Statistics Homework Solutions 230

Statistics Homework Solutions 230 - 230 CHAPTER 10 712 5 30...

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Unformatted text preview: 230 CHAPTER 10 712 5 30 £¨ ¡ 4 79 and X 23 75. ¡ ¨ 143 7 30 A2 R, respectively. ¨ ¡ £¨ 0 729. R ¡ ¡ ¨ From the control chart table, A2 A2 R and X ¢ (c) The upper and lower limits for the X -chart are X ¡ Therefore LCL = 20.258 and UCL = 27.242. 712 5 30 £¨ ¡ 2 08333 and X ¡ ¨ ¡ 62 5 30 A3s, respectively. 23 75. ¨ £¨ ¡ 1 628. s ¡ ¨ From the control chart table, A3 A3 s and X ¢ (d) The upper and lower limits for the X -chart are X ¡ Therefore LCL = 20.358 and UCL = 27.142. 5. The upper and lower limits for the R-chart are D 3 R and D4 R, respectively. ¡ 2 114. 0 131. Therefore LCL = 0 and UCL = 0.277. The variance is in control. ¨ 0 131 and X 1 110. ¡ ¨ ¡ 0 577. R A2 R, respectively. ¨ ¨ From the control chart table, A2 A2 R and X ¡ (b) The upper and lower limits for the X -chart are X ¢ ¡ R 0 and D4 ¡ ¡ From the control chart table, D3 ¨ 3. (a) The sample size is n ¡ Therefore LCL = 1.034 and UCL = 1.186. The process is out of control for the first time on sample 17. 2A2R 3 1 1604, respectively. ¡ £ ¢ ¨ ¡ 1 0596 and X 1 135, respectively. ¨ ¢ ¨ £ ¡ 2A2R 3 A2 R 3 £ ¡ £ The 2σ limits are X 1 085 and X ¡ A2 R 3 ¨ (c) The 1σ limits are X ¡ The process is out of control for the first time on sample 8, where 2 out of the last three samples are below the lower 2σ control limit. 2 4 ¤£ ¡ ¡ The 3σ limits are 10 11 and σX 1. ¡ 5. (a) X has a normal distribution with µ 3 1 , or 7 and 13. ¥¤ 13 . £ ¥ ¡ ¤ ¡ ¡ ¡ £ ©¥ ¤ ¡ The probability that a point plots outside the 3σ limits is the sum of the area to the left of z to the right of z 2. ¡ 2. ¡ ¤ ¡ 11 1 ¡ ¢¥ 4. The z-score for 13 is 13 PX ¤ 11 1 7 ¡ The z-score for 7 is 7 PX ¥ The probability that a point plots outside the 3σ limits is p 4 and the area ¡ ¡ 0 0228 0 0228. ¨ 43 86. 10. 13 . 0. ¡ ¤ £ ¤ ¡ £ ¡¥ ¡ ¡ ¤ £ ¥ 0. ¡ 5 03. So P X 7 ¥ ¤ ¨ ¡ ¡ ¥ 12 03 1 £ ¥ ¨ ¡ ¡ ¤ ¡ ¨ ¤ ¤ 12 03. ¡ 7 12 03. ¨ ¤ The z-score for 7 is 7 ¨ ¡ Therefore m m 1, so m ¡ 13 Now check that P X £ ©¥ ¡ Therefore 0 97 0. ¡ ¥ ¡ The z-score for 13 is 13 m 1. The z-score with an area of 1/6 = 0.1667 to the right is approximately z 0 97. ¨ ¡ ¢¥ ¤ 7 ¡ 1 6, and check that P X 7. PX ¥ PX ¡ 13 13 ¤ ¡ £ Find m so that P X 10, P X ¡ ¡ 1 6. Since m 7 ¢¥ ¡ Since ARL = 6, p PX ¤ ¨ £ The probability that a point plots outside the 3σ limits is p ¡ ¢ ¨ (b) Let m be the required value. Since the shift is upward, m ¥ ¡ 1 0 0228 ¨ 0 0000 ¨ Therefore p The ARL is 1 p ...
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