Asked by gourav
If a power failures occur according to a poisson distribution with an average of 3 failures every twenty weeks,calculate the probability that there will not be more than one failure during a particular week.
Answers
Answered by
MathMate
where μ represents the frequency of occurrence (success) within a given time period (week).
In this case, failure is considered "success".
so
μ
=3 "successes" / 20 weeks
=3/20
By definition of the Poisson distribution,
P(x;μ)=e<sup>-μ</sup>μ<sup>x</sup>/x!
where x∈Z
=P(0;μ)+P(1;μ)
=e<sup>-(3/20)</sup>(3/20)<sup>0</sup>/0! +e<sup>-(3/20)</sup>(3/20)<sup>1</sup>/1!
=0.86+0.13
=0.99
In this case, failure is considered "success".
so
μ
=3 "successes" / 20 weeks
=3/20
By definition of the Poisson distribution,
P(x;μ)=e<sup>-μ</sup>μ<sup>x</sup>/x!
where x∈Z
=P(0;μ)+P(1;μ)
=e<sup>-(3/20)</sup>(3/20)<sup>0</sup>/0! +e<sup>-(3/20)</sup>(3/20)<sup>1</sup>/1!
=0.86+0.13
=0.99
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