Question

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Allegiant Airlines charges a mean base fare of $88. 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 80 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 $35. 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

Given :-

Sample size (n) = 80,

Population standard deviation (sigma) = 35

a)

Population mean(mu) cost per flight is = 88 + 36 = 124

b)

Therefore,

x1 = 124-10 = 114

x2 = 124+10 = 134

So, we have to find P (x1 < X < x2)

Let, z = (X-mu) / (sigma/√n)  

Therefore,

z1= ( 114-124)/(35/√80) = -2.56

z2= (134 -124) / (35/√80) = 2.56

Therefore,

P(114 < X < 134 ) = P( -2.56 < z < 2.56 )

= P(z< 2.56) - P(z < -2.56) --------- (use z-table)

= 0.9948 - 0.0052

P(114 < X < 134 ) = 0.9896

c)

x1= 124-5 = 119

x2 = 124+5 = 129

z1= (119-124) / (35/√80) = -1.28

z2= (129-124) / (35/√80) = 1.28

Therefore,

P(119 < X < 129) = P( -1.28 < z < 1.28)

= P( z <1.28) - P( z < -1.28)

= 0.8997 - 0.1003

P(119 < X < 129) = 0.7994