Deviation In Words 5. Divide by n – 1 to get the sample variance. 6. Find the square root to get the sample standard deviation. In Symbols 2 ( x x ) s2 n 1 ( x x ) 2 s n 1 117 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 42 118 Solution: Finding the Sample Standard Deviation • Determine SSx • n = 10 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 Salary, x 119 Solution: Finding the Sample Standard Deviation Sample Variance ( x x ) 88.5 9.8 • s n 1 10 1 2 2 Sample Standard Deviation 88.5 3.1 • s s 9 2 The sample standard deviation is about 3.1, or $3100. 120 Example: Using Technology to Find the Standard Deviation Sample office rental rates (in dollars per square foot per year) for Miami’s central business district are shown in the table. Use a calculator or a computer to find the mean rental rate and the sample standard deviation. (Adapted from: Cushman & Wakefield Inc.) Office Rental Rates 35.00 33.50 37.00 23.75 26.50 31.25 36.50 40.00 32.00 39.25 37.50 34.75 37.75 37.25 36.75 27.00 35.75 26.00 37.00 29.00 40.50 24.50 33.00 38.00 121 Solution: Using Technology to Find the Standard Deviation Sample Mean Sample Standard Deviation 122 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. 123 Interpreting Standard Deviation: Empirical Rule (68 – 95 – 99.7 Rule) For data with a (symmetric) bell-shaped distribution, the standard deviation has the following characteristics: • About 68% of the data lie within one standard deviation of the mean. • About 95% of the data lie within two standard deviations of the mean. • About 99.7% of the data lie within three standard deviations of the mean. 124 Interpreting Standard Deviation: Empirical Rule (68 – 95 – 99.7 Rule) 99.7% within 3 standard deviations 95% within 2 standard deviations 68% within 1 standard deviation 34% 34% 2.35% 2.35% 13.5% x 3s x 2s 13.5% x s x xs x 2s x 3s 125 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. 126 Solution: Using the Empirical Rule • Because the distribution is bell-shaped, you can use the Empirical Rule. 34% 13.5% 55.87 x 3s 58.58 x 2s 61.29 x s 64 x 66.71 xs 69.42 x 2s 72.13 x 3s 34% + 13.5% = 47.5% of women are between 64 and 69.42 inches tall. 127 Chebychev’s Theorem • The portion of any data set lying within k standard deviations (k > 1) of the mean is at least: 1 1 2 k 1 3 • k = 2: In any data set, at least 1 2 or 75% 2 4 of the data lie within 2 standard deviations of the mean. 1 8 • k = 3: In any data set, at least 1 2 or 88.9% 3 9 of the data lie within 3 standard deviations of the mean. 128 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? 129 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 At least 75% of the population of Florida is between 0 and 88.8 years old. 130 Standard Deviation for Grouped Data Sample standard deviation for a frequency distribution • ( x x ) 2 f s n 1 where n= Σf (the number of entries in the data set) • When a frequency distribution has classes, estimate the sample mean and standard deviation by using the midpoint of each class. 131 Example: Finding the Standard Deviation for Grouped Data You collect a random sample of the number of children per household in a region. Find the sample mean and the sample standard deviation of the data set. Number of Children in 50 Households 1 3 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 4 132 Solution: Finding the Standard Deviation for Grouped Data • First construct a frequency distribution. • Find the mean of the frequency distribution. xf 91 x 1.8 n 50 The sample mean is about 1.8 children. 