Question

You may need to use the appropriate appendix table to answer this question.

The mean cost of domestic airfares in the United States rose to an all-time high of $385 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 $110.

(a)

What is the probability that a domestic airfare is $561 or more? (Round your answer to four decimal places.)

(b)

What is the probability that a domestic airfare is $230 or less? (Round your answer to four decimal places.)

(c)

What is the probability that a domestic airfare is between $300 and $500? (Round your answer to four decimal places.)

(d)

What is the minimum cost in dollars for a fair to be included in the highest 4% of domestic airfares? (Round your answer to the nearest integer.)

Answer 1

Solution :

Given that,

mean = $\mu$ = 385

standard deviation = $\sigma$ = 110

a ) P (x $\geq$   561)

= 1 - P (x $\leq$ 561 )

= 1 - P ( x -  $\mu$/ $\sigma$ ) $\leq$ ( 561 - 385 / 110)

= 1 - P ( z $\leq$ 176 / 110 )

= 1 - P ( z $\leq$ 1.6 )

Using z table

= 1 - 0.9452

= 0.0548

Probability = 0.0548

b ) P( x $\leq$ 230 )

P ( x - $\mu$ / $\sigma$ ) $\leq$ (230- 385 / 110)

P ( z $\leq$ - 155 / 110 )

P ( z $\leq$ - 1.41)

= 0.0793

Probability =0.0793

c ) P (300 < x < 600 )

P ( 300 -385 / 110) < ( x -  $\mu$/ $\sigma$ ) < ( 331 - 385 / 110)

P ( - 85 / 110 < z <  215 / 110 )

P (-0.77 < z < 1.95 )

P ( z < 1.95 ) - P ( z < -0.77)

Using z table

= 0.9744 - 0.2206

= 0.7538

Probability = 0.7538

P( Z > z) = 4%

P(Z > z) = 0.04

1 - P( Z < z) = 0.04

P(Z < z) = 1 - 0.04

P(Z < z) = 0.96

z = 1.75

Using z-score formula,

x = z * $\sigma$ +$\mu$

x = 1.75 * 110  + 385

= 577.5

The minimum cost = 577