Motorola used the normal distribution to determine the probability of defects an

Question

Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production process. Assume a production process produces items with a mean weight of 10 ounces. Calculate the probability of a defect and the expected number of defects for a 1000-unit production run in the following situations. a.The process standard deviation is .15, and the process control is set at plus or minus one standard deviation. Units with weights less than 9.85 or greater than 10.15 ounces will be classified as defects. In a production run of 1000 parts, how many defects would be found (to the nearest whole number)? b.Through process design improvements, the process standard deviation can be reduced to .05. Assume the process control remains the same, with weights less than 9.85 or greater than 10.15 ounces being classified as defects. (to 4 decimals)? In a production run of 1000 parts, how many defects would be found (to the nearest whole number)? c.What is the advantage of reducing process variation?

Explanation / Answer

a)P(defect) = 1 - P(9.85 ? x ? 10.15) = 1 - P(-1 ? z ? 1) = 1 - .6826 = .3174 Expected number of defects = 1000(.3174) = 317.4 b) P(defect) = 1 - P(9.85 ? x ? 10.15) = 1 - P(-3 ? z ? 3) = 1 - .9974 = .0026 Expected number of defects = 1000(.0026) = 2.6 c)the advantage of reducing process variation, thereby causing process control limits to be at a greater number of standard deviations from the mean

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