otorola 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 ounces. a. The process standard deviation is , and the process control is set at plus or minus standard deviation . Units with weights less than or greater than ounces will be classified as defects. What is the probability of a defect (to 4 decimals)? In a production run of 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 . Assume the process control remains the same, with weights less than or greater than ounces being classified as defects. What is the probability of a defect (round to 4 decimals; if necessary)? In a production run of parts, how many defects would be found (to the nearest whole number)? c. What is the advantage of reducing process variation, thereby causing a problem limits
Answer:
a)
for normal distribution z score =(Xμ)/σx 

Let mean= μ= 
8 
std deviation =σ= 
0.100 
probability within specification:
probability = 
P(7.775<X<8.225) 
= 
P(2.25<Z<2.25)= 
0.98780.0122= 
0.9756 
hence probability of a defect =10.9756 =0.0244
number of defects =1000*0.0244=24
b)
probability within specification:
probability = 
P(7.775<X<8.225) 
= 
P(2.5<Z<2.5)= 
0.99380.0062= 
0.9876 
probability of a defect =10.9876 = 0.0124
number of defects =1000*0.0124 =12
c)
As sample SD is decreases, the process control limits increases and probability of producing the defective items is reduced.
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