Instructors manual from Linda Null & Julia Lobur boook
SOLUTION TO EXERCISE 11.1: From Text-Equation 11.1, the mean value of the sample compositions is: 1 31 y y i 21.84% 31 i 1 Since the rtue mixture composition si not known, an estimate of the standard deviation is found from Text_Equation 11.3: 1 31 y i 21.852 2.518 S 31 1 i 1 Since there are less than 50 samples, the variance distribution curve is more likely to be a 2 distribution. Therefore, from Text-Equations 11.11 and 11.12: S 2 N 1 2.518 2 31 1 Lower limit: 2L 2 2
Upper limit: 2U
S 2 N 1 2.518 2 31 1 2 1 2 1
At the 95% confidence level, 0.025 so referring to the 2 distribution tables with 30 degrees of freedom, 2 46.98 and 2 1 16.79 . Hence: 2L 4.05 and 2U 11.33 and so we can say that the actual standard deviation lies between 2.012 and 3.366 with 95% confidence.
SOLUTIONS TO CHAPTER 11 EXERCISES: MIXING AND SEGREGATION
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SOLUTIONS TO CHAPTER 11 EXERCISES: MIXING AND SEGREGATION
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SOLUTIONS TO CHAPTER 11 EXERCISES: MIXING AND SEGREGATION
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SOLUTIONS TO CHAPTER 11 EXERCISES: MIXING AND SEGREGATION
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SOLUTIONS TO CHAPTER 11 EXERCISES: MIXING AND SEGREGATION
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SOLUTIONS TO CHAPTER 11 EXERCISES: MIXING AND SEGREGATION
