Probability

Question

The number of calls coming to a call center follows a Poison distribution. On average 90 calls arrive every hour on Mondays between 10 am and 11 am. The probability that exactly three calls will arrive in a two minutes period is what?

Answer 1

λt = (90 calls/min)[1/30 hr/(2 min)] = 3 calls/(2 min)

 

The distribution needed is Pois(λt) = Pois(3)

 

The PMF is then P(N = n) =(λt)ⁿe^-λt / n! = 3ⁿe^-3 / n!

 

So in a 2 minute time span, P(N = 3) = 3³e^-3/3! = 9/(2e³) ≈ 0.224