According to the MIT Airline Data Project, American Airlines controlled 15.5% of the domestic market in 2010. A random sample of 125 domestic passengers that year was selected. Using the normal approximation to the binomial distribution, what is the probability that more than 12 passengers from this sample were on American Airline flights?
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To calculate the probability that more than 12 passengers from the sample were on American Airlines flights, we need to use the normal approximation to the binomial distribution.
First, we need to calculate the mean (μ) and the standard deviation (σ) of the binomial distribution.
The mean (μ) can be calculated as:
μ = n * p
Where n is the sample size (125 passengers) and p is the probability of success (15.5%).
μ = 125 * 0.155
μ = 19.375
The standard deviation (σ) can be calculated as:
σ = sqrt(n * p * (1 - p))
Where sqrt represents the square root.
σ = sqrt(125 * 0.155 * (1 - 0.155))
σ = sqrt(125 * 0.155 * 0.845)
σ = sqrt(16.0584375)
σ ≈ 4.007
Next, we need to calculate the z-score for the value of 12 passengers (x) using the formula:
z = (x - μ) / σ
z = (12 - 19.375) / 4.007
z ≈ -1.845
Now, we can use a standard normal distribution table or a calculator to find the probability associated with the z-score of -1.845. The probability is the area under the curve to the right of -1.845.
The probability can be calculated as:
P(X > 12) = 1 - P(X ≤ 12)
Using the standard normal distribution table or a calculator, we find that P(X ≤ 12) is approximately 0.0322. Therefore,
P(X > 12) = 1 - P(X ≤ 12)
P(X > 12) = 1 - 0.0322
P(X > 12) ≈ 0.9678
Therefore, the probability that more than 12 passengers from the sample were on American Airlines flights is approximately 0.9678 or 96.78%.
To find the probability that more than 12 passengers from a random sample of 125 were on American Airlines flights, we can use the normal approximation to the binomial distribution.
First, we need to calculate the mean and standard deviation of the binomial distribution. The mean (μ) is equal to n * p, where n is the sample size and p is the probability of success (in this case, the probability of a passenger being on an American Airlines flight). The standard deviation (σ) is equal to sqrt(n * p * (1 - p)).
Given that American Airlines controlled 15.5% of the domestic market in 2010, the probability of a passenger being on an American Airlines flight (p) is 0.155.
μ = n * p = 125 * 0.155 = 19.375
σ = sqrt(n * p * (1 - p)) = sqrt(125 * 0.155 * (1 - 0.155)) = 4.303
Next, we need to calculate the z-score for the desired event, which is more than 12 passengers on American Airlines flights. The z-score formula is z = (x - μ) / σ, where x is the number of passengers and μ is the mean.
For more than 12 passengers, we need to calculate the z-score for x = 12.5 (since we can't have half a passenger).
z = (12.5 - 19.375) / 4.303 = -1.602
Now, we can use a standard normal distribution table or a calculator to find the probability associated with this z-score. The probability of more than 12 passengers on American Airlines flights is equal to 1 minus the cumulative probability up to the z-score.
P(x > 12) = 1 - P(z < -1.602)
Using a standard normal distribution table or a calculator, we find that P(z < -1.602) is approximately 0.0548.
Therefore, the probability that more than 12 passengers from this sample were on American Airlines flights is approximately 1 - 0.0548 = 0.9452, or 94.52%.