Measures of dispersion — variance, standard deviation: Question bank

The vocabulary these traps rely on

Before a single trap, let us pin down every symbol so nothing sneaks up on you.

In Figure s01 the yellow dashed line marks the mean $\overset{x}{ˉ}$ (the balance point), and each coloured stick is a deviation $x_{i} - \overset{x}{ˉ}$ : sticks pointing left are negative, sticks pointing right are positive. Notice with your eye that the left-lengths and right-lengths balance out — that is the picture behind " $\sum (x_{i} - \overset{x}{ˉ}) = 0$ ", and it is why we must square before averaging.

Why $σ^{2} = \overline{x^{2}} - (\overset{x}{ˉ})^{2}$ — derived, not just quoted

The shortcut is not magic. Start from the definition and expand the square inside: $σ^{2} = \frac{1}{n} \sum (x_{i} - \overset{x}{ˉ})^{2} = \frac{1}{n} \sum (x_{i}^{2} - 2 x_{i} \overset{x}{ˉ} + \overset{x}{ˉ}^{2}) .$ Split the one sum into three (summation distributes over $+$ ): $= \overline{x^{2}} \frac{1}{n} \sum x_{i}^{2} - 2 \overset{x}{ˉ} \cdot \overset{x}{ˉ} \frac{1}{n} \sum x_{i} + \frac{1}{n} n \overset{x}{ˉ}^{2} \sum \overset{x}{ˉ}^{2} .$ The middle piece is $2 \overset{x}{ˉ} \cdot \overset{x}{ˉ} = 2 (\overset{x}{ˉ})^{2}$ and the last is $(\overset{x}{ˉ})^{2}$ , so $σ^{2} = \overline{x^{2}} - 2 (\overset{x}{ˉ})^{2} + (\overset{x}{ˉ})^{2} = \overline{x^{2}} - (\overset{x}{ˉ})^{2} .$

Three quick anchors to carry into the traps:

Variance $σ^{2}$ — average squared distance from the centre, never negative.
Standard deviation $σ = σ^{2}$ — the same idea in the data's own units.
The shortcut $σ^{2} = \overline{x^{2}} - (\overset{x}{ˉ})^{2}$ — "mean of squares minus square of mean".

True or false — justify

Each line is a claim. Decide true or false and give the one-sentence reason before revealing. A bare "true"/"false" is worth nothing — the reason is the point.

Variance can be negative if the data has more values below the mean than above.

False. Every term

(x_{i} - \overset{x}{ˉ})^{2}

is a square, so it is

\geq 0

; a sum of non-negative numbers divided by

n > 0

can never be negative, no matter how the data sits around the mean.

If two datasets have the same mean, they must have the same variance.

False. The mean fixes the centre only; the parent note's whole point is that two classes averaging 60% can be tightly bunched or wildly scattered — same mean, different spread.

Standard deviation is always less than or equal to the variance.

False. Only when

σ^{2} \geq 1

. If

σ^{2} = 0.25

then

σ = 0.5 > 0.25

; square-rooting a number below 1 makes it bigger.

Adding 10 to every exam mark raises the standard deviation by 10.

False. Shifting every value slides the whole cloud sideways without changing any gap between points, so both variance and SD are unchanged — variance is shift-invariant.

Doubling every data value doubles the variance.

False. Scaling by

c

multiplies variance by

c^{2}

, so doubling (

c = 2

) multiplies variance by

4

; it doubles the SD, not the variance. See Figure s03 for the stretch picture.

\overline{x^{2}}

and

(\overset{x}{ˉ})^{2}

are just two names for the same number.

False.

\overline{x^{2}}

squares each value then averages;

(\overset{x}{ˉ})^{2}

averages then squares. Their difference is exactly the variance, so they are equal only when variance is

0

(all values identical).

If variance is

0

, the mean, median and mode of the data must all be equal.

True. Variance

0

forces every value to be identical, so a single repeated number is simultaneously the mean, the median and the mode.

Sample variance

s^{2}

is always larger than population variance

σ^{2}

for the same data.

True (for

n > 1

and non-constant data). Both share the same sum of squared deviations, but

s^{2}

divides by the smaller

n - 1

, so it is larger; if all values are equal both are

0

Figure s03 shows why the scaling rule squares $c$ : stretch the number line by $c$ and every deviation stick stretches by $c$ , so every squared stick — and therefore their average, the variance — stretches by $c^{2}$ , while the SD (a length again) stretches by just $c$ .

Spot the error

Each line describes a calculation or statement someone actually made. Find the mistake and state the correct principle.

"I got variance

= 5.6

marks, so the spread is about

5.6

marks."

The error is units. Variance is in marks², not marks; the interpretable spread is

σ = 5.6 \approx 2.37

marks. Report SD, not variance, for a spread in real units.

"I averaged the deviations

(x_{i} - \overset{x}{ˉ})

and got

0

, so this data has no spread."

The error is skipping the square.

\sum (x_{i} - \overset{x}{ˉ}) = 0

for every dataset because positives and negatives cancel (see Figure s01); that zero measures nothing about spread. You must square first.

