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)?

0.0668

Step-by-step explanation:

Mean cost per ticket = u = $ 385

Standard Deviation = = $ 110

Its given that the distribution of domestic airfares is normally distributed. We need to find the probability that a domestic airfare is $550 or more.

i.e we have to find: Probability (Airfares ≥ 550). In symbolic form this can be written as: P(X ≥ 550)

Since, the distribution is normal, we can use the concept of z scores to answer this problem. First we have to convert x = 550 to an equivalent z score.

The formula for the z score is:

Using the values in this formula, we get:

So, P(X ≥ 550) is equivalent to P(z ≥ 1.5)

Using the z table we can find the probability of z score being equal to or greater than 1.5, which comes out to be: 0.0668

So,

P(z ≥ 1.5) = 0.0668

Since,

P(X ≥ 550) = P(z ≥ 1.5), we can conclude:

The probability that a domestic airfare is $550 or more is 0.0668

the answer is 489

step-by-step explanation:

-2x + 7 = 7

subtract 7 from both sides.

-2x = 0

divide both sides by -2.

x = 0