#### Transcript Chapter 2

Chapter 2 Descriptive Statistics © 2012 Pearson Education, Inc. All rights reserved. 1 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 2 of 149 Course Level Expectations • CLE 3136.1.7 Use technologies appropriately to develop understanding of abstract mathematical ideas, to facilitate problem solving, and to produce accurate and reliable models. • CLE 3136.2.1 Understand histograms, parallel box plots, and scatterplots, and use them to compare display data. Larson/Farber 5th ed. 3 Checks for Understanding • 3136.5.1 Use properties of point estimators, including biased/unbiased, and variability. • 3136.5.2 Understand the meaning of confidence level, of confidence intervals, and the properties of confidence intervals. Larson/Farber 5th ed. 4 Section 2.1 Frequency Distributions and Their Graphs © 2012 Pearson Education, Inc. All rights reserved. 5 of 149 Section 2.1 Objectives • Construct frequency distributions • Construct frequency histograms, frequency polygons, relative frequency histograms, and ogives © 2012 Pearson Education, Inc. All rights reserved. 6 of 149 Frequency Distribution Why can’t we do ‘5-1’? 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 This table might represent the ages of people in a room for example © 2012 Pearson Education, Inc. All rights reserved. limits limits 7 of 149 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. •For example, you have a range of ages from 10 to 74 •You decide you want 8 classes © 2012 Pearson Education, Inc. All rights reserved. •The range of data is 74-10 = 64 •64/8 = 8. •Therefore, you have 8 classes with 8 of 149 a width of 8 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. •From the previous example, the minimum data entry is 10 •The next lower limit would be 18 © 2012 Pearson Education, Inc. All rights reserved. •The first upper limit would be 17 •The next would be 17+8 = 25 9 of 149 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. •For our example to continue, we would have to have a number for how many people were each age © 2012 Pearson Education, Inc. All rights reserved. 10 of 149 Example: Constructing a Frequency Distribution The following sample data set lists the prices (in dollars) of 30 portable global positioning system (GPS) navigators. Construct a frequency distribution that has seven classes. 90 130 400 200 350 70 325 250 150 250 275 270 150 130 59 200 160 450 300 130 220 100 200 400 200 250 95 180 170 150 © 2012 Pearson Education, Inc. All rights reserved. 11 of 149 Solution: Constructing a Frequency Distribution 90 130 400 200 350 70 325 250 150 250 275 270 150 130 59 200 160 450 300 130 220 100 200 400 200 250 95 180 170 150 1. Number of classes = 7 (given) 2. Find the class width max min 450 59 391 55.86 #classes 7 7 Round up to 56 © 2012 Pearson Education, Inc. All rights reserved. 12 of 149 Solution: Constructing a Frequency Distribution 3. Use 59 (minimum value) as first lower limit. Add the class width of 56 to get the lower limit of the next class. 59 + 56 = 115 Find the remaining lower limits. © 2012 Pearson Education, Inc. All rights reserved. Lower limit Class width = 56 Upper limit 59 115 171 227 283 339 395 13 of 149 Solution: Constructing a Frequency Distribution The upper limit of the first class is 114 (one less than the lower limit of the second class). Add the class width of 56 to get the upper limit of the next class. 114 + 56 = 170 Find the remaining upper limits. © 2012 Pearson Education, Inc. All rights reserved. Lower limit Upper limit 59 115 171 227 283 339 114 170 226 282 338 394 395 450 Class width = 56 14 of 149 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. Note: you should sum the frequencies. This is the total sample size, and will be used later. Here, the sample size is 30 © 2012 Pearson Education, Inc. All rights reserved. Class Tally Frequency, f IIII 5 115–170 IIII III 8 171–226 IIII I 6 227–282 IIII 5 283–338 II 2 339–394 I 1 395–450 III 3 59–114 15 of 149 Determining the Midpoint Midpoint of a class (Lower class limit) (Upper class limit) 2 Class 59–114 Midpoint 59 114 86.5 2 115–170 115 170 142.5 2 171–226 171 226 198.5 2 © 2012 Pearson Education, Inc. All rights reserved. Frequency, f 5 Class width = 56 8 6 16 of 149 Determining the Relative Frequency Relative Frequency of a class • Portion or percentage of the data that falls in a particular class. •Frequency is how often Class frequency f something • Relative frequency occurs Sample size n Class Frequency, f 59–114 5 115–170 8 171–226 6 © 2012 Pearson Education, Inc. All rights reserved. Relative Frequency 5 0.17 30 8 0.27 30 6 0.2 30 •Relative frequency is how often something occurs relative to the total occurrences (sample size) 17 of 149 Determining the Cumulative Frequency Cumulative frequency of a class • The sum of the frequencies for that class and all previous classes. Class Frequency, f Cumulative frequency 59–114 5 5 115–170 + 8 13 171–226 + 6 19 © 2012 Pearson Education, Inc. All rights reserved. 