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 60 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 $37. Use z-table.

Question

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 60 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 $37. Use z-table.

a. What is the population mean cost per flight?
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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)?

Answer 1

To find the answers to these questions, we need to use the concept of the sampling distribution of the sample mean and the standard normal distribution (z-distribution). Here's how we can solve each part step by step:

a. To find the population mean cost per flight, we simply use the given mean base fare, which is $89.

b. To find the probability that the sample mean will be within $10 of the population mean cost per flight, we need to find the area under the standard normal distribution curve within the interval defined by $89 - $10 = $79 to $89 + $10 = $99. We can do this by converting these values into z-scores using the formula:

z = (x - μ) / (σ / sqrt(n))

where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.

In this case, x = $89, μ = $89, σ = $37, and n = 60. Plugging these values into the formula, we get:

z1 = ($79 - $89) / ($37 / sqrt(60))
z2 = ($99 - $89) / ($37 / sqrt(60))

Now, we can use the z-table to find the cumulative probability associated with each z-score. The probability we want is the difference between these two cumulative probabilities.

c. Similarly, to find the probability that the sample mean will be within $5 of the population mean cost per flight, we repeat the above process with new values of x:

x = $89 and the interval is $89 - $5 = $84 to $89 + $5 = $94.

Calculate new z-scores using the formula:

z1 = ($84 - $89) / ($37 / sqrt(60))
z2 = ($94 - $89) / ($37 / sqrt(60))

Then, find the difference between the cumulative probabilities associated with these z-scores using the z-table.

Using the z-table method, we can find the probabilities for parts b and c by calculating the differences between the cumulative probabilities associated with the respective z-scores.