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Calculating the range is a straightforward way to understand the spread of a data set. Imagine you have a set of numbers, and you want to know how spread out they are. The range gives you a quick and simple insight into this by just focusing on two numbers: the largest and the smallest. Simply subtract the smallest number from the largest number to find the range.
In our example data set \(1, 2, 3, 4, 5\), the largest number is 5, and the smallest is 1. Subtracting these, you get:

• Range: \(5 - 1 = 4\)

This tells us that all numbers fall within a span of 4 units. While the range is easy to calculate, it doesn’t tell us how the data points are distributed between the extremes.

Before calculating standard deviation, finding the mean is necessary. The mean, or average, provides a central value that represents the data set. It is calculated by adding all the numbers together and dividing by the total count.
For the data set \(1, 2, 3, 4, 5\), add all the numbers:

• Sum = \(1 + 2 + 3 + 4 + 5 = 15\)

Now, divide by the number of data points, which is 5:

• Mean: \(\frac{15}{5} = 3\)

The mean provides a point of balance for the set, showing where, on average, numbers lie on the number line. This average is essential for calculating standard deviations be it sample or population since it signifies the center of the data.

The sample standard deviation is a measure of how much our data points deviate from the mean, considering that we are working with a sample of a larger population. This is especially important when your dataset is just a small part of a much larger population, and it adjusts the calculation by using \(n - 1\) (unbiased estimator). Let's explore how it's calculated by starting from the mean, which is 3.
For each data point, calculate its deviation from the mean, square it, sum these squared deviations, divide by \(n-1\) (where \(n\) is 5), and finally take the square root. Here’s how it works for \(1, 2, 3, 4, 5\):

• Deviations squared: \((1-3)^2, (2-3)^2, (3-3)^2, (4-3)^2, (5-3)^2\) equals \(4, 1, 0, 1, 4\)
• Summing gives \(10\)
• Divide by \(n-1\): \(\frac{10}{4} = 2.5\)
• Square root: \(s = \sqrt{2.5} \approx 1.58\)

This yields a sample standard deviation of approximately 1.58, suggesting that sample data points are typically 1.58 units away from the mean.

The population standard deviation is useful when your data set represents the whole population. It tells you how much the data points deviate from the mean on average. Unlike the sample standard deviation, the formula uses \(n\) rather than \(n-1\), because we are not estimating from a subset.
Using the same data points \(1, 2, 3, 4, 5\) and mean 3, let's compute this:

• As before, the squared deviations are \((1-3)^2, (2-3)^2, (3-3)^2, (4-3)^2, (5-3)^2\) resulting in \(4, 1, 0, 1, 4\)
• Sum these to get \(10\)
• Divide by \(n\): \(\frac{10}{5} = 2\)
• Square root: \(\sigma = \sqrt{2} \approx 1.41\)

With a population standard deviation of approximately 1.41, we see a slightly smaller deviation than the sample case, accurately reflecting the entire population's variability.