Lecture 2 Frequency Distributions and Their Graphs PDF

Lecture 2.1 Objectives How to construct a frequency distribution including limits, midpoints, relative frequencies, cumulative...

Lecture 2.1 Objectives How to construct a frequency distribution including limits, midpoints, relative frequencies, cumulative frequencies, and boundaries Lecture 2 How to construct frequency histograms, frequency polygons, relative frequency histograms, and ogives Frequency Distributions and Their Graphs Frequency Distribution Constructing a Frequency Distribution Frequency Distribution Class Frequency, f 1. Decide on the number of classes. A table that shows Class width 1–5 5
Usually between 5 and 20; otherwise, it may be classes or intervals of 6 – 1 = 5 difficult to detect any patterns. 6 – 10 8 data with a count of the 2. Find the class width. number of entries in each 11 – 15 6
Determine the range of the data. class. 16 – 20 8
Divide the range by the number of classes. The frequency, f, of a 21 – 25 5 class is the number of
Round up to the next convenient number. 26 – 30 4 data entries in the class. Lower class Upper class limits limits  Constructing a Frequency Distribution Constructing a Frequency Distribution 3. Find the class limits. 4. Make a tally mark for each data entry in the row of
You can use the minimum data entry as the lower the appropriate class. limit of the first class. 5. Count the tally marks to find the total frequency f
Find the remaining lower limits (add the class for each 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. Example: Constructing a Frequency Solution: Constructing a Frequency Distribution Distribution The following sample data set lists the prices (in 90 130 400 200 350 70 325 250 150 250 dollars) of 30 portable global positioning system (GPS) 275 270 150 130 59 200 160 450 300 130 navigators. Construct a frequency distribution that has 220 100 200 400 200 250 95 180 170 150 seven classes. 90 130 400 200 350 70 325 250 150 250 1. Number of classes = 7 (given) 275 270 150 130 59 200 160 450 300 130 2. Find the class width 220 100 200 400 200 250 95 180 170 150 max − min 450 − 59 391 = = ≈ 55.86 #classes 7 7 Round up to 56  Solution: Constructing a Frequency Solution: Constructing a Frequency Distribution Distribution 3. Use 59 (minimum value) Lower Upper The upper limit of the Lower Upper as first lower limit. Add limit limit first class is 114 (one less limit limit the class width of 56 to Class 59 than the lower limit of the 59 114 Class get the lower limit of the width = 56 115 second class). 115 170 width = 56 next class. 171 Add the class width of 56 171 226 59 + 56 = 115 227 to get the upper limit of 227 282 the next class. Find the remaining 283 283 338 lower limits. 339 114 + 56 = 170 339 394 395 Find the remaining upper 395 450 limits. Solution: Constructing a Frequency Determining the Midpoint Distribution 4. Make a tally mark for each data entry in the row of Midpoint of a class the appropriate class. (Lower class limit) + (Upper class limit) 5. Count the tally marks to find the total frequency f 2 for each class. Class Tally Frequency, f 59 – 114 IIII 5 Class Midpoint Frequency, f 115 – 170 IIII III 8 59 + 114 59 – 114 = 86.5 5 2 171 – 226 IIII I 6 Class width = 56 115 + 170 227 – 282 IIII 5 115 – 170 = 142.5 8 2 283 – 338 II 2 171 + 226 339 – 394 I 1 171 – 226 = 198.5 6 2 395 – 450 III 3  Determining the Relative Frequency Determining the Cumulative Frequency Relative Frequency of a class Cumulative frequency of a class Portion or percentage of the data that falls in a The sum of the frequency for that class and all particular class. previous classes. relative frequency = class frequency f = Class Frequency, f Cumulative frequency Sample size n 59 – 114 5 5 Class Frequency, f Relative Frequency 115 – 170 + 8 13 5 59 – 114 5 ≈ 0.17 30 171 – 226 + 6 19 8 115 – 170 8 ≈ 0.27 30 6 171 – 226 6 = 0.2 30 Expanded Frequency Distribution Graphs of Frequency Distributions Frequency Histogram Relative Cumulative Class Frequency, f Midpoint frequency frequency A bar graph that represents the frequency distribution. 