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Statistics Homework Solutions 231

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

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SECTION 10.2 231 (c) We will find the required value for σ X . The probability that a point plots outside the 3 σ limits is p P X 10 3 σ X P X 10 3 σ X . Since ARL = 6, p 1 6. Since the process mean is 11, P X 10 3 σ X P X 10 3 σ X . Find σ so that P X 10 3 σ X 1 6, and check that P X 10 3 σ X 0. The z -score for 10 3 σ X is 10 3 σ X 11 σ X . The z -score with an area of 1/6 = 0.1667 to the right is approximately z 0 97. Therefore 10 3 σ X 11 σ X 0 97, so σ X 0 4926. Now check that P X 10 3 σ X 0. 10 3 σ X 8 522. The z -score for 8.522 is 8 522 11 0 4926 5 03, so P X 10 3 σ X 0. Therefore σ X 0 4926. Since n 4, σ X σ 4 σ 2. Therefore σ 0 985. (d) Let n be the required sample size. Then σ X 2 n . From part (c), σ X 0 4926. Therefore 2 n 0 4926, so n 16 48. Round up to obtain n 17. 7. 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. (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
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