Question

Allegiant Airlines charges a mean base fare of $89. In addition, the airline charges for making a reservation on its website, checking bags, and inflight beverages. These additional charges average $39 per passenger (Bloomberg Businessweek, October 8 � 14, 2012). Suppose a random sample of 60 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 $40.

a. Allegiant Airline reports a population mean base fare of $89 and a population mean additional fare of $39 per person. What is the population mean cost per flight?
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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

Answer:

a) (μ) = $128

b) 0.9472

c)  0.6671

Step-by-step explanation:

Given that:

Allegiant Airlines charges a mean base fare of $89.

this implies that: mean base fare = $89.

The question proceeds by stating the additional charge on  its website, checking bags, and inflight beverages.

so , additional charges turns out to be = $39 per passenger

Now, Suppose a random sample of 60 passengers is taken

random sample (n) = 60

The population standard deviation of total flight cost is known to be $40

standard deviation (σ) = 40

Question (a) says; we should find the population mean cost per flight

To determine that; we have to consider the total sum(μ) of the mean base fare with the mean additional charges.

Population mean cost per flight (μ)  =  mean base fare + mean additional charges

(μ) = $89 + $39

(μ) = $128

b)

What is the probability the sample mean will be within $10 of the population mean cost per flight (to 4 decimals)?

To determine that; we have:

P(128 - 10 ≤ Х ≤ 128 +10)

= P(118 ≤ Х ≤  138)

= $P[\frac{118-128}{40/\sqrt{60} }\leq \frac{X- \delta }{\alpha /\sqrt{n} }\leq \frac{138-128}{40\sqrt{60} }]$               (where $\delta$  =  μ  and ∝ = σ )

= $P[-1.9365\leq z +1.9365]$

= $P[z\leq 1.9365]-P[z\leq -1.9365]$

Using Excel Command to approach this process, we have;

= 0.9736 - 0.0264

= 0.9472     (to four decimal places)

∴ the probability that the sample mean will be within $10 of the population mean cost per flight = 0.9472

c)

What is the probability the sample mean will be within $5 of the population mean cost per flight (to 4 decimals)?

We wll have to go through the process like the one attempted above in question (b);

So;

P(128 - 5 ≤ Х ≤ 128 + 5)

= P(123 ≤ Х ≤  133)

=  $P[\frac{123-128}{40/\sqrt{60} }\leq \frac{X- \delta }{\alpha /\sqrt{n} }\leq \frac{133-128}{40\sqrt{60} }]$                         (where $\delta$  =  μ  and ∝ = σ )

= $P[-0.9682\leq z +0.9682]$

= $P[z\leq 0.9682]-P[z\leq-0.9682]$

Computing these data in Excel; we have

= 0.8335 -0.1665

= 0.6671         (to 4 decimal places)

∴  the probability that the sample mean will be within $5 of the population mean cost per flight.