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?
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%.