Question

Allegiant Airlines charges a mean base fare of $88. 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 70 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 $39. 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

a) Population mean cost = 88 + 34 = 122

b) P(112 < < 132)

= P((112 - )/() < ( - )/() < (132 - )/() )

= P((112 - 122)/(39/) < Z < (132 -  122)/(39/))

= P(-2.15 < Z < 2.15)

= P(Z < 2.15) - P(Z < -2.15)

= 0.9842 - 0.0158

= 0.9684

c) P(117 < < 127)

= P((117 - )/() < ( - )/() < (127 - )/() )

= P((117 - 122)/(39/) < Z < (127 -  122)/(39/))

= P(-1.07 < Z < 1.07)

= P(Z < 1.07) - P(Z < -1.07)

= 0.8577 - 0.1423

= 0.7154