Frequency Distribution: Mean, Variance, Standard Deviation

1. Frequency Distribution:Mean, Variance, Standard Deviation x f 12 5 13 7 14 6 15 4 16 3 • Given: Number of credit hours a sample of 25 full-time students are taking this semester was collected and is summarized here as a frequency distribution: • Find: a) The values for n, x and x2 using the summations: f, xf and x2 f b) The mean, variance and standard deviation

2. Understanding a Frequency Distribution x f 12 5 13 7 14 6 15 4 16 3 • A sample of 25 data is summarized here as a frequency distribution: • For the above frequency distribution, a) What do the entries x = 12 and f = 5 mean? {The x-value 12 occurred 5 times in the sample} b) If you total the values listed in the x-column, what would this total represent? {It would be the sum of the 5 distinct x-values occurring in the sample, not the sum of all 25 values} • Remember, the x represents the sum of all data values for thesample - this sample has 25 data, not 5 as listed in the x-column

3. Finding the Extensions & Summations x f xf 144 5 12 12 12 12 5 5 5 12 5 12x5=60 122 = 144 144 144x5=720 169 7 x2 x2f 13 13 13 13 7 7 7 13x7=91 13 7 132 = 169 169 169x7=1183 196 6 14 14 14 14 6 6 6 14x6=84 14 6 196 142 = 196 196x6=1176 225 4 15 15 15 15 4 4 4 15 4 15x4=60 225 152 = 225 225x4=900 256 3 16 16 16 16 3 3 3 16 3 16x3=48 162 = 256 256 256x3=768 • Use a table format to find the extensions for each x value and the 3 summations f, xf and x2 f : f = 25 xf = 343 x2f = 4747 The sample size n is f, the sum of the frequencies 1. Find n ; The sample size n = f = 25 2. Find the sum of all data by finding xf ; Find xf for each x The sum of all data = xf = 343 3. Find the sum of all squared data by finding x2f ; First, find x2 for each x ; Second, find x2f for each x The sum of all squared data = x2f = 4747 Notes: Save these 3 summations for future formula work DO NOT find the summations of the x and x2 columns

4. Finding the Sample Mean xf x = f xf x = = f • Formula 2.11 will be used: • Previously determined values: f = 25, xf = 343 343 343 = 13.72 25 25 The sample mean is 13.7 credit hours

5. Finding the Sample Variance (xf)2 (xf)2 x2f - f f s2 = f - 1 ( )2 x2f - ( ) - 41.04 25 25 1.71 s2 = = = = f - 1 25 25 - 1 24 • Formula 2.16 will be used: • Previously determined values: x2f = 4747 xf = 343 f = 25 343 343 4747 4747 The sample variance is 1.71

6. Finding the Sample Standard Deviation s =  s2 s =  s2 =  • The standard deviation is the square root of variance: • Therefore, the standard deviation is: 1.71 = 1.3077 = 1.3 The standard deviation is 1.3 credit hours Notes: 1) The unit of measure for the standard deviation is the unit of the data 2) Use a non-rounded value of variance when calculating the standard deviation

7. Using a Grouped Frequency Distribution Class Interval f 2.50 - 7.50 6 7.50 - 12.50 9 12.50 - 17.50 5 17.50 - 22.50 3 22.50 - 27.50 2 • Given: Twenty-five men were asked, “How much did you spend at the barber shop during your last visit?” The data is summarized using intervals and is listed here as a grouped frequency: • Find: a) The class midpoint for each class b) Estimate the values for n, x and x2 using the summations: f, xf and x2f c) The mean, variance and standard deviation

8. Finding the Class Midpoint Class Interval Midpoint f Class Interval f 2.50 7.50 2.50 - 7.50 6 2.50 - 7.50 6 12.50 7.50 7.50 - 12.50 9 7.50 - 12.50 9 12.50 - 17.50 5 12.50 - 17.50 5 15.0 17.50 - 22.50 3 17.50 - 22.50 3 20.0 22.50 - 27.50 2 22.50 - 27.50 2 25.0 + + 10.00 20.00 = = 2 2 2 2 • Each class interval contains several different data values. In order to use the frequency distribution, a class midpoint must be determined for each class. This center value for the class will be used to approximate the value of each data that belongs to that class. The class midpoints are found by averaging the extreme values for each class: 5.0 5.0 10.0 10.0 • Find: The class midpoints, one class at a time (lower boundary + upper boundary) / 2 2.50 2.50 7.50 7.50 7.50 7.50 12.50 12.50 = 5.0 = 10.0 The midpoint for each class will the be the class’s representative value and be used for finding the extensions

9. Finding the Extensions & Summations x2 x f 25 256 6 5 52 = 25 5 6 5x6=30 25x6=150 x2f xf 100 1009 9 10 102 = 100 10 9 10x9=90 100x9=900 5 56 5 225 2255 5 15 152 = 225 15 5 15x5=75 225x5=1125 109 10 10 400 4003 3 20 202 = 400 20 3 20x3=60 400x3=1200 15 15 155 625 6252 2 25 252 = 625 25 2 25x2=50 625x2=1250 203 20 20 252 25 25 6 9 5 3 2 • Use a table format to find the extensions for each x value and the 3 summations f, xf and x2 f : meaningless totals f = 25 xf = 305 x2f = 4625 1. For x2, multiply each x by itself 2. For xf, multiply each x by its frequency f 3. For x2f, multiply each x2 by its frequency f 4. Find the summations by totaling the columns Notes: Save these 3 summations for future formula work DO NOT find the summations for the x and x2 columns

10. Finding the Sample Mean xf x = f xf x = = f • Formula 2.11 will be used: • Previously determined values: f = 25, xf = 305 305 305 = 12.2 25 25 The sample mean is $12.20

11. Finding the Sample Variance (xf)2 (xf)2 x2f - f f s2 = f - 1 ( )2 x2f - ( ) - 904 25 25 37.6666 37.7 s2 = = = = = f - 1 25 25 - 1 24 • Formula 2.16 will be used: • Previously determined values: x2f = 4625 xf = 305 f = 25 305 305 4625 4625 The sample variance is 37.7

12. Finding the Sample Standard Deviation s =  s2 s =  s2 =  • The standard deviation is the square root of variance: • Therefore, the standard deviation is: 37.6666 = 6.1367 = 6.14 The standard deviation is $6.14 Notes: 1) The unit of measure for the standard deviation is the unit of the data 2) Use a non-rounded value of variance when calculating the standard deviation