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 1000 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 more than four mishandled bags?
To solve this problem, we first need to determine the rate parameter λ of the Poisson distribution. We know that the hotel chain had 2.16 mishandled bags per 1000 customers, so we have:
λ = 2.16/1000 = 0.00216
Next, we need to calculate the probability that the hotel chain will have no more than four mishandled bags in the next 1000 customers. We can calculate this by summing the Poisson probabilities for 0, 1, 2, 3, and 4 mishandled bags:
P(X ≤ 4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)
Using the Poisson probability formula P(X=k) = (e^(-λ) * λ^k) / k!, we have:
P(X=0) = (e^(-0.00216) * 0.00216^0) / 0! = e^(-0.00216) ≈ 0.9978
P(X=1) = (e^(-0.00216) * 0.00216^1) / 1! = 0.00216 * e^(-0.00216) ≈ 0.00216
P(X=2) = (e^(-0.00216) * 0.00216^2) / 2! = (0.00216^2 / 2) * e^(-0.00216) ≈ 0.0005
P(X=3) = (e^(-0.00216) * 0.00216^3) / 3! = (0.00216^3 / 6) * e^(-0.00216) ≈ 0.0001
P(X=4) = (e^(-0.00216) * 0.00216^4) / 4! = (0.00216^4 / 24) * e^(-0.00216) ≈ 0.0000
Therefore,
P(X ≤ 4) = 0.9978 + 0.00216 + 0.0005 + 0.0001 + 0.0000 ≈ 1
So the probability that in the next 1,000 customers, the hotel chain will have no more than four mishandled bags is approximately 1, or 100%.