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Statistical Quality Control IE 3255 Spring 2005 Solution HomeWork #2 1. (a)Stem-and-leaf, No of samples, N = 80 Leaf Unit = 0.10 Stem Leaf Frequency 2 12+ 68 6 13- 3134 12 13+ 776978 28 14- 3133101332423404 (15) 14+ 585669589889695 37 15- 3324223422112232 21 15+ 568987666 12 16- 144011 6 16+ 85996 1 17- 0 (b) Frequency distribution of chemical yield 1 2 3 4 5 6 7 8 9 10 Yield % 12.00 ≤ x <12.67 12.67 ≤ x <13.33 13.33 ≤ x <14.00 14.00 ≤ x <14.67 14.67 ≤ x <15.33 15.33 ≤ x < 16.00 16.00 ≤ x <16.67 16.67 ≤ x < 17.33 17.33 ≤ x <18.00 18.00 ≤ x < 19.00 Total Midpoint Frequency 0 1 4 9 21 22 12 7 4 0 Cumulative Frequency 0 1 5 14 35 57 69 76 80 80 Relative Frequency 0.000 0.013 0.050 0.113 0.263 0.275 0.150 0.088 0.050 0.000 Cum. Rel. Frequency 0.000 0.013 0.063 0.175 0.438 0.713 0.863 0.950 1.000 1.000 13.33 13 13.67 14.33 15 15.67 16.33 17 17.67 18.5 80 80 1 1 Histogram of Batch Viscosity 25 20 Frequency class 15 10 5 0 Viscosity 1 The shape of the histogram is not symmetric, and there appear to be one location of central tendency. The distribution does resemble normal probability distribution. (c)Stem-and-leaf, No of samples, N = 80 Leaf Unit = 0.10 Stem Leaf Frequency 2 12+ 68 6 13- 1334 12 13+ 677789 28 14- 0011122333333444 (15) 14+ 555566688889999 37 15- 1122222222333344 21 15+ 566667889 12 16- 011144 6 16+ 56899 1 17- 0 Median observation rank is (0.5)(N) + 0.5 = (0.5)(80) + 0.5 = 40.5 X0.50 = (14.9 + 14.9)/2 = 14.9 Q1 observation rank is (0.25)(N) + 0.5 = (0.25)(80) + 0.5 = 20.5 Q1 = (14.3 + 14.3)/2 = 14.3 Q3 observation rank is (0.75)(N) + 0.5 = (0.75)(80) + 0.5 = 60.5 Q3 = (15.6 + 15.5)/2 = 15.55 (d) 10th percentile observation rank = (0.10)(N) + 0.5 = (0.10)(80) + 0.5 = 8.5 X0.10 = (13.7 + 13.7)/2 = 13.7 90th percentile observation rank = (0.90)(N) + 0.5 = (0.90)(80) + 0.5 = 72.5 X0.90 = (16.4 + 16.1)/2 = 16.25 (e) Box plot for the chemical process 14.3 14.9 12.6 15.55 17.0 2 2. Consider the viscosity data in Exercise 1. Construct a normal probability plot, a lognormal probability plot, and a Weibull probability plot for these data. Based on the plots, which distribution seems to be the best model for the viscosity data? 3 4 3. A mechatronic assembly is subjected to a final functional test. Suppose that defects occur at random in these assemblies, and that defects occur according to a Poisson distribution with parameter λ = 0.02. (a) What is the probability that an assembly will have exactly one defect? (b) What is the probability that an assembly will have one or more defects? (c) Suppose that you improve the process so that the occurrence rate of defects is cut in half to λ = 0.01. What effect does this have on the probability that an assembly will have one or more defects? Solution This is a Poisson distribution with parameter λ = 0.02, x ~ POI(0.02). (a) e −0.02 (0.02) 1 Pr{ x = 1} = p (1) = = 0.0196 1! (b) e −0.02 (0.02) 0 0! = 1 - 0.9802 = 0.0198 Pr{ x ≥ 1} = 1 − Pr{ x = 0} = 1 − p (0) = 1 − (c) Poisson distribution with parameter λ = 0.01, x ~ POI(0.01). e −0.01 (0.01) 0 Pr{x ≥ 1} = 1 − Pr{x = 0} = 1 − p (0) = 1 − 0! = 1 - 0.9900 = 0.0100 Cutting the rate at which defects occur reduces the probability of one or more defects approximately half, from 0.0198 to 0.0100. 5 4. A production process operates with 2% nonconforming output. Every hour a sample of 50 units of product is taken, and the number of nonconforming units counted. If one or more nonconforming units are found, the process is stopped and the quality control technician must search for the cause of nonconforming production. Evaluate the performance of this decision rule. Solution This is a binomial distribution with parameter p = 0.02 and n = 50. The process is stopped if x ≥ 1. ⎛ 50 ⎞ Pr{x ≥ 1} = 1 − Pr{x < 1} = 1 − Pr{x = 0} = 1 − ⎜⎜ ⎟⎟(0.02) 0 (1 − 0.02) 50−0 = 1 − 0.364 = 0.636 ⎝0⎠ The decision rule means that 63.6% of the samples will have one or more nonconforming units, and the process will be stopped to look for a cause. This is a somewhat difficult operating situation. It will cost the company a lot on down time. 6 5. An inspector is looking for nonconforming welds in the gasoline pipeline between Phoenix and Tucson. The probability that any particular weld will be defective is 0.01. The inspector is determined to keep working until finding three defective welds. If the welds are located 100 ft apart, what is the probability that the inspector will have to walk 5000 ft? What is the probability that the inspector will have to walk more than 5000 ft? 6. A population has a mean µ of 44.3 and a standard deviation σ of 2.1. (a) Analyze the problem with a sketch of the normal curve. (b) What percentage of the measurements is larger than 46? (a) b. P{x > 46} = 1 - P{x < 46} = 1 - P{x < 46} 46 - 44.3 = 1 - P{z < } 2.1 = 1 − Φ (0.8095) = 1 − 0.79103 = 0.20897 = 20.9% 7 7. An electronic component in an automobile has a useful life described by an exponential distribution with failure rate 10-5/h. (a) What is the mean time to failure for this component? (b) What is the probability that this component would fail before its expected life. (c) If failure rate is 10-8/h,compute the probability that the component would fail before its expected life. (d) Compare results from (b) and (c). Solution (a) This is exponential distribution with parameter λ = 10-5 Mean time failure rate = 1/λ = 105 = 10,0000 hours (b) Probability that this component would fail before its expected life. 1/ λ ⎧ 1⎫ P ⎨ x ≤ ⎬ = ∫ λe − λt dt = 1 − e −1 = 0.6321 λ⎭ 0 ⎩ (c) If failure rate, λ = 10-8 , probability that this component would fail before its expected life becomes 1/ λ ⎧ 1⎫ P ⎨ x ≤ ⎬ = ∫ λe − λt dt = 1 − e −1 = 0.6321 λ⎭ 0 ⎩ (d) Comparing b and c, one concludes that the probability that a value of an exponential random variable will be less than its mean is 0.63212, regardless of the value of λ. 8