Problem 27 - Basic Statistics for Business and Economics

Question

Assume that the mean hourly cost to operate a commercial airplane follows the normal distribution with a mean \(\$ 2,100\) per hour and a standard deviation of \(\$ 250 .\) What is the operating cost for the lowest 3 percent of the airplanes?

Step-by-Step Solution

Answer

The operating cost for the lowest 3% of airplanes is approximately $1630.

Step 1: Identify the Problem

We need to find the operating cost for the lowest 3% of airplanes from a normally distributed cost with a mean of $2100 and a standard deviation of $250.

Step 2: Understand the Normal Distribution

The problem states that operating costs follow a normal distribution. Thus, we can use properties of the normal distribution to find corresponding costs for specific percentages or percentiles.

Step 3: Utilize Z-score Formula

The Z-score formula is used to find how many standard deviations an element is from the mean. The formula is: \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the value we are looking for, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

Step 4: Find Z-score for Lowest 3%

Using a Z-table or calculator, find the Z-score that corresponds to the bottom 3%, which is approximately -1.88 (common value for the 3rd percentile in Z-tables).

Step 5: Solve for X (Operating Cost)

Plug the Z-score back into the Z-score formula and solve for \( X \): \[ -1.88 = \frac{X - 2100}{250} \].

Step 6: Calculate the Operating Cost

Rearrange the equation from the previous step: \( X - 2100 = -1.88 \times 250 \). Then \( X = 2100 - 470 = 1630 \). This means the operating cost for the lowest 3% of the airplanes is approximately \( \$1630 \).

Key Concepts

Z-score

The Z-score is a crucial concept in statistics, especially when dealing with normal distributions. It tells us how many standard deviations an element is from the mean of the data set. Essentially, it's a way of standardizing scores on different scales to a common scale.

• Why Use Z-scores? They help in understanding and comparing data from different normal distributions by bringing everything to a standard scale. Thus, comparisons become easier.
• Z-score Formula: To calculate the Z-score, the formula is \( Z = \frac{X - \mu}{\sigma} \). Here, \(X\) is the value we want to analyze, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

For instance, in our problem, we want to find the operating costs that fall into the lowest 3%. By using the Z-score, we discover that this requires finding which value corresponds to a Z-score of approximately -1.88.

Percentile

A percentile provides a clear picture of where a particular value falls within a distribution. It describes the percentage of the data set that exists below a certain point.

• Understanding Percentiles: If a value is in the 3rd percentile, this means that it is higher than 3% of all other values in the distribution, or conversely, 97% of the values exceed it.
• Role in Our Example: To find the operating costs of the lowest 3% of airplanes, we used a percentile to understand which costs are typical for this proportion of the airplanes.

Percentiles are intuitive for grasping how an individual value compares against an overall distribution, especially useful in educational and scientific applications.

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. The lower the standard deviation, the closer the data points tend to be to the mean.

• Importance: In normal distributions, standard deviation determines the spread of the data. It helps identify how much the typical data point differs from the mean.
• Link to the Problem: In this exercise, a standard deviation of $250 means that most operating costs fall within $250 of the mean of $2100. This spread helps us gauge where a certain cost lies in relation to the mean value.

In conclusion, standard deviation is vital for interpreting data, especially when making inferences about a population from a sample, as seen in our problem.