Problem 26 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

In a survey, consumers were asked how many television sets they have in their home. The results are summarized in the following table: $$\begin{array}{lccccc}\hline \text { TVs } & 1 & 2 & 3 & 4 & 5 \\\\\hline \text { Respondents, } \% & 13.9 & 26.5 & 28.6 & 14.8 & 16.2 \\\\\hline\end{array}$$\ Find the average number of TVs in the home of the respondents. What is the standard deviation for these data?

Step-by-Step Solution

Answer

The average number of TVs in the home of the respondents is 3.15, and the standard deviation is approximately 1.2351.

Step 1: Calculate the mean

To find the mean (also called the average), we'll multiply each value in the x array by its corresponding proportion in the p array and then sum up the results. Mean = \(\sum (x_i * p_i)\) Mean = 1*(0.139) + 2*(0.265) + 3*(0.286) + 4*(0.148) + 5*(0.162) After calculating the result, we get: Mean = 3.15 So, the average number of TVs in the home of the respondents is 3.15.

Step 2: Calculate the variance

Now, let's find the variance. The variance is the average of the squared differences from the mean. Variance = \(\sum (x_i - \text{mean})^2 * p_i\) Variance = (1-3.15)^2*(0.139) + (2-3.15)^2*(0.265) + (3-3.15)^2*(0.286) + (4-3.15)^2*(0.148) + (5-3.15)^2*(0.162) After calculating the result, we get: Variance = 1.5275

Step 3: Calculate the standard deviation

Finally, let's find the standard deviation, which is simply the square root of the variance. Standard deviation = \(\sqrt{\text{variance}}\) Standard deviation = \(\sqrt{1.5275}\) After calculating the result, we get: Standard deviation ≈ 1.2351 So, the standard deviation for these data is approximately 1.2351. To summarize, the average number of TVs in the home of the respondents is 3.15, and the standard deviation is approximately 1.2351.

Key Concepts

Mean Calculation

Understanding the mean is crucial for analyzing data sets. The mean, often referred to as the average, represents a central value within a set of numbers. To calculate the mean, each data point is multiplied by its probability or frequency, and these products are then summed together.

In the context of our problem, we have a data set representing the number of television sets in homes, along with the proportion of respondents that have each number of TVs. By multiplying each number of TVs (1 through 5) by their corresponding proportions and summing up these products, we arrive at the mean number of TVs:
Mean = \(1 \times 0.139 + 2 \times 0.265 + 3 \times 0.286 + 4 \times 0.148 + 5 \times 0.162 = 3.15\).

This result tells us that, on average, respondents have about 3.15 television sets in their homes. The mean provides a useful summary of the data but does not give us information about the spread or variability of the data.

Variance Calculation

Variance tells us how much the data are spread out from the mean. It is calculated by taking the average of the squared differences between each data point and the mean. Squaring these differences ensures that negative and positive differences do not cancel each other out and emphasizes larger differences.

In the given exercise, the variance is calculated by subtracting the mean from each number of TVs, squaring the results, multiplying by the corresponding percentages, and then summing up those products:Variance = \((1-3.15)^2\times0.139 + (2-3.15)^2\times0.265 + (3-3.15)^2\times0.286 + (4-3.15)^2\times0.148 + (5-3.15)^2\times0.162 = 1.5275\).

This value of 1.5275 indicates the average squared difference from the mean number of TVs, hence providing insight into the variability of television ownership among the respondents.

Standard Deviation

The standard deviation is a widely used measure of the amount of variation or dispersion of a set of values. It is basically the square root of the variance, providing us with a value that is in the same unit as the data. This makes it more interpretable than the variance.

To find the standard deviation for the number of TVs in the respondents' homes, we calculate the square root of the variance:Standard deviation = \(\sqrt{1.5275} \approx 1.2351\).

A lower standard deviation indicates that the data points are closer to the mean, while a higher standard deviation shows that the data are more spread out. In this survey, a standard deviation of approximately 1.2351 implies that the variability in the number of television sets is relatively moderate. When analyzing data sets, the standard deviation can be particularly useful when comparing the spread of different distributions or understanding the relative risk of an investment.