Customers arrive at bakery at an average of 10 customers per

Customers arrive at bakery at an average of 10 customers per hour. What is the probability that less than 5 customers will arrive in the next hour? Assume Poisson arrivals. a. .0378 b. .0293 c. .2500 d .1126

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
P( X < 5) = P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)   
= e^-10 * 0 ^ 4 / 4! + e^-10 * ^ 3 / 3! + e^-10 * ^ 2 / 2! + e^-10 * ^ 1 / 1! + e^-10 * ^ 0 / 0!
= 0.0293

P(less than 5 customers will arrive in the next hour) = 0.0293