Using R Two machines run simultaneously in a factory. The times at which they need repair servicefollows an exponential distribution. The average time between repairs is about 2 weeks formachine 1, and 2.2 weeks for machine 2. Suppose both machines are put online at the sametime.2(a). [8pts] Using simulation, estimate the probability that the first machine to need repairs ismachine 2.2(b). [8pts] Using simulation, estimate the probability that the time the first machine to needrepairs will be longer than 3 weeks. Make a graph of the distribution of times, and computethis value (Hint: there may be a few approaches, but you may find min() or
Using R Two machines run simultaneously in a factory. The times at which they need repair servicefollows an exponential distribution. The average time between repairs is about 2 weeks formachine 1, and 2.2 weeks for machine 2. Suppose both machines are put online at the sametime.2(a). [8pts] Using simulation, estimate the probability that the first machine to need repairs ismachine 2.2(b). [8pts] Using simulation, estimate the probability that the time the first machine to needrepairs will be longer than 3 weeks. Make a graph of the distribution of times, and computethis value (Hint: there may be a few approaches, but you may find min() or
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
Using R
Two machines run simultaneously in a factory. The times at which they need repair service
follows an exponential distribution. The average time between repairs is about 2 weeks for
machine 1, and 2.2 weeks for machine 2. Suppose both machines are put online at the same
time.
2(a). [8pts] Using simulation, estimate the
machine 2.
2(b). [8pts] Using simulation, estimate the probability that the time the first machine to need
repairs will be longer than 3 weeks. Make a graph of the distribution of times, and compute
this value (Hint: there may be a few approaches, but you may find min() or
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