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.
Customers arrive at bakery at an average of 10 customers per hour. W.pdf
1. 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
