Question

The mean cost of domestic airfares in the United States rose to an all-time high of $385 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 $105. Use Table 1 in Appendix B. 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 $250 or less (to 4 decimals)? c. What if the probability that a domestic airfare is between $310 and $500 (to 4 decimals)? d. What is the cost for the 2% highest domestic airfares? (rounded to nearest dollar) $ or - Select your answer -moreless

Answer 1

Solution:

Given in the question
Mean cost of airfares in the US = $385 per ticket
Standard deviation = $ 105
Assume that domestic airfares are nomally distributed
Solution(a)
We need to calculate that a domestic airfare is $530 or more P(X>=$530),
Here we will use normal distribution table to find probability
We will calculate Z-score, which can be calculated as
Z-score = (X-)/ = (530-385)/105 = 1.38
From Z table we found p-value
P(X>=$530) = 1 - P(X<1.38) = 1- 0.9162 = 0.0838
So there is 8.38% probability that a domestic airfare is $530 or more.
Solution(b)
We need to calculate that a domestic airfare is $250 or less P(X<=$250),
Here we will use normal distribution table to find probability
We will calculate Z-score, which can be calculated as
Z-score = (X-)/ = (250-385)/105 = -1.29
From Z table we found p-value
P(X<=$530) = 0.0985
So there is 9.85% probability that a domestic airfare is $250 or less.
Solution(c)
We need to calculate that a domestic airfare is between $310 and $500 P($310<X<$500),
Here we will use normal distribution table to find probability
We will calculate Z-score, which can be calculated as
Z-score = (X-)/ = (310-385)/105 = -0.71
Z-score = (X-)/ = (500-385)/105 = 1.1
From Z table we found p-value
P($310<X<$500) = P(X<$500) - P(X<$310) = 0.8643 - 0.2389 = 0.6254
So there is 62.54% probability that a domestic airfare is b/w $310 and $500.
Solution(d)
Here given in the question
P-value = 0.98
So Z-score can be found from Z table So Z-score = 2.05375
So Domestic airfare can be found as
X = + Z-score* = 385 + 2.05375*105 = $600.64 or $601