Measures of Spread: Variance and Standard Deviation
Quick Reference
| Field | Value |
|---|---|
| Textbook | Moore, McCabe, Craig, Introduction to the Practice of Statistics (IPS), 9th edition |
| Chapter | 1 (Looking at Data: Distributions) |
| Section | 1.3 (Describing Distributions with Numbers) |
| Subsection | Measuring spread: the standard deviation |
| Pages | Page numbers pending faculty verification. |
| Course | MATH380 |
Try This First
Here are two small classes. Each took a 10-point quiz. Both classes have the same mean score.
- Class A scores: 5, 5, 5, 5, 5
- Class B scores: 1, 3, 5, 7, 9
Before reading any formula, answer these on paper:
- Confirm that both lists have a mean of 5.
- Which class is more “spread out”? Write one sentence saying how you can tell.
- If you had to attach a single number to each class to describe how far the scores sit from 5, what would you try?
Check your thinking
Both lists add to 25, so both have mean $\bar{x} = 5$. Class A has no spread at all (every score equals the mean). Class B fans out from 5 down to 1 and up to 9.
A natural first idea is to measure, for each score, how far it sits from the mean (the deviation), then summarize those distances with a single number. That single number is what the variance and standard deviation produce.
From a Picture to a Number
The mean tells you where the center of a distribution sits. It says nothing about how tightly or loosely the observations cluster around that center. Two distributions can share the same mean and look completely different: one packed near the center, the other widely scattered.
Spread is the missing piece. The standard deviation answers a single question in one number: on a typical observation, how far from the mean does the data sit?
The construction follows four plain steps, and every step has a reason:
- Find the deviation of each observation from the mean. The deviation $x_i - \bar{x}$ is the signed distance from the center. A score above the mean has a positive deviation; a score below has a negative deviation.
- Square each deviation. Squaring does two jobs. It removes the sign (so distances below and above the mean both count as positive spread), and it gives extra weight to observations that sit far out.
- Average the squared deviations. This average is the variance. Averaging keeps the number from simply growing as you collect more data.
- Take the square root. The variance is in squared units (squared points, squared dollars). The square root returns the answer to the original units, giving the standard deviation.
One detail in step 3 surprises almost everyone the first time: the average divides by $n-1$, not $n$. The reason is in the misconceptions section below. For now, treat $n-1$ as the count that makes the sample variance a fair estimate of the spread in the larger population the sample came from.
Prerequisite Hub
How this skill connects to its prerequisites and what it leads to.
Builds on
| Skill | Why it is needed |
|---|---|
| Measures of Center: Mean and Median | The standard deviation measures spread about the mean, so the mean comes first |
| Sigma (Summation) Notation | The variance formula is a sum $\sum (x_i - \bar{x})^2$ over every observation |
| Basic Algebra Review | Squaring deviations and taking a square root require comfort with algebra |
Unlocks
| Skill | How this skill feeds it |
|---|---|
| Density Curves | A density curve has a mean and a standard deviation that describe its center and spread |
| Standardizing and z-Scores | A z-score is measured in units of standard deviations from the mean |
Cross-course prerequisites
None. Everything required is inside MATH380.
Quick Reference
| Quantity | Symbol | Formula | Units |
|---|---|---|---|
| Variance | $s^2$ | $\dfrac{1}{n-1}\displaystyle\sum_{i=1}^{n}(x_i - \bar{x})^2$ | squared units |
| Standard deviation | $s$ | $\sqrt{\dfrac{1}{n-1}\displaystyle\sum_{i=1}^{n}(x_i - \bar{x})^2}$ | original units |
Read both formulas as one process: measure each distance from the mean, square it, average over $n-1$, and (for $s$) take the square root to get back to the original units.
Official Definitions
These definitions are reproduced faithfully from the scope source.
Variance ($s^2$)
The variance $s^2$ of a set of observations is an average of the squares of the deviations of the observations from their mean. In symbols, for $n$ observations $x_1, x_2, \ldots, x_n$, the variance is
$$\boxed{\,s^2 = \frac{1}{n-1}\sum_{i=1}^{n}(x_i - \bar{x})^2\,}$$
Source: Moore, McCabe, Craig, IPS 9th edition, Section 1.3 (Describing Distributions with Numbers); corroborated by OpenStax Introductory Statistics 2e, where variance is the mean of the squared deviations from the mean (key terms).
