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According to the Internal Revenue Service, the mean tax refund for the year 2006 was \(\$ 2,290\). Assume the standard deviation is \(\$ 650\) and that the amounts refunded follow a normal probability distribution. a. What percent of the refunds are more than \(\$ 3,000 ?\) b. What percent of the refunds are more than \(\$ 3,000\) but less than \(\$ 3,500 ?\) c. What percent of the refunds are more than \(\$ 2,500\) but less than \(\$ 3,500 ?\)

Short Answer

Expert verified
a) 13.79%, b) 10.65%, c) 34.07%

Step by step solution

01

Understanding the Normal Distribution

A normal distribution is characterized by its mean and standard deviation. Here, the mean \(\mu\) is \\(2,290\ and the standard deviation \(\sigma\) is \\)650\. The problem asks for percentages, translating to finding areas under specific regions of the normal curve.
02

Calculate the Z-score for $3,000

The Z-score measures how many standard deviations an element is from the mean. The formula is: \[Z = \frac{X - \mu}{\sigma}\]Substituting \(X = 3000\), \(\mu = 2290\), and \(\sigma = 650\), we get: \[Z = \frac{3000 - 2290}{650} = \frac{710}{650} \approx 1.0923\]
03

Find Percent of Refunds More Than $3,000

Using the Z-score from Step 2 (approximately 1.0923), look up this value in the standard normal distribution table to find the corresponding percentile. The cumulative probability (percentile) for \(Z = 1.0923\) is approximately 0.8621, meaning 86.21% of refunds are less than \\(3000\. Thus, \[100\% - 86.21\% = 13.79\%\]of the refunds are more than \\)3000\.
04

Calculate the Z-scores for $3,000 and $3,500

For \$3,500\, the Z-score is: \[Z = \frac{3500 - 2290}{650} = \frac{1210}{650} \approx 1.8615\]
05

Find Percent of Refunds Between $3,000 and $3,500

Using the Z-score from Step 4, look up the value in the standard normal distribution table. The cumulative probability for \(Z = 1.8615\) is approximately 0.9686. To find the percentage between \\(3000\ and \\)3500\:\[96.86\% - 86.21\% = 10.65\%\]
06

Calculate the Z-score for $2,500

For \$2,500\, the Z-score is: \[Z = \frac{2500 - 2290}{650} = \frac{210}{650} \approx 0.3231\]
07

Find Percent of Refunds Between $2,500 and $3,500

Using the Z-scores for \\(2,500\ and \\)3,500\, find the cumulative probabilities: The cumulative probability for \(Z = 0.3231\) is approximately 0.6279. Now, calculate the percentage between \\(2,500\ and \\)3,500\:\[96.86\% - 62.79\% = 34.07\%\]

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Key Concepts

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

Z-score
The Z-score is a fundamental concept in statistics, especially when dealing with normal distributions. It is a numerical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. If a data point's Z-score is 0, it indicates that the data point's score is identical to the mean score. The formula to calculate the Z-score is:\[Z = \frac{X - \mu}{\sigma}\]where:
  • \(X\) is the value to be standardized
  • \(\mu\) is the mean of the distribution
  • \(\sigma\) is the standard deviation
The Z-score tells us how many standard deviations away from the mean a particular data point is. For example, a Z-score of 1.0923 means the value is 1.0923 standard deviations above the mean. This makes it easier to determine the proportion of data points above or below a certain value in the context of a normal distribution. The Z-score is crucial because it helps translate a raw score into a location on the standard normal distribution, which can then be used to find probabilities.
Standard Deviation
Standard deviation is a statistical measurement that quantifies the amount of variation or dispersion in a set of data values. It gives insight into how much the individual data points in a dataset differ from the mean (average) value of the dataset. A small standard deviation indicates that the data points tend to be close to the mean, while a large standard deviation indicates that the data points are spread out over a wider range. The formula for standard deviation is:\[\sigma = \sqrt{\frac{\sum{(X_i - \mu)^2}}{N}}\]where:
  • \(X_i\) represents each data point
  • \(\mu\) is the mean of the data
  • \(N\) is the number of data points
In the context of a normal distribution, standard deviation plays a pivotal role. It helps determine the "spread" of the bell curve and thus significantly impacts the shape of the distribution. In our example, the standard deviation helps us understand how the tax refunds are distributed around the mean of $2,290. It also aids in calculating Z-scores, which further guides us in determining cumulative probabilities associated with certain refund amounts.
Cumulative Probability
Cumulative probability refers to the probability that a random variable takes on a value less than or equal to a specified number. In the context of statistics and the normal distribution, it signifies the area under the probability distribution curve to the left of a given value. To find cumulative probabilities, we rely on the standard normal distribution table, often known as the Z-table. Every Z-score corresponds to a cumulative probability, showing the proportion of values lying below it. For a Z-score of 1.0923, for instance, the cumulative probability is approximately 0.8621. This means about 86.21% of the data lies below this value. Once we have the cumulative probabilities, we can easily find:
  • Percentages more or less than a certain Z-score by subtracting the cumulative probability from 1 (or 100% for percentages).
  • Probabilities between two data points by calculating the difference between their cumulative probabilities.
Employing these techniques allows us to solve real-world problems, such as determining what percentage of tax refunds fall beyond specific cut-off points or within a particular range, as demonstrated in the exercise.

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Most popular questions from this chapter

According to the Insurance Institute of America, a family of four spends between \(\$ 400\) and \(\$ 3,800\) per year on all types of insurance. Suppose the money spent is uniformly distributed between these amounts. a. What is the mean amount spent on insurance? b. What is the standard deviation of the amount spent? c. If we select a family at random, what is the probability they spend less than \(\$ 2,000\) per year on insurance per year? d. What is the probability a family spends more than \(\$ 3,000\) per year?

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?

The net sales and the number of employees for aluminum fabricators with similar characteristics are organized into frequency distributions. Both are normally distributed. For the net sales, the mean is \(\$ 180\) million and the standard deviation is \(\$ 25\) million. For the number of employees, the mean is 1,500 and the standard deviation is \(120 .\) Clarion Fabricators had sales of \(\$ 170\) million and 1,850 employees. a. Convert Clarion's sales and number of employees to \(z\) values. b. Locate the two \(z\) values. c. Compare Clarion's sales and number of employees with those of the other fabricators.

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.

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?

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