Problem

The U.S. Department of Transportation maintains statistics for mishandled bags per 1,000 a...

The U.S. Department of Transportation maintains statistics for mishandled bags per 1,000 airline passengers. In October 2013, Delta mishandled 1.55 bags per 1,000 passengers. What is the probability that in the next 1,000 passengers, Delta will have

a. no mishandled bags?


b. at least one mishandled bag?


c. at least two mishandled bags?

Step-by-Step Solution

Solution 1
The mean number of mishandled bags per 1,000 passengers is 1.55. That is λ=1.55.

The probability mass function of Poisson distribution is,

\(P(X=x)=\frac{e^{-\lambda} \lambda^{x}}{x !}, \quad x=0,1,2, \ldots\)

Here, P(X=x) is the probability that an event x occurs, λ is the parameter, excepted value or mean.

a.

The probability that there is no mishandled bag is,

$$ \begin{aligned} P(X=0) &=\frac{e^{-(1.55)}(1.55)^{0}}{0 !} \\ &=\frac{0.2122 \times 1}{1} \\ &=0.2122 \end{aligned} $$

b.

The probability that there is at least one mishandled bag is,

$$ \begin{aligned} P(X \geq 1) &=1-P(X=0) \\ &=1-\frac{e^{-(1.55)}(1.55)^{0}}{0 !} \\ &=1-\frac{0.2122 \times 1}{1} \\ &=1-0.2122 \\ &=0.7878 \end{aligned} $$

c.

The probability that there are at least two mishandled bags is,

$$ \begin{aligned} P(X \geq 2) &=1-P(X<2) \\ &=1-\{P(X=0)+P(X=1)\} \\ &=1-\left\{\frac{e^{-(1.55)}(1.55)^{0}}{0 !}+\frac{e^{-(1.55)}(1.55)^{1}}{1 !}\right\} \\ &=1-\left\{\frac{0.2122 \times 1}{1}+\frac{0.2122 \times 1.55}{1}\right\} \\ &=1-\{0.2122+0.3290\} \\ &=1-0.5412 \\ &=0.4588 \end{aligned} $$

Add your Solution
Textbook Solutions and Answers Search