Question

Allegiant Airlines charges a mean base fare of $89. In addition, the airline charges for making a reservation on its website, checking bags, and inflight beverages. These additional charges average $36 per passenger. Suppose a random sample of 70 passengers is taken to determine the total cost of their flight on Allegiant Airlines. The population standard deviation of total flight cost is known to be $39. Use z-table. a. What is the population mean cost per flight? $ b. What is the probability the sample mean will be within $10 of the population mean cost per flight (to 4 decimals)? c. What is the probability the sample mean will be within $5 of the population mean cost per flight (to 4 decimals)?

Answer 1

a) The mean of base fare is $89 and the mean of additional fare is $36

The population mean of cost in each flight is calculated as :

Mean cost per flight = mean base fair + mean additional fair = 89 + 36 = $125

b) The population mean is $125 and the sample size is 70. That is, μ = 125 and n = 70. The population standard deviation is provided as $39

The distribution of the mean of the fare of the flight will be approximately normally distributed with the mean, μ = 125 and standard deviation, σ = 39.

Therefore, the probability that the mean fare of the flight is within $10 of population mean cost per flight is computed as:

= P(125 - 10 < X̅ < 125+10)

= P(115 < X̅ < 135)

= P(115-125/39/√70 < z < 135-125/39/√70)

= P( -2.15 < z < 2.15)

= P( z < 2.15) - P( z < -2.15) using z table

= 0.9842 - 0.0158

= 0.9684

c) The probability that the mean fare of the flight is within $5 of population mean cost per flight is computed as :

= P(125 - 5 < X̅ < 125 + 5)

= P(120 < X̅ < 130)

= P(120 - 125/39/√70 < z < 130 - 125/39/√70)

= P( -1.07 < z < 1.07 )

= P( z < 1.07 ) - P ( z < -1.07 ) using z table

= 0.8577 - 0.1423

= 0.7154