Question

The mean cost of domestic airfares in the United States rose to an all-time high of $400 per ticket. Airfares were based on the total ticket value, which consisted of the price charges by the airlines plus any additional taxes and fees. Assume domestic airfares are normally distributed with a standard deviation of $120. Use Table 1 in Appendix B.

a. What is the probability that a domestic airfare is $555 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 $300 and $470 (to 4 decimals)?

d. What is the cost for the 4% highest domestic airfares? [(round to nearest dollar) (more or less)]

Answer 1

the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 400$
standard Deviation ( sd )= 120$

a.
the probability that a domestic airfare is $555 or more
P(X > 555) = (555-400)/120
= 155/120 = 1.2917
= P ( Z >1.2917) From Standard Normal Table
= 0.0982

b.
the probability that a domestic airfare is $250 or less
P(X < 250) = (250-400)/120
= -150/120= -1.25
= P ( Z <-1.25) From Standard Normal Table
= 0.1056

c.
the probability that a domestic airfare is between $300 and $470
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 300) = (300-400)/120
= -100/120 = -0.8333
= P ( Z <-0.8333) From Standard Normal Table
= 0.2023
P(X < 470) = (470-400)/120
= 70/120 = 0.5833
= P ( Z <0.5833) From Standard Normal Table
= 0.7202
P(300 < X < 470) = 0.7202-0.2023 = 0.5178

d.
the cost for the 4% highest domestic airfares
P ( Z > x ) = 0.04
Value of z to the cumulative probability of 0.04 from normal table is 1.8
P( x-u / (s.d) > x - 400/120) = 0.04
That is, ( x - 400/120) = 1.8
--> x = 1.8 * 120+400 = 610.1
the cost is 610$