ReidCh6Sol

# ReidCh6Sol - , 15 10 ) = 0.67 C p = 6 LSL USL-= ) 5 ( 6 70...

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DS 412 Solutions to Ch. 6 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 11 11.2 11.4 11.6 11.8 12 12.2 12.4 1 2 3 4 5 Sample Number Process mean is not in control. Sample mean 3 is above the UCL, sample means 4 and 5 are below the LCL. Control Limits for R-chart: CL = 0.60 UCL = (2.0)(0.60) = 1.20 LCL = (0)(0.45) = 0

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R-Chart 0 0.2 0.4 0.6 0.8 1 1.2 1.4 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 12: C pk = min( σ μ 3 - USL , 3 LSL - ) = min( ) 5 ( 3 80 100 - , ) 5 ( 3 70 80 - ) = min( 15 20
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Unformatted text preview: , 15 10 ) = 0.67 C p = 6 LSL USL-= ) 5 ( 6 70 100-= 1 This process is not capable since the C pk value is less than 1. Looking at the C P index by itself would lead us to believe that the machine is capable. Problem 20: The ten samples have a total of 85 leaky tubes. The proportion of leaky tubes in the population is estimated to be 0.085. σ p = 100 ) 915 . )( 085 . ( = 0.0279 CL = 0.085 UCL = 0.085 + 3(0.0279) = 0.169 LCL = 0.085 - 3(0.0279) = 0.001 Since none of the pbars exceeds 0.12, the process is in control....
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## This note was uploaded on 11/04/2011 for the course DS 412 taught by Professor Eng during the Fall '07 term at S.F. State.

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ReidCh6Sol - , 15 10 ) = 0.67 C p = 6 LSL USL-= ) 5 ( 6 70...

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