Homework 5 Solutions

# Homework 5 Solutions - Homework 5 Solutions BUSQOM 1070,...

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Homework 5 Solutions BUSQOM 1070, Spring 2011 S6.15 (a) The total number defective is 57. 57/1,000 0.057 (0.057)(0.943) 0.0005375 0.023 100 UCL 0.057 3(0.023) 0.057 0.069 0.126 LCL 0.057 3(0.023) 0.057 – 0.069 0.012 0 p p p p    (b) The process is out of control on the third day (of the next 3 days). S6.18 . . . . . . 7 5 3 7 5 3 300 0.040 30 250 7,500 7,500 (1 ) UCL 0.040 3(0.01239) 0.077 ) LCL 0.040 3(0.01239) 0.003 p p p pp pz n n    

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Upper specification Lower specification 6 2,400 1,600 800 1.33 6(100) 600 p C S6.27 min , 33 2,400 1,800 1,800 1,600 min 3(100) 3(100) min [2.00, 0.67] = 0.67 pk USL x x LSL C      The Cp tells us the machine’s variability is acceptable relative to the range of tolerance limits. But Cpk tells us the
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## This note was uploaded on 05/13/2011 for the course BUSORG 1070 taught by Professor Scala during the Spring '11 term at Pittsburgh.

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Homework 5 Solutions - Homework 5 Solutions BUSQOM 1070,...

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