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 0.21, 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 0.07. 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.
(c)	What is the advantage of reducing process variation, thereby causing process control limits to be at a greater number of standard deviations from the mean?
	Reducing the process standard deviation causes a - Select your answer -substantial increasemoderate increasenegligible changemoderate decreasesubstantial decreaseItem 3 in the number of defects.

Answer 1

a)

probability of defect:

probability =

1-P(9.85<X<10.15)

=

1-P(-0.71<Z<0.71)=

1-(0.7611-0.2389)=

0.4778

(try 0.4750 if this comes wrong)

suspected number of defects =np=1000*0.4778 =478

(try 475 if this comes wrong)

b)

probability of defect:

probability =

1-P(9.85<X<10.15)

=

1-P(-2.14<Z<2.14)=

1-(0.9838-0.0162)=

0.0324

(try 0.0322 if this comes wrong)

suspected number of defects =np= 32

c)

Reducing the process standard deviation causes a substantial decrease in the number of defects.