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 16 ounces.

a. The process standard deviation is 0.10 , and the process control is set at plus or minus 1.75 standard deviation . Units with weights less than 15.825 or greater than 16.175 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 0.06 . Assume the process control remains the same, with weights less than 15.825 or greater than 16.175 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)?

Answer 1

a)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	16
std deviation   =σ=	0.1000

probabiltiy of defect =P(X<15.825)+P(X>16.175)=1-P(15.825<X<16.175)=1-P(-1.75 <Z<1.75)=1-(0.9599-0.0401)=0.0802

In a production run of 1000 parts expected defects =1000*0.0802=~80

b)

probabiltiy of defect =P(X<15.825)+P(X>16.175)=1-P(15.825<X<16.175)=1-P(-2.92 <Z<2.92)=1-(0.9982-0.0018)=0.0036

In a production run of 1000 parts expected defects =1000*0.0036=~4