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:
\[ \boxed{0.1845} \]