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. What is the probability that a domestic airfare is $550 or more (to 4 decimals)? What is the probability than a domestic airfare is $250 or less (to 4 decimals)? What if the probability that a domestic airfare is between $300 and $500 (to 4 decimals)? What is the cost for the 3% highest domestic airfares?

Explanation / Answer

Given,

Mean = 385

Standard deviation = 110

( a.)

P ( x 550)

z = ( x – Mean ) / Standard deviation

= (550 – 385)/110

= 1.5

P (z 1.5) = 1 – P (z<1.5)

By using Normal Distribution Table we get,

P (z<1.5 ) = 0.9332

P (z 1.5) = 1 – 0.9332 = 0.0668

Answer: 0.0668

(b.)

P (x 250)

z = ( x – Mean ) / Standard deviation

= (250 – 385)/110

= -1.23

P (z -1.23)

By using Normal Distribution Table we get,

P (z -1.23) = 0.1093

Answer: 0.1093

(c.)

P (300 < x < 500)

z1 = ( x – Mean ) / Standard deviation

= (300 – 385)/110

= -0.77

z2 = ( x – Mean ) / Standard deviation

= (500 – 385)/110

= 1.05

P (-0.77 < z < 1.05)

= P ( z < 1.05 ) – P ( z < -0.77)

By using Normal Distribution Table we get,

P (z<1.05 ) = 0.8531

P (z<-0.77) = 0.2206

P (-0.77 < z < 1.05)

= 0.8531 – 0.2206

= 0.6325

Answer: 0.6325

(d.)

Here first we have to find the z-score,

P (Z>z) = 0.03

By using Normal Table we get,

P (Z>1.88) = 0.03

X = Mean + (z*standard deviation)

   = 385 + (1.88*110)

= 591.8

Answer: 591.8

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