18 of 149 Expanded Frequency Distribution Remember what this is called Class Frequency, f Midpoint Relative Frequency (f/n) 59–114 5 86.5 0.17 5 115–170 8 142.5 0.27 13 171–226 6 198.5 0.2 19 227–282 5 254.5 0.17 24 283–338 2 310.5 0.07 26 339–394 1 366.5 0.03 27 395–450 3 422.5 0.1 f Σf = 30 1 n This is the uppercase Greek Cumulative frequency 30 letter “sigma” and means “to sum” © 2012 Pearson Education, Inc. All rights reserved. 19 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 20 of 149 Class Boundaries Why do this? Class boundaries • The numbers that separate classes without forming Because bars of a histogram must gaps between them. • The distance from the upper limit of the first class to the lower limit of the second class is 115 – 114 = 1. • Half this distance is 0.5. touch, bars begin and end at class boundaries instead of class limits Class Class boundaries Frequency, f 59–114 58.5–114.5 5 115–170 8 171–226 6 • First class lower boundary = 59 – 0.5 = 58.5 • First class upper boundary = 114 + 0.5 = 114.5 © 2012 Pearson Education, Inc. All rights reserved. 21 of 149 Class Boundaries Class 59–114 115–170 171–226 227–282 283–338 339–394 395–450 © 2012 Pearson Education, Inc. All rights reserved. Class boundaries 58.5–114.5 114.5–170.5 170.5–226.5 226.5–282.5 282.5–338.5 338.5–394.5 394.5–450.5 Frequency, f 5 8 6 5 2 1 3 22 of 149 Example: Frequency Histogram Construct a frequency histogram for the Global Positioning system (GPS) navigators. Class Class boundaries 59–114 58.5–114.5 86.5 5 115–170 114.5–170.5 142.5 8 171–226 170.5–226.5 198.5 6 227–282 226.5–282.5 254.5 5 283–338 282.5–338.5 310.5 2 339–394 338.5–394.5 366.5 1 395–450 394.5–450.5 422.5 3 © 2012 Pearson Education, Inc. All rights reserved. Frequency, Midpoint f 23 of 149 Solution: Frequency Histogram (using Midpoints) © 2012 Pearson Education, Inc. All rights reserved. 24 of 149 Solution: Frequency Histogram (using class boundaries) You can see that more than half of the GPS navigators are priced below $226.50. © 2012 Pearson Education, Inc. All rights reserved. 25 of 149 Graphs of Frequency Distributions frequency Frequency Polygon • A line graph that emphasizes the continuous change in frequencies. data values © 2012 Pearson Education, Inc. All rights reserved. 26 of 149 Example: Frequency Polygon Construct a frequency polygon for the GPS navigators frequency distribution. You will use these midpoints to create your frequency polygon Class Midpoint Frequency, f 59–114 86.5 5 115–170 142.5 8 171–226 198.5 6 227–282 254.5 5 283–338 310.5 2 339–394 366.5 1 395–450 422.5 3 © 2012 Pearson Education, Inc. All rights reserved. 27 of 149 Solution: Frequency Polygon The graph should begin and end on the horizontal axis, so extend the left side to one class width before the first class midpoint and extend the right side to one class width after the last class midpoint. Add This And This You can see that the frequency of GPS navigators increases up to $142.50 and then decreases. © 2012 Pearson Education, Inc. All rights reserved. 28 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 29 of 149 Example: Relative Frequency Histogram Construct a relative frequency histogram for the GPS navigators frequency distribution. Class Class boundaries Frequency, f Relative frequency 59–114 58.5–114.5 5 0.17 115–170 114.5–170.5 8 0.27 171–226 170.5–226.5 6 0.2 227–282 226.5–282.5 5 0.17 283–338 282.5–338.5 2 0.07 339–394 338.5–394.5 1 0.03 395–450 394.5–450.5 3 0.1 © 2012 Pearson Education, Inc. All rights reserved. 30 of 149 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 27% of GPS navigators are priced between $114.50 and $170.50. © 2012 Pearson Education, Inc. All rights reserved. 31 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 32 of 149 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. © 2012 Pearson Education, Inc. All rights reserved. 33 of 149 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). © 2012 Pearson Education, Inc. All rights reserved. 34 of 149 Example: Ogive Construct an ogive for the GPS navigators frequency distribution. Class Class boundaries Frequency, f Cumulative frequency 59–114 58.5–114.5 5 5 115–170 114.5–170.5 8 13 171–226 170.5–226.5 6 19 227–282 226.5–282.5 5 24 283–338 282.5–338.5 2 26 339–394 338.5–394.5 1 27 395–450 394.5–450.5 3 30 © 2012 Pearson Education, Inc. All rights reserved. 