In a large corporate computer network, user log-ons to the system can be modeled as a Poisson RV with a mean of 25 log-ons per hour. (20pts) (a) What is the probability that there are no logons in an interval of 6 minutes? (b) What is the probability that the distance between two log-ons be more than one hour?

F (t<0.1) = 0.91791

Step-by-step explanation:

Solution:

- Let X be an exponential RV denoting time t in hours from start of interval to until first log-on that arises from Poisson process with the rate λ = 25 log-ons/hr. Its cumulative density function is given by:

F (t) = 1 - e ^ ( - 25*t) t > 0

A) In this case we are interested in the probability that it takes t = 6/60 = 0.1 hrs until the first log-on. F (t < 0.1 hr), we have:

F (t<0.1) = 1 - e ^ ( - 25*0.1)

F (t<0.1) = 0.91791