Question

The number of phone calls in a given period of time follows a Poisson distribution. Find the mean number of phone calls so that probability of one or more calls exceeds 0.95?​

Answer 1

Answer:

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Answer 2

Given : The number of phone calls in a given period of time follows a Poisson distribution.

To Find :  the mean number of phone calls so that probability of one or more calls exceeds 0.95

Solution:

P(x)  = λˣ  e^(-λ) / x!

λ = Mean

probability of one or more calls exceeds 0.95

=> P(1) + P(2) + .... > 0.95

=> P(0) < 0.05

P(0)  = λ⁰  e^(-λ) / 0!

=> e^(-λ) < 0.05

=> - λ < ln (0.05)

=> - λ < -2.996

=>   λ > 2.996

=>  λ = 3   min value

Minimum  the mean number of phone calls  = 3

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