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 $34 per passenger. Suppose a random sample of 60 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

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 = = 87 + 34 = 121

standard deviation = = 37

n = 60

=   = 121

= / n = 37/ 60 = 4.78

b) P(111 < < 131)  

= P[(111 - 121) / 4.78 < ( - ) / < (131 - 121) /4.78 )]

= P(-2.09 < Z < 2.09)

= P(Z < 2.09) - P(Z < -2.09)

Using z table,  

= 0.9817 - 0.0183

= 0.9634

c) P(116 < < 126)  

= P[(116 - 121) / 4.78 < ( - ) / < (126 - 121) / 4.78 )]

= P(-1.05 < Z < 1.05)

= P(Z < 1.05) - P(Z < -1.05)

Using z table,  

= 0.8531 - 0.1469

= 0.7062