<p> Chapter 10 – Class exercises:</p><p>Problems:</p><p>3. a. n = 20</p><p>A2 = 0.18 = = 3.10 Mean Chart: =± A2 (= 3.1 ± 0.18(0.45) D3 = 0.41 (= 0.45 = 3.1 ± .081 D4 = 1.59 Hence, UCL is 3.181 and LCL is 3.019. All means are within these limits.</p><p>Range Chart: UCL is D4(= 1.59(0.45) = .7155</p><p>LCL is D3(= 0.41(0.45) = .1845 In control since all points are within these limits.</p><p>4. Sample Mean Range Mean Chart: = A (= 79.96 0.58(1.87) 1 79.48 2.6 ± 2 ± 2 80.14 2.3 = 79.96 ± 1.08 3 80.14 1.2 UCL = 81.04, LCL = 78.88</p><p>4 79.60 1.7 Range Chart: UCL = D4(= 2.11(1.87) = 3.95</p><p>5 80.02 2.0 LCL = D3(= 0(1.87) = 0 6 80.38 1.4 [Both charts suggest the process is in control: Neither has any points outside the limits.]</p><p> p(1 p) 6. n = 200 Control Limits = p  2 n</p><p>25 .0096(.9904) p   .0096  .0096  2 13(200) 200  .0096  .0138 Thus, UCL is .0234 and LCL becomes 0. Since n = 200, the fraction represented by each data point is half the amount shown. E.g., 1 defective = .005, 2 defectives = .01, etc. Sample 10 is too large. Omitting that value and re-computing limits with 18 p   .0075 yields 12(200) UCL = .0197 and LCL = 0. 24. Let USL = Upper Specification Limit, LSL = Lower Specification Limit, X = Process mean,  = Process standard deviation</p><p>For process H: X  LSL 15 14.1   .93 3 (3)(.32)</p><p>USL  X 16 15   1.04 3 (3)(.32)</p><p>C pk  min.938, 1.04 .93 .93  1.0, not capable</p><p>For process K: X  LSL 33  30   1.0 3 (3)(1)</p><p>USL  X 36.5  33  1.17 3 (3)(1)</p><p>C pk  min{1.0, 1.17}  1.0 Since 1.0 < 1.33, the process is not capable.</p><p>For process T: X  LSL 18.5 16.5  1.67 3 (3)(0.4)</p><p>USL  X 20.118.5   1.33 3 (3)(0.4)</p><p>C pk  min{1.67, 1.33} 1.33 Since 1.33 = 1.33, the process is capable.</p><p>25. Let USL = Upper Specification Limit, LSL = Lower Specification Limit, X = Process mean,  = Process standard deviation. USL = 90 minutes, LSL = 50 minutes, </p><p>X 1 = 74 minutes, 1 = 4.0 minutes</p><p>X 2 = 72 minutes, 2 = 5.1 minutes For the first repair firm: X  LSL 74  50   2.0 3 (3)(4.0)</p><p>USL  X 90  74   1.333 3 (3)(4.0)</p><p>C pk  min{2.0, 1.33}  1.333 Since 1.333 = 1.333, the firm 1 is capable.</p><p>For the second repair firm: X  LSL 72  50   1.44 3 (3)(5.1)</p><p>USL  X 90  72  1.18 3 (3)(5.1)</p><p>C pk  min{1.44,1.88} 1.18 Since 1.18 < 1.33, the firm 2 is not capable.</p><p>26. Let USL = Upper Specification Limit, LSL = Lower Specification Limit, X = Process mean,  = Process standard deviation.</p><p>USL = 30 minutes, LSL = 45 minutes, X Armand = 38 minutes,  Armand = 3 minutes X Jerry = 37 minutes,  Jerry = 2.5 minutes X Melissa = 37.5 minutes,  Melissa = 2.5 minutes</p><p>For Armand:</p><p>X  LSL 38  30   .89 3 (3)(3)</p><p>USL  X 45  38   .78 3 (3)(3)</p><p>C pk  min{.89, .78}  .78 Since .78 < 1.33, Armand is not capable.</p><p>For Jerry: X  LSL 37  30   .93 3 (3)(2.5)</p><p>USL  X 45  37  1.07 3 (3)(2.5)</p><p>C pk  min{.93, 1.07}  .93 Since .93 < 1.33, Jerry is not capable. For Melissa, sinceUSL  X  X  LSL  7.5, the process is centered, therefore we will use </p><p>Cp to measure process capability.</p><p>USL  LSL 45  30 C    1.39 p 6 (6)(1.8) Since 1.39 > 1.33, Melissa is capable.</p>