Question

The mean cost of domestic airfares in the United States rose to an all-time high of $395 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. Use Table 1 in Appendix B.

Answer 1

Mean, U = $395

Std deviation, s = $110

a) Z-score = (X - U) / s , here X = 560

Z-score = (165 /110) = 1.5

Probability for Z-score 1,5 = 0.93319 i.e. probability of airfare within $560. (Refer standard normal distribution table for Z-score and corresponding p-values)

Hence , the Probability of airfare more than 560 = 1 - 0.93319 = 0.0668

b) Z-score = (265 - 395) / 110 = -1.18

Probability of Z-score , -1.18 = 0.1190 i.e. probability of fare till $265.

c) Here, X = 300 and X1 = 470

Z - score = (300 - 395) / 110 = -0.86, Corresponding p-value = 0.19489

Z-score X1 =  (470 - 395) / 110 = 0.68, Corresponding p-value = 0.75175

Probability of airfare between 300 and 470 = 0.75175 - 0.19489 = 0.5568

d) Cost for 3% highest domestic air fare i.e. p-value > 0.97

For p-value = 0.97, Z value = 1.88

Cost of highest 3% domestic air fare = U + Z*Std deviation = 395 + 1.88 * 110 = 602 dollars or more