Question

Referring to the data presented in the Decision Dilemma, suppose cell phone users receive an average of 3.6 calls per day.

1. what is the probability that a cell phone user receives at least 1 call per day?
2. what is the probability that a cell phone user receives 5 or fewer calls per day?

Answer 1

X ~ poi ( )

Where = 3.6

Poisson probability distribution is

P(X) = e^- * ^X / X!

a)

P(X >= 1) = 1 - P(X = 0)

= 1 - e^-3.6

= 0.9727

b)

P(X <= 5) = P(X = 0) +P(X = 1) +P(X = 2) +P(X = 3) +P(X = 4) +P(X = 5)

= e^-3.6 +e^-3.6 *3.6 + e^-3.6 * 3.6² / 2! + e^-3.6 * 3.6³ / 3! + e^-3.6 * 3.6⁴ / 4! + e^-3.6 * 3.6⁵ / 5!

= 0.8441