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Chapter 7: Problem 25

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

Short Answer

Expert verified
The operating cost for the lowest 3 percent of airplanes is approximately \(\$1630\).

Step by step solution

01

Understand the Problem

We are given that the mean hourly cost to operate a commercial airplane is normally distributed with a mean \( \mu = \\(2100 \) and a standard deviation \( \sigma = \\)250 \). We need to find the cost corresponding to the lowest 3 percent of airplanes. This is a question about finding a specific percentile (3rd percentile) of a normal distribution.
02

Identify the Required Z-Score

Since we want to find the cost that corresponds to the lowest 3 percent, we first need to find the z-score (standard score) corresponding to the 3rd percentile of a standard normal distribution. This can be done using a z-table or calculator, which indicates that a 3rd percentile has a z-score of approximately \( -1.88 \).
03

Use the Z-Score to Find the Operating Cost

With the z-score \( z = -1.88 \), use the formula of the normal distribution to find the operating cost \( x \). The formula is \( x = \mu + z \times \sigma \), where \( \mu = 2100 \) and \( \sigma = 250 \). Substitute the values into the formula: \( x = 2100 + (-1.88) \times 250 = 2100 - 470 = 1630 \).
04

Interpret the Result

The calculated value \( 1630 \) represents the cost per hour at which only 3 percent of airplanes have a lower operating cost. This means that, 3% of commercial airplanes have an operating hourly cost less than or equal to \( \$1630 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Z-Score
The z-score is an important concept in statistics, especially when dealing with normal distributions. A z-score tells you how many standard deviations an element is from the mean. Here's why it's useful:
  • It allows comparisons between different data points in a standard way.
  • If data is normally distributed, the z-score can help us find probabilities and percentiles.
To calculate a z-score, use the formula:\[z = \frac{x - \mu}{\sigma}\]where:
  • \( x \) is the value you're comparing.
  • \( \mu \) is the mean.
  • \( \sigma \) is the standard deviation.
In the context of our problem, finding the 3rd percentile means looking for a negative z-score, because we are dealing with the lowest values. A z-score of -1.88 shows that the operating cost lies 1.88 standard deviations below the mean in our context.
Exploring Percentiles
Percentiles give us a way to understand the relative standing of a set of data points. For any given data set, the nth percentile is the value below which n% of the data fall. Percentiles are useful because:
  • They help us understand the distribution of data.
  • They allow us to compare individual data points to the overall data set.
In terms of practical application, locating the 3rd percentile helps identify the lower 3% of costs in this scenario. If you think of the distribution as a graph, percentiles help locate specific positions on the curve. For example:
  • The 50th percentile (median) splits the data into two equal halves.
  • The 3rd percentile, as in our problem, indicates particularly low values.
This knowledge aids businesses in identifying unusual or unexpected performance patterns.
The Role of Standard Deviation
Standard deviation is key when examining how spread out our data is.A smaller standard deviation means data points are tightly clustered around the mean, while a larger standard deviation means they are more spread out. Standard deviation influences:
  • The calculation of z-scores.
  • The overall shape of a normal distribution graph.
In our airplane example, a standard deviation of \(\$250\) indicates the range where most hourly costs will fall.Knowing the standard deviation allows precise predictions regarding probabilities:
  • A range of \( \mu - \sigma \) to \( \mu + \sigma \) contains about 68% of data.
  • A larger range of \( \mu - 2\sigma \) to \( \mu + 2\sigma \) covers roughly 95% of data.
Understanding standard deviation helps in evaluating how different or unusual a particular cost might be, thus supporting more informed financial decision-making.

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Most popular questions from this chapter

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