Question

The U.S. Department of Transportation maintains statistics for mishandled bags per 1,000 passengers. In 2013, Delta mishandled 2.19 bags per 1,000 passengers. What is the probability that in the next 1000 passengers Delta will have:

a.       No mishandled bags

b.       At least one mishandled bag?

c.       At least two mishandled bags?

Answer 1

Probability mass function of a poisson distribution:

X : Number of mishandled bags per 1000 passengers

Given,

In 2013, Delta mishandled 2.19 bags per 1,000 passengers

Average mishandled bags per 1,000 passengers : = 21.9

X follows a Poission distribution with = 21.9 with Probability mass function

a. Probability of No mishandled bags i.e P(X=0)

P(X=0) = P(0)

Probability of No mishandled bags = 0.1119

b.

Probability of Atleast one mishandled i.e P(X 1)

P(X 1) = 1-P(X<1) = 1 - P(X=0) = 1 - 0.1119 = 0.8881

Probability of Atleast one mishandled = 0.8881

c.

Probability of Atleast two mishandled i.e P(X 2)

P(X 2) = 1-P(X<2) = 1- [P(X=0) +P(X=1)]

P(X=0) +P(X=1) = 0.1119 + 0.2451 = 0.357

1- [P(X=0) +P(X=1)] = 1-0.357 = 0.643

Probability of Atleast two mishandled = 0.643