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Airline Passengers Get Heavier. In response to the increasing weight of airline passengers, the Federal Aviation Administration (FAA) in 2003 told airlines to assume that passengers average 195 pounds in the winter, including clothing and carry-on baggage. But passengers vary, and the FAA did not specify a standard deviation. A reasonable standard deviation is 35 pounds. Weights are not Normally distributed, especially when the population includes both men and women, but they are not very non-Normal. A commuter plane carries 22 passengers. What is the approximate probability that the total weight of the passengers exceeds 4500 pounds? Use the four-step process to guide your work. (Hint: To apply the central limit theorem, restate the problem in terms of the mean weight.)

Step by step solution

01

Understanding the Problem

We want to determine the probability that 22 passengers on a plane have a total weight exceeding 4500 pounds. We know each passenger's average weight is 195 pounds in winter. We will use the Central Limit Theorem to estimate this probability, considering the sample mean.

02

Convert Total Weight Problem to Mean Weight Problem

The total weight exceeding 4500 pounds implies the mean weight per passenger exceeding \( \frac{4500}{22} \). Calculate this mean, \( \bar{x} = \frac{4500}{22} \approx 204.55 \) pounds.

03

Use the Central Limit Theorem

The Central Limit Theorem allows us to approach this as a normal distribution problem. The distribution of the sample mean \( \bar{x} \) can be approximated by a normal distribution with mean \( \mu = 195 \) and standard deviation \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \).

04

Calculate the Standard Deviation of the Sampling Distribution

The standard deviation of the sample mean is \( \sigma_{\bar{x}} = \frac{35}{\sqrt{22}} \), approximately equal to 7.46 pounds.

05

Determine the Z-Score

Calculate the Z-score to determine how many standard deviations \( \bar{x} = 204.55 \) is from the mean \( \mu = 195 \). The Z-score is \( Z = \frac{204.55 - 195}{7.46} \approx 1.28 \).

06

Find Probability from Z-Score

Use a standard normal distribution table or calculator to find the probability for the Z-score. \( P(Z > 1.28) \approx 0.10 \) indicating a 10% probability that the mean exceeds 204.55 pounds.

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Key Concepts

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

Sampling Distribution

When dealing with data from a population, understanding the concept of sampling distribution is essential. Imagine you repeatedly select samples of the same size from a population and calculate the mean weight for each sample. These sample means form their own distribution, known as the sampling distribution of the sample mean.
The beauty of the sampling distribution lies in its properties:

• The mean of the sampling distribution is equal to the mean of the population from which the samples are drawn.
• The shape of the sampling distribution becomes more bell-shaped and normal as you take larger samples, thanks to a key concept called the Central Limit Theorem (CLT).
• Even if the population distribution is not normal, the sampling distribution of the mean will be approximately normal for a large enough sample size.

In our airline passenger example, we took a sample size of 22 passengers to create a sampling distribution, making the problem manageable using the CLT.

Standard Deviation

Standard deviation is a measurement of how spread out the numbers in a data set are. It tells us how much the individual data points differ from the mean value of the data set. A higher standard deviation means the data points are spread out over a larger range of values, while a lower standard deviation indicates that they are closer to the mean.In the context of sampling distribution, we often refer to the standard deviation of the sample means as the standard error. This helps us understand how much the means of different samples will vary from each other:

• The formula for the standard deviation of the sampling distribution (standard error) is given by \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the population standard deviation, and \( n \) is the sample size.
• As you increase the sample size, the standard error decreases, indicating greater precision in estimating the population mean.

In our example, the standard deviation was given as 35 pounds, and with a sample size of 22, the standard error calculated gave us an idea about how reliable our sample mean is in approximating the population mean.

Z-Score

The Z-score is a statistical measurement that describes a data point's position within a distribution relative to the mean, measured in terms of standard deviations. It tells us how many standard deviations a data point, or in our case, the sample mean, is from the population mean.By converting our mean weight problem to a Z-score problem, we simplify the process of finding probabilities:

• The Z-score formula is \( Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} \), where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, and \( \sigma_{\bar{x}} \) is the standard error.
• With the Z-score, we can reference standard normal distribution tables (or use calculators) to determine probabilities and make statistical inferences.
• In our example, a calculated Z-score of approximately 1.28 helped us identify the probability of the average passenger weight exceeding a certain value.

Normal Distribution

A normal distribution is a bell-shaped curve that is symmetric about the mean, depicting the distribution of data points for many types of data. Many natural phenomena and measurements follow this kind of distribution, which makes it extremely useful in statistics. Key features of a normal distribution include:

• The mean, median, and mode of the distribution are all equal.
• About 68% of the data falls within one standard deviation of the mean, 95% within two, and 99.7% within three, a fact known as the empirical rule or 68-95-99.7 rule.

In our example, although individual weights of passengers are not normally distributed, the Central Limit Theorem allows the distribution of the sample mean to be approximated as normal, providing a basis for applying normal probability computations to solve the exercise.