Allegiant Airlines charges a mean base fare of $89. In addition, the airline charges for making a reservation on its website, checking bags, and inflight beverages. These additional charges average $35 per passenger. Suppose a random sample of 50 passengers is taken to determine the total cost of their flight on Allegiant Airlines. The population standard deviation of total flight cost is known to be $40.
b. What is the probability the sample mean will be within $10 of the population mean cost per flight (to 4 decimals)?
c. What is the probability the sample mean will be within $5 of the population mean cost per flight (to 4 decimals)?
To answer these questions, we will use the Central Limit Theorem and the standard deviation of the sample mean formula. The Central Limit Theorem states that if we have a sufficiently large sample size, the distribution of the sample mean will be approximately normal, regardless of the shape of the population distribution.
b. To calculate the probability that the sample mean is within $10 of the population mean cost per flight, we need to find the z-scores for the lower and upper bounds.
First, we need to calculate the standard error of the mean (SE):
SE = population standard deviation / √(sample size)
SE = $40 / √(50)
Next, we find the z-score for the lower bound:
Lower bound z-score = (sample mean - population mean) / SE
Lower bound z-score = (89 - 89) / ($40 / √(50))
Since the sample mean and population mean are the same, the z-score for the lower bound will be 0.
Next, we find the z-score for the upper bound:
Upper bound z-score = (sample mean + $10 - population mean) / SE
Upper bound z-score = (89 + $10 - 89) / ($40 / √(50))
Now, we can use a Z-table or a statistical calculator to find the area between the z-scores of 0 and the upper bound. Subtract this area from 0.5 to find the probability.
c. Similarly, to calculate the probability that the sample mean is within $5 of the population mean cost per flight, we need to find the z-scores for the lower and upper bounds.
Lower bound z-score = (89 - 89) / ($40 / √(50))
Again, the z-score for the lower bound will be 0.
Upper bound z-score = (89 + $5 - 89) / ($40 / √(50))
Using a Z-table or a statistical calculator, find the area between the z-scores of 0 and the upper bound. Subtract this area from 0.5 to find the probability.