664 HW2 Solns

664 HW2 Solns - Statistics 664 Autumn 2010 Homework 2...

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Statistics 664 – Autumn 2010 Homework 2 Solutions 3.12 Here is a normal plot from Minitab 105 100 95 90 85 80 75 99.9 99 95 90 80 70 60 50 40 30 20 10 5 1 0.1 Yield Percent Mean 89.48 StDev 4.158 N 90 AD 0.956 P-Value 0.015 Probability Plot of Yield Normal - 95% CI There do not appear to be any serious departures from normality (although the a test for normality has P-value 0.015, which would suggest that we would reject the hypothesis that the distribution is normal). 3.25 a Using Minitab to do Poisson probability calculations yields the following. Poisson with mean = 0.02 x P( X = x ) 1 0.0196040 b. The probability of 1 or more defects is equal to 1 – probability of 0 defects. From Minitab we find Poisson with mean = 0.02 x P( X = x ) 0 0.980199
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Thus the probability of 1 or more defects is 1 – 0.980199 = 0.019801 c. The probability of 1 or more defects is equal to 1 – probability of 0 defects. From Minitab we find Poisson with mean = 0.01 x P( X = x ) 0 0.990050 Thus, the probability of 1 or more defects is 1 – 0.990050 = 0.00995 Cutting the occurrence rate of defects in half decreased the probability of 1 or more defects (by approximately ½). 3.31 We would model the number of noncomforming units out of the 25 sampled as binomial with n = 25 and probability of “success” p = 0.01. The probability of finding one or more noncomforming units in a sample of 25 is 1 – probability of 0 noncomforming units. From Minitab we find
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664 HW2 Solns - Statistics 664 Autumn 2010 Homework 2...

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