59 – 114 5 86.5 0.17 5 The horizontal scale is quantitative and measures the 115 – 170 8 142.5 0.27 13 data values. 171 – 226 6 198.5 0.2 19 227 – 282 5 254.5 0.17 24 The vertical scale measures the frequencies of the 283 – 338 2 310.5 0.07 26 classes. 339 – 394 1 366.5 0.03 27 Consecutive bars must touch. frequency 395 – 450 3 422.5 0.1 30 Σf = 30 f ∑ =1 n data values  Class Boundaries Class Boundaries Class boundaries Class Frequency, The numbers that separate classes without forming Class boundaries f gaps between them. 59 – 114 58.5 – 114.5 5 The distance from the upper 115 – 170 114.5 – 170.5 8 Class Frequency, limit of the first class to the Class Boundaries f 171 – 226 170.5 – 226.5 6 lower limit of the second 59 – 114 58.5 – 114.5 5 227 – 282 226.5 – 282.5 5 class is 115 – 114 = 1. 115 – 170 8 283 – 338 282.5 – 338.5 2 Half this distance is 0.5. 171 – 226 6 339 – 394 338.5 – 394.5 1 395 – 450 394.5 – 450.5 3 First class lower boundary = 59 – 0.5 = 58.5 First class upper boundary = 114 + 0.5 = 114.5 Solution: Frequency Histogram Example: Frequency Histogram (using Midpoints) Construct a frequency histogram for the global positioning system (GPS) navigators. Class Frequency, Class boundaries Midpoint f 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  Solution: Frequency Histogram Graphs of Frequency Distributions (using class boundaries) Frequency Polygon A line graph that emphasizes the continuous change in frequencies. frequency data values You can see that more than half of the GPS navigators are priced below $226.50. Example: Frequency Polygon Solution: Frequency Polygon Construct a frequency polygon for the GPS navigators The graph should begin and end on the frequency distribution. horizontal axis, so extend the left side to Class Midpoint Frequency, f one class width before 59 – 114 86.5 5 the first class midpoint and extend 115 – 170 142.5 8 the right side to one 171 – 226 198.5 6 class width after the 227 – 282 254.5 5 last class midpoint. 283 – 338 310.5 2 339 – 394 366.5 1 You can see that the frequency of GPS navigators increases 395 – 450 422.5 3 up to $142.50 and then decreases.  Graphs of Frequency Distributions Example: Relative Frequency Histogram Relative Frequency Histogram Construct a relative frequency histogram for the GPS Has the same shape and the same horizontal scale as navigators frequency distribution. the corresponding frequency histogram. Class Frequency, Relative Class boundaries f frequency The vertical scale measures the relative frequencies, 59 – 114 58.5 – 114.5 86.5 0.17 not frequencies. 115 – 170 114.5 – 170.5 142.5 0.27 171 – 226 170.5 – 226.5 198.5 0.2 frequency relative 227 – 282 226.5 – 282.5 254.5 0.17 283 – 338 282.5 – 338.5 310.5 0.07 339 – 394 338.5 – 394.5 366.5 0.03 data values 395 – 450 394.5 – 450.5 422.5 0.1 Solution: Relative Frequency Histogram Graphs of Frequency Distributions Cumulative Frequency Graph or Ogive A line graph that displays the cumulative frequency of each class at its upper class boundary. The upper boundaries are marked on the horizontal axis. The cumulative frequencies are marked on the 6.5 18.5 30.5 42.5 54.5 66.5 78.5 90.5 vertical axis. cumulative frequency From this graph you can see that 20% of GPS navigators are priced between $170.50 and $226.50. data values  Constructing an Ogive Constructing an Ogive 1. Construct a frequency distribution that includes 4. Connect the points in order from left to right. cumulative frequencies as one of the columns. 5. The graph should start at the lower boundary of the 2. Specify the horizontal and vertical scales. first class (cumulative frequency is zero) and should
The horizontal scale consists of the upper class end at the upper boundary of the last class boundaries. (cumulative frequency is equal to the sample size).
