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 14 ounces. Use Table 1 in Appendix B. The process standard deviation is 0.15, and the process control is set at plus or minus 1 standard deviation. Units with weights less than 13.85 or greater than 14.15 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 the nearest whole number)? Through process design improvements, the process standard deviation can be reduced to 0.05. Assume the process control remains the same, with weights less than 13.85 or greater than 14.15 ounces being 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 the nearest whole number)? What is the advantage of reducing process variation

Answer 1

  probability of a defect =P(X<13.85)+P(X>14.15)=1-P(13.85<X<14.15)

=1-P((13.85-14)/0.2<Z<(14.15-14)/0.2)=1-(0.7734-0.2266)=0.4532

In a production run of 1000 parts, number of defects =np=1000*0.4532 =453

with improved std deviaiton

probability of a defect =1-P(-2.14<Z<2.14)=1-(0.9838-0.0162)=0.0324

In a production run of 1000 parts ; number of defects =1000*0.0324 =32

  It can substantially reduce the number of defects.