Statistics Homework Solutions 231

Statistics Homework Solutions 231 - 231 SECTION 10.2 (c) We...

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Unformatted text preview: 231 SECTION 10.2 (c) We will find the required value for σX . ¤ ¡ ¥ ¥ ¢ ¡ ¡ ¤ ¤ ¡ £ £ ¥ 0 4926. ¨ ¡ 10 3σX 0. ¥ 5 03, so P X ¡ ¤ ¡ ¨ ¡ ¡ 11 0 4926 ¨ £ ©¥ ¡ ¡¥ ¢ ¡ £ ¢ ¡ ¡ ¤ ¨ 0. ¤ ¡ ¥ ¢ ¨ £ ¥ ¡ ¡ ¡ ¢ ¨ ¤ ¨ σ 2. £ ¡ 4 ¤£ ¨ ¡ 0 985. ¨ 0 4926, so n ¨ ¡ n 16 48. ¡ ¡ ¡ ¡ ¤£ ¨ ¡ ¡ Round up to obtain n ¡ 0 4926. Therefore 2 n. ¤£ ¡ ¡ From part (c), σX 2 ¨ ¤ ¤ ¡ σ (d) Let n be the required sample size. Then σX 17. The probability of a false alarm on any given sample is 0 0027, and the probability that there will not be a false alarm on any given sample is 0.9973. ¨ 7. 3σ X . 0 4926. 4, σX Therefore σ ¡ 0 97, so σX 3σX 0. 3σX . 11 σX . The z-score with an area of 1/6 = 0.1667 to the right is 8 522. The z-score for 8.522 is 8 522 Therefore σX Since n 10 3σX 10 ¡ 3σX 11 σX 3σX 10 PX 10 ¥ 3σX Now check that P X 10 ¡ Therefore 10 ¤ The z-score for 10 3σX is 10 approximately z 0 97. ¡ 1 6, and check that P X ¡ 3σX ¢¥ 10 3σX 10 PX ¤ Find σ so that P X 3σX ¡ 1 6. Since the process mean is 11, P X 10 ¢ Since ARL = 6, p PX ¥ The probability that a point plots outside the 3σ limits is p (a) The probability that there will be no false alarm in the next 50 samples is 0 9973 50 0 874. Therefore the probability that there will be a false alarm within the next 50 samples is 1 0 874 0 126. ¨ ¡ ¨ ¨ ¡ ¨ ¡ (b) The probability that there will be no false alarm in the next 100 samples is 0 9973 100 0 763. Therefore the probability that there will be a false alarm within the next 50 samples is 1 0 763 0 237. ¡ ¨ 256 37 ¡ ¨ ¡ ¨ ¨ ¡ ¨ ¨ ¡ ¡ ¨ ¡ ¨ ¡ 1 961 and X A3s, respectively. 199 816. ¨ ¨ ¡ 1 287. s ¡ ¨ ¡ From the control chart table, A3 A3 s and X ¢ ¡ (b) The upper and lower limits for the X -chart are X Therefore LCL = 197.292 and UCL = 202.340. The process is in control. A3 s 3 ¡ £ ¢ 198 975 and X 200 657, respectively. ¨ ¨ ¡ A3 s 3 ¨ ¨ 1 970. ¡ 0 030 and B4 1 961. Therefore LCL = 0.0588 and UCL = 3.863. The variance is in control. £ ¡ (c) The 1σ limits are X ¡ ¨ ln0 5. Solving for n yields n 6. The upper and lower limits for the S-chart are B 3 s and B4 s, respectively. From the control chart table, B3 s 0 582. ¨ ¡ 9. (a) The sample size is n 0 5, so n ln 0 9973 ¨ (d) Let n be the required number. Then 0 9973n ¨ (c) The probability that there will be no false alarm in the next 200 samples is 0 9973 200 257. ...
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This note was uploaded on 12/20/2011 for the course STA 3163 taught by Professor Mattgilg during the Fall '11 term at UNF.

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