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 4.06 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 customer, the hotel chain will have no more than three mishandled bags?
To calculate the probability that the hotel chain will have no more than three mishandled bags in the next 1,000 customers, we can use the Poisson probability formula:
P(X <= 3) = e^(-λ) * (λ^0 / 0!) + e^(-λ) * (λ^1 / 1!) + e^(-λ) * (λ^2 / 2!) + e^(-λ) * (λ^3 / 3!)
Where λ is the average number of mishandled bags per 1,000 customers, which is 4.06.
P(X <= 3) = e^(-4.06) * (4.06^0 / 0!) + e^(-4.06) * (4.06^1 / 1!) + e^(-4.06) * (4.06^2 / 2!) + e^(-4.06) * (4.06^3 / 3!)
Calculating each term:
P(X <= 3) = 0.0170 + 0.0690 + 0.1404 + 0.1908
P(X <= 3) = 0.4172
Therefore, the probability that the hotel chain will have no more than three mishandled bags in the next 1,000 customers is 0.4172 or 41.72%.