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Chapter 17: Problem 8

The sale prices (in \(\$ 1000\) s) for eight houses on a certain road are: \(\$ 820, \$ 930, \$ 780, \$ 950, \$ 3540, \$ 680, \$ 920,\) \(\$ 900 .\) Find the mean and standard deviation of the (a) Eight houses. (b) Seven similar houses (leave out the top-priced house).

Short Answer

Expert verified
Answer: (a) The mean and standard deviation for all eight houses are 1065 and 634.77 respectively. (b) For the seven similar houses excluding the highest-priced house, the mean and standard deviation are 710 and 504.48 respectively.

Step by step solution

01

Calculate the mean for all eight houses

First, let's add the prices of all eight houses: \(820 + 930 + 780 + 950 + 3540 + 680 + 920 + 900 = 8520.\) Now, divide the total by the number of houses (8): Mean = \(\frac{8520}{8} = 1065.\)
02

Calculate the mean for the seven similar houses

First, remove the highest-priced house ($3540) from the total prices, and then add the remaining seven prices: \(820 + 930 + 780 + 950 + 680 + 920 + 900 = 4980.\) Now, divide the total by the number of houses (7): Mean = \(\frac{4980}{7} = 710.\)
03

Calculate the standard deviation for all eight houses

The formula for the standard deviation is: Standard Deviation = \(\sqrt{\frac{\sum(x_i - \text{Mean})^2}{N}}.\) Now, subtract the mean from each house's price, square the result, and add all these squared differences together: SD = \(\sqrt{\frac{(820-1065)^2+(930-1065)^2+\cdots+(900-1065)^2}{8}}.\) SD = \(\sqrt{\frac{60025+18225+...11449}}{8}}.\) SD = \(\sqrt{\frac{3226824}{8}}= 634.77.\)
04

Calculate the standard deviation for the seven similar houses

Now, we will find the standard deviation for the seven similar houses. First, subtract the mean from each house's price, square the result, and add all these squared differences together: SD = \(\sqrt{\frac{(820-710)^2+(930-710)^2+\cdots+(900-710)^2}{7}}.\) SD = \(\sqrt{\frac{12100+48400+...36100}}{7}}.\) SD = \(\sqrt{\frac{1786010}{7}}= 504.48.\) Thus, the mean and standard deviation of the prices for (a) all eight houses are 1065 and 634.77 respectively, while (b) for the seven similar houses, they are 710 and 504.48 respectively.

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Key Concepts

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

Descriptive Statistics
Descriptive statistics is a branch of statistics that deals with the quantitative description and summary of data. It helps us make sense of the raw data by converting it into useful information. In this exercise, descriptive statistics involve computing the mean and standard deviation.
  • The **mean**, often referred to as the average, is a measure that summarizes a set of numbers by a single value. It is found by adding all the numbers together and then dividing by the total count. For the exercise, the mean gives an idea of the average house price.

  • The **standard deviation** is a statistic that measures the dispersion or spread of a set of values. A low standard deviation indicates that the values tend to be close to the mean, whereas a high standard deviation indicates a wider range of values.
In the given solutions, we calculated both these measures for all eight houses and again for seven houses, excluding the highest-priced house. This helps in understanding how one extremely high-priced house can affect the overall statistics.
Data Analysis
Data analysis is the process of inspecting, cleaning, and modeling data with the aim of discovering useful information. In our context, data analysis allows us to look beyond simple numbers and uncover insights, such as trends in house prices.
  • Calculating the **mean** and **standard deviation** is part of preliminary data analysis. These statistics help summarize the data set and provide an overview of the "average" and the "spread" of the house prices.

  • By excluding the outlier, a particularly high-priced property, we perform a simple yet effective analytical step that demonstrates the concept of robust statistics. This gives us a better understanding of the underlying distribution of more typical house prices.
With this exercise, we clearly see how data analysis can reveal significant patterns, such as how outliers can skew statistical measures like the mean and standard deviation.
House Prices
House prices often serve as an excellent example for teaching statistical concepts because they can vary significantly depending on many factors. The exercise addresses the influence of a single, much higher price on the overall statistics.
  • The **mean house price** calculated from all eight house prices shows the effect of having one very high price in the dataset. At $1065k, the mean is significantly lifted by the single expensive property valued at $3540k.

  • After removing the highest-priced house, the mean of the remaining seven houses drops to $710k, illustrating how the typical house price appears much lower when not skewed by an outlier.
This example highlights the importance of carefully analyzing property data, considering factors like location, property size, and economic conditions, all of which can impact the overall statistics encountered when studying house prices.
Outliers
Outliers are data points that differ significantly from other observations in the dataset. They can heavily influence some statistical measures, potentially leading to misleading results.
  • In this exercise, the house priced at $3540k is an outlier because it is substantially higher than the other prices in the dataset. As a result, it raises the overall mean, skewing the data.

  • To mitigate the effect of outliers, statistical analysis often involves calculating separate statistics excluding these anomalies. This gives a more accurate representation of the typical dataset without the influence of extreme values.
Understanding outliers is crucial in data analysis as they can reveal interesting patterns or errors in the data. However, they must be treated with caution to ensure that they reflect genuine data phenomena rather than data collection or input errors.

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