The vertical scale measures cumulative frequencies. 3. Plot points that represent the upper class boundaries and their corresponding cumulative frequencies. Example: Ogive Solution: Ogive Construct an ogive for the GPS navigators frequency distribution. Class Frequency, Cumulative Class boundaries f frequency 59 – 114 58.5 – 114.5 86.5 5 115 – 170 114.5 – 170.5 142.5 13 171 – 226 170.5 – 226.5 198.5 19 6.5 18.5 30.5 42.5 54.5 66.5 78.5 90.5 227 – 282 226.5 – 282.5 254.5 24 283 – 338 282.5 – 338.5 310.5 26 From the ogive, you can see that about 25 GPS navigators cost 339 – 394 338.5 – 394.5 366.5 27 $300 or less. The greatest increase occurs between $114.50 and 395 – 450 394.5 – 450.5 422.5 30 $170.50.  Graphing Quantitative Data Sets Stem-and-leaf plot Each number is separated into a stem and a leaf. Lecture 2.2 Similar to a histogram. Still contains original data values. 26 More Graphs and Displays Data: 21, 25, 25, 26, 27, 28, 2 1 5 5 6 7 8 30, 36, 36, 45 3 0 6 6 4 5. Example: Constructing a Stem-and-Leaf Solution: Constructing a Stem-and-Leaf Plot Plot The following are the numbers of text messages sent 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 last month by the cellular phone users on one floor of a 139 139 122 78 133 126 123 145 121 134 124 119 132 133 124 college dormitory. Display the data in a stem-and-leaf 129 112 126 148 147 plot. 155 159 144 129 105 145 126 116 130 114 122 112 112 142 126 The data entries go from a low of 78 to a high of 159. 118 118 108 122 121 109 140 126 119 113 117 118 109 109 119 Use the rightmost digit as the leaf. 139 139 122 78 133 126 123 145 121 134 124 119 132 133 124 129 112 126 148 147
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.  Solution: Constructing a Stem-and-Leaf Graphing Quantitative Data Sets Plot Dot plot Include a key to identify Each data entry is plotted, using a point, above a the values of the data. 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 From the display, you can conclude that more than 50% of the cellular phone users sent between 110 and 130 text messages.. Example: Constructing a Dot Plot Solution: Constructing a Dot Plot Use a dot plot organize the text messaging data. 155 159 144 129 105 145 126 116 130 114 122 112 112 142 126 118 118 108 122 121 109 140 126 119 113 117 118 109 109 119 155 159 144 129 105 145 126 116 130 114 122 112 112 142 126 139 139 122 78 133 126 123 145 121 134 124 119 132 133 124 118 118 108 122 121 109 140 126 119 113 117 118 109 109 119 129 112 126 148 147 139 139 122 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. 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...  Graphing Qualitative Data Sets Example: Constructing a Pie Chart Pie Chart The numbers of earned degrees conferred (in thousands) A circle is divided into sectors that represent in 2007 are shown in the table. Use a pie chart to categories. organize the data. (Source: U.S. National Center for Educational Statistics) The area of each sector is proportional to the Type of degree Number frequency of each category. (thousands) Associate’s 728 Bachelor’s 1525 Master’s 604 First professional 90 Doctoral 60. Solution: Constructing a Pie Chart Solution: Constructing a Pie Chart Find the relative frequency (percent) of each category. Type of degree Frequency, f Relative frequency Construct the pie chart using the central angle that corresponds to each category. Associate’s 728 728 ≈ 0.24
To find the central angle, multiply 360º by the 3007 Bachelor’s 1525 category's relative frequency. 1525 ≈ 0.51 3007
