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

Answer:

a) P=0.998

b) P=0.879

c) P=0.268

Step-by-step 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:

$z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{10}{25/\sqrt{60}}=\dfrac{10}{3.2275}=3.1$

$P(|X-\mu|$

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:

$z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{5}{25/\sqrt{60}}=\dfrac{5}{3.2275}=1.55$

$P(|X-\mu|$

c) the probability the sample mean will exceed $100

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

$z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{100-98}{25/\sqrt{60}}=\dfrac{2}{3.2275}=0.62$

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