hwsol3 - IE 360 Homework Assignment #3 Solutions 4-8 4-28...

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Homework Assignment #3 Solutions 4-8 4-28 4-36 a& b only 4-54 4-62 4-86 4-8 a) 25 . 0 25 . 1 25 . 1 ) 8 . 74 ( 8 . 74 6 . 74 8 . 74 6 . 74 = = = < x dx X P b) P(X < 74.8 or X > 75.2) = P(X < 74.8) + P(X > 75.2) because the two events are mutually exclusive. The result is 0.25 + 0.25 = 0.50. c) 750 . 0 ) 6 . 0 ( 25 . 1 25 . 1 25 . 1 ) 3 . 75 7 . 74 ( 3 . 75 7 . 74 3 . 75 7 . 74 = = = = < < x dx X P 4-28 a) 887 . 2 ) ( , 333 . 8 3 ) 1205 ( 1 . 0 1 . 0 ) 1205 ( ) ( 1205 05 . 0 1 . 0 ) ( 1210 1200 3 1210 1200 2 1210 1200 2 1210 1200 = = = - = - = = = = X V Therefore x dx x X V x dx x X E x σ b) Clearly, centering the process at the center of the specifications results in the greatest proportion of cables within specifications. 5 . 0 1 . 0 1 . 0 ) 1205 1200 ( ) 1205 1195 ( 1205 1200 1205 1200 = = = < < = < < x dx X P X P 4-36 X = time to fill out form X~U(1.5,2.2) a) min 85 . 1 2 ) 2 . 2 5 . 1 ( ) ( = + = X E 2 2 min 0408 . 0 12
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This note was uploaded on 05/01/2008 for the course IE 360 taught by Professor Chen during the Spring '08 term at University of Louisville.

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hwsol3 - IE 360 Homework Assignment #3 Solutions 4-8 4-28...

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