Question

Allegiant Airlines charges a mean base fare of $89. In addition, the airline charges for making a reservation on its website, checking bags, and inflight beverages. These additional charges average $35 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 $40  Use z-table.

1.) What is the population mean cost per flight?

2.) What is the probability the sample mean will be within $10 of the population mean cost per flight (to 4 decimals)?

3.) 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 = = 89 + 35 = 124

standard deviation = = 40

n = 80

=   = 128

= / n = 40/ 80 = 4.47

a) P(118 < < 138)  

= P[(118 - 128) / 4.47 < ( - ) / < (138 - 128) / 4.47 )]

= P(-2.24 < Z < 2.24)

= P(Z < 2.24) - P(Z < -2.24)

Using z table,  

= 0.9875 - 0.0125

= 0.9750

b) P(123 < < 133)  

= P[(123 - 128) / 4.47 < ( - ) / < (133 - 128) / 4.47 )]

= P(-1.12 < Z < 1.12)

= P(Z < 1.12) - P(Z < -1.12)

Using z table,  

= 0.8686 - 0.1314

= 0.7372