Question

A major hotel chain keeps a record of the number of mishandled bags per 1,000 customers. In a recent year, the hotel chain had 3.5 mishandled bags per 1,000 customers. Assume that the number of mishandled bags has a Poisson distribution. What is the probability that in the next 1,000 customers, the chain will have exactly 2 mishandled bags?

Answer 1

To find the probability of exactly 2 mishandled bags when the average number of mishandled bags per 1,000 customers is 3.5, we can use the Poisson probability formula:

\[
P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}
\]

where:
- \( k \) is the number of occurrences (in this case, the number of mishandled bags),
- \( \lambda \) is the average number of occurrences (the mean),
- \( e \) is the base of the natural logarithm (approximately equal to 2.71828).

Here, we have:
- \( \lambda = 3.5 \)
- \( k = 2 \)

Plugging in these values gives:

\[
P(X = 2) = \frac{3.5^2 \cdot e^{-3.5}}{2!}
\]

Calculating \( 3.5^2 \):

\[
3.5^2 = 12.25
\]

Calculating \( 2! \):

\[
2! = 2
\]

Now, we need to calculate \( e^{-3.5} \):

Calculating \( e^{-3.5} \) using a calculator or exponential function:

\[
e^{-3.5} \approx 0.0301974
\]

Now substituting these values into the Poisson formula:

\[
P(X = 2) = \frac{12.25 \cdot 0.0301974}{2}
\]

Calculating the numerator:

\[
12.25 \cdot 0.0301974 \approx 0.36901845
\]

Now divide by 2:

\[
P(X = 2) \approx \frac{0.36901845}{2} \approx 0.184509225
\]

Thus, the probability that in the next 1,000 customers, the hotel chain will have exactly 2 mishandled bags is approximately:

\[
\boxed{0.1845}
\]