35 of 149 Solution: Ogive 6.5 18.5 30.5 42.5 54.5 66.5 78.5 90.5 From the ogive, you can see that about 25 GPS navigators cost $300 or less. The greatest increase occurs between $114.50 and $170.50. © 2012 Pearson Education, Inc. All rights reserved. 36 of 149 Section 2.1 Summary • Constructed frequency distributions • Constructed frequency histograms, frequency polygons, relative frequency histograms and ogives © 2012 Pearson Education, Inc. All rights reserved. 37 of 149 Assignment • Page 47-51 1-6 Orally • 7-43 Odd Larson/Farber 5th ed. 38 Chapter 2 Quiz 1(20 points) • Using the information from Question number 34 on page 50, created an expanded Frequency Distribution Chart. • It should look like this: Class Larson/Farber 5th ed. Frequency, f Midpoint Relative Frequency (f/n) Cumulative frequency 39 Answers to Quiz 1 Class Frequency, f Midpoint Relative Frequency (f/n) 2456-2542 7 2499 .28 7 2543-2629 3 2586 .12 10 2630-2716 2 2673 .08 12 2717-2803 4 2760 .16 16 2804-2888 9 2847 .36 25 Larson/Farber 5th ed. Cumulative frequency 40 Section 2.2 More Graphs and Displays © 2012 Pearson Education, Inc. All rights reserved. 41 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 42 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 2 3 1 5 5 6 7 8 0 6 6 4 5 43 of 149 Example: Constructing a Stem-and-Leaf Plot The following are the numbers of text messages sent last week by the cellular phone users on one floor of a college dormitory. Display the data in a stem-and-leaf plot. 155 159 118 118 139 139 129 112 144 108 122 126 129 122 78 148 105 145 126 116 130 114 122 112 112 142 126 121 109 140 126 119 113 117 118 109 109 119 133 126 123 145 121 134 124 119 132 133 124 147 © 2012 Pearson Education, Inc. All rights reserved. 44 of 149 Solution: Constructing a Stem-and-Leaf Plot 155 159 118 118 139 139 129 112 144 108 122 126 129 122 78 148 105 145 126 116 130 114 122 112 112 142 126 121 109 140 126 119 113 117 118 109 109 119 133 126 123 145 121 134 124 119 132 133 124 147 • 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. © 2012 Pearson Education, Inc. All rights reserved. 45 of 149 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. © 2012 Pearson Education, Inc. All rights reserved. 46 of 149 Graphing Quantitative Data Sets We did a version of this on the 1st day of school Dot plot • Each data entry is plotted, using a point, above a horizontal axis. Data: 21, 25, 25, 26, 27, 28, 30, 36, 36, 45 26 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 © 2012 Pearson Education, Inc. All rights reserved. 47 of 149 Example: Constructing a Dot Plot Use a dot plot organize the text messaging data. 155 159 118 118 139 139 129 112 144 108 122 126 129 122 78 148 105 145 126 116 130 114 122 112 112 142 126 121 109 140 126 119 113 117 118 109 109 119 133 126 123 145 121 134 124 119 132 133 124 147 • 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. © 2012 Pearson Education, Inc. All rights reserved. 48 of 149 Solution: Constructing a Dot Plot 155 159 118 118 139 139 129 112 144 108 122 126 129 122 78 148 105 145 126 116 130 114 122 112 112 142 126 121 109 140 126 119 113 117 118 109 109 119 133 126 123 145 121 134 124 119 132 133 124 147 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. This is commonly called an “outlier” © 2012 Pearson Education, Inc. All rights reserved. 49 of 149 Graphing Qualitative Data Sets Pie Chart • A circle is divided into sectors that represent categories. • The area of each sector is proportional to the frequency of each category. • Therefore you will need Relative Frequency © 2012 Pearson Education, Inc. All rights reserved. 50 of 149 Example: Constructing a Pie Chart The numbers of earned degrees conferred (in thousands) in 2007 are shown in the table. Use a pie chart to organize the data. (Source: U.S. National Center for Educational Statistics) Type of degree Associate’s Bachelor’s Master’s First professional Doctoral © 2012 Pearson Education, Inc. All rights reserved. Number (thousands) 728 1525 604 90 60 51 of 149 Solution: Constructing a Pie Chart • Find the relative frequency (percent) of each category. Type of degree Frequency, f Associate’s 728 Bachelor’s 1525 Master’s 604 First professional Doctoral 728 0.24 3007 1525 0.51 3007 604 0.20 3007 90 90 0.03 3007 60 60 0.02 3007 Σf = 3007 © 2012 Pearson Education, Inc. All rights reserved. Relative frequency 52 of 149 Solution: Constructing a Pie Chart • Construct the pie chart using the central angle that corresponds to each category. To find the central angle, multiply 360º by the category's relative frequency. For example, the central angle for associate’s degrees is 360º(0.24) ≈ 86º © 2012 Pearson Education, Inc. All rights reserved. 