hw3solutions - IEOR 130 Solutions to HW 3 Spring 2007 Prof...

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IEOR 130 Solutions to HW 3 Spring, 2007, Prof. Leachman 1. (a) UCL - LCL = 6* σ / n**0.5 95.4 - 79.0 = 6* σ / 5**0.5 -> σ = 6.11 C p = (USL - LSL) / 6* σ = (105.0 - 65.0) / 6*6.11 = 1.09 μ = (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 R-bar = d 2 * σ = (2.326)(6.11) = 14.21 Î LCL = 0, UCL = d 4 *R-bar = (2.11)(14.21) = 29.98 (c) Suppose μ -> 92 = 87.2 + 4.8 = μ + 0.786* σ Prob of Type 2 error = Prob { LCL <= X-bar <= UCL | Exp X-bar = 87.2 + 0.786* σ } = Prob { (-3* σ )/(n**0.5) <= X-bar - μ <= (3* σ )/(n**0.5) | Exp X-bar = 87.2 + 0.786* σ } = Prob { (-3* σ )/(n**0.5) - 0.786* σ <= X-bar - μ - 0.786* σ <= (3* σ )/(n**0.5) - 0.786* σ | Exp X-bar = 87.2 + 0.786* σ } = Prob { -3 - 0.786*(5**0.5) <= { X-bar - μ - 0.786* σ } / [ σ /(5**0.5)] <= 3 - 0.786*(5**0.5) | Exp X-bar = μ + 0.786* σ } = 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) }
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= 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 out-of-spec. 2. (a) The value of C pk is determined as follows: Scrap rate = 1.2% -> Prob { Z <= (USL – μ ) / σ } = 0.988 -> Z = 2.26 from the table at back of notes. Hence (USL – μ ) / σ = 2.26. Now C pk = (USL – μ ) / 3* σ = 2.26 / 3 = 0.753.
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