M119L06-page9 - 139.24 100 12.2 148.84 83 4.8 23.04 100...

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(Lesson 6: Measures of Spread or Variation; 3-3) 3.23 Solution to Example 2 Observe that these are the same five scores as in Example 1, but we are treating them as sample data this time. The computations in the table are exactly the same as in Example 1, although we now use x instead of μ to denote the mean of the data. Data x ( ) Step 2 Deviations: x x ( ) values Step 3 Squared Deviations: x x ( ) 2 values 80 7.8 60.84 76 11.8
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Unformatted text preview: 139.24 100 12.2 148.84 83 4.8 23.04 100 12.2 148.84 Step 1 : x = 87.8 points Sum = 520.8 Do Steps 4, 5 . Step 4 : VAR, or s 2 = the "tilted" average of the squared deviations = 520.8 4 = 130.2 square points Here, we differ from Example 1 in that the sum from Column 3, 520.8, is divided not by 5, but by 4 (i.e., n 1 ). Step 5 : SD, or = VAR = 130.2 11.4 points...
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This note was uploaded on 12/29/2011 for the course MATH 119 taught by Professor Kim during the Fall '09 term at SUNY Stony Brook.

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