For example, the central angle for cars is Master’s 604 604 ≈ 0.20 360(0.24) ≈ 86º 3007 First professional 90 90 ≈ 0.03 3007 Doctoral 60 ≈ 0.02 60 3007 3007..  Solution: Constructing a Pie Chart Solution: Constructing a Pie Chart Relative Relative Central Type of degree Frequency, f frequency Central angle Type of degree frequency angle Associate’s 728 0.24 360º(0.24)≈86º Associate’s 0.24 86º Bachelor’s 0.51 184º Bachelor’s 1525 0.51 360º(0.51)≈184º Master’s 0.20 72º 360º(0.20)≈72º First professional 0.03 11º Master’s 604 0.20 Doctoral 0.02 7º First professional 90 0.03 360º(0.03)≈11º From the pie chart, you can see that most fatalities in motor Doctoral 60 0.02 360º(0.02)≈7º vehicle crashes were those involving the occupants of cars... Measures of Central Tendency Measure of central tendency A value that represents a typical, or central, entry of a data set. Measures of Central Tendency Most common measures of central tendency:
Mean
Median
Mode..  Measure of Central Tendency: Mean Example: Finding a Sample Mean Mean (average) The prices (in dollars) for a sample of roundtrip flights The sum of all the data entries divided by the number from Chicago, Illinois to Cancun, Mexico are listed. of entries. What is the mean price of the flights? Sigma notation: Σx = add all of the data entries (x) 872 432 397 427 388 782 397 in the data set. Σx Population mean: µ = N Σx Sample mean: x= n.. Solution: Finding a Sample Mean Measure of Central Tendency: Median 872 432 397 427 388 782 397 Median The sum of the flight prices is The value that lies in the middle of the data when the Σx = 872 + 432 + 397 + 427 + 388 + 782 + 397 = 3695 data set is ordered. Measures the center of an ordered data set by dividing To find the mean price, divide the sum of the prices by the number of prices in the sample it into two equal parts. If the data set has an Σx 3695
odd number of entries: median is the middle data x= = ≈ 527.9 entry. n 7
even number of entries: median is the mean of The mean price of the flights is about $527.90. the two middle data entries...  Example: Finding the Median Solution: Finding the Median The prices (in dollars) for a sample of roundtrip flights 872 432 397 427 388 782 397 from Chicago, Illinois to Cancun, Mexico are listed. Find the median of the flight prices. First order the data. 872 432 397 427 388 782 397 388 397 397 427 432 782 872 There are seven entries (an odd number), the median is the middle, or fourth, data entry. The median price of the flights is $427... Example: Finding the Median Solution: Finding the Median The flight priced at $432 is no longer available. What is 872 397 427 388 782 397 the median price of the remaining flights? First order the data. 872 397 427 388 782 397 388 397 397 427 782 872 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...  Measure of Central Tendency: Mode Example: Finding the Mode Mode The prices (in dollars) for a sample of roundtrip flights The data entry that occurs with the greatest frequency. from Chicago, Illinois to Cancun, Mexico are listed. If no entry is repeated the data set has no mode. Find the mode of the flight prices. If two entries occur with the same greatest frequency, 872 432 397 427 388 782 397 each entry is a mode (bimodal)... Solution: Finding the Mode Example: Finding the Mode 872 432 397 427 388 782 397 At a political debate a sample of audience members was asked to name the political party to which they belong. Ordering the data helps to find the mode. Their responses are shown in the table. What is the 388 397 397 427 432 782 872 mode of the responses? Political Party Frequency, f The entry of 397 occurs twice, whereas the other Democrat 34 data entries occur only once. Republican 56 Other 21 The mode of the flight prices is $397. Did not respond 9..  Solution: Finding the Mode Comparing the Mean, Median, and Mode Political Party Frequency, f All three measures describe a typical entry of a data Democrat 34 set. Republican 56 Advantage of using the mean: Other 21
The mean is a reliable measure because it takes Did not respond 9 into account every entry of a data set. Disadvantage of using the mean: 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.
