Calculating Standard Deviation

From Variance to Standard Deviation

Variance summarizes squared distances from the mean. It is useful for measuring spread, but those squared distances are not expressed on the same scale as the observations themselves. Standard deviation brings the summary back to a typical distance on the original measurement scale: it is found by taking the square root of the variance.

The calculation depends on whether the data describe an entire population or a sample. Population variance divides the sum of squared deviations by the population size \(N\). Sample variance uses the \(n-1\) convention. Once the appropriate variance has been calculated, the corresponding standard deviation is its nonnegative square root.

Definition: For a population with mean \(\mu\) and variance \(\sigma^2\), the population standard deviation is \(\sigma=\sqrt{\sigma^2}\). For a sample with mean \(\bar{x}\) and sample variance \(s^2\), the sample standard deviation is \(s=\sqrt{s^2}\). In either case, standard deviation is a measure of the typical distance of observations from their mean, expressed in the original units of the data.

The square root is the nonnegative root. Variance cannot be negative because it is a sum of squared deviations divided by a positive number. Therefore, its square root is a real, nonnegative number. A standard deviation of zero occurs only when every observation is exactly equal to the mean, so there is no spread in the data.

Calculating a Population Standard Deviation

When the data include every member of the population of interest, calculate \(\sigma\) by first finding \(\sigma^2\) and then taking its square root. In the formula, \(x_i\) is the \(i\)th population value, and \(N\) is the number of values in the population.

\sigma^2=\frac{\sum_{i=1}^{N}(x_i-\mu)^2}{N} \qquad\text{and}\qquad \sigma=\sqrt{\sigma^2}

Worked Example: Standard Deviation of a Population

Suppose, hypothetically, that a small population consists of the scores \(1, 3, 5,\) and \(7\) points. Find its population standard deviation.

Step 1: Find the population mean. The population has \(N=4\) values, so

\mu=\frac{1+3+5+7}{4}=\frac{16}{4}=4\text{ points}.

Step 2: Find the population variance. The deviations from the mean are \(-3,-1,1,\) and \(3\) points. Squaring and adding gives \(9+1+1+9=20\). Divide by \(N=4\):

\sigma^2=\frac{(1-4)^2+(3-4)^2+(5-4)^2+(7-4)^2}{4} =\frac{9+1+1+9}{4} =\frac{20}{4}=5.

Step 3: Take the square root.

\sigma=\sqrt{5}=2.236\text{ points}.

The population’s standard deviation is about \(2.236\) points. In this population, scores are typically about \(2.236\) points from their mean of \(4\) points. This is a summary of spread, not a claim that each score is exactly \(2.236\) points away.

Calculating a Sample Standard Deviation

For a sample, the standard deviation is the square root of the sample variance. Under the \(n-1\) convention, the sample variance is calculated by summing squared deviations from the sample mean and dividing by \(n-1\). Do not divide by \(n\) when calculating \(s^2\) under this convention; once the correct sample variance is found, take its square root.

s^2=\frac{\sum_{i=1}^{n}(x_i-\bar{x})^2}{n-1} \qquad\text{and}\qquad s=\sqrt{s^2}

Worked Example: Standard Deviation of a Sample

Suppose a hypothetical sample contains four quiz scores: \(4, 6, 8,\) and \(10\) points. Find the sample standard deviation using the \(n-1\) convention.

Step 1: Calculate the sample mean.

\bar{x}=\frac{4+6+8+10}{4}=\frac{28}{4}=7\text{ points}.

Step 2: Calculate the sample variance. The deviations from \(7\) are \(-3,-1,1,\) and \(3\), so the sum of squared deviations is \(9+1+1+9=20\). Since \(n=4\), divide by \(n-1=3\):

s^2=\frac{(4-7)^2+(6-7)^2+(8-7)^2+(10-7)^2}{4-1} =\frac{9+1+1+9}{3} =\frac{20}{3}=6.6667.

Step 3: Take the square root of the sample variance.

s=\sqrt{\frac{20}{3}}=\sqrt{6.6667}\approx2.582\text{ points}.

The sample standard deviation is about \(2.582\) points. In this sample, scores are typically about \(2.582\) points from the sample mean of \(7\) points. The sample variance’s denominator is \(3\), not \(4\); taking a square root does not change which variance formula is used.

Why the Square Root Gives a Typical Distance

A standard deviation is built from distances to the mean, but it is not calculated by simply averaging the absolute distances. The variance squares each deviation before combining them. Taking the square root of the variance returns the summary to the original measurement scale, while retaining the effect of the squared deviations used in the variance calculation.

