University:
California State University, Northridge
Course:
SOM 306 | Operations Management
Academic year:

2024
Views:

293

Pages:

5
Author:

Franco Cannon

Chapter 6 - Solutions Problem 1: a. Mean of sample 1 = (5.8+5.9+6.0+6.1)/4 = 5.95 Mean of sample 2 = (6.2+6.0+5.9+5.9)/4 = 6 Mean of sample 3 = (6.1+5.9+6.0+5.8)/4 = 5.95 Mean of sample 4 = (6.0+5.9+5.0+6.1)/4 = 5.975 b. The mean of the sampling distribution is the average of the sample means x = Mean = (5.95+6+5.95+5.975)/4 = 5.97 The standard deviation of the sampling distribution is computed as σ n . The population n standard deviation σ can be estimated from the 4 samples using the equation: ∑ (x i =1 i − x) 2 n −1 where n = 16, and X = 5.97, therefore σ = 0.1138 (Note: that in this case the standard deviation is not given, but estimated from the data collected. We did not cover this method in the class so if the standard deviation is not given use the alternative method.) The standard deviation of the sampling distribution of the sample means is equal to 0.0569 σ , where σ = 0.1138, and n = 4 (i.e., the number of observations which is estimated using n in each sample). c. Center Line (CL) = x = 5.97 σ = 5.97 + 3 (0.0569) = 6.14 UCL = x + 3 n σ LCL = x - 3 = 5.97 - 3 (0.0569) = 5.80 n Problem 4: Sample 1 2 3 4 5 6 7 8 9 10 1 16.40 15.97 15.91 16.20 15.87 15.43 16.43 15.50 16.13 15.68 Sample 2 3 16.11 15.90 16.10 16.20 16.00 16.04 16.21 15.93 16.21 16.34 15.49 15.55 16.21 15.99 15.92 16.12 16.21 16.05 16.43 16.20 4 15.78 15.81 15.92 15.95 16.43 15.92 16.00 16.02 16.01 15.97 Mean: Control limits for X-bar chart: Mean 16.05 16.02 15.97 16.07 16.21 15.60 16.16 15.89 16.10 16.07 Range 0.62 0.39 0.13 0.28 0.56 0.49 0.44 0.62 0.20 0.75 16.01 0.45 , CL = 16.01 UCL = 16.01 + (0.73)(0.45) = 16.34 LCL = 16.01 – (0.73)(0.45) = 15.68 X-bar Chart 16,40 16,20 16,00 15,80 15,60 15,40 15,20 1 2 3 4 5 6 7 8 9 10 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 1,20 1,00 0,80 0,60 0,40 0,20 0,00 1 2 3 4 5 6 7 Sample Number 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 that do not fall in this range. Problem 6: a. Control limits for X-bar chart: CL = 12.00 UCL = 12.00 + (0.48)(0.60) = 12.29 LCL = 12.00 – (0.48)(0.60) = 11.71 X-bar Chart 12,4 12,2 12 11,8 11,6 11,4 11,2 11 1 2 3 Sample Number Process mean is not in control 4 5 Control Limits for R-chart: CL = 0.60 UCL = (2.0)(0.60) = 1.20 LCL = (0)(0.45) = 0 R-Chart 1,4 1,2 1 0,8 0,6 0,4 0,2 0 1 2 3 4 5 Sample Number R-Chart is in control, but since X-bar chart is not in control, then the process is not in control. Problem 7: 6 = 0.06 100 0.06(1 − 0.06) UCL = 0.06 +3 = 0.22 20 CL = P = LCL = 0.06 -3 0.06(1 − 0.06) = 0 (rounded to zero since the LCL value is negative) 20 Problem 8: 18 = 0.09 200 (0.09)(0.91) σp= = 0.064 20 UCL = 0.09 + 3(0.064) = 0.282 LCL = 0.09 – 3(0.064) = - 1.02 = 0 a. CL = p = b. P-Chart 0,35 0,3 0,25 0,2 0,15 0,1 0,05 0 1 2 3 Sample Number Process is out of control Problem 10: CL = 53/12 = 4.42 errors UCL = 4.42 + 3 4.42 = 10.73 LCL = 4.42 - 3 4.42 = - 1.89 = 0 4