Question

"Ryanair charges a mean base fare of $59. In addition, the airline charges for checking in at the airport, carry-on baggage, checked baggage, and inflight food/beverage. These additional charges average $39 per passenger. Suppose a random sample of 60 passengers is taken to determine the total cost of their flight on Ryanair. The population standard deviation of total flight cost is known to be $25."

a. What is the probability the sample mean will be within $10 of the population mean cost per flight?
b. What is the probability the sample mean will be within $5 of the population mean cost per flight?
c. What is the probability the sample mean will exceed $100?

Answer 1

a) P=0.998

b) P=0.879

c) P=0.268

Explanation:

The mean charges per passenger are:

`\mu=59+39=98`

The standard deviation is known to be σ=$25.

If the sample size is 60, we have to calculate:

a) the probability the sample mean will be within $10 of the population mean cost per flight.

This can be calculated from the z-value for a margin or error of 10 from the mean:

`z=(X-\mu)/(\sigma/√(n))=(10)/(25/√(60))=(10)/(3.2275)=3.1`

`P(|X-\mu|<10)=P(|z|<3.1)=0.998`

b) the probability the sample mean will be within $5 of the population mean cost per flight.

This can be calculated from the z-value for a margin or error of 5 from the mean:

`z=(X-\mu)/(\sigma/√(n))=(5)/(25/√(60))=(5)/(3.2275)=1.55`

`P(|X-\mu|<5)=P(|z|<1.55)=0.879`

c) the probability the sample mean will exceed $100

Again we will calculate the z-score for X=100.

`z=(X-\mu)/(\sigma/√(n))=(100-98)/(25/√(60))=(2)/(3.2275)=0.62`

`P(X>100)=P(z>0.62)=0.268`