Question

A store typically has 5 customers arrive per hour, and arrivals follow a Poisson distribution. What is the likelihood that a. no customers will arrive in an hour? b. less than three customers arrive in an hour? c. five or more customers arrive in an hour?

Answer 1

Mean/Expected number of events of interest: λ =                5 / hour

poisson probability distribution

P(X=x) = e-λλx/x!

a)

P ( X =    0   ) = e^-5*5^0/0!=   0.0067

b)

P ( X =    1   ) = e^-5*5^1/1!=   0.0337
P ( X =    2   ) = e^-5*5^2/2!=   0.0842

P(X<3) = P(x=0) + P(X=1) + P(X=2) = 0.0067+0.0337 + 0.0842 = 0.1247

c)

P(X≥5) = 1 - P(X≤4) = 1 - 0.4405 = 0.5595