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 14 ounces.

a. The process standard deviation is 0.10, and the process control is set at plus or minus 2 standard deviation s. Units with weights less than 13.8 or greater than 14.2 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 (round to the nearest whole number)?

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 13.8 or greater than 14.2 ounces being classified as defects. What is the probability of a defect (round to 4 decimals; if necessary)?

In a production run of 1000 parts, how many defects would be found (to the nearest whole number)?

Explanation / Answer

a)probability of a defect =P(X<13.8 or X>14.2)=1-P(13.8<X<14.2)=1-P((13.8-14)/0.1<Z<(14.2-14)/0.1)

=1-P(-2<Z<2) =1-(0.9772-0.0228)=0.0455

b) number of expected defects =np =1000*0.0455 =~46

b)

probability of a defect =P(X<13.8 or X>14.2)=1-P(13.8<X<14.2)=1-P((13.8-14)/0.08<Z<(14.2-14)/0.08)

=1-P(-2.5<Z<2.5) =1-(0.0062-0.9938)=0.0124

number of expected defects =np =1000*0.0124 =~12

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