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 Σf = 50 Σ(xf )= 91 133 Solution: Finding the Standard Deviation for Grouped Data • Determine the sum of squares. x f xx ( x x )2 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 f ( x x )2 f 145.40 134 Solution: Finding the Standard Deviation for Grouped Data • Find the sample standard deviation. x 2 x ( x x )2 ( x x ) f 145.40 s 1.7 n 1 50 1 ( x x )2 f The standard deviation is about 1.7 children. 135 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 136 Section 2.5 Measures of Position 137 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) 138 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. 139 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 Solution: • Q2 divides the data set into two halves. Lower half Upper half 5 7 9 10 11 13 14 15 16 17 18 18 20 21 37 Q2 140 Solution: Finding Quartiles • The first and third quartiles are the medians of the lower and upper halves of the data set. Lower half Upper half 5 7 9 10 11 13 14 15 16 17 18 18 20 21 37 Q1 Q2 Q3 About one fourth of the employees scored 10 or less, about one half scored 15 or less; and about three fourths scored 18 or less. 141 The number of nuclear power plants in the top 15 nuclear power-producing countries in the world are listed. Find the first, second, and third quartiles of the data set. What can you conclude? National Institute of Standards and Technology ( N 1) Pth Rank n 100 ( N 1)Qth Rank n 4 ( N 1) Dth Rank n 10 The 2-quantile is called the median The 3-quantiles are called tertiles or terciles → T The 4-quantiles are called quartiles → Q The 5-quantiles are called quintiles → QU The 6-quantiles are called sextiles → S The 10-quantiles are called deciles → D The 12-quantiles are called duo-deciles → Dd The 20-quantiles are called vigintiles → V The 100-quantiles are called percentiles → P The 1000-quantiles are called permilles → Pr . Slide 2- 143 Interquartile Range Interquartile Range (IQR) • The difference between the third and first quartiles. • IQR = Q3 – Q1 144 Example: Finding the Interquartile Range Find the interquartile range of the test scores. Recall Q1 = 10, Q2 = 15, and Q3 = 18 Solution: • IQR = Q3 – Q1 = 18 – 10 = 8 The test scores in the middle portion of the data set vary by at most 8 points. 145 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 146 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. Box Whisker Minimum entry Whisker Q1 Median, Q2 Q3 Maximum entry 147 Example: Drawing a Box-and-Whisker Plot Draw a box-and-whisker plot that represents the 15 test scores. Recall Min = 5 Q1 = 10 Q2 = 15 Q3 = 18 Max = 37 Solution: 5 10 15 18 37 About half the scores are between 10 and 18. By looking at the length of the right whisker, you can conclude 37 is a possible outlier. 148 Percentiles and Other Fractiles Fractiles Quartiles Deciles Percentiles Summary Divides data into 4 equal parts Divides data into 10 equal parts Divides data into 100 equal parts Symbols Q1, Q2, Q3 D1, D2, D3,…, D9 P1, P2, P3,…, P99 149 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) 150 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. 151 The Standard Score Standard Score (z-score) • Represents the number of standard deviations a given value x falls from the mean μ. value - mean x • z standard deviation 152 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. 153 Solution: Comparing z-Scores from Different Data Sets • Forest Whitaker z x • Helen Mirren z x 45 43.7 0.15 8.8 0.15 standard deviations above the mean 61 36 2.17 11.5 2.17 standard deviations above the mean 154 Solution: Comparing z-Scores from Different Data Sets z = 0.15 z = 2.17 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. 