Question

Messages arrive to a computer server according to a poisson distribution with a mean rate of 10 per hour. determine the length of an interval of time such that the probability that no messages arrive during this interval is 0.90.

Answer 1

Assuming that the Poisson distribution is applicable:
The Poisson distribution is f(k) = λ^k * e^(-λ) / k!
Where:λ is the expected number of circumstances of the occurrence during interval t ;

In the problem, r = 10 calls/hour. Where it is also = 1/360 calls/sec , therefore λ = t/360 calls during interval t ;
f(k) is the probability of exactly "k" occurrences;

So based from the problem: f = 0.9 ; k = 0 ; λ = t/360

0.9 = (t/360)^0 * e^(-t/360) / 0!

0.9 = e^(-t/360)

t = -360 ln 0.9 = 37.93 s
it is approximated: 38 s is the interval with 0.9 probability of no calls received