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A system consists of two subsystems connected in series,

Statistics for Engineers and Scientists | 4th Edition | ISBN: 9780073401331 | Authors: William Navidi ISBN: 9780073401331 38

Solution for problem 10E Chapter 4.12

Statistics for Engineers and Scientists | 4th Edition

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Statistics for Engineers and Scientists | 4th Edition | ISBN: 9780073401331 | Authors: William Navidi

Statistics for Engineers and Scientists | 4th Edition

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Problem 10E

A system consists of two subsystems connected in series, as shown in the following schematic illustration. Each subsystem consists of two components connected in parallel. The AB subsystem fails when both A and B have failed. The CD subsystem fails when both C and D have failed. The system fails as soon as one of the two subsystems fails. Assume that the lifetimes of the components, in months, have the following distributions: A: Exp(l), B: Exp(0.l), C: Exp(0.2), D: Exp(0.2).

a. Generate, by simulation, a large number (at least 1000) of system lifetimes.

b. Estimate the mean system lifetime.

c. Estimate the median system lifetime.

d. Estimate the probability that the system functions for more than 6 months.

e. Estimate the 90th percentile of system lifetimes.

f. Estimate the probability that the AB subsystem fails before the CD subsystem does.

Step-by-Step Solution:

Step 1of 4:

  A system consist of two subsystems connected in series. Each subsystem consist of two components connected in parallel. Subsystem  1 and 2 fails if both components of the system fails. The system fails if one of two subsystem fails. The lifetime of components in months have distributed as A: exp(1) , B: exp(0.1) ,C: exp(0.2), D: exp (0.2).

               

The system fails if one of two subsystem fails. Let X1 , X2 and Y1,Y2 denote the lifetime of components A,B, C, D respectively. And L1 and L2 are the lifetime of both subsystem. So the lifetime of the system can be represented as  L= Min (L1,L2).

Step 2 of 4:

We have to generate a sample of 1000 system lifetime.

We can generate the sample of 1000 lifetime of each component of the system in minitab as

Calc- Random Data- Distribution(exponential) - Parameters(mean,standard deviation(with sample size))- Ok (store data in column C1 , C2 , C3, C4).

Find the lifetime of system 1 as Max( x1,x2), and lifetime of system 2 as Max( y1, y2).

Generated sample will be different from ours.

                         

(b) We have to estimate the mean system lifetime.

      We can find the mean of the system lifetime as

Stat - Basic statistics - Display  Descriptive statistics-  Assign variable - Statistics(mean) - ok.

         E(L)  2.069 months

Therefore the mean system lifetime is  2.069 months.

(c)  we  have to estimate the median of the system lifetime.

 

          Median 1.67 months.

Therefore the estimated median of the system lifetime is 1.67 months.

Step 3 of 4

Chapter 4.12, Problem 10E is Solved
Step 4 of 4

Textbook: Statistics for Engineers and Scientists
Edition: 4
Author: William Navidi
ISBN: 9780073401331

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