Variance and Its Relationship to Standard Deviation

Variance: The Quantity Behind Standard Deviation

In Interpreting Standard Deviation Correctly, you learned that standard deviation describes a typical distance of observations from their mean and is reported in the same units as the data. This tutorial looks at the calculation beneath that interpretation: variance. Variance is built from squared deviations, and standard deviation is the square root of variance.

The key trade-off is straightforward. Squaring deviations makes variance useful as a measure of spread and a building block for later statistical methods, but it gives variance squared units. Taking the square root returns the measure to the units of the original data, which is usually more natural to explain in context.

Definition: Variance summarizes the spread of observations around their mean by combining their squared deviations from that mean. Sample variance is written \(s^2\), and population variance is written \(\sigma^2\).

The superscript \(2\) in \(s^2\) and \(\sigma^2\) is part of the notation and also reminds us that the units are squared. For a sample, divide the sum of squared deviations by \(n-1\). For a population, divide by the population size \(N\). Standard deviation is the square root of the corresponding variance.

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

s=\sqrt{s^2} \qquad\text{and}\qquad \sigma=\sqrt{\sigma^2}

These formulas distinguish the sample from the population just as \(s\) and \(\sigma\) did in Standard Deviation Using a Calculator. Use \(s^2\) when the observations are a sample and \(\sigma^2\) when they are the entire population of interest. In particular, sample variance uses \(n-1\), not \(n\).

Why Are the Units Squared?

Suppose a sample records distances in centimeters. Each deviation \(x-\bar{x}\) is in centimeters. Squaring a deviation gives square centimeters, and adding those squared deviations still gives square centimeters. Dividing by the unit-free count \(n-1\) leaves the sample variance in square centimeters. Taking the square root gives the sample standard deviation in centimeters.

The same reasoning applies to other units. If observations are measured in minutes, variance is in minutes squared and standard deviation is in minutes. If the observations are weights in grams, variance is in grams squared and standard deviation is in grams. The variance’s units are not a mistake: they follow directly from squaring the deviations.

Key relationship: Variance is in squared units because it is calculated from squared deviations. Standard deviation is the square root of variance, so it returns to the original units of the observations.

Standard deviation is therefore usually easier to interpret in a sentence about the data. A standard deviation of 2.1 centimeters can describe a typical distance from the mean in centimeters. A variance of 4.3 square centimeters is mathematically meaningful, but it does not describe a typical distance of 4.3 centimeters. Keep the two quantities distinct: variance is a squared-spread summary; standard deviation is the square-root measure in the original units.

Variance is not simply the ordinary average of the squared deviations in every setting. For a population, the formula divides by \(N\), so it is the mean of the population’s squared deviations from \(\mu\). For a sample, the AP Statistics sample variance formula divides by \(n-1\). That denominator is part of the sample-variance convention.

Worked Examples: Calculating and Interpreting Variance

Worked Example: Leaf Lengths in a Sample

A fictional sample of four leaves has lengths of 5, 7, 8, and 10 centimeters. Find the sample variance and sample standard deviation. Explain which measure is easier to interpret in context.

Find the sample mean. The lengths sum to \(5+7+8+10=30\) centimeters. Thus, \(\bar{x}=30/4=7.5\) centimeters.

Find the squared deviations. The deviations from 7.5 are \(-2.5,-0.5,0.5,\) and \(2.5\) centimeters. Their squares are \(6.25,0.25,0.25,\) and \(6.25\) square centimeters, which sum to \(13\) square centimeters.

Calculate the sample variance and standard deviation. The sample size is \(n=4\), so the denominator is \(n-1=3\):

s^2=\frac{13}{4-1}=\frac{13}{3}\approx4.333\text{ cm}^2 \qquad s=\sqrt{\frac{13}{3}}\approx2.082\text{ cm}

Check the squared-deviation sum another way. The squared lengths sum to \(5^2+7^2+8^2+10^2=238\). Subtract \(n\bar{x}^2=4(7.5^2)=225\), giving \(238-225=13\), the same squared-deviation sum. Dividing by 3 gives \(s^2=13/3\approx4.333\text{ cm}^2\); taking its square root gives \(s\approx2.082\text{ cm}\).

