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 chain will have at least one mishandled bags?
To find the probability that the chain will have at least one mishandled bag in the next 1,000 customers, we can use the Poisson probability formula:
P(X >= 1) = 1 - P(X = 0)
where X is the number of mishandled bags in 1,000 customers.
Given that the average number of mishandled bags per 1,000 customers is 2.16, we can calculate the probability of having zero mishandled bags in 1,000 customers using the Poisson distribution formula:
P(X = 0) = (e^(-λ) * λ^0) / 0!
where λ is the average number of mishandled bags per 1,000 customers.
Plugging in the values:
P(X = 0) = (e^(-2.16) * 2.16^0) / 0!
P(X = 0) = (e^(-2.16) * 1) / 1
P(X = 0) = e^(-2.16)
Now, we can find the probability of having at least one mishandled bag in the next 1,000 customers:
P(X >= 1) = 1 - e^(-2.16)
P(X >= 1) ≈ 0.8847
Therefore, the probability that in the next 1,000 customers, the hotel chain will have at least one mishandled bag is approximately 0.8847 or 88.47%.