A website receives an average of four visits per ten minutes. Let X_t be the num

Question

A website receives an average of four visits per ten minutes. Let X_t be the number of visits after t minutes. Use a Poisson process to model the number of visits. (a) Identify the distribution (by name and parameter value(s)) of the number of visits in the next two minutes. (b) The website receives advertising revenue after 25,000 visits. Identify the distribution (by name and parameter value(s)) of the time until the website begins to receive revenue. (c) Compute the expected time until the website receives revenue. Give your answer in days, rounded to two places after the decimal. (d) Express the probability of the website receiving revenue in at most 40 days using an appropriate CDF (you do not have to compute the probability). (e) Compute the probability that at least five minutes pass until the next visit. (f) What is the distribution of X_15?

Explanation / Answer

average number of visits per minute = 4/10 =0.4 clicks per minute

(a) Distribution is poission distribution with parameter t = 0.4t = 0.4 * 2 = 0.8 clicks per minute

Where P( Xt) = e-t (t)X /X! = e-0.8 0.8X /X!

b) Website receive revenue after 25000 visits

so average time taken in getting 25000 visits = 25000/0.4= 62500 minutes

For sufficiently large values of , (say >1,000), theNormal( = ,2 = ) Distribution is an excellent approximation to the Poisson() Distribution.

So, here distribution can be said normal distribution where Mean () = 62500 minutes = 43.40 days and standerddeviation = sqrt( 43.40) = 6.58

c) E(time website receives revenue) = 62500 minutes = 43.40 days

d) Here we get it as a normal distribution so we have to calculate that time must be at most 40 days

so P( t< 40 days) = 1 - (z) = 1 - [ (40 - 43.40)/6.58]

where (z) is the cdf of the given probability distribution

P( t< 40 days) = 1 - (- 0.51)

(e) In the next five minutes , there must be 2 clicks so

P( x =0; = 2) = e -2 20/ 0! = 0.1353

(f) X15= e-t (t)X /X! = e-6 6X /X!

[here = 0.4 and t = 15 minutes ;t = 6 clicks]

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