The mean cost of domestic airfares in the United States rose to an all-time high

Question

The mean cost of domestic airfares in the United States rose to an all-time high of $385 per ticket (Bureau of Transportation Statistics website, November 2, 2012). Airfares were based on the total ticket value, which consisted of the price charged by the airlines plus any additional taxes and fees. Assume domestic airfares are normally distributed with a standard deviation of $110.

a. What is the probability that a domestic airfare is $550 or more (to 4 decimals)?

b. What is the probability than a domestic airfare is $250 or less (to 4 decimals)?

c. What if the probability that a domestic airfare is between $300 and $500 (to 4 decimals)?

d. What is the cost for the 3% highest domestic airfares? $

Explanation / Answer

Normal Distribution
Mean ( u ) =385
Standard Deviation ( sd )=110
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X => 550) = (550-385)/110
= 165/110 = 1.5
= P ( Z >1.5) From Standard Normal Table
= 0.0668
b)
P(X <= 250) = (250-385)/110
= -135/110= -1.2273
= P ( Z <-1.2273) From Standard Normal Table
= 0.1099                  
c)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 300) = (300-385)/110
= -85/110 = -0.7727
= P ( Z <-0.7727) From Standard Normal Table
= 0.21984
P(X < 500) = (500-385)/110
= 115/110 = 1.0455
= P ( Z <1.0455) From Standard Normal Table
= 0.85209
P(300 < X < 500) = 0.85209-0.21984 = 0.6323                  
d)
P ( Z > x ) = 0.03
Value of z to the cumulative probability of 0.03 from normal table is 1.88
P( x-u/ (s.d) > x - 385/110) = 0.03
That is, ( x - 385/110) = 1.88
--> x = 1.88 * 110+385 = 591.91                  
                  

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