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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 $250 or less (to 4 decimals)?

Respuesta :

Answer:

P [ Z ≤ 250 ] = 0.1099

Step-by-step explanation:

We have:

Normal distribution

μ = 385 (mean of population

σ = 110 (standard deviation of population)

Z = 250  The our critical value

Therefore:

× [ Z ≤ 250 ] =  (Z - μ ) ÷ σ         ⇒  × [ Z ≤ 250 ] =  ( 250 - 385 )÷110

× [ Z ≤ 250 ] = - 1.2272

As the critical point has 4 decimal and Z table only give three we need interpolate hence from points 1.22 and 1.23

                      × (values)               Probabilty (from z table)

                         -1.22                           0,1112

                    × = - 1.227                           Uknown

                         - 1.23                          0.1093

Diferences :

1.22 - 1.23 = 0.01                    0.1112 - 0.1093  = 0.0019

So  using rule of three:

                                                                     0.01       ⇒   0.0019

Diference between (1.22-1.227) =            0.007    ⇒       ? (α)

α = 0.00133

This value must be subtracted from the probability associated to the point 1.22 which is 0.1112

0.1112 - 0.00133 = 0.10987

P [ Z ≤ 250 ] = 0.10987        ⇒    P [ Z ≤ 250 ] = 0.1099

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