eBook Motorola used the normal distribution to determine the probability of defe

Question

eBook 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 15 ounces. a. The process standard deviation is 0.10, and the process control is set at plus or minus 1.25 standard deviation s. Units with weights less than 14.875 or greater than 15.125 ounces will be dlassified as defects. What is the probability of a defect (to 4 decimals)? -2112 In a production run of 1000 parts, how maniy defects would be found (round to the nearest whole number)? 211 b. Through process design improvements, the process standard deviation can be reduced to 0.08. Assume the process control remains the same, with weights less than 14.875 or greater than 15.125 ounces being dlassified as defects. What is the probability of a defect (round to 4 decimals: if necessary)? 11883 In a production run of 1000 parts, how many defects would be found (to the nearest whole number)? 119 C. What is the advantage of reducing process variation, thereby causing a problern limits to be at a greater number of standard deviations from the mean? It can substantially reduce the number of defects Hide Feedback Partially Correct

Explanation / Answer

a) Z scores of the test are -1.25 and 1.25

So, P(defect) = P(z > 1.25) or P(z < -1.25)

= 1 - P(z < 1.25) + P(z < -1.25)

= 1 - 0.8944 + 0.1056

= 0.2112

b) Number of defects in 1000 = 1000x0.2112 = 211

c) Standard deviation changed from 0.10 to 0.08

So, new z score = 1.25x0.10/0.08 = 1.56

P(defect) = 1 - P(z < 1.5625) + P(z < -1.5625)

= 1 - 0.9409 + 0.0591

= 0.1182

d) Number of defects = 0.1182x1000 = 118

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

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