a roofing company receives on average 4 calls per hour. In any given hour, what is the probability that it will receive at least 1 call?

Question

I need to see how this will be calculated in order to get the answer

Answer 1

Patricia,

 

The Poisson distribution does a nice job of modeling discrete events such as calls in a period of time like an hour.

 

P (k) = (λ^k) e^(-λ) / k!   for k = 0, 1, 2, ...  where

 

P(k)  = probability of k events (calls) in an hour.

   e   = Euler constant (2.718282).

   λ   = constant that represents both the mean value and standard deviation of this distribution (given as 4 calls per hr).

   k!  = factorial of k

 

Then  P(k ≥ 1) = 1 - P(k =0)

 

and P(k=0) = (4^o) e^(-4) / 0! = (1/1) e^(-4) = 0.0183   ,  since there is only one way of getting no calls in an hour 0! =1

 

Then  P(k ≥ 1) = 1 - P(k =0) 1 - 0.0183 = 0.9817 is the probability of receiving at least 1 call in an hour.