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.

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 

The 8th term is 512