Joey L.
asked • 06/26/20Poisson Distribution Question
A commercial washing machine has non-repairable motor with a constant failure rate of 0.08 failures per year. The service organization has purchased 2 spare motors. If the design life of the washing machine is 7 years, what is the probability that the 2 spares will be adequate?
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1 Expert Answer
Tom K. answered • 06/26/20
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.08 * 7 = .56
Two spares are adequate if there are 2 or less failures in the 7 years.
P(x) = e^-.56(.56)^x/x!
P(0) + P(1) + P(2) = e^-.56(1 + .56 + .56^2/2) = 0.980651720815645
Alternatively, you can use a package or calculator that will return the result.
For example, in Excel, =poisson.dist(2,.56,1) generates the same value. (The final 1 says cumulative; if 0, this would return the probability of exactly 2.)
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Terman F.
The SMp work "GenDPD" is currently available from: http://www.mpijournal.org/pdf/2020-03/MPI-2020-03-p491.pdf; and will help medical physics community and others to understand: 1) How the binomial distribution (BD) and Poisson distribution (PD) were derived; 2) What really the BD and PD are; 3) How one should use the BD; 4) The PD is not a new probabilistic function (PF), but the own BD with simplifications valid for some values of BD parameters, and changes of variable and parameters; and 5) Given the essential condition for a PF is not satisfied in the PD for some values of its parameter, also we can say that: The PD is not a PF. For these reasons, the PD(x;µ) could be replaced with the BD(x;Xmax,p), where Xmax is the possible outcomes of a stochastic process, Xmax=n and p= µ/Xmax.01/24/21