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To calculate the mean of a set of numbers, you first need to find the total sum of those numbers. Let's consider the sample values: 1.3, 7.0, 3.6, 4.1, and 5.0. Add them together to get the sum:

• 1.3 + 7.0 + 3.6 + 4.1 + 5.0 = 21.0

The next step is to divide this sum by the total number of values, which is 5 in this case:

• Mean, \( \bar{x} = \frac{21.0}{5} = 4.2 \)

Therefore, the mean \( \bar{x} \) is 4.2. The mean of a data set is a central value that represents the data, providing an average or midpoint around which the data values are spread. This central tendency gives you an idea of what a typical value might be in your data set.

Once you have the mean, you can calculate each data point’s deviation from the mean. Deviation indicates how much a specific value differs from this average value. Let's find the deviation for each of our sample values by subtracting the mean (4.2) from each value:

• 1.3 - 4.2 = -2.9
• 7.0 - 4.2 = 2.8
• 3.6 - 4.2 = -0.6
• 4.1 - 4.2 = -0.1
• 5.0 - 4.2 = 0.8

These deviations show how much each value strays from the mean. Some are negative, meaning the actual data point is less than the mean. Others are positive, indicating it is more than the mean. Recognizing these deviations is essential for understanding the variability within your data.

An interesting property of deviations from the mean lies in their summation. If you add up all of these individual deviations, the result is always zero. This is because the deviations above the mean (positive) and below the mean (negative) balance each other out. Let’s add our deviations to illustrate this:

• -2.9 + 2.8 - 0.6 - 0.1 + 0.8 = 0

The sum, as expected, is zero. This characteristic is a fundamental concept in statistics. It shows that the mean truly is a central point and that deviations are naturally spread around it. Understanding this balance helps in demonstrating why the mean is an important measure of central tendency in statistical analysis. The sum of zero implies there's no total loss or gain when you regard deviations, confirming their balance around the mean.