Question

Consider a Poisson probability distribution with

lambda

equals5.3

.

Determine the following probabilities.

​a) exactly

5

occurrences

​b) more than

6

occurrences

​c)

3

or fewer occurrences

Answer 1

a) P(X = 5) = e^-5.3 * 5.3⁵ / 5! = 0.1740

b) P(X > 6) = 1 - P(X < 6)

                   = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6))

                   = 1 - (e^-5.3 * 5.3⁰ / 0! + e^-5.3 * 5.3¹ / 1! + e^-5.3 * 5.3² / 2! + e^-5.3 * 5.3³ / 3! + e^-5.3 * 5.3⁴ / 4! + e^-5.3 * 5.3⁵ / 5! + e^-5.3 * 5.3⁶ / 6!)

                   = 1 - 0.7171

                   = 0.2829

c) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

                  = e^-5.3 * 5.3⁰ / 0! + e^-5.3 * 5.3¹ / 1! + e^-5.3 * 5.3² / 2! + e^-5.3 * 5.3³ / 3!

                  = 0.2254