Answer:
a)0.067
b)0.111
c)0.612
d)$687.28
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = $385
Standard Deviation, σ = $110
We are given that the distribution of domestic airfares is a bell shaped distribution that is a normal distribution.
Formula:
a) P(domestic airfare is $550 or more)
P(x > 550)
Calculation the value from standard normal z table, we have,
b) P(domestic airfare is $250 or less)
Calculating the value from the standard normal table we have,
![P( x \leq 250) = 0.111 = 11.1\%](https://tex.z-dn.net/?f=P%28%20x%20%5Cleq%20250%29%20%3D%200.111%20%3D%2011.1%5C%25)
c))P(domestic airfare is between $300 and $500)
![P(300 \leq x \leq 500) = 61.2\%](https://tex.z-dn.net/?f=P%28300%20%5Cleq%20x%20%5Cleq%20500%29%20%3D%2061.2%5C%25)
d) P(X=x) = 0.03
We have to find the value of x such that the probability is 0.03.
P(X > x)
Calculation the value from standard normal z table, we have,
Hence, the domestic fares must be $687.28 or greater for them to lie in the highest 3%.