Suppose that the number of phone calls in an hour is modeled

Suppose that the number of phone calls in an hour is modeled by the Poisson distribution with 2
calls per hour on average. Let Y be a random variable for the number of phone calls in 4 hours. Find:

(a) P(3 < Y 6)
(b) P(Y > 5)
(c) P(Y 9)

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)
Mean rate is 2 calls per hour on average
and the number of phone calls in 4 hours is = 2*4 = 8
P( X = 4 ) = e ^-8 * 8^4 / 4! = 0.0573
P( X = 5 ) = e ^-8 * 8^5 / 5! = 0.0916

P( 3 < X < 6) = P(X=4) + P(X=5) = 0.0573+0.0916 = 0.1489

b)
P( X < = 5) = P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)   
= e^-8 * 8 ^ 5 / 5! + e^-8 * 5 ^ 4 / 4! + e^-8 * ^ 3 / 3! + e^-8 * ^ 2 / 2! + e^-8 * ^ 1 / 1! + e^-8 * ^ 0 / 0!
= 0.1912
P( X > 5) = 1 -P ( X <= 5) = 1 - 0.1912 = 0.8088

c)

P( X < 9) = P(X=8) + P(X=7) + P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= e^-8 * 5 ^ 8 / 8! + e^-8 * ^ 7 / 7! + e^-8 * ^ 6 / 6! + e^-8 * ^ 5 / 5! + e^-8 * ^ 4 / 4! + e^-8 * ^ 3 / 3! + e^-8 * ^ 2 / 2! + e^-8 * ^ 1 / 1! + e^-8 * ^ 0 / 0!
= 0.5925
P( X > = 9 ) = 1 - P (X < 9) = 0.4075