Motorola used the normal distribution to determine the probability of defects an
ID: 3059978 • Letter: M
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 15 ounces.
The process standard deviation is 0.1, and the process control is set at plus or minus 2 standard deviations. Units with weights less than 14.8 or greater than 15.2 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 0.08. Assume the process control remains the same, with weights less than 14.8 or greater than 15.2 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?
- 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
1)
probability of a defect =P(X>14.8)+P(X<15.2)=1-P(14.8<X<15.2) =1-P(-2<Z<2) =1-(0.9772-0.0228)=0.0455
numbr of defects =1000*0.0455 =~46
2)
probability of a defect =P(X>14.8)+P(X<15.2)=1-P(14.8<X<15.2) =1-P(-2.5<Z<2.5) =1-(0.9938-0.0062)=0.0124
numbr of defects =1000*0.0124 =~12
It can substantially reduce the number of defects
for normal distribution z score =(X-)/ here mean= = 15 std deviation == 0.1000Related Questions
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