Question

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 1

Answer:

a)

for normal distribution z score =(X-μ)/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)=

0.9878-0.0122=

0.9756

hence probability of a defect =1-0.9756 =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)=

0.9938-0.0062=

0.9876

probability of a defect =1-0.9876 = 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.