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Unformatted text preview: IEOR 130 Solutions to HW 3 Spring, 2007, Prof. Leachman 1. (a) UCL  LCL = 6* / n**0.5 95.4  79.0 = 6* / 5**0.5 > = 6.11 C p = (USL  LSL) / 6* = (105.0  65.0) / 6*6.11 = 1.09 = (95.4 + 79.0) / 2 = 87.2 which is closer to USL than to LSL C pk = (105.0  87.2) / 3*6.11 = 0.97 (b) n=5 > d 3 = 0, d 4 = 2.11, d 5 = 2.326 Rbar = d 2 * = (2.326)(6.11) = 14.21 LCL = 0, UCL = d 4 *Rbar = (2.11)(14.21) = 29.98 (c) Suppose > 92 = 87.2 + 4.8 = + 0.786* Prob of Type 2 error = Prob { LCL <= Xbar <= UCL  Exp Xbar = 87.2 + 0.786* } = Prob { (3* )/(n**0.5) <= Xbar  <= (3* )/(n**0.5)  Exp Xbar = 87.2 + 0.786* } = Prob { (3* )/(n**0.5)  0.786* <= Xbar   0.786* <= (3* )/(n**0.5)  0.786*  Exp Xbar = 87.2 + 0.786* } = Prob { 3  0.786*(5**0.5) <= { Xbar   0.786* } / [ /(5**0.5)] <= 3  0.786*(5**0.5)  Exp Xbar = + 0.786* } = Phi( 3  0.786*(5**0.5) )  Phi( 3  0.786*(5**0.5) ) = Phi(1.243)  Phi(4.757) ~ Phi(1.243) = 0.893 The probability the shift is not detected in the first 5 lots is (0.893)**5 = 0.568 (d) The probability of falling below LCL after the mean shifts upward to 92.0 is negligible. Hence the Prob of scrap = Prob { X > 105  X ~ N( 92, 6.11**2) } = Prob { Z > (105  92) / 6.11 } = Prob { Z > 2.128 } = 0.0165 We expect 1.65% of the die output will be defective because the etch dimension is outofspec. 2. (a) The value of C pk is determined as follows: Scrap rate = 1.2% > Prob { Z <= (USL ) / } = 0.988 > Z = 2.26 from the table at back of notes....
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This note was uploaded on 04/01/2008 for the course IEOR 131 taught by Professor Leachman during the Spring '08 term at University of California, Berkeley.
 Spring '08
 Leachman

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