Standard deviation ($s$)
The standard deviation $s$ is the square root of the variance $s^2$:
$$\boxed{\,s = \sqrt{\frac{1}{n-1}\sum_{i=1}^{n}(x_i - \bar{x})^2}\,}$$
The standard deviation measures spread about the mean and should be used only when the mean is chosen as the measure of center. It is zero only when there is no spread, that is, when all observations have the same value.
Source: Moore, McCabe, Craig, IPS 9th edition, Section 1.3 (Describing Distributions with Numbers); corroborated by OpenStax Introductory Statistics 2e, where the standard deviation is the square root of the variance and measures how far values are from the mean (key terms).
Two facts that follow directly from the definition
| Property | Statement | Why it holds |
|---|---|---|
| $s \geq 0$ always | The standard deviation is never negative | It is the square root of an average of squares, and squares are never negative |
| $s = 0$ means no spread | $s = 0$ if and only if every observation equals the mean | Each squared deviation must be zero, so each $x_i = \bar{x}$ |
The scope lists no separate named theorem for this node, so there is no theorem-statement block. The two properties above are immediate consequences of the definition itself.
Worked Examples
Every example states a prediction before computing, then checks the prediction against the result.
Worked Example 1: The two quiz classes from the opener
Return to Class B: scores 1, 3, 5, 7, 9, with mean $\bar{x} = 5$.
Predict first. The scores reach 4 points below and 4 points above the mean, so a “typical distance from 5” should land somewhere between 2 and 4. Hold that prediction.
Step 1: deviations $x_i - \bar{x}$.
| $x_i$ | $x_i - \bar{x}$ | $(x_i - \bar{x})^2$ |
|---|---|---|
| 1 | $-4$ | 16 |
| 3 | $-2$ | 4 |
| 5 | $0$ | 0 |
| 7 | $2$ | 4 |
| 9 | $4$ | 16 |
Step 2: sum the squared deviations. $$\sum_{i=1}^{5}(x_i - \bar{x})^2 = 16 + 4 + 0 + 4 + 16 = 40$$
Step 3: divide by $n - 1 = 5 - 1 = 4$ to get the variance. $$s^2 = \frac{40}{4} = 10 \text{ (squared points)}$$
Step 4: take the square root to get the standard deviation. $$s = \sqrt{10} \approx 3.16 \text{ points}$$
Check the prediction. The predicted range was 2 to 4 points; the result, about 3.16 points, sits inside it. For Class A (all scores 5), every deviation is 0, so $s^2 = 0$ and $s = 0$, matching the fact that a list with no spread has standard deviation zero.
Worked Example 2: Daily high temperatures
A town recorded high temperatures (in degrees Fahrenheit) on six days: 68, 70, 72, 65, 74, 71.
Predict first. The values run from 65 to 74, a range of 9 degrees, so a typical distance from the center should be a few degrees, not 9 and not a fraction of one.
Step 1: find the mean. $$\bar{x} = \frac{68 + 70 + 72 + 65 + 74 + 71}{6} = \frac{420}{6} = 70 \text{ degrees}$$
Step 2: deviations and squared deviations.
| $x_i$ | $x_i - 70$ | $(x_i - 70)^2$ |
|---|---|---|
| 68 | $-2$ | 4 |
| 70 | $0$ | 0 |
| 72 | $2$ | 4 |
| 65 | $-5$ | 25 |
| 74 | $4$ | 16 |
| 71 | $1$ | 1 |
Step 3: sum and divide by $n - 1 = 5$. $$\sum (x_i - \bar{x})^2 = 4 + 0 + 4 + 25 + 16 + 1 = 50$$ $$s^2 = \frac{50}{5} = 10 \text{ (squared degrees)}$$
Step 4: square root. $$s = \sqrt{10} \approx 3.16 \text{ degrees}$$
Check the prediction. About 3.16 degrees is a few degrees, smaller than the full range of 9, which is exactly what a typical distance from the center should be. The deviations summed to 0 ($-2 + 0 + 2 - 5 + 4 + 1 = 0$), a useful arithmetic check that the deviations were computed correctly.
Worked Example 3: Units and why the square root matters
Suppose a small data set of package weights (in kilograms) has variance $s^2 = 0.25$.
Predict first. A variance of 0.25 is in squared kilograms, which is hard to interpret directly. The square root should give a plain number of kilograms.
$$s = \sqrt{0.25} = 0.5 \text{ kilograms}$$
The variance, 0.25, is in squared kilograms and cannot be compared to a weight. The standard deviation, 0.5 kilograms, is in the same units as the data, so a reader can say “weights typically sit about half a kilogram from the mean.” This is the entire reason the final step takes a square root: it restores the original units.
