Answer:
0.0668
Step-by-step explanation:
Mean cost per ticket = u = $ 385
Standard Deviation = [tex]\sigma[/tex] = $ 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:
[tex]z=\frac{x-u}{\sigma}[/tex]
Using the values in this formula, we get:
[tex]z=\frac{550-385}{110}=1.5[/tex]
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