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.

1. The process standard deviation is 0.1, and the process control is set at plus or minus 1.5 standard deviations. 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 0 decimals)?
   
   
2. 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 (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)?
   

Answer 1

Answer a)

Probability of Defect = 1 - Probability of No Defect

So, first we will find probability of no defect

P (-1.5<Z<1.5 ) =  P(Z<1.5) − P (Z<-1.5)

P (Z<1.5) = 0.9332

P (Z<-1.5) = 1 - P (Z<1.5)

P (Z<-1.5) = 1 - 0.9332 = 0.0668

P (-1.5<Z<1.5 ) =  0.9332 - 0.0668 = 0.8664

Probability of No Defect = 0.8664

Probability of Defect = 1 - Probability of No Defect = 1 - 0.8664

Probability of Defect = 0.1336

The expected number of defects is the product of the probability and the number of units = 0.1336*1000 = 133.6

The expected number of defects is the product of the probability and the number of units is 134

Answer b)

We will use same procedure as shown and Part a, but this time we will use σ = 0.05

P (-3<Z<3 ) =  P(Z<3) − P (Z<-3)

P (Z<3) = 0.9987

P (Z<-3) = 1 - P (Z<3)

P (Z<-3) = 1 - 0.9987 = 0.0013

P (-3<Z<3) =  0.9987 - 0.0013 = 0.9974

Probability of No Defect = 0.9974

Probability of Defect = 1 - Probability of No Defect = 1 - 0.9974

Probability of Defect = 0.0026

The expected number of defects is the product of the probability and the number of units = 0.0026*1000 = 2.6

The expected number of defects is the product of the probability and the number of units is 3