## Question

The number of messages sent to a bulletin board is a Poisson random variable with a mean of five messages per hour.

1. What is the probability that more than 2 messages are received between 14:00 and 15:00?
1. How many hits would you expect in two-hour time period?
2. Calculate the probability of 14 or more messages in a two-hour period.
3. Would it be unusual to receive 14 messages in a two-hour period? Explain briefly.

X follows Poisson distribution: P(X=x)=(lambda^xe^-lambda)/(x!),x>=0.

1. Since mean is five message per hour then lambda=5.

Probability that more than 2 messages received is equal to 1 minus probability that at most 2 messages are received:

P(X>2)=1-P(X<=2)=1-P(X=0)-P(X=1)-P(X=2)=

=1-(lambda^0e^-lambda)/(0!)-(lambda^1e^-lambda)/(1!)-(lambda^2e^-lambda)/(2!)=1-(5^0e^-5)/(0!)-(5^1e^-5)/(1!)-(5^2e^-5)/(2!)~~0.87534798.

1. Since it is expected to receive 5 message per hour then expected number of message per 2-hour period is 2*5=10.

2. Probability of receiving 14 or more messages is equal to 1 minus probability of receiving less than 14 messages:

P(X>=14)=1-P(X<14)=1-(X=0)-P(X=1)-P(X=2)-(X=3)-P(X=4)-

-P(X=5)-(X=6)-P(X=7)-P(X=8)-

-(X=9)-P(X=10)-P(X=11)-(X=12)-P(X=13)=

=1-(lambda^0 e^-lambda)/(0!)-(lambda^1e^-lambda)/(1!)-(lambda^2 e^-lambda)/(2!)-(lambda^3 e ^-lambda)/(3!)-(lambda^4e^-lambda)/(4!)-(lambda^5e^-lambda)/(5!)-(lambda^6e^-lambda)/(6!)-(lambda^7e^-lambda)/(7!)-(lambda^8e^-lambda)/(8!)-

-(lambda^9e^-lambda)/(9!)-(lambda^10e^-lambda)/(10!)-(lambda^11e^-lambda)/(11!)-(lambda^12e^-lambda)/(12!)-(lambda^13e^-lambda)/(13!)=1-(10^0e^-10)/(0!)-(10^1e^-10)/(1!)-

-(10^2e^-10)/(2!)-(10^3e^-10)/(3!)-(10^4e^-10)/(4!)-(10^5e^-10)/(5!)-(10^6e^-10)/(6!)-(10^7e^-10)/(7!)-

-(10^8e^-10)/(8!)-(10^9e^-10)/(9!)-(10^10e^-10)/(10!)-(10^11e^-10)/(11!)-(10^12e^-10)/(12!)-(10e^13e^-10)/(13!)~~0.1355356.

3. Since P(X=14)=(10^14e^-10)/(14!)~~0.0520771 then it would be quite unlikely (approximately 5%) to receive 14 messages in a 2-hour period.