Question

The average number of phone calls per minute into the switchboard of a company between 10.00 am and 2.30 pm is 2.5. Find the probability that during one particular minute there will be 

(i) no phone at all 

(ii) exactly 3 calls 

(iii) at least 5 calls.

Answer 1

Let X be the Poisson variable denoting the number of phone calls per minute.

P (X = 1) = (e^-2.5(2.5))/1! = (0.08208) (2.5) = 0.2052 

Using the above values and P (X = 0) and P (X = 3) from the previous subdivisions in (A) we get, 

P(X ≥ 5) = 1 – [0.1336 + 0.2138 + 0.2565 + 0.2052 + 0.08208] 

= 1 – 0.89118 

= 0.10882