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Suppose the Internal Revenue Service reported that the mean tax refund for the year 2017 was \(\$ 2,800 .\) Assume the standard deviation is \(\$ 450\) and that the amounts refunded follow a normal probability distribution. a. What percent of the refunds are more than \(\$ 3,100 ?\) b. What percent of the refunds are more than \(\$ 3,100\) but less than \(\$ 3,500 ?\) c. What percent of the refunds are more than \(\$ 2,250\) but less than \(\$ 3,500 ?\)

Short Answer

Expert verified
a) 25.14% b) 19.20% c) 82.94%

Step by step solution

01

Understand the Problem

We are given a normal distribution for tax refunds with a mean of \(\\( 2,800\) and a standard deviation of \(\\) 450\). We'll find the percentages for ranges above and between specified amounts.
02

Calculate the Z-Score for $3,100

To find the percentage above \(\$ 3,100\), we calculate the Z-score using the formula: \[Z = \frac{{X - \mu}}{{\sigma}}\]where \(X = 3,100\), \(\mu = 2,800\), and \(\sigma = 450\).\[Z = \frac{{3,100 - 2,800}}{{450}} = \frac{{300}}{{450}} \approx 0.67\]
03

Find the Percentage Above $3,100

Using the Z-score table, find the probability corresponding to \(Z = 0.67\). This is approximately 0.7486. The percentage above \(\$ 3,100\) is:\[100\% - 74.86\% = 25.14\%\]
04

Calculate the Z-Scores for $3,100 and $3,500

Repeat the Z-score calculation for \(X = 3,500\):\[Z = \frac{{3,500 - 2,800}}{{450}} = \frac{{700}}{{450}} \approx 1.56\]We already have \(Z = 0.67\) for \(\$ 3,100\) from Step 2.
05

Find the Percentage Between $3,100 and $3,500

Using the Z-score table:- The probability for \(Z = 1.56\) is approximately 0.9406.- The probability for \(Z = 0.67\) is 0.7486 (from Step 3).The percentage between \(\\( 3,100\) and \(\\) 3,500\) is:\[94.06\% - 74.86\% = 19.20\%\]
06

Calculate the Z-Score for $2,250

Calculate the Z-score for \(X = 2,250\):\[Z = \frac{{2,250 - 2,800}}{{450}} = \frac{{-550}}{{450}} \approx -1.22\]
07

Find the Percentage Between $2,250 and $3,500

Using the Z-score table:- The probability for \(Z = -1.22\) is approximately 0.1112.- The probability for \(Z = 1.56\) is 0.9406 (from Step 5).The percentage between \(\\( 2,250\) and \(\\) 3,500\) is:\[94.06\% - 11.12\% = 82.94\%\]
08

Conclusion

We have calculated the percentages for refunds in these specified ranges using the normal distribution and Z-score method.

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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 when working with normal distributions. It allows us to understand how far a specific value is from the mean of the distribution in terms of standard deviations. To compute the Z-score, we use the formula:
  • \[ Z = \frac{X - \mu}{\sigma} \]
Here, \( X \) is the value for which we want to find the Z-score, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

The Z-score tells us:
  • How many standard deviations a particular value is from the mean.
  • If the Z-score is positive, the value is above the mean; if negative, it is below.
In the context of the IRS tax refund example, calculating the Z-score for amounts like \( \$3,100 \) helps determine the position of these amounts within the normal distribution of all tax refunds.
Standard Deviation
Standard deviation is a key measure in statistics that describes the amount of variation or dispersion in a set of values.
It tells us how much the individual data points typically deviate from the mean of the data set.
A smaller standard deviation indicates that the data points are close to the mean, while a larger one suggests more spread out data.

In a normal distribution:
  • About 68% of data falls within one standard deviation of the mean.
  • Approximately 95% is within two standard deviations.
  • Nearly all (99.7%) lies within three standard deviations.

In our example with tax refunds, the standard deviation is set at \( \\(450\).
This gives us insight into how much individual refunds typically vary from the average refund of \( \\)2,800\).
Understanding standard deviation helps us better grasp the overall spread and reliability of the values within a dataset.
Probability Calculation
Probability calculations using the normal distribution involve finding the likelihood of a particular range of values occurring.
This is where Z-scores and standard deviation play a crucial role.
By calculating Z-scores, we transform specific data values into a standard form, making it easier to find probabilities using a standard normal distribution table or calculator.

To find probabilities:
  • First, calculate the Z-score of the values of interest.
  • Next, use the Z-score to find the corresponding probability from a Z-table.
  • Interpret the result: for values greater than a certain point, subtract the Z-score probability from 1.

In the refund problem, we determined the probabilities of refunds being over \( \\(3,100 \), between \( \\)3,100 \) and \( \\(3,500 \), and between \( \\)2,250 \) and \( \$3,500 \).
These calculations help in making informed predictions about the distribution of tax refunds. It shows how likely different refund amounts are and supports decision-making processes.

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Most popular questions from this chapter

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.

Explain what is meant by this statement: "There is not just one normal probability distribution but a 'family' of them."

A normal population has a mean of 80.0 and a standard deviation of \(14.0 .\) a. Compute the probability of a value between 75.0 and 90.0 . b. Compute the probability of a value of 75.0 or less. c. Compute the probability of a value between 55.0 and 70.0 .

Many retail stores offer their own credit cards. At the time of the credit application, the customer is given a \(10 \%\) discount on the purchase. The time required for the credit application process follows a uniform distribution with the times ranging from 4 minutes to 10 minutes. a. What is the mean time for the application process? b. What is the standard deviation of the process time? c. What is the likelihood a particular application will take less than 6 minutes? d. What is the likelihood an application will take more than 5 minutes?

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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