The number of calls that come into a small mailorder company

The number of calls that come into a small mail-order company follows a Poisson distribution. Currently, these calls are serviced by a single operator. The manager knows from past experience that an additional operator will be needed if the rate of calls exceeds 28 per hour. The manager observes that 9 calls came into the mail-order company during a randomly selected 15-minute period. If the rate of calls is actually 28 per hour, what is the probability that 9 or more calls will come in during a given 15-minute period?

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
rate of calls is actually 28 per 60 imns ;
mean for 15 mins will be = 15*28/60 = 7 mins

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^-7 * 3 ^ 8 / 8! + e^-7 * ^ 7 / 7! + e^-7 * ^ 6 / 6! + e^-7 * ^ 5 / 5! + e^-7 * ^ 4 / 4! + e^-7 * ^ 3 / 3! + e^-7 * ^ 2 / 2! + e^-7 * ^ 1 / 1! + e^-7 * ^ 0 / 0!
= 0.7291

P( X > = 9 ) = 1 - P (X < 9) = 0.2709