Question

otorola used the normal distribution to determine the probability of defects and the number of defects...

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

Homework Answers

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.

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