For the major airlines, 21% of all flights depart at least 15 minutes late with a standard deviation of 5%. Currently, 24% of all American Airlines flights depart at least 15 minutes late. If American wants to be in the top 10% of all major airlines with the FEWEST late departures, to what level must American cut its late departures? (please round your answer to 1 decimal place)
I need this answer.
To find the level that American Airlines must cut its late departures in order to be in the top 10% of all major airlines, we first need to calculate the z-score associated with the given percentage.
A z-score represents how many standard deviations a particular value is from the mean. It can be calculated using the formula:
z = (x - μ) / σ
Where:
- z is the z-score
- x is the value
- μ is the mean
- σ is the standard deviation
In this case, we are given that the mean (μ) of all major airlines' late departures is 21% and the standard deviation (σ) is 5%.
Now, we can plug in the values into the formula:
z = (x - 21%) / 5%
Rearranging the formula to solve for x:
x = (z * 5%) + 21%
To be in the top 10% of all major airlines with the fewest late departures, we need to find the z-score associated with the top 10%.
The z-score associated with the top 10% is found by subtracting 10% from 100% (total percentage), which gives us 90%.
Using a standard normal distribution table or a z-score calculator, we find that the z-score corresponding to 90% is approximately 1.28.
Plugging this value into the equation, we can solve for x:
x = (1.28 * 5%) + 21%
x = 6.4% + 21%
x = 27.4%
Therefore, American Airlines must cut its late departures to at least 27.4% in order to be in the top 10% of all major airlines with the fewest late departures.