Question

Allegiant Airlines charges a mean base fare of $70. In addition, the airline charges for making a reservation on its website, checking bags, and inflight beverages. These additional charges average $34 per passenger (Bloomberg Businessweek, October 8-14 , 2012). Suppose a random sample of 100 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. What is the probability that the sample mean will be within $5 of the population mean cost per flight? Express your answer as a number with 4 decimal places.

Answer 1

1)

expected mean cost =70+34=104

for normal distribution z score =(X-μ)/σx
here mean=       μ=	104
std deviation   =σ=	40.000
sample size       =n=	100
std error=σx̅=σ/√n=	4.000

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

probability =P(99<X<109)=P((99-104)/4)<Z<(109-104)/4)=P(-1.25<Z<1.25)=0.8944-0.1056=0.7888