Question

Allegiant Airlines charges a mean base fare of $85. In addition, the airline charges for making a reservation on its website, checking bags, and inflight beverages. These additional charges average $34 per passenger. Suppose a random sample of 80 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 $36. 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

solution:-

a. the population mean cost per flight

=> 85 + 34 = 119

b. the probability the sample mean will be within $10 of the population mean

=> P((119-10) < x < (119+10))

=> P(109 < X < 129)

=> P((109-119)/(36/sqrt(80)) < z < (129-119)/(36/sqrt(80)))

=> P(Z < 2.48 < Z < -2.48)

=> P(z < 2.48) - P(z < -2.48)

=> P(z < 2.48) - (1-P(z < 2.48))

=> 0.9934 - (1-0.9934)

=> 0.9868

c. the probability the sample mean will be within $5 of the population mean

=> P((119-5) < x < (119+5))

=> P(114 < X < 124)

=> P((114-119)/(36/sqrt(80)) < z < (124-119)/(36/sqrt(80)))

=> P(-1.24 < z < 1.24)

=> P(z < 1.24) - P(z < -1.24)

=> P(z < 1.24) - (1-P(z < 1.24))

=> 0.8925 - (1-0.8925)

=> 0.7850