Question

The exponential distribution is frequently applied

to the waiting times between successes in a Poisson

process. If the number of calls received per hour

by a telephone answering service is a Poisson random

variable with parameter λ = 6, we know that the time,

in hours, between successive calls has an exponential

distribution with parameter β =1/6. What is the probability

of waiting more than 15 minutes between any

two successive calls?

Answer #1

probability that there is 00 call in 1515 min. Therefore, for the Poisson parameter, λt=3/2λt=3/2 where t=0.25t=0.25.

Based on the Poisson distribution formula p(X=x)=λxe−λx!p(X=x)=λxe−λx!,

the probability that zero call arrives in the next 15 minutes is p(X=0)=(3/2)0e−3/20!p(X=0)=(3/2)0e−3/20! which is e−3/2e−3/2

Is this the correct way of doing it? I have been searching similar questions online and I found someone's answer on reddit using exponential distribution and I dont know if that's correct because the answer is different from mine.

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