The learning goal
For 2, 4, 4 and 6 cm, the mean is 4 cm and the squared deviations sum to 8 cm². Treating these values as the whole population gives variance 2 cm² and standard deviation √2≈1.414 cm. The conventional sample formulas give variance 8/3 cm² and standard deviation about 1.633 cm. State the reporting scope before choosing a calculator output.
Put the population or sample in the question
The denominator follows the task. To describe exactly these four objects as the population of interest, use N=4. To treat them as a sample for estimating a larger population, the conventional sample-variance formula uses n−1=3.
OpenStax distinguishes these calculations. This exercise does not prove estimation properties or present invented numbers as a real random sample. If the question omits the scope, ask for it or state the definition adopted.
Every length and variation below is an original KitNelo teaching example, not a measurement or learning-effect experiment. Two observations happen to equal 4; keep both. The median exercise’s observation-count checks help distinguish repeated values from duplicate records.
| Question | Definition | Denominator here |
|---|---|---|
| Describe only these four objects | Population variance and standard deviation | 4 |
| Treat four objects as a sample of a larger population | Conventional sample variance and standard deviation | 3 |
| The target or sampling task is unspecified | Clarify before combining answers | Undetermined |
Carry the values through to the squared-deviation sum
Use the same mean for every observation. The lengths total 16 cm, so the mean is 16÷4=4 cm.
The deviations are −2, 0, 0 and 2 cm, which sum to zero. Adding signed deviations and declaring no spread would allow opposite directions to cancel. Squaring gives 4, 0, 0 and 4 cm², with a sum of 8 cm². Both calculations start from that same intermediate value.
| Observation | Length/cm | Deviation/cm | Squared deviation/cm² |
|---|---|---|---|
| 1 | 2 | −2 | 4 |
| 2 | 4 | 0 | 0 |
| 3 | 4 | 0 | 0 |
| 4 | 6 | 2 | 4 |
| Total | 16 | 0 | 8 |
How the same 8 produces two standard deviations
Divide the squared-deviation sum before taking the square root. For the population, σ²=8÷4=2 cm² and σ=√(8÷4)=√2≈1.414 cm.
For the sample, s²=8÷(4−1)=8/3 cm² and s=√(8/3)≈1.633 cm. The sample result is larger here because the same positive sum is divided by a smaller positive number. It does not describe a second, less stable measurement or replace the population answer with a universally correct one.
Preserve parentheses. √8÷3 is about 0.943: it takes the square root before dividing by 3, unlike √(8÷3). Retain the definition, observation count and unit alongside the decimal.
| Scope | Squared-deviation sum | Variance | Standard deviation |
|---|---|---|---|
| Population, N=4 | 8 cm² | 2 cm² | √2≈1.414 cm |
| Sample, n=4 | 8 cm² | 8/3≈2.667 cm² | √(8/3)≈1.633 cm |
Locate the mistake in “a standard deviation of 2 centimeters”
The population variance has numerical value 2, but its unit is cm². A square root produces a standard deviation in cm. NIST explains this change of units.
The average absolute deviation here is (2+0+0+2)÷4=1 cm. That is a different quantity, not a replacement for the requested standard deviation. Check the name and definition of a software output first.
An editor-designed faulty report and its repair
The four objects have mean 4 cm and standard deviation 2 cm.
Treating the four objects as the whole population: mean 4 cm, population variance 2 cm², population standard deviation √2≈1.414 cm, N=4.
Test the reasoning by shifting and scaling all values
Predict what happens when 10 cm is added to every length, or every length is multiplied by 2. Then recompute the mean and squared-deviation sum. Every row below uses the population definition.
Adding a constant moves the values and their mean equally, leaving deviations unchanged. Multiplying all values by 2 multiplies every deviation by 2 and its square by 4. Variance multiplies by 4 and standard deviation by 2. Changing one observation does not meet that whole-data scaling condition.
Identify the observations, scope, denominator and unit in a report. Preserve working values such as 8/3 and √(8/3), then format the final answer. Consult the intermediate-versus-final rounding exercise when checking display rules.
Standard deviation alone cannot establish normality, a successful learning method or small-sample representativeness. Real data require missing-value and unusual-point checks; squaring increases the contribution of observations further from the mean.
| Data/cm | Mean/cm | Variance/cm² | Standard deviation/cm |
|---|---|---|---|
| 2, 4, 4, 6 | 4 | 2 | √2≈1.414 |
| 12, 14, 14, 16 | 14 | 2 | √2≈1.414 |
| 4, 8, 8, 12 | 8 | 8 | √8≈2.828 |
Have you completed the exercise?
- Explain why both observations equal to 4 remain.
- Recover the squared-deviation sum 8 cm² without looking at the decimal answer.
- Label population denominator 4 and sample denominator 3 separately.
- Keep the square root over the whole quotient and distinguish cm² from cm.
- Explain the effect of whole-data shifts and scaling on deviations.
Common questions
Can one value have a standard deviation?
A one-member population has population standard deviation 0. The conventional sample formula divides by n−1, which is zero when n=1, so it cannot calculate a sample standard deviation there. One value also does not support a useful sample estimate of spread.
Can the mean 4 alone determine the standard deviation?
No. The values 4, 4, 4, 4 also have mean 4, but population standard deviation 0. You need the observations or enough information to recover their squared-deviation sum and count.
