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

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IE 360 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 ) 5 . 1 2 . 2 ( ) ( = - = X V
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b) 7143 . 0 ) 5 . 0 )( 7 . 0 / 1 ( ) 7 . 0 / 1 ( ) 7 . 0 / 1 ( ) 5 . 1 2 . 2 ( 1 ) 2 ( 2 5 . 1 2 5 . 1 2 5 . 1 = = = = - = < x dx dx X P 4-54 X=fill volume of can X~N(12.4, 0.1 2 ) a) P(X < 12) = P(Z < 1 . 0 4 . 12 12 - ) = P(Z < - 4) 0 b) P(X < 12.1) = P Z < - 121 12 4 01 . . . = P(Z < - 3) = 0.00135 and P(X > 12.6) = P Z > - 12 6 12 4 01 . . . = P(Z > 2) = 0.02275. Therefore, the proportion of cans scrapped is 0.00135 + 0.02275 = 0.0241, or 2.41% c) P(12.4 - x < X < 12.4 + x) = 0.99. Therefore, < < - 1 . 0 1 . 0 x Z x P = 0.99 Consequently, < 1 . 0 x Z P = 0.995 and x = 0.1(2.58) = 0.258. The limits are (12.142, 12.658). 4-62 X= diameter of dot X~N(0.002, 0.0004 2 ) a) P(X > 0.0026) = - > 0004 . 0 002 . 0 0026 . 0 Z P = P(Z > 1.5)
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