Full Question
The number of messages that arrive at a Web site is a Poisson distributed random variable with a mean of 6 messages per hour.
a. What is the probability that 6 messages are received in 1 hour?
b. What is the probability that 10 messages are received in 1.5 hours?
c. What is the probability that fewer than 2 messages are received in 0.5 hour?
Answer and Explanation
Given
λ = 6 per hour
Poisson Probability P(X = k) = (λ^k e^-λ)/k!
a. K = 6
P(X = 6) = (6^6 e^-6)/6!
P(X = 6) = 0.160623141047980
P(X = 6) = 0.1606--------- Approximated
b.
If 6 messages are received on average per hour then the number of messages received on average per 1.5 hours is
λ = 6 *1.5
λ = 9
For k = 10
P(X = 10) = (9^10 e^-9)/10!
P(X = 10) = 0.118580076008570
P(X=10) = 0.1186 ---------- Approximated
c.
If 6 messages are received on average per hour then the number of messages received on average per 0.5 hours is
λ = 6 *0.5
λ = 3
For messages fewer than 2 means than k = 0 or k = 1
For k = 0
P(X = 0) = (3^0 e^-3)/0!
P(X = 0) = 0.049787068367863
P(X = 0) = 0.0498 ------_--- Approximated
For X = 1
P(X = 1) = (3^1 e^-3)/1!
P(X = 1) = 0.149361205103591
P(X = 1) = 0.1494 ---------- Approximated
P(X <2) = P(X=0) + P(X=1)
P(X<2) = 0.0498 + 0.1494
P(X<2) = 0.1996
