Question 25: Motorola used the normal distribution to determine the probability

Question

Question 25: 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. Please include work.

a. The process standard deviation is 0.1, and the process control is set at plus or minus 2.25 standard deviations. Units with weights less than 9.775 or greater than 10.225 ounces will be classified as defects. What is the probability of a defect (to 4 decimals)?

In a production run of 1000 parts, how many defects would be found (to 0 decimals) ____?

b. Through process design improvements, the process standard deviation can be reduced to 0.09. Assume the process control remains the same, with weights less than 9.775 or greater than 10.225 ounces being classified as defects. What is the probability of a defect (rounded to 4 decimals; getting the exact answer, although not necessary, will require Excel)?

In a production run of 1000 parts, how many defects would be found (to 0 decimals)____?

c. What is the advantage of reducing process variation?

- Select answer -It can substantially reduce the number of defects or It may slightly reduce the number of defects or It has no effect on the number of defects

Explanation / Answer

a) Start with the given +/- 1 standard deviation values = 9.775 and 10.225. Now, find the probability of defects ...

P(X < 9.775) = P[z < (9.775 - 10) / 0.1) = P(z < -2.25)
From the Standard Normal table ...

P(z < -2.25) = .012224

Since the Normal is perfectly symmetric, P(X > 10.225) = P(z > 2.25) will also equal 0.012224

So the total probability of defects = 2* .0122244 = .024449

Number defects in a 1000 unit production = 1000 x 0.024449 = 24.449 defects

(b) P(X < 9.775) = P[z < (9.775 - 10) / 0.09) = P(z < -2.5)
From the Standard Normal table ...

P(z < -2.5) = .00621

Since the Normal is perfectly symmetric, P(X > 10.225) = P(z > 2.25) will also equal 0.00621

So the total probability of defects = 2* .00621 = .012419

Number defects in a 1000 unit production = 1000 x 0.012419 = 12.419 defects

(c) Answer would be - It can substantially reduce the number of defects as reduction in variance reduces the variability in the output and data is less sparsed from the average value. Thus Defects are reduced.

Answer

TY!

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.