53 of 149 Solution: Constructing a Pie Chart Type of degree Relative Frequency, f frequency Central angle Associate’s 728 0.24 360º(0.24)≈86º Bachelor’s 1525 0.51 360º(0.51)≈184º 604 0.20 360º(0.20)≈72º First professional 90 0.03 360º(0.03)≈11º Doctoral 60 0.02 360º(0.02)≈7º Master’s © 2012 Pearson Education, Inc. All rights reserved. 54 of 149 Solution: Constructing a Pie Chart Relative frequency Central angle Associate’s 0.24 86º Bachelor’s 0.51 184º Master’s 0.20 72º First professional 0.03 11º Doctoral 0.02 7º Type of degree From the pie chart, you can see that over one half of the degrees conferred in 2007 were bachelor’s degrees. © 2012 Pearson Education, Inc. All rights reserved. 55 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 56 of 149 Example: Constructing a Pareto Chart In a recent year, the retail industry lost $36.5 billion in inventory shrinkage. Inventory shrinkage is the loss of inventory through breakage, pilferage, shoplifting, and so on. The causes of the inventory shrinkage are administrative error ($5.4 billion), employee theft ($15.9 billion), shoplifting ($12.7 billion), and vendor fraud ($1.4 billion). Use a Pareto chart to organize this data. (Source: National Retail Federation and Center for Retailing Education, University of Florida) © 2012 Pearson Education, Inc. All rights reserved. 57 of 149 Chapter 2 Quiz 2(20 points) • Using the information from Question number 32 on page 50, created an expanded Frequency Distribution Chart. • It should look like this: Class Larson/Farber 5th ed. Frequency, f Midpoint Relative Frequency (f/n) Cumulative frequency 58 Answers to Quiz 2 Class Frequency, f Midpoint Relative Frequency (f/n) 32-35 3 36.5 .125~.13 3 36-39 9 40.5 .375~.38 12 40-43 8 44.5 .33 20 44-47 3 48.5 .125~.13 23 48-51 1 51 .04 24 Larson/Farber 5th ed. Cumulative frequency 59 Solution: Constructing a Pareto Chart Cause $ (billion) Admin. error 5.4 Employee theft 15.9 Shoplifting 12.7 Vendor fraud Millions of dollars Causes of Inventory Shrinkage 1.4 20 15 10 5 0 Employee Theft Shoplifting Admin. Error Cause Vendor fraud From the graph, it is easy to see that the causes of inventory shrinkage that should be addressed first are employee theft and shoplifting. © 2012 Pearson Education, Inc. All rights reserved. 60 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 61 of 149 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) © 2012 Pearson Education, Inc. All rights reserved. 62 of 149 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. © 2012 Pearson Education, Inc. All rights reserved. 63 of 149 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. © 2012 Pearson Education, Inc. All rights reserved. 64 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 65 of 149 Example: Constructing a Time Series Chart The table lists the number of cellular telephone subscribers (in millions) for the years 1998 through 2008. Construct a time series chart for the number of cellular subscribers. (Source: Cellular Telecommunication & Internet Association) © 2012 Pearson Education, Inc. All rights reserved. 66 of 149 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. © 2012 Pearson Education, Inc. All rights reserved. 67 of 149 Solution: Constructing a Time Series Chart The graph shows that the number of subscribers has been increasing since 1998, with greater increases recently. © 2012 Pearson Education, Inc. All rights reserved. 68 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 69 of 149 Assignment • Page 60 1-4 orally • Page 60-63 5-33 odd Larson/Farber 5th ed. 70 Chapter 2 Quiz 3 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. Lower limit 59 115 Upper limit 114 170 What is the class width in this chart?? What are the class boundaries for the first class in this chart? What is the midpoint of the first class in this chart? What do you call a chart which has classes, frequency, midpoints, relative frequency and cumulative frequency? What does this ∑ symbol mean? What is on the “y” axis on a frequency histogram? What is on the “x” axis on a frequency histogram? On the “x” axis of a frequency histogram, you may use two methods to indicate different classes. What are they? Does the first bar of a frequency histogram touch the “Y” axis? True/False: A stem and leaf plot is a qualitative graph of data. True/False: A pie chart is a qualitative graph of data. True/False: A Pareto Graph has all bars touching each other, and is ordered greatest to least. A ___________ uses a number line with marks above it to represent data values. Each entry in one data set corresponds to one entry in a second data set on a _____________ Data set is composed of quantitative entries taken at regular intervals over a period of time in a _________ _______ _________. Larson/Farber 5th ed. 71 Section 2.3 Measures of Central Tendency © 2012 Pearson Education, Inc. All rights reserved. 