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 customers, the hotel chain will have at least two mishandled bags?
To solve this problem, we first need to calculate the average number of mishandled bags per 1,000 customers, which is given as 4.06. This average rate is also the rate parameter for the Poisson distribution.
Let X be the number of mishandled bags in the next 1,000 customers. We want to find P(X ≥ 2).
The probability mass function of a Poisson distribution is given by:
P(X = k) = (λ^k * e^(-λ)) / k!, where λ is the average rate of occurrence.
First, calculate the probability of having 0 or 1 mishandled bag in the next 1,000 customers:
P(X = 0) = (4.06^0 * e^(-4.06)) / 0! ≈ 0.016
P(X = 1) = (4.06^1 * e^(-4.06)) / 1! ≈ 0.065
Now, calculate the probability of having at least 2 mishandled bags:
P(X ≥ 2) = 1 - P(X = 0) - P(X = 1) ≈ 1 - 0.016 - 0.065 ≈ 0.919
Therefore, the probability that in the next 1,000 customers, the hotel chain will have at least two mishandled bags is approximately 0.919 or 91.9%.