Greatly affected by outliers (a data entry that is far removed from the other entries in the data set)... Example: Comparing the Mean, Median, Solution: Comparing the Mean, Median, and Mode and Mode Ages in a class Find the mean, median, and mode of the sample ages of 20 20 20 20 20 20 21 a class shown. Which measure of central tendency best 21 21 21 22 22 22 23 describes a typical entry of this data set? Are there any 23 23 23 24 24 65 outliers? Ages in a class Mean: Σx 20 + 20 +... + 24 + 65 x= = ≈ 23.8 years 20 20 20 20 20 20 21 n 20 21 21 21 22 22 22 23 Median: 21 + 22 23 23 23 24 24 65 = 21.5 years 2 Mode: 20 years (the entry occurring with the greatest frequency)..  Solution: Comparing the Mean, Median, Solution: Comparing the Mean, Median, and Mode and Mode Sometimes a graphical comparison can help you decide which measure of central Mean ≈ 23.8 years Median = 21.5 years Mode = 20 years tendency best represents a data set. 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. In this case, it appears that the median best describes the data set... Weighted Mean Example: Finding a Weighted Mean Weighted Mean You are taking a class in which your grade is The mean of a data set whose entries have varying determined from five sources: 50% from your test weights. mean, 15% from your midterm, 20% from your final exam, 10% from your computer lab work, and 5% from Σ( x ⋅ w) your homework. Your scores are 86 (test mean), 96 x = where w is the weight of each entry x (midterm), 82 (final exam), 98 (computer lab), and 100 Σw (homework). What is the weighted mean of your scores? If the minimum average for an A is 90, did you get an A?..  Solution: Finding a Weighted Mean Mean of Grouped Data Source Score, x Weight, w x·w Test Mean 86 0.50 86(0.50)= 43.0 Mean of a Frequency Distribution Midterm 96 0.15 96(0.15) = 14.4 Approximated by Final Exam 82 0.20 82(0.20) = 16.4 Σ( x ⋅ f ) Computer Lab 98 0.10 98(0.10) = 9.8 x= n = Σf Homework 100 0.05 100(0.05) = 5.0 n Σw = 1 Σ(x·w) = 88.6 where x and f are the midpoints and frequencies of a class, respectively Σ(x ⋅w) 8 8.6 x = = = 8 8.6 Σw 1 Your weighted mean for the course is 88.6. You did not get an A... Example: Find the Mean of a Frequency Solution: Find the Mean of a Frequency Distribution Distribution Use the frequency distribution to approximate the mean Class Midpoint, x Frequency, f (x·f) number of minutes that a sample of Internet subscribers 7 – 18 12.5 6 12.5·6 = 75.0 spent online during their most recent session. 19 – 30 24.5 10 24.5·10 = 245.0 31 – 42 36.5 13 36.5·13 = 474.5 Class Midpoint Frequency, f 43 – 54 48.5 8 48.5·8 = 388.0 7 – 18 12.5 6 55 – 66 60.5 5 60.5·5 = 302.5 19 – 30 24.5 10 67 – 78 72.5 6 72.5·6 = 435.0 31 – 42 36.5 13 79 – 90 84.5 2 84.5·2 = 169.0 43 – 54 48.5 8 n = 50 Σ(x·f) = 2089.0 55 – 66 60.5 5 67 – 78 72.5 6 Σ( x ⋅ f ) 2089 x= = ≈ 41.8 minutes 79 – 90 84.5 2 n 50..  Range Range The difference between the maximum and minimum Lecture 2.4 data entries in the set. The data must be quantitative. Range = (Max. data entry) – (Min. data entry) Measures of Variation.. Example: Finding the Range Solution: Finding the Range A corporation hired 10 graduates. The starting salaries Ordering the data helps to find the least and greatest for each graduate are shown. Find the range of the salaries. starting salaries. 37 38 39 41 41 41 42 44 45 47 Starting salaries (1000s of dollars) minimum maximum 41 38 39 45 47 41 44 41 37 42 Range = (Max. salary) – (Min. salary) = 47 – 37 = 10 The range of starting salaries is 10 or $10,000...  Deviation, Variance, and Standard Example: Finding the Deviation Deviation Deviation A corporation hired 10 graduates. The starting salaries The difference between the data entry, x, and the for each graduate are shown. Find the deviation of the mean of the data set. starting salaries. Population data set: Starting salaries (1000s of dollars)
Deviation of x = x – µ 41 38 39 45 47 41 44 41 37 42 Sample data set: Solution: First determine the mean starting salary.
