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 0.30, 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 0.10. 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 anser to four decimal places.

Answer 1

Solution:

(a) z score for 9.85 = (9.85 - 10)/0.30

= -0.15/ 0.30

z score for 9.85 = -0.5

z score for 10.15 = (10.15 - 10)/0.30

= 0.15/0.30

z score for 10.15 = 0.5

Now

P(Defect) = P(X < 9.85 or X > 10.15)

= P(z < -0.5) + P(z >0.5)

= [1−P ( Z<0.5 )]+ [ 1−P ( Z<0.5 )]

= [ 1- 0.6915] + [1- 0.6915]

= 0.3085+0.3085

= 0.6170

Therefore P(Defect ) = 0.6170

(b) here

z score for 9.85 = (9.85 - 10)/0.10

z score for 9.85 = -0.15/0.10

z score for 9.85 = -1.5

Now

z score for 10.15 = (10.15 - 10)/0.10

z score for 10.15 = 0.15/0.10

z score for 10.15 = 1.5

Therefore,

P(Defect) = P(X < 9.85 or X > 10.15)

= P(z < -1.5) + P(z > 1.5)

= [1−P ( Z<1.5 )] + [ 1−P ( Z<1.5 )]

= [1-0.9332] + [ 1- 0.9332]

= 0.0668+ 0.0668

= 0.1336

Therefore  P(Defect) = 0.1336