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 6 ounces.

The process standard deviation is 0.1, and the process control is set at plus or minus 2.25 standard deviations. Units with weights less than 5.775 or greater than 6.225 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)

Through process design improvements, the process standard deviation can be reduced to 0.09. Assume the process control remains the same, with weights less than 5.775 or greater than 6.225 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)?

What is the advantage of reducing process variation?

Explanation / Answer

as z=(X-mean)/std deviation

hence P(5.775<X<6.225) =P((5.775-6)/0.1<Z<(6.225-6)/0.1)=P(-2.25<Z<2.25) =0.9878-0.0122=0.9756

hence probabilty of defect =1-0.9756 =0.0245

therefore number of defects =np=1000*0.0245=24.5

for std deviation =0.09

P(5.775<X<6.225) =P((5.775-6)/0.09<Z<(6.225-6)/0.09)=P(-2.5<Z<2.5) =0.9938-0.0062=0.9876

hence probabilty of defect =1-0.9876 =0.0124

therefore number of defects =np=1000*0.0124=12.4

therefore as we reduce process variation, number of defects will reduce as more and more parts will be in pass region.

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