Deviation of x = x – x Σx 415 µ= = = 41.5 N 10.. Deviation, Variance, and Standard Solution: Finding the Deviation Deviation Determine the Salary ($1000s), x Deviation: x – µ Population Variance deviation for each 41 41 – 41.5 = –0.5 data entry. 38 38 – 41.5 = –3.5 Σ( x − µ ) 2 2 Sum of squares, SSx 39 39 – 41.5 = –2.5 σ = N 45 45 – 41.5 = 3.5 47 47 – 41.5 = 5.5 Population Standard Deviation 41 41 – 41.5 = –0.5 44 44 – 41.5 = 2.5 Σ( x − µ ) 2 2 41 41 – 41.5 = –0.5 σ = σ = 37 37 – 41.5 = –4.5 N 42 42 – 41.5 = 0.5 Σx = 415 Σ(x – µ) = 0..  Finding the Population Variance & Finding the Population Variance & Standard Deviation Standard Deviation In Words In Symbols In Words In Symbols 1. Find the mean of the Σx 5. Divide by N to get the Σ( x − µ ) 2 µ= σ2 = population data set. N population variance. N 2. Find deviation of each x–µ 6. Find the square root to get Σ( x − µ ) 2 entry. the population standard σ= deviation. N 3. Square each deviation. (x – µ)2 4. Add to get the sum of SSx = Σ(x – µ)2 squares... Example: Finding the Population Solution: Finding the Population Standard Deviation Standard Deviation A corporation hired 10 graduates. The starting salaries Determine SSx Salary, x Deviation: x – µ Squares: (x – µ)2 for each graduate are shown. Find the population N = 10 41 41 – 41.5 = –0.5 (–0.5)2 = 0.25 38 38 – 41.5 = –3.5 (–3.5)2 = 12.25 variance and standard deviation of the starting salaries. 39 39 – 41.5 = –2.5 (–2.5)2 = 6.25 Starting salaries (1000s of dollars) 45 45 – 41.5 = 3.5 (3.5)2 = 12.25 41 38 39 45 47 41 44 41 37 42 47 47 – 41.5 = 5.5 (5.5)2 = 30.25 Recall µ = 41.5. 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..  Solution: Finding the Population Deviation, Variance, and Standard Standard Deviation Deviation Population Variance Sample Variance Σ( x − µ ) 2 88.5 2 2Σ( x − x ) 2 σ = = ≈ 8.9 s = N 10 n −1 Population Standard Deviation Sample Standard Deviation 2 2 Σ( x − x ) 2 σ = σ = 8.85 ≈ 3.0 s= s = n −1 The population standard deviation is about 3.0, or $3000... Example: Finding the Sample Standard Solution: Finding the Sample Standard Deviation Deviation The starting salaries are for the Chicago branches of a Determine SSx Salary, x Deviation: x – µ Squares: (x – µ)2 corporation. The corporation has several other branches, n = 10 41 41 – 41.5 = –0.5 (–0.5)2 = 0.25 38 38 – 41.5 = –3.5 (–3.5)2 = 12.25 and you plan to use the starting salaries of the Chicago 39 39 – 41.5 = –2.5 (–2.5)2 = 6.25 branches to estimate the starting salaries for the larger 45 45 – 41.5 = 3.5 (3.5)2 = 12.25 population. Find the sample standard deviation of the 47 47 – 41.5 = 5.5 (5.5)2 = 30.25 starting salaries. 41 41 – 41.5 = –0.5 (–0.5)2 = 0.25 Starting salaries (1000s of dollars) 44 44 – 41.5 = 2.5 (2.5)2 = 6.25 41 38 39 45 47 41 44 41 37 42 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..  Solution: Finding the Sample Standard Interpreting Standard Deviation Deviation Sample Variance Standard deviation is a measure of the typical amount an entry deviates from the mean. Σ( x − x ) 2 88.5 2 The more the entries are spread out, the greater the s = = ≈ 9.8 n −1 10 − 1 standard deviation. Sample Standard Deviation 88.5 s= s2 = ≈ 3.1 9 The sample standard deviation is about 3.1, or $3100... Interpreting Standard Deviation: Interpreting Standard Deviation: Empirical Rule (68 – 95 – 99.7 Rule) Empirical Rule (68 – 95 – 99.7 Rule) 99.7% within 3 standard deviations For data with a (symmetric) bell-shaped distribution, the 95% within 2 standard deviations standard deviation has the following characteristics: 68% within 1 standard deviation About 68% of the data lie within one standard deviation of the mean. About 95% of the data lie within two standard 34% 34% deviations of the mean. About 99.7% of the data lie within three standard 2.35% 2.35% 13.5% 13.5% deviations of the mean. x − 3s x − 2s x −s x x +s x + 2s x + 3s..  