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A recent report in USA Today indicated a typical family of four spends \(\$ 490\) per month on food. Assume the distribution of food expenditures for a family of four follows the normal distribution, with a mean of \(\$ 490\) and a standard deviation of \(\$ 90 .\) a. What percent of the families spend more than \(\$ 30\) but less than \(\$ 490\) per month on food? b. What percent of the families spend less than \(\$ 430\) per month on food? c. What percent spend between \(\$ 430\) and \(\$ 600\) per month on food? d. What percent spend between \(\$ 500\) and \(\$ 600\) per month on food?

Short Answer

Expert verified
a. 50%, b. 25.14%, c. 63.74%, d. 34.50%

Step by step solution

01

Understand the Problem

We have a normal distribution characterized by a mean \( \mu = 490 \) and a standard deviation \( \sigma = 90 \). We need to find the percentage of families in different expenditure ranges concerning this normal distribution.
02

Calculate Z-Scores

A Z-score indicates how many standard deviations an element is from the mean. The formula for a z-score is \( Z = \frac{X - \mu}{\sigma} \) where \( X \) is the value concerned.- For \( X = 30 \): \( Z = \frac{30 - 490}{90} = -5.11 \)- For \( X = 430 \): \( Z = \frac{430 - 490}{90} = -0.67 \)- For \( X = 490 \): \( Z = \frac{490 - 490}{90} = 0 \)- For \( X = 500 \): \( Z = \frac{500 - 490}{90} = 0.11 \)- For \( X = 600 \): \( Z = \frac{600 - 490}{90} = 1.22 \)
03

Use the Z-Table or Calculator

The Z-table or a standard normal distribution calculator gives us the probability that a score is less than a given Z-score. Get these probabilities:- Probability \( Z < -5.11 \) is practically 0 since it's way off the normal limits.- Probability \( Z < -0.67 \) is approximately 0.2514.- Probability \( Z < 0 \) is 0.5.- Probability \( Z < 0.11 \) is approximately 0.5438.- Probability \( Z < 1.22 \) is approximately 0.8888.
04

Calculate Percentages for Each Question

**a. Percent of families spending more than \( 30 \) but less than \( 490 \):**The percentage can be found by subtracting the probability of spending less than \( 30 \) from less than \( 490 \):\[0.5 - 0 = 0.5\] which corresponds to 50%.**b. Percent of families spending less than \( 430 \):**This is just the probability of \( Z < -0.67 \):\[0.2514\] which corresponds to 25.14%.**c. Percent spending between \( 430 \) and \( 600 \):**Find the difference between the probabilities:\[0.8888 - 0.2514 = 0.6374\] which corresponds to 63.74%.**d. Percent spending between \( 500 \) and \( 600 \):**Difference between probabilities \( Z < 1.22 \) and \( Z < 0.11 \):\[0.8888 - 0.5438 = 0.3450\] which corresponds to 34.50%.

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Key Concepts

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

Z-Score
A z-score is a valuable statistical tool that helps us understand where a particular data point stands in a normal distribution. Essentially, it represents the number of standard deviations a data point is from the mean. Calculating the z-score is simple with the formula:
  • \( Z = \frac{X - \mu}{\sigma} \)
Here, \( X \) is the data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. For instance, a z-score of 0 indicates the data point is exactly at the mean. A positive z-score means it's above the mean, while a negative z-score means it's below. For example, the z-score for a family spending \(430 is calculated as follows:
  • \( Z = \frac{430 - 490}{90} = -0.67 \)
This implies that \)430 is 0.67 standard deviations below the mean of $490. Understanding z-scores allows for determining the probability of a value lying within a specific range.
Standard Deviation
Standard deviation is a key concept in statistics, providing a measure of the dispersion or spread of a set of data values around their mean. In our context, the standard deviation is $90, which means that a typical family's food expenditure is likely to vary by $90 above or below the mean of $490.
Here's a breakdown:
  • Low Standard Deviation: Indicates that the data points are closely clustered around the mean.
  • High Standard Deviation: Suggests that the data points are spread out over a larger range of values.
Being aware of the standard deviation enables us to determine how typical or atypical certain data points are within the context of the entire dataset. For example, if a family's expenditure is $600, they are further from the average, as indicated by their relatively high z-score.
Probability
Probability tells us how likely an event is to occur. In the context of normal distribution, we can calculate the probability of a data point falling within a certain range using a z-score table or normal distribution calculator. It's expressed as a percentage but can also be a fraction or decimal. For example, to find the probability of a family's expenses being between \(430 and \)600, we look at their respective z-scores:
  • Z-score for \(430: \( -0.67 \)
  • Z-score for \)600: \( 1.22 \)
Using a z-table, we find:
  • Probability that expenses are less than \(430 is approximately 25.14%.
  • Probability that expenses are less than \)600 is approximately 88.88%.
The difference, 63.74%, shows the probability of a family's monthly spending falling between these values. Probability is essential in understanding and interpreting statistical data, especially in real-world scenarios like budgeting for a family's monthly expenses.
Mean
In statistics, the mean is a measure of central tendency and is simply the average of a set of numbers. In the problem here, the mean food expenditure for a family of four is $490.
This mean is indicative of what is typical for this group:
  • Calculate the mean by adding all the observed data points and dividing by the number of points.
  • The mean represents the central point of a data set.
In the context of a normal distribution, the mean is the peak of the bell curve. It helps in understanding where most families' expenditures lie. For this exercise, having a mean of $490 means this is the central value around which the family expenditures are distributed. Knowing the mean enables easier calculation and understanding of other statistical measures like standard deviation and z-scores.

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Most popular questions from this chapter

A uniform distribution is defined over the interval from 2 to \(5 .\) a. What are the values for \(a\) and \(b\) ? b. What is the mean of this uniform distribution? c. What is the standard deviation? d. Show that the total area is 1.00 . e. Find the probability of a value more than 2.6 . f. Find the probability of a value between 2.9 and 3.7 .

The Kamp family has twins, Rob and Rachel. Both Rob and Rachel graduated from college 2 years ago, and each is now earning \(\$ 50,000\) per year. Rachel works in the retail industry, where the mean salary for executives with less than 5 years' experience is \(\$ 35,000\) with a standard deviation of \(\$ 8,000 .\) Rob is an engineer. The mean salary for engineers with less than 5 years' experience is \(\$ 60,000\) with a standard deviation of \(\$ 5,000\). Compute the \(z\) values for both Rob and Rachel and comment on your findings.

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The monthly sales of mufflers in the Richmond, Virginia, area follow the normal distribution with a mean of 1,200 and a standard deviation of \(225 .\) The manufacturer would like to establish inventory levels such that there is only a 5 percent chance of running out of stock. Where should the manufacturer set the inventory levels?

A normal distribution has a mean of 50 and a standard deviation of \(4 .\) Determine the value below which 95 percent of the observations will occur.

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