W2s - WS # 2: 1016-351: Coverage - 1.3, 1.4: Dr. Chulmin...

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WS solution for Probability (KIM) Page 1 WS # 2: 1016-351: Coverage - 1.3, 1.4: Dr. Chulmin Kim KEY 1. For a sample of size 4, if 1 2 3 10, 8, and 6, x x x x x x         then the sample variance is equal to _____ 72 _____. Use the property that the sum of all deviations always is zero, we can find 4 4 xx  . So the sample variance is [10 2 +(-8) 2 +(-6) 2 +4 2 ]/(4-1)=72. 2. Calculate the values of the sample mean and sample standard deviation for the following observations on fracture strength by the computational formulas and calculator. 128 131 142 168 87 93 105 114 96 98 The sample mean, 11 (1,162) 116.2. 10 i x x x n The sample standard deviation, 2 2 2 () 140,992 10 25.75 19 i i x x n s n  3. Please refer to the following data. 86, 93, 45, 49, 51, 44, 77, 74, 92, 80
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This note was uploaded on 12/07/2010 for the course MTH 1016.351 taught by Professor Dr.chulminkim during the Winter '10 term at RIT.

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W2s - WS # 2: 1016-351: Coverage - 1.3, 1.4: Dr. Chulmin...

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