Question

Allegiant Airlines charges a mean base fare of $86. In addition, the airline charges for making a reservation on its website, checking bags, and inflight beverages. These additional charges average $36 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 $37. 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 = = 86 + 36 = 122

standard deviation = = 37

n = 70

=   = 122

= / n = 37 / 70 = 4.42

b) P(112 < < 132)  

= P[(112 - 122) / 4.42 < ( - ) / < (132 - 122) /4.42 )]

= P(-2.26 < Z < 2.26)

= P(Z < 2.26) - P(Z < -2.26)

Using z table,  

= 0.9881 - 0.0119

= 0.9762

c) P(117 < < 127)  

= P[(117 - 122) / 4.42 < ( - ) / < (127 - 122) /4.42 )]

= P(-1.13 < Z < 1.13)

= P(Z < 1.13) - P(Z < -1.13)

Using z table,  

= 0.8708 - 0.1292

= 0.7416