Question

Messages arrive at an electronic mail server at the average rate of 4 messages every 5 minutes. Their number is modeled by a Binomial counting process.

(a) What frame length makes the probability of a new message arrival during a given frame equal 0.05?

(b) Suppose that 50 messages arrived during some 1-hour period. Does this indicate that the arrival rate is on the increase? Use frames computed in (a).

Answer 1

Average number. q. moils cuining in mail bone is said to follow bingmical distribution with trean of 4 in the minute interval. This i 5 minute internal consists of infinite humba of brone intervals of small length (very small) and probability that mail assives in this integral is very small. Honce as h-sos and pho we can use Poisson appronimation to binomial distribution. Hince rate of assuals in 5 minute interval is d=bp = 15 minute Hence number of anivals is in one house is hence 4x12=u8 [hour. Now, t= 48 hour When humben g ainsirals is distributed as Puisson, interestinal time ir distributed as caberential with in tavaniyal time of GO -|-25. The distribution of this exponential vasicable B given by P[Tat] dett, de 1.25 48 [Tnt] = drettet - Vedt | édt.

:-P (Tst) - ett The probabibility that time frame fro. which new message aires i at (interoviral time) = 0.05 = log est _ lug oor => -dt = log oor - 11.25 (t) = -2.9957 => t = -2.9957 = 2.3966. -1-25

52 Probability that there are so ansivcis in GO minuts is equvi valent to saying that gap between Assoreds is 60=12 P[ T >2.396) = 6.65 but here time gap is 1.2 minutes. Hance we can say that so ansin als in 1 minute does not indicate decrease in late parameter (significantly). level of significance here is o.or. Critical region is (2.3966, a) where (2.3966, ) B gap between assinals