Question

Allegiant Airlines charges a mean base fare of $85. In addition, the airline charges for making a reservation on its website, checking bags, and inflight beverages. These additional charges average $34 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.

b. What is the probability the sample mean will be within $10 of the population mean cost per flight (to 4 decimals)? 0.9641

c. What is the probability the sample mean will be within $5 of the population mean cost per flight (to 4 decimals)?

Answer 1

b)

Here, μ = 119, σ = 4.6614, x1 = 109 and x2 = 129. We need to compute P(109<= X <= 129). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (109 - 119)/4.6614 = -2.15
z2 = (129 - 119)/4.6614 = 2.15

Therefore, we get
P(109 <= X <= 129) = P((129 - 119)/4.6614) <= z <= (129 - 119)/4.6614)
= P(-2.15 <= z <= 2.15) = P(z <= 2.15) - P(z <= -2.15)
= 0.9842 - 0.0158
= 0.9684

c)

Here, μ = 119, σ = 4.6614, x1 = 114 and x2 = 124. We need to compute P(114<= X <= 124). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (114 - 119)/4.6614 = -1.07
z2 = (124 - 119)/4.6614 = 1.07

Therefore, we get
P(114 <= X <= 124) = P((124 - 119)/4.6614) <= z <= (124 - 119)/4.6614)
= P(-1.07 <= z <= 1.07) = P(z <= 1.07) - P(z <= -1.07)
= 0.8577 - 0.1423
= 0.7154