Cars arrive at a car wash randomly and independently the pro

Cars arrive at a car wash randomly and independently; the probability of an arrival is the same for any two time intervals of equal length. The mean arrival rate is 15 cars per hour.

1) What is the probability that 12 or more cars will arrive during any given hour of operation?

2) What is the probability that 5 or fewer cars will arive during any given 20 minutes of operation?

3) What is the probability that more than 5 but less than 10 cars will arrive during any given 40 minutes of operation?

4) What is the probaility that no arrivals happen in any given 10 minutes of operation?

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)
P( X < 12) = P(X=11) + P(X=10) + P(X=9) + P(X=8) + P(X=7) + P(X=6) + P(X=5) + ..... + P(X=1)+ P(X=0)   
= e^-15 * 2 ^ 11 / 11! + e^-15 * ^ 10 / 10! + e^-15 * ^ 9 / 9! + e^-15 * ^ 8 / 8! + e^-15 * ^ 7 / 7! + e^-15 * ^ 6 / 6! + e^-15 * ^ 5 / 5! + e^-15 * ^ 4 / 4! + e^-15 * ^ 3 / 3! + e^-15 * ^ 2 / 2! + ...... e^-15 * ^0/ 0!
= 0.1848
P( X > = 12 ) = 1 - P (X < 12) = 0.8152

b)
15 Cars per 60 mins, For 40 mins of approach = 20*15/60 = 5 Cars

P( X < = 5) = P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)   
= e^-5 * 5 ^ 5 / 5! + e^-5 * 5 ^ 4 / 4! + e^-5 * ^ 3 / 3! + e^-5 * ^ 2 / 2! + e^-5 * ^ 1 / 1! + e^-5 * ^ 0 / 0!   
= 0.616

c)

15 Cars per 60 mins, For 20 mins of approach = 40*15/60 = 10 Cars
P( X = 6 ) = e ^-10 * 10^6 / 6! = 0.0631
P( X = 7 ) = e ^-10 * 10^7 / 7! = 0.0901
P( X = 8 ) = e ^-10 * 10^8 / 8! = 0.1126
P( X = 9 ) = e ^-10 * 10^9 / 9! = 0.1251

P( 5 < X < 10 ) = P( X = 6 ) + P( X = 7 ) + P( X = 8 )+ P( X = 9 ) = 0.0631 + 0.0901 + 0.1126 + 0.1251 =0.3909
d)
15 Cars per 60 mins, For 10 mins of approach = 10*15/60 = 2.5 Cars
P( X = 0 ) = e ^-2.5 * 2.5^0 / 0! = 0.0821