Question

Allegiant Airlines charges a mean base fare of $84. 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 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

Answer:

Given,

n = 50

standard deviation = 40

= /sqrt(n)

substitute values

= 40/sqrt(50)

= 5.66

a)

To determine population mean cost per flight

Mean = 84 + 34

= 118

Mean = 118

b)

To determine the probability the sample mean will be within $10 of the population mean cost per flight

Here x = 108,

z = (108 - 118)/40/sqrt(50)

= -10 / 5.66

= - 1.77

x = 128 ,

z = (128 - 118) / 40/sqrt(50)

= 10 / 5.66

= 1.77

P(- 1.77 < z < 1.77) = P(z < 1.77) - P(z < -1.77)

= 0.9616 - 0.0384

= 0.9232

Probability = 0.9232

c)

To determine the probability the sample mean will be within $5 of the population mean cost per flight

Here x = 113

z = (113 - 118)/40/sqrt(50)

= - 5 / 5.66

= - 0.88

x = 123

z = (123 - 118)/40/sqrt(50)

= 5/5.66

= 0.88

P(-0.88 < z < 0.88) = P(z < 0.88) - P(z < -0.88)

= 0.8106 - 0.1894

= 0.6212

Probability = 0.6212