Question

The mean cost of domestic airfares in the United States rose to an all-time high of $380 per ticket. 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 $120.

a. What is the probability that a domestic airfare is $530 or more (to 4 decimals)?

b. What is the probability that a domestic airfare is $260 or less (to 4 decimals)?

c. What if the probability that a domestic airfare is between $310 and $470 (to 4 decimals)?

d. What is the cost for the 3% highest domestic airfares? (rounded to nearest dollar)

Answer 1

Here, we will use the z statistics to find out the values using mean () = $380 and Standard deviation () = $120 and

z= (x-)/

a) For airfare x>$530 the z value will be = 530 - 380/120 = 1.25

so the respective probability of z value 1.25 will be 0.3944(from z table)

but this is the probability of fare between 380 to 530. So to find out the fare more than 530 we have to subtract it from 0.5 and it will give us 0.1056(0.5-0.3944).

so the required probability will be 0.1056

b) For airfare x<$260 the z value = 260-380/120 = -1 or 1(ignoring sign)

so the respective probability for z = 1 will be 0.3413 (from z table) but this is the probability for fare between 260 to 380

so for actual less than 260 we have to subtract it from 0.5 that will be 0.1587(0.5-0.3413).

so the required probability will be 0.1587

c) For airfare between $310<x> $470 we have to find out two z value and its respective probability and then we have to add them to find out total

so for x>310 z = 310-380/120 = 0.583

and for x<470 z= 470-380/120 = 0.75

now respective probabilities will be for z= 0.583 is 0.2190 and for 0.75 is 0.2734 (from z table)

so the total will be 0.2190+0.2734 = 0.4924

d) Here 3% highest fare means we have to find out that fare whose chance is 0.03 in normal distribution or you can say that whose area lies in the extreme right is 0.03.

so actually we have to find out x, but here we will take the probability by subtracting it from 0.5 because we have to find out that how far the point is from the mean in the distribution and that will be the answer.

here 0.5-0.03 will give 0.47, when you will search in the z table you will find that probability on the z value of 1.88 that is 0.4699 close to 0.47

taking z = 1.88 and mean as 380 and standard deviation as 120 we can find the x

so, 1.88 = x-380/120

x = (1.88 x 120) + 380 = $605.6 approx

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