72 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 73 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 74 of 149 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 Pronounced “X Bar” always used for sample means © 2012 Pearson Education, Inc. All rights reserved. Lowercase Greek letter “mu” always used for the mean of 75 of 149 the population Example: Finding a Sample Mean The prices (in dollars) for a sample of round-trip flights from Chicago, Illinois to Cancun, Mexico are listed. What is the mean price of the flights? 872 432 397 427 388 782 397 © 2012 Pearson Education, Inc. All rights reserved. 76 of 149 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. © 2012 Pearson Education, Inc. All rights reserved. 77 of 149 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. © 2012 Pearson Education, Inc. All rights reserved. 78 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 79 of 149 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. © 2012 Pearson Education, Inc. All rights reserved. 80 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 81 of 149 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. © 2012 Pearson Education, Inc. All rights reserved. 82 of 149 Measure of Central Tendency: Mode Mode • The data entry that occurs with the greatest frequency. • A data set can have one mode, more than one mode, or no mode. • If no entry is repeated the data set has no mode. • If two entries occur with the same greatest frequency, each entry is a mode (bimodal). © 2012 Pearson Education, Inc. All rights reserved. 83 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 84 of 149 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. © 2012 Pearson Education, Inc. All rights reserved. 85 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 86 of 149 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. © 2012 Pearson Education, Inc. All rights reserved. 87 of 149 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). © 2012 Pearson Education, Inc. All rights reserved. 88 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 89 of 149 Solution: Comparing the Mean, Median, and Mode Ages in a class Mean: Median: Mode: 20 20 20 20 20 20 21 21 21 21 22 22 22 23 23 23 23 24 24 65 x 20 20 ... 24 65 x 23.8 years n 20 21 22 21.5 years 2 20 years (the entry occurring with the greatest frequency) © 2012 Pearson Education, Inc. All rights reserved. 90 of 149 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. © 2012 Pearson Education, Inc. All rights reserved. 91 of 149 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. © 2012 Pearson Education, Inc. All rights reserved. 92 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 93 of 149 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? © 2012 Pearson Education, Inc. All rights reserved. 94 of 149 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. © 2012 Pearson Education, Inc. All rights reserved. 95 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 96 of 149 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. © 2012 Pearson Education, Inc. All rights reserved. ( x f ) x n 97 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 98 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 99 of 149 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. © 2012 Pearson Education, Inc. All rights reserved. 100 of 149 The Shape of Distributions Uniform Distribution (rectangular) • All entries or classes in the distribution have equal or approximately equal frequencies. • Symmetric. © 2012 Pearson Education, Inc. All rights reserved. 101 of 149 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. © 2012 Pearson Education, Inc. All rights reserved. 102 of 149 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. © 2012 Pearson Education, Inc. All rights reserved. 103 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 104 of 149 Assignment • Page 72 1-4 orally • Page 72-77 5-57 odd Larson/Farber 5th ed. 105 Chapter 2 Quiz 4 (5 points) 1. T/F –In a uniform distribution, the mean, median and mode are all the same 2. T/F -If two entries occur with the same least frequency, each entry is a mode (bimodal). 3. T/F –An advantage of using the mean as a measure of central tendency is that it is not affected by outliers. 