Question

Suppose that Motorola uses the normal distribution to determine the probability of defects and the number of defects in a particular production process. Assume that the production process manufactures items with a mean weight of 10 ounces. Calculate the probability of a defect and the suspected number of defects for a 1,000-unit production run in the following situations.

(a)	The process standard deviation is 0.15, and the process control is set at plus or minus one standard deviation. Units with weights less than 9.85 or greater than 10.15 ounces will be classified as defects. If required, round your answer to four decimal places.

b)	Through process design improvements, the process standard deviation can be reduced to 0.05. Assume that the process control remains the same, with weights less than 9.85 or greater than 10.15 ounces being classified as defects. If required, round your answer to four decimal places.

Answer 1

X : Weight of the item

(a) X follows normal distribution with mean 10 ounces and standard deviation 0.15

Unit is classified as defect if weight is less than 9.85 or greater than 10.15; i.e A unit is classified as non-defective if weight is between 9.85 and 10.15 i.e 9.85 X 10.15

Probability of a non-defect = P(9.85 X 10.15) ;

P(9.85 X 10.15) = P(X 10.15) - P(X 9.85)

P(X 10.15)

Z-score for 10.15 = (10.15 - mean)/standard deviation = (10.15 - 10) / 0.15 = 1

From standard normal tables , P(Z1) = 0.8413

P(X 10.15)=P(Z1) = 0.8413

P(X 9.85)

Z-score for = (9.85 - mean)/standard deviation = (9.85 - 10) / 0.15 = -1

From standard normal tables , P(Z-1) = 0.1587

P(X 9.85) = P(Z-1) = 0.1587

P(9.85 X 10.15) = P(X 10.15) - P(X 9.85) = 0.8413-0.1587=0.6826

Probability of a non-defect = P(9.85 X 10.15) = 0.6826

Probability of a defect = 1-Probability of a non-defect = 1-0.6826 = 0.3174

Probability of a defect = 0.3174

Suspected number of defects for a 1,000 unit production run = 1000 x Probability of a defect = 1000*0.3174=317.4

Suspected number of defects for a 1,000 unit production run = 317.4

(b) With improved process, the process standard deviation reduced to 0.05; Therefore ,

X follows normal distribution with mean 10 ounces and standard deviation 0.05

Unit is classified as defect if weight is less than 9.85 or greater than 10.15; i.e A unit is classified as non-defective if weight is between 9.85 and 10.15 i.e 9.85 X 10.15

Probability of a non-defect = P(9.85 X 10.15) ;

P(9.85 X 10.15) = P(X 10.15) - P(X 9.85)

P(X 10.15)

Z-score for 10.15 = (10.15 - mean)/standard deviation = (10.15 - 10) / 0.05 = 0.15/0.05 = 3

From standard normal tables , P(Z3) = 0.9987

P(X 10.15)=P(Z3) = 0.9987

P(X 9.85)

Z-score for = (9.85 - mean)/standard deviation = (9.85 - 10) / 0.05 = -3

From standard normal tables , P(Z-3) = 0.0013

P(X 9.85) = P(Z-3) = 0.0013

P(9.85 X 10.15) = P(X 10.15) - P(X 9.85) = 0.9987-0.0013=0.9974

Probability of a non-defect = P(9.85 X 10.15) = 0.9974

Probability of a defect = 1-Probability of a non-defect = 1-0.9974 = 0.0026

Probability of a defect = 0.0026

Suspected number of defects for a 1,000 unit production run = 1000 x Probability of a defect = 1000 x 0.0026=2.6

Suspected number of defects for a 1,000 unit production run = 2.6