Interpret the two summaries. The sample variance of leaf lengths is about \(4.333\) square centimeters. That reports spread in squared units, so it is not a typical distance of 4.333 centimeters. The sample standard deviation is about \(2.082\) centimeters. In context, the sampled leaf lengths typically differ from their sample mean of 7.5 centimeters by about 2.082 centimeters.

Worked Example: A Population of Package Weights

A fictional small workshop checks every package in one batch. The four package weights are 4, 6, 8, and 10 grams. Treating these four packages as the entire population of interest, find the population variance and population standard deviation.

State the population mean. The weights sum to \(4+6+8+10=28\) grams. Therefore, \(\mu=28/4=7\) grams.

Calculate deviations and squares. The deviations from 7 grams are \(-3,-1,1,\) and \(3\) grams. The squared deviations are \(9,1,1,\) and \(9\) square grams, with total \(20\) square grams.

Use the population denominator. Since the list includes all \(N=4\) packages in this population, divide by 4, not by \(4-1\):

\sigma^2=\frac{20}{4}=5\text{ g}^2 \qquad \sigma=\sqrt{5}\approx2.236\text{ g}

Check the squared-deviation sum another way. The squared weights sum to \(4^2+6^2+8^2+10^2=216\). Subtract \(N\mu^2=4(7^2)=196\), giving \(216-196=20\). Thus, \(\sigma^2=20/4=5\text{ g}^2\) and \(\sigma=\sqrt{5}\approx2.236\text{ g}\), matching the direct calculation.

Interpret in context. The population variance is 5 square grams, while the population standard deviation is about 2.236 grams. The standard deviation is easier to describe: these package weights typically differ from their population mean of 7 grams by about 2.236 grams. The variance records spread too, but its squared units make it less direct as a description of package weights.

Worked Example: Changing Meters to Centimeters

Three fictional trail segments have lengths of 2, 3, and 4 meters. Find the sample variance and standard deviation in meters, then express the same results in centimeters. What happens to the numerical value of each measure?

Calculate in meters. The sample mean is \(\bar{x}=(2+3+4)/3=3\) meters. The deviations are \(-1,0,\) and \(1\) meter, so the squared deviations sum to \(1+0+1=2\) square meters. With \(n-1=2\):

s^2=\frac{2}{2}=1\text{ m}^2 \qquad s=\sqrt{1}=1\text{ m}

Check in meters another way. The squared lengths sum to \(2^2+3^2+4^2=29\), and \(n\bar{x}^2=3(3^2)=27\). Their difference is \(29-27=2\), so \(s^2=2/2=1\text{ m}^2\) and \(s=1\text{ m}\).

Convert the observations to centimeters. The lengths are 200, 300, and 400 centimeters, with mean 300 centimeters. Their deviations are \(-100,0,\) and \(100\) centimeters, whose squares sum to \(20{,}000\) square centimeters. Therefore \(s^2=20{,}000/2=10{,}000\text{ cm}^2\), and \(s=\sqrt{10{,}000}=100\text{ cm}\).

Check in centimeters another way. The squared lengths sum to \(200^2+300^2+400^2=290{,}000\). Subtract \(n\bar{x}^2=3(300^2)=270{,}000\), giving \(20{,}000\). This confirms \(s^2=10{,}000\text{ cm}^2\) and \(s=100\text{ cm}\).

Compare the units and values. The standard deviation is the same distance described in different units: \(1\) meter equals \(100\) centimeters. The variance converts by the square of the unit conversion: \(1\text{ m}^2=10{,}000\text{ cm}^2\). This example shows why it is essential to report units and why standard deviation’s original-unit scale is often more intuitive.

