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 2.16 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 hotel chain will have no mishandled bags?
To calculate the probability that in the next 1,000 customers there will be no mishandled bags, we can use the Poisson probability formula:
P(X = k) = (e^(-λ) * λ^k) / k!
Where:
- P(X = k) is the probability that exactly k mishandled bags occur
- λ is the average number of mishandled bags per 1,000 customers
- k is the number of mishandled bags (in this case, k = 0)
Given that λ = 2.16 mishandled bags per 1,000 customers, we need to first convert this to the average number of mishandled bags per customer:
λ = 2.16 / 1000 = 0.00216 mishandled bags per customer
Now we can plug in the values to the formula:
P(X = 0) = (e^(-0.00216) * 0.00216^0) / 0!
P(X = 0) = (e^(-0.00216) * 1) / 1
P(X = 0) = e^(-0.00216)
Using a calculator, we find that e^(-0.00216) ≈ 0.997845288
Therefore, the probability that in the next 1,000 customers, the hotel chain will have no mishandled bags is approximately 0.9978 or 99.78%.