Standard Deviation and Variance

The same average can hide very different data

Two processes produce values 10, 10, 10, 10, 10 and 6, 8, 10, 12, 14. Both have a mean of 10. The first has no observed spread; the second varies around that mean.

Variance summarizes squared deviations from the mean. Standard deviation is its square root, restoring the original measurement units. Neither requires the data to be normally distributed just to calculate it.

Decide whether these data are a sample or the population

For a complete population of N values:

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

For a sample of n values used to estimate population variability:

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

The denominator depends on the goal, not on whether the dataset feels “large.” If you describe all five readings as the complete set of interest, divide by 5. If you use those readings to estimate a process variance, the usual sample formula divides by 4.

The sample mean is estimated from the same observations. Once it is fixed, the deviations sum to zero, leaving n − 1 freely varying deviations. Under independent sampling with finite variance, this correction makes s² an unbiased estimator of population variance. It does not make s itself exactly unbiased.

Worked example from raw data

Use 6, 8, 10, 12, 14. The mean is 50/5 = 10.

x	x − x̄	(x − x̄)²
6	−4	16
8	−2	4
10	0	0
12	2	4
14	4	16
Total	0	40

For the sample calculation:

$s^2=\frac{40}{4}=10,\qquad s=\sqrt{10}=3.1623$

For the population calculation:

$\sigma^2=\frac{40}{5}=8,\qquad\sigma=\sqrt{8}=2.8284$

If the readings are in minutes, the variance is in minutes squared and the SD is in minutes. Keep full precision while calculating and round the final result.

Reproduce both versions with the standard deviation calculator. Make sure its sample/population setting matches your purpose.

What standard deviation tells you

SD measures spread around the mean using squared distances. It is not the average absolute distance. In the example, the mean absolute deviation is (4 + 2 + 0 + 2 + 4)/5 = 2.4, which differs from both SD values.

For a genuinely normal model, about 68% of observations lie within one population SD of the mean and about 95% within two. These percentages are not guaranteed for arbitrary data. A skewed or multimodal distribution can have a perfectly valid SD without obeying that rule.

An SD of zero means every recorded value is equal. It does not prove the underlying process has no variation; limited measurement resolution or a tiny sample may hide it.

SD and standard error have different jobs

Suppose a sample of 25 waits has s = 10 minutes.

SD = 10 minutes: observed variability among individual waits.
Estimated standard error = 10/√25 = 2 minutes: estimated sampling variability of the sample mean, assuming independent observations from the same population.

Taking more observations can improve the precision of the mean without making individual waits less variable. If n becomes 100 and variability remains similar, SE is about 1 minute. Do not substitute SE for SD when describing the spread of individual values.

Outliers, robustness, and scale

Squaring deviations makes SD sensitive to extreme observations. Check whether an unusual reading is a data-entry error, a valid rare event, or evidence that several processes were mixed. Do not delete it merely because it increases SD.

For skewed data, consider reporting median and interquartile range alongside a distribution plot. These summarize a different aspect of variability; they are not interchangeable estimates of the same quantity.

Adding a constant changes the mean but not SD. Multiplying every value by a constant a multiplies SD by |a| and variance by a². Converting minutes to seconds multiplies SD by 60 and variance by 3,600.

Practice

1. Calculate sample variance and SD for 2, 4, 4, 6.

Show the calculation

The mean is 4. Squared deviations are 4, 0, 0, 4; their sum is 8. Sample variance is 8/3 = 2.6667, and sample SD is √(8/3) = 1.6330. Dividing by 4 instead gives population variance 2 and population SD 1.4142.

2. A sample has SD 12 and n = 36. What is the estimated SE of its mean? What sample size would halve it, assuming comparable variability and independent observations?

Show the explanation

SE = 12/√36 = 2. Halving it to 1 requires quadrupling n to 144. Doubling n reduces SE by a factor of √2, not by a factor of 2.