Example: Using the Empirical Rule Solution: Using the Empirical Rule Because the distribution is bell-shaped, you can use 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. 34% + 13.5% = 47.5% of women are between 59.06 and 64.3 inches tall... Chebychev’s Theorem Example: Using Chebychev’ ’s Theorem The portion of any data set lying within k standard The age distribution for Florida is shown in the deviations (k > 1) of the mean is at least: histogram. Apply Chebychev’s Theorem to the data 1 using k = 2. What can you conclude? 1− 2 k k = 2: In any data set, at least 1 3 1− = or 75% 22 4 of the data lie within 2 standard deviations of the mean. k = 3: In any data set, at least 1 8 1− = or 88.9% 32 9 of the data lie within 3 standard deviations of the mean...  Solution: Using Chebychev’ ’s Theorem Standard Deviation for Grouped Data Sample standard deviation for a frequency distribution Σ( x − x ) 2 f where n= Σf (the number of s= n −1 entries in the data set) k = 2: µ – 2σ = 39.2 – 2(24.8) = -10.4 (use 0 since age can’t be negative) When a frequency distribution has classes, estimate the µ + 2σ = 39.2 + 2(24.8) = 88.8 sample mean and standard deviation by using the At least 75% of the population of Florida is between 0 and 88.8 years old. midpoint of each class... Example: Finding the Standard Deviation Solution: Finding the Standard Deviation for Grouped Data for Grouped Data You collect a random sample of the Number of Children in First construct a frequency distribution. 50 Households number of children per household in 1 3 1 1 1 x f xf a region. Find the sample mean and Find the mean of the frequency 0 10 0(10) = 0 1 2 2 1 0 the sample standard deviation of the 1 1 0 0 0 distribution. 1 19 1(19) = 19 2 7 2(7) = 14 data set. 1 5 0 3 6 Σxf 91 3 0 3 1 1 x= = ≈ 1.8 3 7 3(7) =21 1 1 6 0 1 n 50 4 2 4(2) = 8 3 6 6 1 2 5 1 5(1) = 5 The sample mean is about 1.8 children. 2 3 0 1 1 6 4 6(4) = 24 4 1 1 2 2 Σf = 50 Σ(xf )= 91 0 3 0 2 4..  Solution: Finding the Standard Deviation Solution: Finding the Standard Deviation for Grouped Data for Grouped Data Determine the sum of squares. Find the sample standard deviation. x f x−x ( x − x )2 ( x − x )2 f x−x ( x − x )2 ( x − x )2 f Σ( x − x ) 2 f 145.40 0 10 0 – 1.8 = –1.8 (–1.8)2 = 3.24 3.24(10) = 32.40 s= = ≈ 1.7 1 19 1 – 1.8 = –0.8 (–0.8)2 = 0.64 0.64(19) = 12.16 n −1 50 − 1 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 The standard deviation is about 1.7 children. 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 Σ( x − x ) 2 f = 145.40.. Example: Comparing Variation in Coefficient of Variation Different Data Sets Coefficient of Variation (CV) The table shows the population Describes the standard deviation of a data set as a heights (in inches) and weights (in percent of the mean. pounds) of the members of a Population data set: basketball team. Find the
CV = σ 100% coefficient of variation for the µ heights and the weighs. Then compare the results. Sample data set:
s CV = 100% x..  Solution: Comparing Variation in Solution: Comparing Variation in Different Data Sets Different Data Sets The mean weight is µ ≈ 187.8 pounds with a standard The mean height is µ ≈ 72.8 inches with a standard deviation of σ ≈ 17.7 pounds. The coefficient of deviation of σ ≈ 3.3 inches. The coefficient of variation variation for the weights is for the heights is σ σ CVweight = 100% CVheight = 100% µ µ 17.7 3.3 = 100% = 100% 187.8 72.8 ≈ 9.4% ≈ 4.5% The weights (9.4%) are more variable than the heights (4.5%)... Quartiles Fractiles are numbers that partition (divide) an ordered data set into equal parts. Quartiles approximately divide an ordered data set Lecture 2.5 into four equal parts.
