Emails arrive at the server of a company at the rate of 20 per hour. It is assum

Question

Emails arrive at the server of a company at the rate of 20 per hour. It is assumed that a Poisson process is a good model for the arrivals of the emails. What is the probability (to 2 decimal places) that the time between two consecutive emails is more than two minutes? Consider the total waiting time (in minutes) for three consecutive emails to arrive. Which of the following is a plot of the density function for this random variable? Find to 2 d.p. the probability that the waiting time in (b) is greater than 5 minutes. Answer:

Explanation / Answer

(A)
email rate 20 per hour
email rate per 2 mins = (20/60*3) = 1 per 2 mins

Here we need to find, not more than 1 mail arrived in 2 mins
P(X<=1) = P(X=0) + P(X=1)
P(X=0) = (1)^0*(e)^(-1)/0! = 0.3679
P(X=1) = (1)^1*(e)^(-1)/1! = 0.3679

P(X<=1) = 0.3679 + 0.3679 = 0.7358

(B)
Option II, as time passes waiting time goes on decreasing and density goes on increasing.

(C)
email rate per 5 mins = (20/60*5) = 5/3 = 1.667

Here we need to find, less than 3 mail arrived in 5 mins
P(X<3) = P(X=0) + P(X=1) + P(X=2)
P(X=0) = (1.667)^0*(e)^(-1.667)/0! = 0.1888
P(X=1) = (1.667)^1*(e)^(-1.667)/1! = 0.3148
P(X=2) = (1.667)^2*(e)^(-1.667)/2! = 0.2623

P(X<=1) = 0.1888 + 0.3148 + 0.2623 = 0.7659

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