Problem 20 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

A study of the records of 85,000 apartment units in the greater Boston area revealed the following data: $$\begin{array}{llllll}\hline \text { Year } & 2002 & 2003 & 2004 & 2005 & 2006 \\ \hline \text { Average Rent, } \$ & 1352 & 1336 & 1317 & 1308 & 1355 \\\\\hline\end{array}$$ Find the average of the average rent for the \(5 \mathrm{yr}\) in question. What is the standard deviation for these data?

Step-by-Step Solution

Answer

The average of the average rents for the 5-year period is 1333.6, and the standard deviation is approximately 20.82.

Step 1: Find the mean of the average rents

To find the mean of the average rents for the given data, add up all the average rents for each year, and then divide by the total number of years. Mean = \(\frac{1352+1336+1317+1308+1355}{5}\)

Step 2: Calculate the mean

Mean = \(\frac{1352+1336+1317+1308+1355}{5}\) = \(\frac{6668}{5} \) = 1333.6

Step 3: Calculate the deviations from the mean

Next, we will calculate the deviations from the mean for each year by subtracting the mean from each average rent: \(x_1 - \mu = 1352 - 1333.6 = 18.4\) \(x_2 - \mu = 1336 - 1333.6 = 2.4\) \(x_3 - \mu = 1317 - 1333.6 = -16.6\) \(x_4 - \mu = 1308 - 1333.6 = -25.6\) \(x_5 - \mu = 1355 - 1333.6 = 21.4\)

Step 4: Calculate the square of deviations

Now, we will calculate the square of deviations from the mean for each year: \((x_1 - \mu)^2 = (18.4)^2 = 338.56\) \((x_2 - \mu)^2 = (2.4)^2 = 5.76\) \((x_3 - \mu)^2 = (-16.6)^2 = 275.56\) \((x_4 - \mu)^2 = (-25.6)^2 = 655.36\) \((x_5 - \mu)^2 = (21.4)^2 = 457.96\)

Step 5: Calculate the sum of squared deviations

Add up all squared deviations: Sum of squared deviations = \(338.56 + 5.76 + 275.56 + 655.36 + 457.96 = 1733.20\)

Step 6: Calculate the standard deviation

Standard deviation = \(\sqrt{\frac{1}{N-1}\sum_{i=1}^N (x_i - \mu)^2}\) Standard deviation = \(\sqrt{\frac{1}{5-1}\times 1733.20} = \sqrt{\frac{1}{4}\times 1733.20} = \sqrt{433.30}\) Standard deviation ≈ 20.82 The average of the average rents for the 5-year period is 1333.6, and the standard deviation is approximately 20.82.

Key Concepts

Mean Calculation

When analyzing data, the mean provides a quick snapshot of what is typical. To find the mean of a set of numbers, you add them together and then divide by how many numbers there are. In the exercise, you add up each year's average rent—1352, 1336, 1317, 1308, and 1355—and then divide this total by 5. This operation results in a mean average rent of 1333.6 dollars for the 5 years.

• Add together the total rents: 1352 + 1336 + 1317 + 1308 + 1355 = 6668
• Divide this sum by the number of years: \(\frac{6668}{5} = 1333.6\)

This mean calculation is important because it simplifies the dataset into one easy-to-read figure, making it easier to quickly understand the general trend of the rents over these years.

Standard Deviation

The standard deviation is a critical statistic that tells us how much variation, or "spread," there is in a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates a wider spread.
For the rents data in question, after calculating the mean (1333.6), we find how far each rent is from that mean. We square each of these distances (deviations) to remove negative numbers and then find their average by dividing the sum by the number of values minus one, which applies Bessel's correction for sample data. Finally, we take the square root to bring it back to the original unit (dollars):

• Calculate each deviation: 1352 - 1333.6 = 18.4, and so on
• Square each deviation: \((18.4)^2 = 338.56\) and so forth
• Sum these squared deviations and divide: \(\sqrt{\frac{1733.20}{4}} = \sqrt{433.30}\)
• The resulting standard deviation is about 20.82 dollars.

This standard deviation helps to understand how consistent the rent values are relative to their mean value.

Data Analysis

Data analysis involves breaking down complex sets of data into smaller, understandable chunks to discern patterns and insights. In this exercise, we are analyzing the average rent over a 5-year period—specifically looking at how average rents change annually. By calculating both the mean and the standard deviation, we can derive trends and variability in rent values over time.

Insights from Mean and Standard Deviation

• The mean gives us a single figure summarizing the central tendency of the rent data.
• The standard deviation provides a sense of the spread and gives context to whether rents were stable or varied greatly around the mean.

Yet, mean and standard deviation are only the starting points. They can assist in further analysis such as detecting trends in rental price changes. For instance, a decreasing mean might indicate an overall decline in rent, while a high standard deviation could show that rent prices are volatile. Understanding these concepts allows one to draw deeper insights from rental data or any averaged numerical data set.