A division within a transportation company is responsible fo

A division within a transportation company is responsible for performing maintenance on the fleet of trucks used by the company. Trucks in need of repair arrive to the maintenance division according to a Poisson process having a rate of 0.3 trucks per hour. Suppose we begin observing the maintenance division at some point in time. What is the probability that 3 trucks arrive during the first 90 minutes? If 2 trucks arrive between 10 a.m. and noon, what is the probability that 4 trucks arrive between 8 a.m. and noon? If 2 trucks arrive between 10 a.m. and noon, what is the probability that 3 trucks arrive between 8 a.m. and 10 a.m.?

Solution

Possion Distribution
PMF of P.D is = f ( k ) = e- x / x!
Where   
= parameter of the distribution.
x = is the number of independent trials
a)
0.3 trucks per hours
mean rate for 90 minuets is 0.3+0.15 is 0.45
P( X = 3 ) = e ^-0.45 * 0.45^3 / 3! = 0.0097

b)
Mean rate= 2 trucks arrive in b/w 10 AM to NOON, i.e 1 Per hour
Mean rate for time 8 AM to Noon are = 1*4 = 4
P( X = 4 ) = e ^-4 * 4^4 / 4! = 0.1954

c)
Mean rate for time 8 AM to 10 AM are = 1*2 = 2
P( X = 3 ) = e ^-2 * 2^3 / 3! = 0.1804