The calculation can be seen directly from the variance formula. For a population, the variance is the average of the squared deviations \(x_i-\mu\); its square root is therefore the square root of that average. For a sample under the \(n-1\) convention, the same reasoning applies with deviations from \(\bar{x}\) and divisor \(n-1\). In both cases, the result summarizes the scale of the deviations around the mean.

\sqrt{\frac{\sum_{i=1}^{N}(x_i-\mu)^2}{N}} =\sigma \qquad\text{and}\qquad \sqrt{\frac{\sum_{i=1}^{n}(x_i-\bar{x})^2}{n-1}} =s

This is why standard deviation is commonly interpreted as a typical distance from the mean. “Typical” is an interpretation of the overall spread, not a promise that every observation lies that distance from the mean. Some values can be closer and others farther away. The standard deviation also is not, in general, the average of the absolute deviations.

Worked Example: Interpreting a Standard Deviation in Context

A hypothetical sample of five package-delivery times, measured in minutes, is \(3, 5, 6, 8,\) and \(8\). Calculate its sample standard deviation and interpret it.

Step 1: Find the sample mean.

\bar{x}=\frac{3+5+6+8+8}{5}=\frac{30}{5}=6\text{ minutes}.

Step 2: Calculate the sample variance. The deviations from \(6\) are \(-3,-1,0,2,\) and \(2\). Their squared values add to \(9+1+0+4+4=18\). With \(n=5\), divide by \(n-1=4\):

s^2=\frac{(3-6)^2+(5-6)^2+(6-6)^2+(8-6)^2+(8-6)^2}{5-1} =\frac{18}{4}=4.5.

Step 3: Take the square root.

s=\sqrt{4.5}\approx2.121\text{ minutes}.

The sample standard deviation is about \(2.121\) minutes. The delivery times in this sample are typically about \(2.121\) minutes from their mean of \(6\) minutes. The actual absolute distances are \(3,1,0,2,\) and \(2\) minutes, so they are not all equal to the standard deviation. Their average absolute distance is \(8/5=1.6\) minutes, which is a different summary.

A Reliable Calculation Process

A careful order of operations prevents common errors. Keep intermediate values unrounded where possible, especially the variance, and round the final standard deviation to a sensible number of significant figures.

Identify the data scope.
Decide whether the values represent a complete population or a sample. This determines whether to use \(N\) or \(n-1\) in the variance calculation.

Find the appropriate mean and variance.
Use \(\mu\) and \(\sigma^2\) for a population, or \(\bar{x}\) and \(s^2\) for a sample. Follow the corresponding variance formula.

Take the nonnegative square root.
Calculate \(\sqrt{\sigma^2}\) or \(\sqrt{s^2}\), retaining precision until the final result.

Interpret in context.
State the standard deviation with the original measurement units and describe it as a typical distance from the relevant mean.

Common Mistakes and Why This Summary Matters

Standard deviation condenses the spread around the mean into a single number that can be described on the data’s original scale. That makes the result easier to interpret in context than a variance alone. Its meaning still depends on the data being summarized: a standard deviation describes spread around the mean, so it should be reported with the variable and units rather than as a context-free number.

Using variance as if it were standard deviation. The variance is not the final answer when standard deviation is requested. Take its square root.
Using the wrong variance denominator. For a population, divide the squared-deviation sum by \(N\). For a sample using the \(n-1\) convention, divide by \(n-1\) before taking the square root.
Taking a square root too early. Find the variance from the complete sum of squared deviations, then take its square root. Rounding intermediate calculations too aggressively can slightly change the final result.
Calling it the exact average distance. Standard deviation is a typical distance summary based on squared deviations. It is not generally the arithmetic mean of the absolute distances from the mean.
Leaving out context. A result such as \(2.121\) does not explain what is varying. State, for example, that delivery times in the sample are typically about \(2.121\) minutes from their mean.

Key takeaway: Standard deviation is the nonnegative square root of variance: \(\sigma=\sqrt{\sigma^2}\) for a population and \(s=\sqrt{s^2}\) for a sample. Interpret it as a typical distance from the mean, in the original units of the observations.

Check Your Understanding

Try these questions before reviewing the worked examples.

A complete population has variance \(16\) square units. What is its population standard deviation, and how would you describe it?
A sample variance calculated using the \(n-1\) convention is \(12.25\). Find the sample standard deviation.
Explain why a sample variance calculation using the \(n-1\) convention does not divide by \(n\) before taking the square root.
In a hypothetical sample of temperatures measured in degrees Celsius, the sample standard deviation is \(3.4\). Write an interpretation in context, without claiming every observation is exactly \(3.4\) degrees from the mean.
Why is standard deviation not generally the same as the mean absolute distance from the mean?

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