First quartile, Q1: About one quarter of the data fall on or below Q1. Measures of Position
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...  Example: Finding Quartiles Solution: Finding Quartiles The first and third quartiles are the medians of the The number of nuclear power plants in the top 15 lower and upper halves of the data set. nuclear power-producing countries in the world are listed. Find the first, second, and third quartiles of the Lower half Upper half data set. 6 7 8 10 11 15 17 18 18 19 20 31 54 59 104 7 18 11 6 59 17 18 54 104 20 31 8 10 15 19 Q1 Q2 Q3 Solution: Q2 divides the data set into two halves. Lower half Upper half About one fourth of the countries have 10 or less, 6 7 8 10 11 15 17 18 18 19 20 31 54 59 104 about one half have 18 or less; and about three fourths have 31 or less. Q2.. Interquartile Range Example: Finding the Interquartile Range Interquartile Range (IQR) Find the interquartile range of the data set. The difference between the third and first quartiles. Recall Q1 = 10, Q2 = 18, and Q3 = 31 IQR = Q3 – Q1 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...  Box-and-Whisker Plot Drawing a Box-and-Whisker Plot Box-and-whisker plot 1. Find the five-number summary of the data set. Exploratory data analysis tool. 2. Construct a horizontal scale that spans the range of Highlights important features of a data set. the data. Requires (five-number summary): 3. Plot the five numbers above the horizontal scale.
Minimum entry 4. Draw a box above the horizontal scale from Q1 to Q3 and draw a vertical line in the box at Q2.
First quartile Q1 5. Draw whiskers from the box to the minimum and
Median Q2 maximum entries.
Third quartile Q3 Whisker Box Whisker
Maximum entry Minimum Maximum entry Q1 Median, Q2 Q3 entry.. Example: Drawing a Box-and-Whisker Percentiles and Other Fractiles Plot Draw a box-and-whisker plot that represents the 15 data Fractiles Summary Symbols set. Quartiles Divides data into 4 equal Q1, Q2, Q3 parts Min = 6, Q1 = 10, Q2 = 18, Q3 = 31, Max = 104, Deciles Divides data into 10 equal D1, D2, D3,…, D9 Solution: parts Percentiles Divides data into 100 equal P1, P2, P3,…, P99 parts About half the scores are between 10 and 31. By looking at the length of the right whisker, you can conclude 104 is a possible outlier...  Example: Interpreting Percentiles Solution: Interpreting Percentiles The ogive represents the cumulative frequency The 62nd percentile distribution for SAT test corresponds to a test score scores of college-bound of 1600. students in a recent year. What This means that 62% of the test score represents the 62nd students had an SAT score percentile? How should you of 1600 or less. interpret this? (Source: College Board).. Example: Comparing z-Scores from The Standard Score Different Data Sets Standard Score (z-score) In 2009, Heath Ledger won the Oscar for Best Represents the number of standard deviations a given Supporting Actor at age 29 for his role in the movie The value x falls from the mean µ. Dark Knight. Penelope Cruz won the Oscar for Best Supporting Actress at age 34 for her role in Vicky value - mean x−µ Cristina Barcelona. The mean age of all Best z= = standard deviation σ 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-score that corresponds to the ages of Ledger and Cruz. Then compare your results...  Solution: Comparing z-Scores from Solution: Comparing z-Scores from Different Data Sets Different Data Sets Heath Ledger x−µ 29 − 49.5 1.49 standard z= = ≈ −1.49 deviations above σ 13.8 the mean Penelope Cruz Both z-scores fall between −2 and 2, so neither score x−µ 34 − 39.9 0.42 standard would be considered unusual. Compared with other z= = ≈ −0.42 deviations below Best Supporting Actor winners, Heath Ledger was σ 14.0 the mean relatively younger, whereas the age of Penelope Cruz was only slightly lower than the average age of other Best Supporting Actress winners...