Worked Example: How an Extreme Value Affects Variance

A fictional sample of three sensor readings is 10, 11, and 12 units. A fourth reading, 30 units, is later verified and added to the sample. Calculate the sample variance before and after adding it, and explain the change.

Before adding 30. The mean is \((10+11+12)/3=11\) units. The squared deviations are \(1,0,\) and \(1\), summing to 2. Thus \(s^2=2/(3-1)=1\text{ unit}^2\), and \(s=1\) unit.

Check the original calculation. The squared readings sum to \(100+121+144=365\). Subtract \(n\bar{x}^2=3(11^2)=363\), leaving 2. Dividing by 2 gives variance \(1\text{ unit}^2\) and standard deviation 1 unit.

After adding 30. The new mean is \((10+11+12+30)/4=63/4=15.75\) units. The deviations are \(-5.75,-4.75,-3.75,\) and \(14.25\). Their squared deviations sum to \(272.75\) square units. The new sample variance is \(272.75/(4-1)=90.9167\text{ unit}^2\), rounded, and the sample standard deviation is \(\sqrt{90.9167}\approx9.535\) units.

Check the updated calculation. The squared readings sum to \(100+121+144+900=1265\). Subtract \(n\bar{x}^2=4(15.75^2)=992.25\), giving \(1265-992.25=272.75\). Then \(272.75/3\approx90.9167\text{ unit}^2\), and \(\sqrt{90.9167}\approx9.535\) units, as before.

Explain the change. The added reading is far from the other three, so it greatly increases their squared deviations from the new mean. Variance and standard deviation are both non-resistant summaries: an extreme value can substantially affect them. As in Choosing IQR or Standard Deviation to Describe Spread, inspect the distribution and unusual values when deciding whether standard deviation is an appropriate spread summary.

Common Mistakes and AP Exam Tips

Confusing variance with standard deviation. \(s^2\) is the sample variance; \(s\) is the sample standard deviation. Do not call a variance a typical distance. The typical-distance interpretation belongs to standard deviation.
Using the wrong units. If observations are measured in centimeters, report \(s\) in centimeters and \(s^2\) in square centimeters. Do not label variance in the original units.
Using the wrong denominator. The sample variance formula divides by \(n-1\); the population variance formula divides by \(N\). Identify whether the data are a sample or the entire population before choosing the formula.
Forgetting to take the square root. The sum of squared deviations divided by the appropriate denominator gives variance. Take its square root to obtain standard deviation.
Treating variance as resistant. Because deviations are squared, large deviations can contribute much more than small ones. An unusual value can have a strong effect on variance and standard deviation.
Comparing numerical variances without checking units. Variance changes with the square of a unit conversion. State the units and compare like with like; standard deviations are expressed in the original units.

For a complete response, show whether the calculation is for a sample or a population, use the corresponding formula and denominator, and label the result with squared units. If the question asks what spread means in context, also report standard deviation in the original units and interpret it as a typical distance from the mean.

Key takeaway: Variance summarizes squared deviations from the mean, so its units are squared. Standard deviation is the square root of variance and returns to the original units, making it the more natural measure for describing typical distance from the mean.

Check Your Understanding

Use the distinction between variance and standard deviation, and pay attention to whether the values describe a sample or an entire population.

A sample is measured in minutes and has \(s^2=16\). What is its sample standard deviation, and what are the units of each measure?
For a population of 8 observations, the squared deviations from the population mean sum to 56 square meters. Find \(\sigma^2\) and \(\sigma\), including units.
Why is it incorrect to say that a variance of \(9\text{ cm}^2\) means observations typically differ from the mean by 9 centimeters?
A sample of distances is converted from meters to centimeters. How does the standard deviation’s numerical value change? How does the variance’s numerical value change?
Why can adding one extreme observation substantially affect variance?