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%.
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?
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