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Unformatted text preview: Statistics 664 – Autumn 2010 Homework 6 Solutions 1. In Homework 4 we found that the center line and control limits for the obtained the x and R control charts are x chart center line = 10.9 UCL = 47.53 LCL = –25.73 R chart center line = 63.5 UCL = 134.3 LCL = 0 (Note: An alternative answer is to argue that if the process is in control the mean should be 0. This should be the center line of the x chart. In this case, the control limits would be LCL = –36.63 and UCL = 36.63.) When we change from samples of size 5 to samples of size 3 our control limits change to new x chart center line = 10.9 (same as previously) UCL = 10.9 + A 2, n=3 [ d 2, n = 3 δ 2, ν =5 ] R old = 10.9 + 1.023[1.693/2.326]63.5 = 10.9 + 47.3 = 58.2 LCL = 10.9 – A 2, n=3 [ d 2, n = 3 δ 2, ν =5 ] R old = 10.9 – 1.023[1.693/2.326]63.5 = 10.9 – 47.3 = –36.4 (Note: An alternative answer is to argue that if the process is in control the mean should be 0. This should be the center line of the x chart. In this case, the new control limits would be LCL = –47.3 and UCL = 47.3. ) new R chart center line = [ d 2, n = 3 δ 2, ν =5 ] R old = [1.693/2.326]63.5 = 46.2 UCL = D 4, n=3 [ d 2, n = 3 δ 2, ν =5 ] R old = 2.574[1.693/2.326]63.5 = 119.0 LCL = D 3, n=3 [ d 2, n = 3 δ 2, ν =5 ] R old = 0[1.693/2.326]63.5 = 0 2. Problem 6.40 The probability that a sample mean (based on a sample of size n = 5) will fall between the control limits of 96 and 104 if the process is normally distributed with true mean 98 and standard deviation 8 is...
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 Spring '10
 BillNotz
 Statistics, Normal Distribution, Standard Deviation, UCL

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