Question

Motorola 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 10 ounces.

(a) The process standard deviation is 0.18, and the process control is set at plus or minus one standard deviation. Units with weights less than 9.82 or greater than 10.18 ounces will be classified as defects. (Round your answer to the nearest integer.)

Calculate the probability of a defect. (Round your answer to four decimal places.)

Calculate the expected number of defects for a 1,000-unit production run. (Round your answer to the nearest integer.) defects

(b) Through process design improvements, the process standard deviation can be reduced to 0.06. Assume the process control remains the same, with weights less than 9.82 or greater than 10.18 ounces being classified as defects.

Calculate the probability of a defect. (Round your answer to four decimal places.)

Calculate the expected number of defects for a 1,000-unit production run. (Round your answer to the nearest integer.) defects

(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 no change in the number of defects.

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

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

Answer #1

**SOLUTION:**

Let X be the weight of any given item. X has a normal
distribution with mean _{} and
standard deviation

a) the probability of a defect is the probability that a randomly selected unit weighs less than 9.79 or greater than 10.21 ounces

^{Get the Z scores}

ans: the probability of a defect is 0.3173

the probability that a randomly selected item is defective is 0.3173

Let Y be the number of units out of 1000 which are defective. We can say that Y has a Binomial distribution with parameters, number of trials (number of units) n=1000 and success probability ( the probability that a randomly selected item is defective) p=0.3173

We know the the expectation of Y is

ans: the expected number of defects for a 1,000-unit production run is 317

b) the process standard deviation can be reduced to 0.07, that is if X is the weight of any given item. X has a normal distribution with mean and standard deviation

the probability of a defect is the probability that a randomly selected unit weighs less than 9.79 or greater than 10.21 ounces

^{get
the Z-score.}

^{}

^{}

^{}

^{}

^{}

ans: the probability of a defect is 0.0027

the probability that a randomly selected item is defective is 0.0027

Let Y be the number of units out of 1000 which are defective. We can say that Y has a Binomial distribution with parameters, number of trials (number of units) n=1000 and success probability ( the probability that a randomly selected item is defective) p=0.0027

We know the the expectation of Y is

ans: the expected number of defects for a 1,000-unit production run is 3

c) We can see that the number of defects has reduced due to the decrease in the standard deviation

ans: Reducing the process standard deviation causes a substantial reduction in the number of defects.

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