Question

Allegiant Airlines charges a mean base fare of $87. In addition, the airline charges for making a reservation on its website, checking bags, and inflight beverages. These additional charges average $37 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 $38. Use z-table.

a. What is the population mean cost per flight?

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

• Here let X be the random variable denoting the cost per flight
• Airlines mean base fare is $87 and the additional charges are $37 per passenger.
• Hence the total mean cost per flight is $87+$37=$124
• So Here X~Normal(124,38²/50)
• The probability that the sample mean will within $10 of the population mean cost per flight
• P(114<Xbar<134)
• =P(<Z<)
• =P(<Z<)
• =P(-1.86<Z<1.86)
• =P(Z<1.86)-[1-P(Z<1.86)]
• =0.96856-[1-0.96856](i.e the p value associated with the z value in the normal distribution table)
• =0.93712
• c)
• To find  probability that the sample mean will within $5 of the population mean cost per flight.
• P(119<Xbar<129)
• =P(<Z<)
• =P(<Z<)
• P(-0.930<Z<0.930)
• P(Z<0.930)-[1-P(Z<0.930)]
• 0.82381-[1-0.82381](i.e the p value associated with the z value in the normal distribution table)
• 0.64762