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

you didn't post any answer choices, so its not 100% clear what your teacher is looking for

however, log(15/7) is equivalent to log(15) - log(7)

i'm using the rule log(x/y) = log(x) - log(y)