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 $38 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 $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

Solution :

Given that ,

a) mean = = 88 + 38 = 126

standard deviation = = 39

n = 80

=   = 126

= / n = 39/ 80 = 4.36

b) P(116 < < 136)  

= P[(116 - 126) / 4.36 < ( - ) / < (136 - 126) /4.36 )]

= P(- 2.29 < Z < 2.29)

= P(Z < 2.29) - P(Z < - 2.29)

Using z table,  

= 0.9890 - 0.0110

= 0.9780

c) P(121 < < 131)  

= P[(121 - 126) / 4.36 < ( - ) / < (131 - 126) /4.36 )]

= P(- 1.15 < Z < 1.15)

= P(Z < 1.15) - P(Z < - 1.15)

Using z table,  

= 0.8749 - 0.1251

= 0.7498