Problem 2 The number of messages posted in a bulletin board system is a Poisson
ID: 3321071 • Letter: P
Question
Problem 2
The number of messages posted in a bulletin board system is a Poisson random variable with a mean of 10 messages per hour.
What is the probability that less than two messages are posted in one hour?
Suppose that no message has been posted for 3 hours. Find the probability that another hour will elapse before the next message arrives.
Problem 3
Hits on a high-volume website are assumed to follow a Poisson distribution with a mean of 10,000 hits per day. What is the probability of more than 10,150 hits in a day? Approximate the question by using a normal distribution.
Explanation / Answer
Problem #2.
10 messages per hour i.e. lambda = 10
P(X = 2) = e^(-lambda) * lambda^x / x!
= e^(-10) * 10^2/2!
= 0.00227
As per second condition, we want to know no message arrives in 4 hours.
lambda = 40 messages per 4 hour
P(X = 0) = e^(-40) * 10^0/0! = 0.0000
Problem #3
lambda = 10000
P(X > 10,150) = 1 - P(X < 10,150)
P(X < 10150) = e^(10000) * 10000^10150/10150!
= 0.933568998 (excel formula =POISSON.DIST(10150,10000,TRUE))
P(X > 10,150) = 1 - 0.933568998 = 0.0664
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