Question

Check My Work 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 $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 $37. Use z-table. a. What is the population mean cost per flight? $ 120 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)? Hide Feedback Partially Correct

Answer 1

Solution :

Given that,

mean = $\mu$ = 120

standard deviation = $\sigma$ = 37

n = 70

$\mu$_{$\bar x$} = $\mu$ = 120

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 37/ $\sqrt$ 70 = 4.42

b) P(110 < $\bar x$ < 130)  

= P[(110 - 120) / 4.42 < ($\bar x$ - $\mu$ _{$\bar x$}) / $\sigma$ _{$\bar x$} < (130 - 120) / 4.42)]

= P(-2.26 < Z < 2.26)

= P(Z < 2.26) - P(Z < -2.26)

Using z table,  

= 0.9881 - 0.0119  

= 0.9762

c) P(115 < $\bar x$ < 125)  

= P[(115 - 120) / 4.42 < ($\bar x$ - $\mu$ _{$\bar x$}) / $\sigma$ _{$\bar x$} < (125 - 120) / 4.42)]

= P(-1.13 < Z < 1.13)

= P(Z < 1.13) - P(Z < -1.13)

Using z table,  

= 0.8708 - 0.1292

= 0.7416