Question
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?
0.0254
0.0687
0.0256
0.0172
0.0025
0.0254
0.0687
0.0256
0.0172
0.0025
Answers
GPT 3.5
The probability of having no mishandled bags in a Poisson distribution with 4.06 mishandled bags per 1000 customers is given by the formula:
P(X = 0) = (e^(-λ) * λ^0) / 0!
Where λ is the average number of mishandled bags per 1000 customers.
In this case, λ = 4.06. Thus, the probability can be calculated as:
P(X = 0) = (e^(-4.06) * 4.06^0) / 0!
P(X = 0) = (e^(-4.06) * 1) / 1
P(X = 0) = e^(-4.06)
P(X = 0) = 0.0172
Therefore, the probability that in the next 1,000 customers, the hotel chain will have no mishandled bags is 0.0172 or 1.72%.
Hence, option "0.0172" is the correct answer.
P(X = 0) = (e^(-λ) * λ^0) / 0!
Where λ is the average number of mishandled bags per 1000 customers.
In this case, λ = 4.06. Thus, the probability can be calculated as:
P(X = 0) = (e^(-4.06) * 4.06^0) / 0!
P(X = 0) = (e^(-4.06) * 1) / 1
P(X = 0) = e^(-4.06)
P(X = 0) = 0.0172
Therefore, the probability that in the next 1,000 customers, the hotel chain will have no mishandled bags is 0.0172 or 1.72%.
Hence, option "0.0172" is the correct answer.