# Refer back to the series-parallel system configuration

## Problem 81E Chapter 2

Probability and Statistics for Engineering and the Sciences | 9th Edition

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Problem 81E

Refer back to the series-parallel system configuration introduced in Example 2.35, and suppose that there are only two cells rather than three in each parallel subsystem [in Figure 2.14(a), eliminate cells 3 and 6, and renumber cells 4 and 5 as 3 and 4]. Using P(Ai) = .9, the probability that system lifetime exceeds t0 is easily seen to be .9639. To what value would .9 have to be changed in order to increase the system lifetime reliability from .9639 to .99? [Hint: LetP(Ai) = .9 , express system reliability in terms of p, and then le? ?= ? . Reference example 2.35 Each day, Monday through Friday, a batch of components sent by a first supplier arrives at a certain inspection facility. Two days a week, a batch also arrives from a second supplier. Eighty percent of all supplier 1’s batches pass inspection, and 90% of supplier 2’s do likewise. What is the probability that, on a randomly selected day, two batches pass inspection? We will answer this assuming that on days when two batches are tested, whether the first batch passes is independent of whether the second batch does so. Figure 2.13 displays the relevant information. Reference 2.13 Reference 2.14

Step-by-Step Solution:

Answer : Step 1 : We have new system series-parallel system as Event A =ievent that lifetime of cell exceed t . 0 We assume all cell as identical and their failure or success that is lifespan system independent of each other. Given P(Ai) = .9, the probability that system lifetime exceeds t0 is easily seen to be .9639. Let P(A ) =iP Then the probability of the life system exceed t . 0 So P(life system exceed t ) =1-P(both subsystem lives are t ) 0 0 P(life system exceed t ) 01-P(both subsystem lives are t ) 0 P(life system exceed t ) 01-P[P(subsystem A has life t )P( su0system B has life t )] 0 Since,subsystems A and B are identical and independent of each other, P(system life t ) = 1 [P(subsystem life t )] 2 0 0 Since A and B are independent and identical of each other, Then, 2 P(system life > t )0=1 [P(subsystem life t )] 0 Because P(A)P(B) = P(A) P(B) for independent event. Here P(A) = P(B) So P(A)P(B)= P(A) P(B) P(A)P(B) = P(A) P(A) P(A)P(B) = P[(A) ] 2 P(system life > t )0=1 [P(subsystem life > t )] 0 2 2 P(system life > t )0= 1-[1-[P(cell 1 has life > t )0(P(cell 1 has life > t )]] 0 Because cell1 and cell2 are independent and...

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##### ISBN: 9780321629111

The answer to “Refer back to the series-parallel system configuration introduced in Example 2.35, and suppose that there are only two cells rather than three in each parallel subsystem [in Figure 2.14(a), eliminate cells 3 and 6, and renumber cells 4 and 5 as 3 and 4]. Using P(Ai) = .9, the probability that system lifetime exceeds t0 is easily seen to be .9639. To what value would .9 have to be changed in order to increase the system lifetime reliability from .9639 to .99? [Hint: LetP(Ai) = .9 , express system reliability in terms of p, and then le? ?= ? . Reference example 2.35 Each day, Monday through Friday, a batch of components sent by a first supplier arrives at a certain inspection facility. Two days a week, a batch also arrives from a second supplier. Eighty percent of all supplier 1’s batches pass inspection, and 90% of supplier 2’s do likewise. What is the probability that, on a randomly selected day, two batches pass inspection? We will answer this assuming that on days when two batches are tested, whether the first batch passes is independent of whether the second batch does so. Figure 2.13 displays the relevant information. Reference 2.13 Reference 2.14” is broken down into a number of easy to follow steps, and 202 words. This textbook survival guide was created for the textbook: Probability and Statistics for Engineering and the Sciences, edition: 9. This full solution covers the following key subjects: system, batch, supplier, reference, cells. This expansive textbook survival guide covers 18 chapters, and 1582 solutions. The full step-by-step solution to problem: 81E from chapter: 2 was answered by , our top Statistics solution expert on 05/06/17, 06:21PM. Since the solution to 81E from 2 chapter was answered, more than 433 students have viewed the full step-by-step answer. Probability and Statistics for Engineering and the Sciences was written by and is associated to the ISBN: 9780321629111.

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