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 $37 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.

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

solution:-

a) the population mean μ = 84 + 37 = 121

b) here given that sample mean within 10

=> P(111 < X < 131) = P((111-121)/(39/sqrt(70)) < Z < (131-121)/(39/sqrt(70)))

= P(-2.15 < Z < 2.15)

= P(Z < 2.15) - P(Z < -2.15)

= 0.9842 - 0.0158

= 0.9684

c) here given that sample mean within 5

P(116 < X < 126) = P((116-121)/(39/sqrt(70)) < Z < (126-121)/(39/sqrt(70)))

= P(-1.07 < Z < 1.07)

= P(Z < 1.07) - P(Z < -1.07)

= 0.8577 - 0.1423

= 0.7154