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 $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 $40. Use z-table.

a. What is the population mean cost per flight?
$ 120

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 per flight =86+34 =120

b)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	120
std deviation   =σ=	40.000
sample size       =n=	70
std error=σx̅=σ/√n=	4.7809

probability the sample mean will be within $10 of the population mean cost per flight:

probability =

P(110<X<130)

=

P(-2.09<Z<2.09)=

0.9817-0.0183=

0.9634

c)

probability the sample mean will be within $5 of the population mean cost per flight :

probability =

P(115<X<125)

=

P(-1.05<Z<1.05)=

0.8531-0.1469=

0.7062