4. In a symmetrical distribution, the mean, median and mode are all the same 5. In a “Skewed Left” distribution, the median is to the ______ of the mean 6. In a “Skewed Right” distribution, the mean is to the _____ of the median • Quiz continues on next slide Larson/Farber 5th ed. 106 Chapter 2 Quiz 4 Continued (5 points) Use the frequency distribution to approximate the mean number of minutes that a sample of Internet subscribers spent online during their most recent session. Class Midpoint Frequency, f 7 – 18 12.5 8 19 – 30 24.5 11 31 – 42 36.5 16 43 – 54 48.5 10 55 – 66 60.5 8 67 – 78 72.5 5 79 – 90 84.5 2 © 2012 Pearson Education, Inc. All rights reserved. Quiz continues on next slide 107 of 149 Chapter 2 Quiz 4 Continued (5 points) Use the scores earned by the student below, calculate his weighted grade. Did he earn an A? (assume the teacher rounds grades to whole numbers). Source Score, x Weight, w Test Mean 90 0.50 Midterm 90 0.15 Final Exam 98 0.20 Computer Lab 98 0.10 Homework 100 0.05 © 2012 Pearson Education, Inc. All rights reserved. 108 of 149 Section 2.4 Measures of Variation © 2012 Pearson Education, Inc. All rights reserved. 109 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 110 of 149 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) © 2012 Pearson Education, Inc. All rights reserved. 111 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 112 of 149 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. © 2012 Pearson Education, Inc. All rights reserved. 113 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 114 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 115 of 149 Solution: Finding the Deviation • Determine the deviation for each data entry. Notice that the sum of the deviations is zero. This makes it hard to use the information without manipulation © 2012 Pearson Education, Inc. All rights reserved. Deviation ($1000s) Salary ($1000s), x 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 Σx = 415 42 – 41.5 = 0.5 Σ(x – μ) = 0 116 of 149 Deviation, Variance, and Standard Deviation By squaring the deviations, we can get usable information which relates to central tendency Population Variance ( x ) • N 2 2 Sum of squares, SSx Population Standard Deviation 2 ( x ) 2 • N © 2012 Pearson Education, Inc. All rights reserved. Greek lower case sigma –Sigma squared is the standard symbol for Pop variance Sigma is the standard symbol for Pop Standard Deviation 117 of 149 Finding the Population Variance & Standard Deviation In Words 1. Find the mean of the population data set. 2. Find the deviation of each entry. In Symbols x N x–μ 3. Square each deviation. (x – μ)2 4. Add to get the sum of squares. SSx = Σ(x – μ)2 © 2012 Pearson Education, Inc. All rights reserved. 118 of 149 Finding the Population Variance & Standard Deviation In Words 5. Divide by N to get the population variance. 6. Find the square root of the variance to get the population standard deviation. © 2012 Pearson Education, Inc. All rights reserved. In Symbols 2 ( x ) 2 N ( x ) 2 N 119 of 149 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. © 2012 Pearson Education, Inc. All rights reserved. 120 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 121 of 149 Solution: Finding the Population Standard Deviation Population Variance ( x ) 88.5 8.9 • N 10 2 2 We don’t use this value much Population Standard Deviation • 2 8.85 3.0 The population standard deviation is about 3.0, or $3000. © 2012 Pearson Education, Inc. All rights reserved. 122 of 149 Deviation, Variance, and Standard Deviation Sample Variance ( x x ) • s n 1 2 2 Everything is basically the “same” except that we use n-1 instead of N. This is considered a more conservative value Sample Standard Deviation • 2 ( x x ) s s2 n 1 © 2012 Pearson Education, Inc. All rights reserved. 123 of 149 Finding the Sample Variance & Standard Deviation In Words In Symbols x n 1. Find the mean of the sample data set. x 2. Find the 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 © 2012 Pearson Education, Inc. All rights reserved. 124 of 149 Finding the Sample Variance & Standard Deviation In Words 5. Divide by n – 1 to get the sample variance. 6. Find the square root of the variance to get the sample standard deviation. © 2012 Pearson Education, Inc. All rights reserved. In Symbols 2 ( x x ) s2 n 1 ( x x ) 2 s n 1 125 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 126 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 127 of 149 Solution: Finding the Sample Standard Deviation Sample Variance These were the same data values as before ( x x ) 88.5 9.8 Notice that the • s n 1 10 1 2 2 Sample Standard Deviation 88.5 3.1 • s s 9 2 sample standard deviation is slightly larger. In simple terms, we aren’t quite as accurate The sample standard deviation is about 3.1, or $3100. © 2012 Pearson Education, Inc. All rights reserved. 