Common Misconceptions
the variance divides by $n$, the way an ordinary average does. The sample variance divides the sum of squared deviations by $n-1$, not $n$. The reason is that the deviations $x_i - \bar{x}$ are measured from the sample mean $\bar{x}$, which was itself computed from the same data. The deviations are forced to sum to zero, so once $n-1$ of them are known, the last one is determined. Only $n-1$ of the deviations are free to vary, and dividing by $n-1$ corrects the systematic underestimate that dividing by $n$ would produce. Predict-then-check: for the list 2, 4, 6 with mean 4, the squared deviations are 4, 0, 4, summing to 8. Dividing by $n-1=2$ gives $s^2 = 4$; dividing by $n=3$ would give the smaller, biased value $8/3 \approx 2.67$.
the deviations can be averaged directly without squaring. The deviations $x_i - \bar{x}$ always sum to zero, so their plain average is always zero and carries no information about spread. Squaring is what prevents the positive and negative deviations from canceling. Predict-then-check: for the temperatures 68, 70, 72, 65, 74, 71 in Worked Example 2, the raw deviations were $-2, 0, 2, -5, 4, 1$, which sum to 0. The average deviation is 0 for every data set, which is why the construction squares first.
a larger standard deviation means a larger mean. The mean locates the center; the standard deviation measures spread about that center. They are independent. Predict-then-check: the list 100, 100, 100 has a large mean (100) but standard deviation 0, while the list 1, 5, 9 has a small mean (5) but standard deviation $\sqrt{16} = 4$. A small center can carry a large spread, and a large center can carry no spread.
the variance and the standard deviation are interchangeable numbers. They describe the same spread but are in different units. The variance is in squared units; the standard deviation is in the original units. Reporting a variance of 10 “points” is a unit error; it is 10 squared points, and the standard deviation $\sqrt{10} \approx 3.16$ is the value in points.
Practice Problems
A data set has every observation equal to 12. Without computing a sum, state the variance and the standard deviation, and give the one-sentence reason.
Find the variance and the standard deviation of the data set 4, 8, 6, 2. Show each step.
Two data sets each have mean 50. Set P: 50, 50, 50, 50. Set Q: 47, 49, 51, 53. Without finishing every arithmetic step, decide which set has the larger standard deviation, then compute $s$ for Set Q to confirm.
A student computes the standard deviation of 3, 5, 7 like this:
“Mean is 5. Deviations are $-2, 0, 2$. Average of the deviations is $\dfrac{-2 + 0 + 2}{3} = 0$, so the standard deviation is 0.”
Identify the error and give the correct standard deviation.
Start with the data set 10, 14, 18 and find its standard deviation. Then add 100 to every value to get 110, 114, 118, and find the new standard deviation. Explain, using the definition, why the two standard deviations are equal even though the means differ by 100.
Mastery Checklist
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Mental Model
Think of the standard deviation as a typical distance from the mean. Picture the mean as a stake driven into the middle of the data. Each observation is tied to that stake by a string whose length is its deviation. Some strings reach above the stake, some below, but length is what matters. Squaring throws away the up-or-down direction and keeps the length (and gives extra weight to the long strings). Averaging the squared lengths and then taking a square root gives one representative string length: the standard deviation. A data set with all strings of length zero (every value at the stake) has standard deviation zero.
Connections
Looking back:
- Measures of Center: Mean and Median supplies the mean that the deviations are measured from
- Sigma (Summation) Notation gives the $\sum$ that the variance formula relies on
Looking ahead:
- Density Curves describe an idealized distribution by its mean and standard deviation
- Standardizing and z-Scores report each value as a number of standard deviations from the mean
Real-world connections:
- Investment risk is often reported as the standard deviation of returns
- Manufacturing quality control flags parts whose measurements sit too many standard deviations from the target
- Test-score reports use standard deviations to compare an individual to a group
Resources
| Resource | What it covers | Link |
|---|---|---|
| IPS 9th edition, Section 1.3 (Describing Distributions with Numbers) | The official definition of variance and standard deviation as measures of spread | Moore, McCabe, Craig, IPS 9e, Ch. 1, Sec. 1.3 |
| OpenStax Introductory Statistics 2e, 2.7 Measures of the Spread of the Data | Worked development of the standard deviation with additional examples | openstax.org/books/introductory-statistics-2e/pages/2-7-measures-of-the-spread-of-the-data |
| Previous | Up | Next |
|---|---|---|
| Measures of Center: Mean and Median | Skills Index | Density Curves |
Last updated: 2026-06-16