# ch06 - Chapter 6 Problem 2 20 19.875 20 = 19.96 3 0.2 UCL =...

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Chapter 6 Problem 2: CL = x = 3 20 875 . 19 20 + + = 19.96 UCL = x + 3 n σ = 19.96 + 3 4 2 . 0 = 20.26 LCL = x - 3 n σ = 19.96 - 3 4 2 . 0 = 19.66 Problem 4: Sample Sample 1 2 3 4 Mean Range 1 16.40 16.11 15.90 15.78 16.05 0.62 2 15.97 16.10 16.20 15.81 16.02 0.39 3 15.91 16.00 16.04 15.92 15.97 0.13 4 16.20 16.21 15.93 15.95 16.07 0.28 5 15.87 16.21 16.34 16.43 16.21 0.56 6 15.43 15.49 15.55 15.92 15.60 0.49 7 16.43 16.21 15.99 16.00 16.16 0.44 8 15.50 15.92 16.12 16.02 15.89 0.62 9 16.13 16.21 16.05 16.01 16.10 0.20 10 15.68 16.43 16.20 15.97 16.07 0.75 Mean: 16.01 0.45 x = 16.01 R = 0.45 Control limits for X-bar chart : CL = 16.01 UCL = 16.01 + (0.73)(0.45) = 16.34 LCL = 16.01 – (0.73)(0.45) = 15.68

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X-bar Chart 15.20 15.40 15.60 15.80 16.00 16.20 16.40 1 2 3 4 5 6 7 8 9 10 Sample Number Control Limits for R-chart: CL = 0.45 UCL = (2.28)(0.45) = 1.03 LCL = (0)(0.45) = 0 R-Chart 0.00 0.20 0.40 0.60 0.80 1.00 1.20 1 2 3 4 5 6 7 8 9 10 Sample Number The sample mean for sample 6 is below the LCL. The process mean is not in control. b. The process is not capable of meeting the design standards. Design standards dictate that fill levels range between 16.3 ounces and 15.7. There are nine observations (22.5%) that do not fall in this range. Alternatively, the process standard deviation can be estimated from the data to be 0.251 ounces. The process capability index would also show the process is not capable of meeting the specification.ot
Problem 6: a.

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