128 of 149 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.) © 2012 Pearson Education, Inc. All rights reserved. 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 38.00 129 of 149 Using the Calculator • • • • • Select “Stat” Select Edit Clear L1 (list 1) Enter the data into L1 Select “Stat” and highlight “Calc” to display Calc menu • Select 1: “1-Var Stats” and press “2nd L1” Enter Larson/Farber 5th ed. 130 Chapter 2 Quiz 5 • Do all parts of problem 61 on page 78 • Show all work • 4 points for each part (16 points total) Larson/Farber 5th ed. 132 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. © 2012 Pearson Education, Inc. All rights reserved. 133 of 149 Interpreting Standard Deviation: Empirical Rule (68 – 95 – 99.7 Rule) For data with a (symmetric) bell-shaped distribution, the standard deviation has the following characteristics: • About 68% of the data lie within one standard deviation of the mean. • About 95% of the data lie within two standard deviations of the mean. • About 99.7% of the data lie within three standard deviations of the mean. This may also be © 2012 Pearson Education, Inc. All rights reserved. referred to as a “normal distribution” 134 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. x x s x 2s x 3s 135 of 149 Example: Using the Empirical Rule In a survey conducted by the National Center for Health Statistics, the sample mean height of women in the United States (ages 20-29) was 64.3 inches, with a sample standard deviation of 2.62 inches. Estimate the percent of the women whose heights are between 59.06 inches and 64.3 inches. © 2012 Pearson Education, Inc. All rights reserved. 136 of 149 Solution: Using the Empirical Rule • Because the distribution is bell-shaped, you can use the Empirical Rule. 34% + 13.5% = 47.5% of women are between 59.06 and 64.3 inches tall. © 2012 Pearson Education, Inc. All rights reserved. 137 of 149 Chebychev’s Theorem The empirical rule works on bell (normal) distributions. This works on ALL distributions • 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. © 2012 Pearson Education, Inc. All rights reserved. 138 of 149 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? © 2012 Pearson Education, Inc. All rights reserved. 139 of 149 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. © 2012 Pearson Education, Inc. All rights reserved. 140 of 149 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 the sample standard deviation by using the midpoint of each class. This is the equation for Sample standard deviation: © 2012 Pearson Education, Inc. All rights reserved. ( x x ) s n 1 2 141 of 149 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. © 2012 Pearson Education, Inc. All rights reserved. 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 4 142 of 149 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. © 2012 Pearson Education, Inc. All rights reserved. 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 )= 91 143 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 144 of 149 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. What can I do with this information? Remember that the sample mean is 1.8 children. If we can assume that the distribution is normal, then about 68 percent of the sample lies within .1 children and 3.5 children © 2012 Pearson Education, Inc. All rights reserved. 145 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 146 of 149 Assignment • Page 90 1-10 orally • Page 90-97 11-47 odd Larson/Farber 5th ed. 147 Chapter 2 Quiz 6 (10 points) • Using the data from question 53 on page 76: 1. Calculate the sample standard deviation (3 points) 2. What is the range of beds which would include 68% of the sample? (3 points) 3. What is the range of beds which would include 95% of the beds? (3 points) Larson/Farber 5th ed. 148 Section 2.5 Measures of Position © 2012 Pearson Education, Inc. All rights reserved. 149 of 149 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) © 2012 Pearson Education, Inc. All rights reserved. 150 of 149 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. © 2012 Pearson Education, Inc. All rights reserved. 151 of 149 Example: Finding Quartiles The number of nuclear power plants in the top 15 nuclear power-producing countries in the world are listed. Find the first, second, and third quartiles of the data set. 7 18 11 6 59 17 18 54 104 20 31 8 10 15 19 Solution: • Q2 divides the data set into two halves. Lower half Upper half 6 7 8 10 11 15 17 18 18 19 20 31 54 59 104 Q2 © 2012 Pearson Education, Inc. All rights reserved. 152 of 149 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 6 7 8 10 11 15 17 18 18 19 20 31 54 59 104 Q1 Q2 Q3 About one fourth of the countries have 10 or fewer nuclear power plants; about one half have 18 or fewer; and about three fourths have 31 or fewer. © 2012 Pearson Education, Inc. All rights reserved. 