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Refer to the Real Estate data, which report information on homes sold in the Denver, Colorado, area during the last year. a. The mean selling price (in \(\$$ thousands) of the homes was computed earlier to be \)\$ 221.10,\( with a standard deviation of \)\$ 47.11 .\( Use the normal distribution to estimate the percent of homes selling for more than \)\$ 280.0 .$ Compare this to the actual results. Does the normal distribution yield a good approximation of the actual results? b. The mean distance from the center of the city is 14.629 miles with a standard deviation of 4.874 miles. Use the normal distribution to estimate the number of homes 18 or more miles but less than 22 miles from the center of the city. Compare this to the actual results. Does the normal distribution yield a good approximation of the actual results?

Short Answer

Expert verified
a. About 10.56% of homes sell for more than $280,000. b. About 17.96% of homes are between 18 and 22 miles from the city center.

Step by step solution

01

Convert Price to Z-Score for Part a

To find the percentage of homes selling for more than \(\$280.0\), we first convert the selling price to a z-score. We use the formula for z-score: \[ z = \frac{X - \mu}{\sigma} \]where \(X = 280.0\), \(\mu = 221.10\), and \(\sigma = 47.11\). Calculating this gives:\[ z = \frac{280.0 - 221.10}{47.11} \approx 1.25 \]
02

Find Probability for Part a

Using the z-score of 1.25, we refer to the standard normal distribution table to find the probability that corresponds to a z-score of 1.25, which is about 0.8944. This probability represents the area to the left of the z-score.
03

Calculate Percentage for Part a

To determine the percentage of homes selling for more than \(\\(280.0\), subtract the probability 0.8944 from 1:\[ 1 - 0.8944 = 0.1056 \]Thus, approximately 10.56% of homes are estimated to sell for more than \(\\)280.0\).
04

Convert Distance Values to Z-Scores for Part b

For homes 18 or more miles but less than 22 miles from the city center, we convert these distances to z-scores using:\[ z_{18} = \frac{18 - 14.629}{4.874} \approx 0.69 \]\[ z_{22} = \frac{22 - 14.629}{4.874} \approx 1.51 \]
05

Find Probabilities for Part b

Refer to the standard normal distribution table for the probabilities corresponding to z-scores 0.69 and 1.51. The probability for 0.69 is about 0.7549 and for 1.51 is about 0.9345.
06

Calculate Probability Range for Part b

The percentage of homes between 18 and less than 22 miles is found by subtracting the probability of 0.7549 from 0.9345:\[ 0.9345 - 0.7549 = 0.1796 \]This indicates approximately 17.96% of homes are estimated to lie in this range.
07

Compare to Actual Results

Compare the estimated percentages (10.56% and 17.96%) to the actual distribution of homes from the data. If the percentages are close, the normal distribution yields a good approximation. Evaluate how the estimates match the actual data to determine accuracy.

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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 helpful tool for determining how far an individual data point is from the mean in terms of standard deviations. It's like a translator, turning raw scores into a standardized form so we can use the normal distribution table. The formula to calculate a z-score is \( z = \frac{X - \mu}{\sigma} \), where \(X\) is the data value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Imagine you want to know how well a selling price or distance is aligned with the average values. If the z-score is positive, the data point is above the mean; if negative, it's below. For example, if a house price in our data has a z-score of 1.25, it means the price is 1.25 standard deviations above the average price. Comparing z-scores across different datasets makes it easy to see how data varies from the expected norms.
standard deviation
Standard deviation is a key measure in statistics, providing insight into data spread or variability. It tells how much the individual data points deviate from the mean. A small standard deviation means data points are close to the mean, while a large one indicates more spread.

The formula is given by \( \sigma = \sqrt{\frac{1}{N}\sum_{i=1}^{N}(X_i - \mu)^2} \), where \(N\) is the number of data points, \(X_i\) are the values, and \(\mu\) is the mean.
- **High variability**: Wider range of values, hence a larger standard deviation.- **Low variability**: Data points clustered around the mean implies a smaller standard deviation.

In the context of real estate, knowing this variability helps predict future pricing and identify unusual selling prices or property distances from the city center. It’s like the compass in understanding how much the properties differ from the expected average.
probability estimation
Probability estimation involves calculating the likelihood of an event occurring within a specific range of the normal distribution. Once you identify the z-score of a particular value, you can use the standard normal distribution table to estimate the probability of a data point falling above or below a certain threshold.

Here's how you might apply this in different scenarios:
  • Estimate how many homes lie within a certain distance range from the city center.
  • Predict the number of homes selling for over a set price point.

By looking at the area under the curve of a normal distribution graph, we can infer probabilities. In Part a of our exercise, finding homes selling for over $280,000 involves determining the area to the right of the corresponding z-score on the graph. Subtracting the area to the left (found through the z-score probability) from 1 gives the sought-after probability.

Therefore, probability estimation is about harnessing the power of the normal distribution to make educated guesses about data behavior and tendencies.

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Most popular questions from this chapter

A recent study of the hourly wages of maintenance crew members for major airlines showed that the mean hourly salary was \(\$ 20.50,\) with a standard deviation of \(\$ 3.50 .\) Assume the distribution of hourly wages follows the normal probability distribution. If we select a crew member at random, what is the probability the crew member earns: a. Between \(\$ 20.50\) and \(\$ 24.00\) per hour? b. More than \(\$ 24.00\) per hour? c. Less than \(\$ 19.00\) per hour?

A normal distribution has a mean of 80 and a standard deviation of 14. Determine the value above which 80 percent of the values will occur.

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.

List the major characteristics of a normal probability distribution.

Shaver Manufacturing, Inc., offers dental insurance to its employees. A recent study by the human resource director shows the annual cost per employee per year followed the normal probability distribution, with a mean of \(\$ 1,280\) and a standard deviation of \(\$ 420\) per year. a. What fraction of the employees cost more than \(\$ 1,500\) per year for dental expenses? b. What fraction of the employees cost between \(\$ 1,500\) and \(\$ 2,000\) per year? c. Estimate the percent that did not have any dental expense. d. What was the cost for the 10 percent of employees who incurred the highest dental expense?

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