Problem 33 - Calculus Early Transcendentals: Pearson New International Edition

Question

In statistics we define the mean \(\bar{x}\) and the variance \(s^{2}\) of a sequence of numbers \(x_{1}, x_{2}, \ldots, x_{n}\) by $$ \bar{x}=\frac{1}{n} \sum_{i=1}^{n} x_{i}, \quad s^{2}=\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2} $$ Find \(\bar{x}\) and \(s^{2}\) for the sequence of numbers \(2,5,7,8,9,10,14\).

Step-by-Step Solution

Answer

\(\bar{x} \approx 7.857\), \(s^2 \approx 12.410\)

Step 1: Calculate the Mean

First, find the total number of observations, \( n \). Here, the sequence has 7 numbers: \( n = 7 \). To find the mean, \( \bar{x} \), add all the numbers in the sequence and divide by \( n \). The sum of the sequence is:\[2 + 5 + 7 + 8 + 9 + 10 + 14 = 55\]Now, divide the sum by \( n \) to find the mean:\[\bar{x} = \frac{55}{7} \approx 7.857\]

Step 2: Calculate Each Deviation from the Mean

Calculate the deviation of each number from the mean, \( \bar{x} \). This involves subtracting the mean from each number in the sequence:- \( 2 - 7.857 = -5.857 \)- \( 5 - 7.857 = -2.857 \)- \( 7 - 7.857 = -0.857 \)- \( 8 - 7.857 = 0.143 \)- \( 9 - 7.857 = 1.143 \)- \( 10 - 7.857 = 2.143 \)- \( 14 - 7.857 = 6.143 \)

Step 3: Square Each Deviation

Square each deviation calculated in the previous step to eliminate negative values:- \((-5.857)^2 = 34.321 \)- \((-2.857)^2 = 8.163 \)- \((-0.857)^2 = 0.735 \)- \((0.143)^2 = 0.020 \)- \((1.143)^2 = 1.307 \)- \((2.143)^2 = 4.592 \)- \((6.143)^2 = 37.735 \)

Step 4: Calculate the Variance

Find the variance, \( s^2 \), by taking the average of the squared deviations. Add up all the squared deviations:\[34.321 + 8.163 + 0.735 + 0.020 + 1.307 + 4.592 + 37.735 = 86.873\]Now divide by \( n \):\[s^2 = \frac{86.873}{7} \approx 12.410\]

Key Concepts

Understanding the Mean

In statistics, the mean, often referred to as the average, is a measure that represents the central value of a data set. To calculate the mean, you add up all the numbers in your sequence and then divide by the total count of numbers. This gives an idea of the overall trend or central point within your data.

The mean is beneficial because:

• It provides a simple, single value representing an entire data set.
• It's a measure everybody can relate to, often used in everyday contexts like calculating average scores or average spending.

In our exercise, we started by calculating the mean for the sequence of numbers: 2, 5, 7, 8, 9, 10, and 14. By adding these numbers together, we got a total of 55. Dividing 55 by the number of items in the sequence, which is 7, gives us a mean of approximately 7.857. This shows that on average, the values in the sequence linger around this number. Understanding the mean helps to summarize large data sets effectively.

Delving into Variance

Variance is a statistical concept that measures how much the numbers in a data set differ from the mean. Unlike the mean, which provides a central value, the variance helps us understand the spread or dispersion of the data points.

Calculating variance involves several steps:

• First, find the deviation of each number from the mean. Deviations are the difference between each data point and the mean.
• Next, square each of these deviations. Squaring ensures all values are positive, and it amplifies larger deviations, emphasizing greater dispersions.
• Finally, find the average of these squared deviations by summing them up and dividing by the number of data points.

In our exercise, the squared deviations were totaled to 86.873 and divided by 7 to obtain a variance of approximately 12.410. This value indicates the degree of variation around the mean. A high variance suggests a wide spread of numbers, while a low variance indicates they are tightly clustered around the mean.

Exploring Standard Deviation

Standard deviation is closely related to the variance; it's the square root of the variance. It provides a more interpretable measure of spread in the same units as the original data, making it easier to understand at a glance.

Some key points about standard deviation include:

• It's widely used to determine how much variation exists from the average (mean).
• It helps in comparing different data sets that might have the same mean but different spreads.

Though the original exercise focused on calculating variance, knowing that the standard deviation is simply the square root of variance makes it straightforward to find. For example, given our variance of 12.410, the standard deviation would be approximately \(\sqrt{12.410} \approx 3.52\). This tells us that, on average, the values in our data set vary about 3.52 units from the mean. Understanding standard deviation is crucial for a deeper insight into the data's variability.