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91Ó°ÊÓ

Recently the United States Department of Agriculture issued a report (http://www.cnpp. usda.gov/sites/default/files/CostoffoodMar2015.pdf) indicating a family of four spent an average of about \(\$ 890\) per month on food. Assume the distribution of food expenditures for a family of four follows the normal distribution, with a standard deviation of \(\$ 90\) per month. a. What percent of the families spend more than \(\$ 430\) but less than \(\$ 890\) per month on food? b. What percent of the families spend less than \(\$ 830\) per month on food? c. What percent spend between \(\$ 830\) and \(\$ 1,000\) per month on food? d. What percent spend between \(\$ 900\) and \(\$ 1,000\) per month on food?

Short Answer

Expert verified
a) 50%, b) 25.14%, c) 63.74%, d) 34.5%.

Step by step solution

01

Understanding the Problem

We are given the average monthly expenditure on food for a family of four as \( \\( 890 \) with a standard deviation of \( \\) 90 \), and a normal distribution is assumed. We need to find the percentages for various spending ranges using the concept of z-scores and the normal distribution table.
02

Calculate Z-score for Part a

For expenditure greater than \( \\( 430 \) and less than \( \\) 890 \), calculate the z-scores. \(z_1 = \frac{430 - 890}{90} = \frac{-460}{90} \approx -5.11\) and \(z_2 = \frac{890 - 890}{90} = 0\).
03

Determine Probability for Part a

Using the standard normal distribution table, find the probability for \( z_1 \leq z < z_2 \).- Probability for \( z = -5.11 \) is approximately 0.- Probability for \( z = 0 \) is 0.5.Thus, the percentage is \( 50\%.\)
04

Calculate Z-score for Part b

For expenditure less than \( \$ 830 \), calculate the z-score.\( z = \frac{830 - 890}{90} = \frac{-60}{90} \approx -0.67 \).
05

Determine Probability for Part b

Using the standard normal distribution table, find the probability for \( z < -0.67 \).The probability for \( z = -0.67 \) is around 0.2514.Thus, the percentage is \( 25.14\%.\)
06

Calculate Z-scores for Part c

For expenditure between \( \\( 830 \) and \( \\) 1000 \), calculate z-scores.\( z_1 = \frac{830 - 890}{90} \approx -0.67 \) and \( z_2 = \frac{1000 - 890}{90} \approx 1.22 \).
07

Determine Probability for Part c

Using the standard normal distribution table, find the probability for \( z_1 < z < z_2 \).- Probability for \( z = -0.67 \) is 0.2514.- Probability for \( z = 1.22 \) is 0.8888.Thus, the percentage is \( 88.88\% - 25.14\% = 63.74\%.\)
08

Calculate Z-scores for Part d

For expenditure between \( \\( 900 \) and \( \\) 1000 \), calculate z-scores.\( z_1 = \frac{900 - 890}{90} \approx 0.11 \) and \( z_2 = \frac{1000 - 890}{90} \approx 1.22 \).
09

Determine Probability for Part d

Using the standard normal distribution table, find the probability for \( z_1 < z < z_2 \).- Probability for \( z = 0.11 \) is 0.5438.- Probability for \( z = 1.22 \) is 0.8888.Thus, the percentage is \( 88.88\% - 54.38\% = 34.5\%.\)

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Key Concepts

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

Z-Scores
In statistics, a z-score is a way of describing a value's relation to the mean of a group of values. It's a simple method for comparing an individual data point to the overall distribution. The z-score is calculated by subtracting the mean from the individual data point and then dividing by the standard deviation. This results in a standardized score that can help you understand how many standard deviations a value is away from the mean.
For example, if the mean expenditure on food is $890 per month and the standard deviation is $90, a family spending $830 can be calculated as:
  • Subtract the mean: $830 - $890 = -$60
  • Divide by the standard deviation: -$60 ÷ $90 ≈ -0.67
This means the family's spending is 0.67 standard deviations below the mean.
Standard Deviation
Standard deviation is a fundamental concept in statistics that describes how much variation exists in a set of data. It provides insight into the spread of the data around the mean and is a critical component in calculating z-scores. The larger the standard deviation, the more spread out the data points are.
In the context of our problem, the standard deviation of $90 indicates how scattered the monthly food expenditures are for different families compared to the average expenditure of $890.
  • A small standard deviation implies most families spend close to the average.
  • A large standard deviation means spending varies widely among families.
Using standard deviation helps statisticians determine the probability of an event occurring within a certain range, especially when dealing with normal distribution.
Probability Distribution
Probability distribution is a statistical function that describes possible outcomes of a statistical experiment and the likelihood of each outcome. When dealing with food expenditures, assigning probabilities to different spending ranges reveals how common or rare certain spending habits are within the population.
The normal distribution, often called the bell curve, is a common example of a probability distribution. It's symmetric and centered around the mean. Most values cluster around this mean, with probabilities tapering off symmetrically as you move away.
Calculating the probability of spending between specific amounts (like $830 to $1000) involves understanding where these values lie in the distribution curve and using z-scores to find the respective probabilities in a normal distribution table.
Normal Distribution Table
A normal distribution table, or z-table, is a handy tool for finding probabilities and proportions associated with z-scores in a standard normal distribution. This table displays the cumulative probability of a score being below a given z-score.
For instance, to find out how many families spend less than $830, you would calculate the z-score and look this up in the table. For a z-score of -0.67, the table shows a probability of 0.2514, meaning about 25.14% of families spend less than $830 on food each month.
  • Values in the central part of the table correspond to areas under the curve from the left up to the z-score.
  • By combining table values, you can determine probabilities for any range of values on the distribution.
Understanding how to read and interpret this table is essential to analyzing data with normal distribution, offering a clear picture of relative positions and likelihoods of occurrences.

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Most popular questions from this chapter

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