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The U.S. Bureau of Labor Statistics reports that the average annual expenditure on food and drink for all families is \(\$ 5700\) (Money, December 2003 ). Assume that annual expenditure on food and drink is normally distributed and that the standard deviation is \(\$ 1500\) a. What is the range of expenditures of the \(10 \%\) of families with the lowest annual spending on food and drink? b. What percentage of families spend more than \(\$ 7000\) annually on food and drink? c. What is the range of expenditures for the \(5 \%\) of families with the highest annual spending on food and drink?

Short Answer

Expert verified
a) Less than $3780; b) 19.23% spend more than $7000; c) More than $8167.5

Step by step solution

01

Understand the Problem

We have a normally distributed dataset representing annual expenditures on food and drinks by families. The mean (average) of this distribution is given as \( \mu = 5700 \) dollars, and the standard deviation is \( \sigma = 1500 \) dollars. We need to find specific percentiles and ranges using the normal distribution properties.
02

Find the 10th Percentile for Part (a)

To find the lowest 10% expenditure, we need to determine the value below which 10% of the data falls. This is a percentile problem. The 10th percentile corresponds to a \( z \)-score that we find using a standard normal distribution table or calculator. The \( z \)-score for the 10th percentile is approximately \( -1.28 \). We will use the formula: \[ X = \mu + z \sigma \]Substituting the values, we get:\[ X = 5700 + (-1.28)(1500) \]Calculate \( X \) to find the 10 percent threshold for expenditures.
03

Calculate the 10th Percentile Expenditure

Substitute the values:\[ X = 5700 + (-1.28)(1500) = 5700 - 1920 = 3780 \]Thus, the bottom 10% of families spend less than \( \$ 3780 \) annually.
04

Find Percentage of Families Spending More Than $7000 for Part (b)

We want to find the percentage of families spending more than \( \\( 7000 \). First, calculate the \( z \)-score for \( \\) 7000 \) using the formula:\[ z = \frac{X - \mu}{\sigma} = \frac{7000 - 5700}{1500} \]Calculate \( z \) and use a standard normal distribution table to find the corresponding percentage.
05

Calculate the Percentage Greater Than $7000

Substitute the values:\[ z = \frac{7000 - 5700}{1500} = \frac{1300}{1500} \approx 0.87 \]Using a standard normal distribution table, a \( z \)-score of \( 0.87 \) indicates approximately 80.77% of families spend less than \( \\( 7000 \). Therefore, 100% - 80.77% = 19.23% of families spend more than \( \\) 7000 \) annually.
06

Find the 95th Percentile for Part (c)

For the top 5% of spenders, we need to find the 95th percentile because 5% of the data should fall above this point. The \( z \)-score for the 95th percentile is approximately \( 1.645 \). Use the same formula to find the expenditure value:\[ X = \mu + z \sigma \]Substitute to find \( X \).
07

Calculate the 95th Percentile Expenditure

Substitute the values:\[ X = 5700 + (1.645)(1500) = 5700 + 2467.5 = 8167.5 \]So, the top 5% of families spend more than \( \$ 8167.5 \) annually.

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Key Concepts

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

Understanding Percentiles
Percentiles are a way to understand and interpret data by dividing a dataset into 100 equal parts. Each part represents 1% of the data. For example, the 10th percentile is the value below which 10% of the data falls.
In this exercise, we wanted to find the annual expenditure on food and drink that falls into the 10th percentile. Knowing this helps us identify the lowest 10% of spenders in the population. It's a statistical measure that gives insight into the distribution and spread of data within a given sample.
Percentiles are crucial in various fields for making decisions and understanding position relative to the rest of the dataset.
Calculating Z-scores
The concept of a z-score is fundamental in statistics. It tells us how many standard deviations a data point is from the mean. In the context of a normal distribution, z-scores help us find percentiles and probabilities. Computing a z-score requires a specific formula:
  • Subtract the mean from the data point.
  • Divide the result by the standard deviation.
For instance, in the problem, the z-score was used to identify both the 10th and 95th percentiles. These z-scores come from standard normal distribution tables, allowing us to find the probabilities associated with these percentiles.
Z-scores help transform data into a form that is more comprehensible, aiding in comparing different datasets or values easily.
The Role of Standard Deviation
Standard deviation is a key concept when dealing with distributions. It measures the average distance of each data point from the mean, which provides insight into the dispersion or spread of a dataset. A smaller standard deviation means data points are closer to the mean, while a larger standard deviation indicates more spread.
In our example, the standard deviation of $1500 tells us the average variance in annual spending on food and drink from the mean spending of $5700.
  • The standard deviation allowed us to calculate z-scores.
  • Understanding this helps delineate the percentage of families spending below or above certain thresholds.
Standard deviation helps in understanding variability, which is essential in statistical analysis and predictions.
Basics of Statistical Analysis
Statistical analysis involves collecting and analyzing data to identify patterns and trends. It encompasses various techniques and tools, and in this problem, we focus on the analysis of normally distributed data.
This involves checking the structure of a dataset - finding the mean, understanding the normal distribution, and calculating percentiles and z-scores. These methods help to describe, infer, and predict the dataset's characteristics. By doing this analysis, we gain insight into how a subset of a population behaves, make forecasts, or drive informed decisions.
  • Statistical analysis is crucial for making data-driven decisions.
  • It allows us to handle and interpret large datasets effectively.
Understanding statistical analysis is essential for fields like economics, social sciences, and business, where data drives decisions.

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Most popular questions from this chapter

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Is lack of sleep causing traffic fatalities? A study conducted under the auspices of the National Highway Traffic Safety Administration found that the average number of fatal crashes caused by drowsy drivers each year was 1550 ( Business Week , January 26,2004 ). Assume the annual number of fatal crashes per year is normally distributed with a standard deviation of 300 . a. What is the probability of fewer than 1000 fatal crashes in a year? b. What is the probability that the number of fatal crashes will be between 1000 and 2000 for a year? c. For a year to be in the upper \(5 \%\) with respect to the number of fatal crashes, how many fatal crashes would have to occur?

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