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A factory uses three production lines to manufacture cans

Probability and Statistics for Engineers and the Scientists | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye ISBN: 9780321629111 32

Solution for problem 91E Chapter 2

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Probability and Statistics for Engineers and the Scientists | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Problem 91E

A factory uses three production lines to manufacture cans of a certain type. The accompanying table gives percentages of nonconforming cans, categorized by type of nonconformance, for each of the three lines during a particular time period. During this period, line 1 produced 500 nonconforming cans, line 2 produced 400 such cans, and line 3 was responsible for 600 nonconforming cans. Suppose that one of these 1500 cans is randomly selected. a. ?What is the probability that the can was produced by line 1? That the reason for nonconformance is a crack? b. ?If the selected can came from line 1, what is the probability that it had a blemish? c. ?Given that the selected can had a surface defect, what is the probability that it came from line 1?

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Problem 91E Answer: Step1: We have A factory uses three production lines to manufacture cans of a certain type. The accompanying table gives percentages of nonconforming cans, categorized by type of non conformance, for each of the three lines during a particular time period. Line 1 Line 2 Line 3 Blemish 15 12 20 Crack 50 44 40 Pull-Tab Problem 21 28 24 Surface Defect 10 8 15 Other 4 8 2 During this period, line 1 produced 500 nonconforming cans, line 2 produced 400 such cans, and line 3 was responsible for 600 nonconforming cans. Suppose that one of these 1500 cans is randomly selected. We need to find, a. What is the probability that the can was produced by line 1 That the reason for nonconformance is a crack b. If the selected can came from line 1, what is the probability that it had a blemish c. Given that the selected can had a surface defect, what is the probability that it came from line 1 Step2: a). Total cans = 500+400+600 = 1500. Nonconforming cans produced by line 1 is 500 Now, The probability that the can was produced by line 1 is given by P(the can was produced by line 1 =Nonconforming cans produced by line 1 500 total cans = 1500 = 0.3333 Consider, Line 1 Line 2 Line 3 Blemish 15 12 20 Crack 50 44 40 Pull-Tab Problem 21 28 24 Surface Defect 10 8 15 Other 4 8 2 Total 100 100 101 1).Cracked lines in Line 1 =crack(line× Nonconforming cans produced by line 1 total(line 1) = 50 × 500 100 = 0.50 × 500 = 250 crack(line 2) 2).Cracked lines in Line 2 =total(line 2)onconforming cans produced by line 2 = 44 × 400 100 = 0.44 × 400 = 176 crack(line 3) 3).Cracked lines in Line 3 =total(line 3)onconforming cans produced by line 3 = 40 × 600 100 = 0.40 × 600 = 240 Now, The reason for nonconformance is a crack is given by (Cracked lines in Line 1)+(Cracked lines in Line 2)+(Cracked lines in Line 3) = total cans = 250+176+240 1500 = 1500 = 0.444 Step3: b). Let us assume that Blemish = A and Line 1 = B Now, If the selected can came from line 1,then the probability that it had a blemish is given by P(AB) P(A/B) = P(B) Where, P(A B) = ( Blemish(line× Nonconforming cans produced by line 1)/total cans total(line 1) = ( 100 × 500)/1500 = (0.15×500)/1500 = 75/1500 = 0.05. And P(B) = Nonconforming cans produced by line 1 total cans = 500 1500 = 0.3333 Consider, P(AB) P(A/B) = P(B) = 0.05 0.3333 = 0.1515. Therefore,the probability that it had a blemish is 0.1515. Step4: c). Let us assume that Line 1 = A and Surface defect = B Now, Given that the selected can had a surface defect, the probability that it came from line 1 is given by P(AB) P(A/B) = P(B) Where, P(A B) = ( surface defect(l× Nonconforming cans produced by line 1)/total cans total(line 1) = ( 10 × 500)/1500 100 = (0.10×500)/1500 = 50/1500 = 0.0333 Therefore, P(A B) = 0.0333. And P(B) = (surface defect lines in Line 1)+(surface defect lines in Line 2)+(surface defect lines in Line 3) total cans 1).Surface defect in Line 1 = surface defect(li× Nonconforming cans produced by line 1 total(line 1) = 100 × 500 = 0.10 × 500 = 50 surface defect(line 2) 2).Surface defect in Line 2 = total(line 2)× Nonconforming cans produced by line 2 8 = 100× 400 = 0.08 × 400 = 32 3).Surface defect in Line 3 = surface defect(li× Nonconforming cans produced by line 3 total(line 3) = 15 × 600 100 = 0.15 × 600 = 90 Therefore, P(B) = (surface defect lines in Line 1)+(surface defect lines in Line 2)+(surface defect lines in Line 3) 50+32+90 total cans = 1500 172 = 1500 = 0.1146. Consider, P(AB) P(A/B) = P(B) 0.0333 = 0.1146 = 0.2904 Therefore, Given that the selected can had a surface defect, the probability that it came from line 1 is 0.2904.

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Chapter 2, Problem 91E is Solved
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Textbook: Probability and Statistics for Engineers and the Scientists
Edition: 9
Author: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye
ISBN: 9780321629111

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A factory uses three production lines to manufacture cans