Question

Suppose that you have an online blog, and you have observed that the number of visits...

Suppose that you have an online blog, and you have observed that the number of visits to your blog follows a Poisson distribution at a rate of 10 visits per day.

(a) What is the probability that there will be no hits in the next 6 hours?

(b) Suppose that there is no visit in the past 6 hours. What is the probability that you will have to wait for at least 3 more hours till the next visit?

(c) What is the probability your blog will have at least 2 visits in the next 12 hours?

(d) What is the length of a time interval within which the probability of no visits is 0.1?

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Answer #1

a) expected number of visits in next 6 hours =10*6/24 =2.5

therefore probability that there will be no hits =P(X=0)=e-2.5*2.50/0! =0.0821

b)

expected number of visits in next 3 hours =10*3/24 =1.25

probability that you will have to wait for at least 3 more hours till the next visit

=P(no visit in next 3 hours )=e-1.25*1.250/0! =0.2865

c)

expected number of visits in next 12 hours =10*12/24 =5

probability your blog will have at least 2 visits in the next 12 hours =P(X>=2)=1-P(X<=1)

=1-(e-5*50/0!+e-5*51/1!)

=1-0.6446

=0.3554

d)

let time interval is a hours ; expected number of visits =10a/24

P(

therefore P(n visits )=e-10a/24 =0.1

taking log on both sides:

a =-ln(0.1)/(10/24) =5.5626 Hours

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