Question

The mean cost of domestic airfares in the United States rose to an all-time high of $390 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 $115. Use Table 1 in Appendix B. What is the probability that a domestic airfare is $560 or more (to 4 decimals)?

Answer 1

The probability that a domestic airfare is $560 or more is found by calculating the z-score of 1.4783 and then subtracting the table value from 1, resulting in a probability of 0.0694 to four decimal places.

Step-by-step explanation:

The question is asking about the probability that a domestic airfare is $560 or more given that the mean cost of domestic airfares in the United States is $390 and the standard deviation is $115. Since domestic airfares are normally distributed, we can use the z-score formula to find this probability.

The z-score is calculated using the formula z = (X - μ) / σ, where X is the value we are looking for ($560), μ is the mean ($390), and σ is the standard deviation ($115).

Calculating the z-score gives us:
z = ($560 - $390) / $115 ≈ 1.4783

We would then look up this z-score value in a standard normal distribution table to find the probability that a score is above this value. However, typically, tables provide the area under the curve to the left of the z-score, so we would subtract this value from 1 to find the probability that a score is above (to the right of) this z-score.

Assuming we find the corresponding value for z = 1.4783 in the table and it is 0.9306, the probability that an airfare costs $560 or more is:
1 - 0.9306 = 0.0694 (to 4 decimal places).