153 of 149 Interquartile Range Interquartile Range (IQR) • The difference between the third and first quartiles. • IQR = Q3 – Q1 © 2012 Pearson Education, Inc. All rights reserved. 154 of 149 Example: Finding the Interquartile Range Find the interquartile range of the data set. 7 18 11 6 59 17 18 54 104 20 31 8 10 15 19 Recall Q1 = 10, Q2 = 18, and Q3 = 31 Solution: • IQR = Q3 – Q1 = 31 – 10 = 21 The number of power plants in the middle portion of the data set vary by at most 21. © 2012 Pearson Education, Inc. All rights reserved. 155 of 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 © 2012 Pearson Education, Inc. All rights reserved. 156 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. Median, Q2 Q3 Maximum entry 157 of 149 Example: Drawing a Box-and-Whisker Plot Draw a box-and-whisker plot that represents the data set. 7 18 11 6 59 17 18 54 104 20 31 8 10 15 19 Min = 6, Q1 = 10, Q2 = 18, Q3 = 31, Max = 104, Solution: About half the data values are between 10 and 31. By looking at the length of the right whisker, you can conclude 104 is a possible outlier. © 2012 Pearson Education, Inc. All rights reserved. 158 of 149 Percentiles and Other Fractiles Fractiles Quartiles Deciles Percentiles Summary Divide a data set into 4 equal parts Divide a data set into 10 equal parts Divide a data set into 100 equal parts © 2012 Pearson Education, Inc. All rights reserved. Symbols Q1, Q2, Q3 D1, D2, D3,…, D9 P1, P2, P3,…, P99 159 of 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 62nd percentile? How should you interpret this? (Source: College Board) © 2012 Pearson Education, Inc. All rights reserved. 160 of 149 Solution: Interpreting Percentiles The 62nd percentile corresponds to a test score of 1600. This means that 62% of the students had an SAT score of 1600 or less. © 2012 Pearson Education, Inc. All rights reserved. 161 of 149 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 © 2012 Pearson Education, Inc. All rights reserved. 162 of 149 Example: Comparing z-Scores from Different Data Sets In 2009, Heath Ledger won the Oscar for Best Supporting Actor at age 29 for his role in the movie The Dark Knight. Penelope Cruz won the Oscar for Best Supporting Actress at age 34 for her role in Vicky Cristina Barcelona. The mean age of all Best Supporting Actor winners is 49.5, with a standard deviation of 13.8. The mean age of all Best Supporting Actress winners is 39.9, with a standard deviation of 14.0. Find the z-scores that correspond to the ages of Ledger and Cruz. Then compare your results. © 2012 Pearson Education, Inc. All rights reserved. 163 of 149 Solution: Comparing z-Scores from Different Data Sets • Heath Ledger z x 29 49.5 1.49 13.8 1.49 standard deviations below the mean • Penelope Cruz z x 34 39.9 0.42 14.0 © 2012 Pearson Education, Inc. All rights reserved. 0.42 standard deviations below the mean 164 of 149 Solution: Comparing z-Scores from Different Data Sets About 95 percent of the data should fall between -2 and 2. Both z-scores fall between –2 and 2, so neither score would be considered unusual. Compared with other Best Supporting Actor winners, Heath Ledger was relatively younger, whereas the age of Penelope Cruz was only slightly lower than the average age of other Best Supporting Actress winners. © 2012 Pearson Education, Inc. All rights reserved. 165 of 149 Section 2.5 Summary • • • • • Determined the quartiles of a data set Determined the interquartile range of a data set Created a box-and-whisker plot Interpreted other fractiles such as percentiles Determined and interpreted the standard score (z-score) © 2012 Pearson Education, Inc. All rights reserved. 166 of 149 Assignment • Page 107 7-13 odd, 31,33, 37-50 Larson/Farber 5th ed. 167 Group Project (30 points) • 5 pts • 5 pts • 5 pts • 5 pts • • • Gather a sample of both men’s and women’s shoe sizes Create an Expanded Frequency chart using 5 classes Create a Frequency Histogram (using midpoints for x) Calculate the mean, median, and mode of your data Calculate the sample standard deviation of your data If your sample IS normally distributed, then: Using your mean and sample deviation, and the Empirical Rule: 5 pts Between what two shoe sizes would we find 68% of the shoe sizes? 5 pts Between what two shoe sizes would we find 95% of the shoe sizes? -Or • If your sample is NOT normally distributed, then use Chebychev’s theorem to find: Between what two shoe sizes would we find 75% of the shoe sizes? 5 pts Between what two shoe sizes would we find 88.9% of the shoe sizes? 5 pts Larson/Farber 5th ed. 168