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hmk10_Q2_soln_f08

# hmk10_Q2_soln_f08 - 12 = ± = ± All points are between...

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ENGRD2700 Fall 2008 Solutions for Homework 10, Exercise 2, Problems from Devore 7e, Chapter 16. 5. a. P(point falls outside the limits when σ μ μ 5 . 0 + = ) + = + < < - - = σ μ μ σ μ σ μ 5 . 3 3 1 0 0 0 when n X n P ( 29 n Z n P 5 . 3 5 . 3 1 - < < - - - = ( 29 0301 . 9699 . 1 882 . 1 12 . 4 1 = - = < < - - = Z P . b. - = + < < - - σ μ μ σ μ σ μ 0 0 0 3 3 1 when n X n P ( 29 n Z n P + < < + - - = 3 3 1 ( 29 2236 . 24 . 5 76 . 1 = < < - - = Z P c. ( 29 ( 29 9292 . 47 . 1 47 . 7 1 2 3 2 3 1 = - < < - - = - < < - - - Z P n Z n P 6. The limits are ( 29 ( 29 80 . 00 . 13 5 6 . 3 00 . 13 ± = ± , from which LCL = 12.20 and UCL = 13.80. Every one of the 22 x values is well within these limits, so the process appears to be in control with respect to location. 7. 95 . 12 = x and 526 . = s , so with 940 . 5 = a , the control limits are 70 . 13 , 20 . 12 75 . 95 . 12 5 940 . 526 . 3 95 . 12 = ± = ± . Again, every point ( 29 x is between these limits, so there is no evidence of an out-of-control process. 8. 336 . 1 = r and 325 . 2 5 = b , yielding the control limits
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Unformatted text preview: . 12 = ± = ± . All points are between these limits, so the process again appears to be in control with respect to location. 16. 5264 . = s , 940 . 5 = a , LCL = 0 (since n = 5) and UCL = ( 29 ( 29 0996 . 1 5732 . 5264 . 940 . 940 . 1 5172 . 3 5264 . 2 = + =-+ . The largest s i is s 9 = .963, so all standard deviations fall between the control limits. Also, 3364 . 1 = r , c 5 = .864, and b 5 = 2.325, so the LCL is 0 (since n = 5) and UCL = 1.3364 + 3(.864)(1.3364)/2.325 = 2.826. The largest r i is r 9 = 2.4, so all ranges fall within the control limits....
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