Suppose the number of calls per hour to an answering service follows a poisson process with rate 4.
a.what is the probability that fewer than 2 calls came in the first hour?

Question

02-05-2018
Mathematics

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Suppose the number of calls per hour to an answering service follows a poisson process with rate 4.
a.what is the probability that fewer than 2 calls came in the first hour?

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mathmate mathmate · Answer 1 · 2018-05-02T14:48:47+02:00

With a mean of λ , the probability mass distribution (pmf) is given by
[tex]P(k,\lambda)=\lambda^k*e^(-\lambda)/k![/tex]

for λ = 4, and k<2 (i.e. k=0 or 1)

P(k<2, λ )=P(k=1, λ ) + P(k=1, λ )
[tex]=4^0*e^(-4)/4!+4^1*e^(-4)/1![/tex]
[tex]=4^0*e^(-4)/4!+4^1*e^(-4)/1![/tex]
=0.01832+0.07326
=0.09158 (to the fifth place of decimal)

Note: Poisson processes have no memory, so 2 calls in first hour has the same probability as 2 calls in any other hour.

Suppose the number of calls per hour to an answering service follows a poisson process with rate 4. a.what is the probability that fewer than 2 calls came in the first hour?

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Suppose the number of calls per hour to an answering service follows a poisson process with rate 4.
a.what is the probability that fewer than 2 calls came in the first hour?