hmewrk_06_sol - 3.43 = = = 0.6 3 3 i b) Line 1: UCL 3.52...

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Solution Homework #6: Process Control 2.008 Design and Manufacturing II Spring 2004 Problem 1 You are responsible for two different manufacturing lines, both of which produce arms for Karl´s Mechanical Squid. The Upper Specification Limit of the line is 3.52 and the Lower Specification Limit is 3.18. Line 1 produces arms with a µ of 3.35 and an σ of 0.07. Line 2 produces arms with a µ of 3.43 and an σ of 0.05. a) For Line 1: what is the C p ? What is the C pk ? For Line 2: what is the C p ? What is the C pk ? b) Which line produces a greater percentage of parts within the specification limits? What are the exact percentages for each line? Solution a) Line 1: 3.52 3.18 C P = = 0 . 8 1 60 .07 i C Pk z min 3.35 3.18 = = = 0 . 8 1 3 3 i σ Line 2: 3.52 3.18 C P = = 1 .
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Unformatted text preview: 3.43 = = = 0.6 3 3 i b) Line 1: UCL 3.52 3.35 z UCL = = = 2.43 0.07 LCL 3.18 3.35 z LCL = = = 2.43 0.07 Using the table on slide 17 lecture 15 (Quality): P(z<z UCL ) = 0.9925 Using the table on slide 16 lecture 15 (Quality): P(z<z LCL ) = 0.0075 Percentage of parts within the specification limits: P=(0.9925-0.0075)*100% = 98.5 % Line 2: UCL 3.52 3.43 z UCL = = = 1.8 0.05 LCL 3.18 3.43 z LCL = = = 5 0.05 Using the table on slide 17 lecture 15 (Quality): P(z<z UCL ) = 0.9641 P(z<z LCL =-5) 0 Percentage of parts within the specification limits: P=0.9641*100% = 96.41 % Line 1 produces a greater percentage of parts within the specification limits!...
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This note was uploaded on 02/23/2012 for the course MECHANICAL 2.008 taught by Professor Jung-hoonchun during the Spring '04 term at MIT.

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