An airline estimates that 97% of people booked on their flights actually show up. The airline books 204 people on a flight for which the maximum number is 200.The probability that the number of people who show up will exceed the capacity of the plane is closest to ....Group of answer choices0.14000.74860.85890.25140.8701
Question
An airline estimates that 97% of people booked on their flights actually show up. The airline books 204 people on a flight for which the maximum number is 200.The probability that the number of people who show up will exceed the capacity of the plane is closest to ....Group of answer choices0.14000.74860.85890.25140.8701
Solution
This is a binomial problem where we are trying to find the probability of more than 200 people showing up out of 204 booked, given that the probability of a person showing up is 0.97.
The binomial distribution can be approximated by a normal distribution when the sample size is large, which is the case here.
First, we need to calculate the mean (μ) and standard deviation (σ) of the binomial distribution.
The mean (μ) is np, where n is the number of trials (people booked) and p is the probability of success (people showing up).
μ = np = 204 * 0.97 = 197.88
The standard deviation (σ) is sqrt(np(1-p)), where sqrt is the square root function.
σ = sqrt(np(1-p)) = sqrt(204 * 0.97 * 0.03) = sqrt(5.8828) = 2.426
Next, we need to standardize the value we are interested in (200 people) using the formula for a z-score, which is (X - μ) / σ, where X is the value we are interested in.
However, to avoid a continuity correction error, we use 200.5 instead of 200.
Z = (200.5 - 197.88) / 2.426 = 1.08
Finally, we need to find the probability that the z-score is more than 1.08. Looking this up in a standard normal distribution table or using a calculator, we find that the probability is approximately 0.1400.
So, the probability that the number of people who show up will exceed the capacity of the plane is closest to 0.1400.
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