Question

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?

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