Question

The mean cost of domestic airfares in the United States rose to an all-time high of $385 per ticket (Bureau of Transportation Statistics website, November 2, 2012). Airfares were based on the total ticket value, which consisted of the price charged by the airlines plus any additional taxes and fees. Assume domestic airfares are normally distributed with a standard deviation of $110.

a. What is the probability that a domestic airfare is $550 or more (to 4 decimals)?

b. What is the probability than a domestic airfare is $250 or less (to 4 decimals)?

c. What if the probability that a domestic airfare is between $300 and $500 (to 4 decimals)?

d. What is the cost for the 3% highest domestic airfares?

Answer 1

Concepts and reason

The concept of normal distribution is used to solve this problem.

The normal distribution function is used to find the probability of the continuous variable when the data is more or less symmetric. The probability of being less than or more than some value can be calculated by calculating the area of the curve to the left of that value.

Fundamentals

A continuous random variable is said to follow a normal distribution if the probability density function can be written in the form,

$f\left( x \right) = \left\{ \begin{array}{l}\\\frac{1}{{\sigma \sqrt {2\pi } }}{e^{ - {{\left( {\frac{{x - \mu }}{\sigma }} \right)}^2}}}{\rm{ }} - \infty \le x \le \infty \\\\0{\rm{ Otherwise}}\\\end{array} \right.$

The value of $\mu$ is the mean and $\sigma$ is the standard deviation of the distribution. The probability of the normal distribution can be defined as:

$P(x > \alpha ) = \int\limits_\alpha ^\infty {\frac{1}{{\sigma \sqrt {2\pi } }}} {e^{ - {{\left( {\frac{{x - \mu }}{\sigma }} \right)}^2}}}dx$

To calculate the probability for a random variable that follows $N\left( {\mu ,{\sigma ^2}} \right)$ , convert the variable to a standard normal variable by the transformation,

$z = \frac{{x - \mu }}{\sigma }$

The probability can be obtained by the formula:

$\begin{array}{c}\\P\left( {x \le a} \right) = P\left( {\frac{{x - \mu }}{\sigma } \le \frac{{a - \mu }}{\sigma }} \right)\\\\ = \Phi \left( {\frac{{a - \mu }}{\sigma }} \right)\\\end{array}$

$\begin{array}{c}\\P\left( {x \ge a} \right) = P\left( {\frac{{x - \mu }}{\sigma } \ge \frac{{a - \mu }}{\sigma }} \right)\\\\ = 1 - P\left( {\frac{{x - \mu }}{\sigma } \le \frac{{a - \mu }}{\sigma }} \right)\\\\ = 1 - \Phi \left( {\frac{{a - \mu }}{\sigma }} \right)\\\end{array}$

Use the formula where the interval is $a < x < b$ as:

$\begin{array}{c}\\P\left( {a < x < b} \right) = P\left( {\frac{{a - \mu }}{\sigma } \le \frac{{x - \mu }}{\sigma } \le \frac{{b - \mu }}{\sigma }} \right)\\\\ = P\left( {\frac{{a - \mu }}{\sigma } \le z \le \frac{{b - \mu }}{\sigma }} \right)\\\\ = \Phi \left( {\frac{{b - \mu }}{\sigma }} \right) - \Phi \left( {\frac{{a - \mu }}{\sigma }} \right)\\\end{array}$

The $= {\rm{NORMDIST}}\left( {} \right)$ function used in Excel to find the probabilities.

(a)

The mean $\left( \mu \right)$ of the domestic airfares in the United States is $385 per ticket and the standard deviation $\left( \sigma \right)$ is $110. Consider the domestic airfares as a random variable $\left( x \right)$ which follows $N\left( {\mu = 385,\sigma = 110} \right)$ . The probability that the domestic airfare is $550 or more is calculated as,

$\begin{array}{c}\\P\left( {x \ge 550} \right) = P\left( {\frac{{x - \mu }}{\sigma } \ge \frac{{550 - 385}}{{110}}} \right)\\\\ = P\left( {z \ge 1.5} \right)\\\\ = 1 - \Phi (1.5)\\\end{array}$

Use Excel to calculate the probability for the Z-Value 1.5.

The screenshot of the formula used is shown below:

So,

$\begin{array}{c}\\P\left( {x \ge 550} \right) = 1 - 0.9331928\\\\ = 0.0668072\\\\ \approx 0.0668\\\end{array}$

(b)

The calculation of the probability that the domestic airfare is $250 or less is,

$\begin{array}{c}\\P\left( {x \le 250} \right) = p\left( {\frac{{x - \mu }}{\sigma } \le \frac{{250 - 385}}{{110}}} \right)\\\\ = P\left( {z \le - 1.23} \right)\\\\ = \Phi \left( { - 1.23} \right)\\\\ = 1 - \Phi \left( {1.23} \right)\\\end{array}$

Use Excel to calculate the probability for the Z-Value 1.23. The screenshot of the formula used is shown below:

Thus,

$\begin{array}{c}\\P\left( {x \le 250} \right) = 1 - 0.8906514\\\\ = 0.1093486\\\\ \approx 0.1093\\\end{array}$

(c)

The calculation of the probability that the domestic airfares lie between $300 and $500 is,

$\begin{array}{c}\\P\left( {300 \le x \le 500} \right) = p\left( {\frac{{300 - 385}}{{110}} \le \frac{{x - \mu }}{\sigma } \le \frac{{500 - 385}}{{110}}} \right)\\\\ = P\left( { - 0.77 \le z \le 1.05} \right)\\\\ = \Phi \left( {1.05} \right) - \Phi \left( { - 0.77} \right)\\\\ = \Phi \left( {1.05} \right) - \left( {1 - \Phi \left( {0.77} \right)} \right)\\\end{array}$

Use Excel to calculate the probability for the Z-Value 1.05 and 0.77. The screenshot of the formula used is shown below:

So, the probability is calculated as,

$\begin{array}{c}\\P\left( {300 \le x \le 500} \right) = 0.8531409 - \left( {1 - 0.7793501} \right)\\\\ = 0.8531409 - 0.2206499\\\\ = 0.632491\\\\ \approx 0.6325\\\end{array}$

(d)

To find the cost of the 3% of the highest domestic airfares, find the value of $y$ , which is the new cost where $P\left( {x > y} \right) = 3\%$ . The calculation of the cost for the 3% of the highest domestic airfares is,

$\begin{array}{c}\\P\left( {x > y} \right) = 3\% \\\\P\left( {\frac{{x - \mu }}{\sigma } > \frac{{y - \mu }}{\sigma }} \right) = 0.03\\\\P\left( {z > \frac{{y - \mu }}{\sigma }} \right) = 0.03\\\\P\left( {z < \frac{{y - \mu }}{\sigma }} \right) = 0.97\\\end{array}$

$\begin{array}{l}\\\Phi \left( {\frac{{y - \mu }}{\sigma }} \right) = 0.97\\\\\\\end{array}$

Use Excel to find the inverse value of Z for the value 0.97. The screenshot of the formula is used shown below:

So,

$\begin{array}{c}\\\frac{{y - \mu }}{\sigma } = 1.88\\\\y = 1.88\sigma + \mu \\\\y = \left( {1.88 \times 110} \right) + 385\\\\y = 591.8\\\end{array}$

Ans: Part a

The probability that the domestic airfare is $550 or more is approximately 0.0668.