To calculate the probability that in the next 1,000 customers there will be no mishandled bags, we can use the Poisson probability formula:
P(X = k) = (e^(-λ) * λ^k) / k!
Where:
P(X = k) is the probability of getting k mishandled bags
e is the base of the natural logarithm (approximately 2.71828)
λ is the average number of mishandled bags per 1,000 customers
k is the number of mishandled bags we are interested in (in this case, k = 0)
Given that the hotel chain had 4.06 mishandled bags per 1,000 customers, we can calculate the average number of mishandled bags in 1,000 customers:
λ = 4.06/1000 = 0.00406
Now, we can plug the values into the formula and calculate the probability of having 0 mishandled bags in the next 1,000 customers:
P(X = 0) = (e^(-0.00406) * 0.00406^0) / 0!
P(X = 0) = (e^(-0.00406) * 1) / 1
P(X = 0) = e^(-0.00406)
Using a calculator, we find that e^(-0.00406) is approximately 0.995946
Therefore, the probability that in the next 1,000 customers, the hotel chain will have no mishandled bags is approximately 0.995946 or 99.59%.
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 no mishandled bags?
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