A telemarketer makes 5 phone calls per hour and is able to make a sale on 40% of all the contacts.
Q1: What is the probability of making exactly 3 sales in 1 hour?
Which distribution, binomial or Poisson, is more suitable? I would go for binomial, but Poisson seems correct as well.
Let X be the number of sales made per hour.
- Binomial: X~B(n=5, p=0.4)
P(X=3) = 5C3 x (0.4)^3 x (0.6)^2 = 0.2304
Justification: 2 outcomes. Independent Bernoulli experiments.
- Poisson: In one hour, 5 phone calls were made and 5 x 0.4 = 2 sales per hour.
X~Po(lambda = 2)
P(X=3) = (e^-2)(3^2)/2! = 0.1804
Justification: On average, 2 sales are made per hour.
Q2: What is the probability of making at most 3 sales in two hours? Answer obtained using binomial: 0.618 Answer obtained using Poisson: 0.433
If both distributions can be used, then I would expect to obtain the same answer.