155 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) 156 Over the same period, the lowest-income fifth saw a decrease in real income of 7.4 percent. Between 1979 and 2009, the top 5 percent of American families saw their real incomes increase 72.7 percent, according to Census data. Chapter 2: Descriptive Statistics Elementary Statistics: Picturing the World Fifth Edition by Larson and Farber . Slide 4- 160 Find the class width: A. 3 Class 1– 5 Frequency, f 21 6 – 10 11 – 15 16 – 20 16 28 13 B. 4 C. 5 D. 19 . Slide 2- 161 Find the class width: A. 3 Class 1– 5 Frequency, f 21 6 – 10 11 – 15 16 – 20 16 28 13 B. 4 C. 5 D. 19 (6 1)or (11 6)or (10 5) . Slide 2- 162 Estimate the frequency of the class with the greatest frequency. 60 Ages of Concert Attendees A. 28 B. 21 C. 58 D. 53 Frequency 50 40 30 20 10 0 18 28 38 48 58 Age . Slide 2- 163 Estimate the frequency of the class with the greatest frequency. 60 Ages of Concert Attendees A. 28 B. 21 C. 58 D. 53 Frequency 50 40 30 20 10 0 18 28 38 48 58 Age . Slide 2- 164 What is the maximum data entry? Key: 3 | 8 = 38 3 8 9 A. 96 4 0 2 7 B. 38 5 1 1 4 8 C. 9 6 3 3 3 8 9 9 D. 41 7 0 0 1 1 2 4 7 8 8 8 8 9 8 2 2 3 4 7 7 8 9 9 9 1 1 4 5 6 . Slide 2- 165 What is the maximum data entry? Key: 3 | 8 = 38 3 8 9 A. 96 4 0 2 7 B. 38 5 1 1 4 8 C. 9 6 3 3 3 8 9 9 D. 41 7 0 0 1 1 2 4 7 8 8 8 8 9 8 2 2 3 4 7 7 8 9 9 9 1 1 4 5 6 Slide 2- 166 The heights (in inches) of a sample of basketball players are shown: 76 79 81 78 82 78 Find the mean. A. 78.5 B. 79 C. 474 D. 78 . Slide 2- 167 The heights (in inches) of a sample of basketball players are shown: 76 79 81 78 82 78 Find the mean. A. 78.5 B. 79 C. 474 D. 78 . Slide 2- 168 The heights (in inches) of a sample of basketball players are shown: 76 79 81 78 82 78 Find the median. A. 78.5 B. 79 C. 79.5 D. 78 . Slide 2- 169 The heights (in inches) of a sample of basketball players are shown: 76 79 81 78 82 78 Find the median. A. 78.5 B. 79 C. 79.5 D. 78 . Slide 2- 170 The heights (in inches) of a sample of basketball players are shown: 76 79 81 78 82 78 Find the mode. A. 78.5 B. 79 C. 79.5 D. 78 . Slide 2- 171 The heights (in inches) of a sample of basketball players are shown: 76 79 81 78 82 78 Find the mode. A. 78.5 B. 79 C. 79.5 D. 78 . Slide 2- 172 The heights (in inches) of a sample of basketball players are shown: 76 79 81 78 82 78 Find the standard deviation. A. 2.2 B. 6 C. 2 D. 4.8 . Slide 2- 173 The heights (in inches) of a sample of basketball players are shown: 76 79 81 78 82 78 Find the standard deviation. A. 2.2 B. 6 C. 2 D. 4.8 . Slide 2- 174 The mean annual automobile insurance premium is $950, with a standard deviation of $175. The data set has a bell-shaped distribution. Estimate the percent of premiums that are between $600 and $1300. A. 68% B. 75% C. 95% D. 99.7% . Slide 2- 175 600 950 2(175) x 950 2(175) 1300 The mean annual automobile insurance premium is $950, with a standard deviation of $175. The data set has a bell-shaped distribution. Estimate the percent of premiums that are between $600 and $1300. A. 68% B. 75% C. 95% D. 99.7% . Slide 2- 176 Use the box-and-whisker plot to identify the first quartile. 10 | 18 | | | | 24 26 | | | | 30 | | 10 12 14 16 18 20 22 24 26 28 30 A. 10 B. 18 C. 24 D. 26 . Slide 2- 177 Use the box-and-whisker plot to identify the first quartile. 10 | 18 | | | | 24 26 | | | | 30 | | 10 12 14 16 18 20 22 24 26 28 30 A. 10 B. 18 C. 24 D. 26 . Slide 2- 178 The mean annual automobile insurance premium is $950, with a standard deviation of $175. Find the z-score that corresponds to a premium of $1250. A. 1.13 B. –1.13 C. 1.71 D. –1.71 . Slide 2- 179 The mean annual automobile insurance premium is $950, with a standard deviation of $175. Find the z-score that corresponds to a premium of $1250. A. 1.13 B. –1.13 C. 1.71 z x 1250 950 1.71 175 D. –1.71 . Slide 2- 180