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 9 ounces. a. The process standard deviation is 0.1, and the process control is set at plus or minus 1.25 standard deviations. Units with weights less than 8.875 or greater than 9.125 ounces will be classified as defects. What is the probability of a defect (to 4 decimals)? In a production run of 1000 parts, how many defects would be found (to 0 decimals)? b. Through process design improvements, the process standard deviation can be reduced to 0.08. Assume the process control remains the same, with weights less than 8.875 or greater than 9.125 ounces being classified as defects. What is the probability of a defect (rounded to 4 decimals; getting the exact answer, although not necessary, will require Excel)? In a production run of 1000 parts, how many defects would be found (to 0 decimals)? C. What is the advantage of reducing process variation? Select your answer It can substantially reduce the number of defects It may slightly reduce the number of defects It has no effect on the number of defects

Explanation / Answer

Solution:

a). Given in the question Mean = 9

Z = 1.25

It is a two tail test so from z table probability of defects is (0.1056+0.1056) = 0.2112or 21.12%

In a production run of 1000 parts, defects would be 1000*0.2112 = 211.2 or 212

b). Z value for other process after reducing the standard deviation

Z = (9.125 - 9)/0.08 = 0.125/0.08 = 1.5625

It is a two tail test so from z table probability of defects is (0.0594+0.0594) = 0.1188 or 11.88%

In a production run of 1000 parts, defects would be 1000*0.1188 = 118.8 or 119

By reducing the process variation from 0.1 to 0.08 we found that no. of defects decrease from 212 to 119 from which we can say that it can substantially reduce the number of defects.

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