Question

Allegiant Airlines charges a mean base fare of $86. In addition, the airline charges for making a reservation on its website, checking bags, and inflight beverages. These additional charges average $35 per passenger. Suppose a random sample of 50 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 $40. 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 population mean cost per flight = mean base fare + additional charges
= 86 + 35 = 121

b)

Here, μ = 121, σ = 5.6569, x1 = 111 and x2 = 131. We need to compute P(111<= X <= 131). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (111 - 121)/5.6569 = -1.77
z2 = (131 - 121)/5.6569 = 1.77

Therefore, we get
P(111 <= X <= 131) = P((131 - 121)/5.6569) <= z <= (131 - 121)/5.6569)
= P(-1.77 <= z <= 1.77) = P(z <= 1.77) - P(z <= -1.77)
= 0.9616 - 0.0384
= 0.9232

c)

Here, μ = 121, σ = 5.6569, x1 = 116 and x2 = 126. We need to compute P(116<= X <= 126). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (116 - 121)/5.6569 = -0.88
z2 = (126 - 121)/5.6569 = 0.88

Therefore, we get
P(116 <= X <= 126) = P((126 - 121)/5.6569) <= z <= (126 - 121)/5.6569)
= P(-0.88 <= z <= 0.88) = P(z <= 0.88) - P(z <= -0.88)
= 0.8106 - 0.1894
= 0.6212