# 1 Frequency Distribution: Mean, Variance, Standard Deviation Given: Number of credit hours a sample of 25 full- time students are taking this semester.

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1 Frequency Distribution: Mean, Variance, Standard Deviation 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 x 2 using the summations: f, xf and x 2 f x f 125 137 146 154 163 b)The mean, variance and standard deviation

2 Understanding a Frequency Distribution A sample of 25 data is summarized here as a frequency distribution: x f 125 137 146 154 163 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 the sample - this sample has 25 data, not 5 as listed in the x-column

3 1. Find n ; Finding the Extensions & Summations Use a table format to find the extensions for each x value and the 3 summations f, xf and x 2 f : x 12 13 14 15 16 5 7 6 4 f 3 The sample size n is f, the sum of the frequencies The sample size n = f = 25 f = 25 2. Find the sum of all data by finding xf ; The sum of all data = xf = 343 12x5=60 13x7=91 14x6=84 15x4=60 16x3=48 12 13 14 15 16 5 7 6 4 3 12 5 13 7 14 6 15 4 16 3 xf = 343 3. Find the sum of all squared data by finding x 2 f ; x2x2 x2fx2f 144x5=720 169x7=1183 196x6=1176 225x4=900 256x3=768 12 2 = 144 13 2 = 169 14 2 = 196 15 2 = 225 16 2 = 256 12 13 14 15 16 12 13 14 15 16 x 2 f = 4747 Notes:Save these 3 summations for future formula work DO NOT find the summations of the x and x 2 columns xf Find xf for each x First, find x 2 for each x ; Second, find x 2 f for each x The sum of all squared data = x 2 f = 4747 144 169 196 225 256 144 169 196 225 256 5 7 6 4 3 5 7 6 4 3

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

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

6 Finding the Sample Standard Deviation The standard deviation is the square root of variance: s = s 2 Therefore, the standard deviation is: s = s 2 = = 1.3077 The standard deviation is 1.3 credit hours = 1.3 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 1.71

7 Using a Grouped Frequency Distribution Find:a)The class midpoint for each class b)Estimate the values for n, x and x 2 using the summations: f, xf and x 2 f c)The mean, variance and standard deviation 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: Class Intervalf 2.50 - 7.506 7.50 - 12.509 12.50 - 17.505 17.50 - 22.503 22.50 - 27.502

8 Class Intervalf 2.50 - 7.50 7.50 - 12.50 12.50 - 17.50 17.50 - 22.50 22.50 - 27.50 6 9 5 3 2 MidpointClass Intervalf 2.50 - 7.50 7.50 - 12.50 12.50 - 17.50 17.50 - 22.50 22.50 - 27.50 6 9 5 3 2 12.507.50 Finding the Class Midpoint Find:The class midpoints, one class at a time (lower boundary + upper boundary) / 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: The midpoint for each class will the be the classs representative value and be used for finding the extensions 2.50 = 10.00 2 = 5.0 = 20.00 2 = 10.0 7.50 2 + 2.50 7.50 2 + 12.50 15.0 20.0 25.0 5.0 10.0 5.0 10.0

9 5 2 = 25 10 2 = 100 15 2 = 225 20 2 = 400 25 2 = 625 25x6=150 100x9=900 225x5=1125 400x3=1200 625x2=1250 5x6=30 10x9=90 15x5=75 20x3=60 25x2=50 xf 1. For x 2, multiply each x by itself 2. For xf, multiply each x 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 x 2 columns Finding the Extensions & Summations Use a table format to find the extensions for each x value and the 3 summations f, xf and x 2 f : x2fx2f x2x2 x 5 10 15 20 25 6 9 5 3 f 2 5 10 15 20 25 5 10 15 20 25 3. For x 2 f, multiply each x 2 by its frequency f 25 6 100 9 225 5 400 3 625 2 25 100 225 400 625 6 9 5 3 2 5 6 10 9 15 5 20 3 25 2 6 9 5 3 2 5 10 15 20 25 f = 25 xf = 305 x 2 f = 4625 meaningless totals

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

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

12 Finding the Sample Standard Deviation The standard deviation is the square root of variance: s = s 2 Therefore, the standard deviation is: s = s 2 = = 6.1367 The standard deviation is \$6.14 = 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 37.6666

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