"For

2, 4, 4, 6, 9

I used

\overline{x^{2}} = (\overset{x}{ˉ})^{2} = 25

, so variance

= 25 - 25 = 0

The error is confusing

\overline{x^{2}}

with

(\overset{x}{ˉ})^{2}

. Here

\overline{x^{2}} = 153/5 = 30.6

while

(\overset{x}{ˉ})^{2} = 25

, so variance

= 30.6 - 25 = 5.6

, not

0

"My data is the entire population of 5 students, so I divide the sum of squares by

n - 1 = 4

The error is applying Bessel's correction to a full population. Divide by

n - 1

only when the data is a sample estimating a bigger population; a whole population uses

n

"I computed the mean as

5

, then used a different rounded mean

5.2

when finding deviations."

The error is using an inconsistent centre. Every deviation must be measured from the same

\overset{x}{ˉ}

you actually computed, or the sum of deviations won't cancel and the variance is corrupted.

"Variance came out as

- 3

, which tells me the data trends downward."

The error is that variance is impossible to make negative — a negative result means an arithmetic slip (likely

(\overset{x}{ˉ})^{2} > \overline{x^{2}}

, which can never truly happen). Variance carries no direction/trend information at all.

"Two SDs are

4

cm and

9

cm, so together the combined SD is

4 + 9 = 13

cm."

The error is adding SDs directly. Standard deviations do not add; even variances only add under independence, and you never combine spreads by simple addition of SDs.

Why questions

State the reason in your own words first, then check.

Why do we square deviations instead of taking absolute values?

Squaring is smooth/differentiable (good for calculus and least-squares), penalises large deviations more heavily, and links cleanly to geometry and the normal distribution — advantages the kink of

∣ x ∣

lacks.

Why is

\overline{x^{2}}

always at least as large as

(\overset{x}{ˉ})^{2}

Because their difference is the variance (Figure s02's green gap), and variance is a sum of squares divided by

n

, hence

\geq 0

; so

\overline{x^{2}} - (\overset{x}{ˉ})^{2} \geq 0

always.

Why does dividing by

n - 1

(instead of

n

) make the sample variance bigger?

The sample mean

\overset{x}{ˉ}

is chosen to minimise the sum of squared deviations, so those deviations are on average slightly too small; dividing by the smaller

n - 1

inflates the result to remove that downward bias.

Why does variance ignore adding a constant but square the effect of multiplying?

Sliding every point by

+ c

leaves all pairwise distances unchanged, so spread is untouched; multiplying by

c

stretches every distance by

c

, and since variance uses squared distances they stretch by

c^{2}

(Figure s03).

Why do we bother with SD at all when variance already measures spread?

Variance lives in squared units (marks², cm²) which are hard to interpret; taking the square root returns SD to the original units so a "spread of

2.37

marks" makes sense.

Why can the range be a poor spread measure compared to variance?

Range uses only the two extreme points and ignores everything in between, so a single outlier controls it; variance uses every value's distance from the centre.

Why is

σ / \overset{x}{ˉ}

(the coefficient of variation) sometimes preferred over

σ

alone?

SD of

5

cm is large for a beetle but tiny for a building; dividing by the mean gives a unitless relative spread — see Coefficient of Variation — that lets you compare scales fairly.

Edge cases

The boundary and degenerate scenarios. If you never met these, you don't fully own the definition.

What is the variance of a single data point

x_{1} = 7

(as a population)?

Zero. The mean equals

7

, the only deviation is

0

, so

σ^{2} = 0/1 = 0

; one point has no spread.

What is the sample variance of a single data point?

Undefined. The formula divides by

n - 1 = 0

, and a single observation gives no information about spread around a population — you cannot estimate it.

Data is

5, 5, 5, 5

. What are the mean, variance and SD?

Mean

= 5

, variance

= 0

, SD

= 0

. All values identical means every deviation is

0

; this is the unique case where variance hits its minimum.

Data is

0, 0, 0, 10

. Is the variance zero because most values are zero?

No. The mean is

2.5

, so every point deviates from it; three deviations of

- 2.5

and one of

+ 7.5

give a positive variance. "Mostly zeros" is not "all identical".

If every value is negative, say

- 2, - 4, - 6

, can the variance still be positive?

Yes. Variance measures spread, not sign; deviations from the mean

- 4

are

2, 0, - 2

, whose squares give a positive variance. Negative data never forces negative or zero variance.

Two datasets:

{10, 10, 10}

and

{9, 10, 11}

. Which has larger SD and why?

The second. The first has all-identical values so SD

= 0

; the second has genuine spread around the mean

10

, giving a positive SD.

n \to \infty

for a fixed population, does the gap between the

n

-divisor and

(n - 1)

-divisor variance matter?

No — the two estimates converge, since

\frac{n}{n - 1} \to 1

; Bessel's correction only matters noticeably for small samples.

What happens to the SD if you append one point exactly equal to the current mean?

The mean is unchanged but the sum of squared deviations is unchanged while

n

grows, so the variance and SD shrink slightly — adding a point right at the centre dilutes the average spread.

Recall One-line self-test before you leave

Say aloud: variance is never negative, is measured in squared units, is unchanged when you shift all data, and is multiplied by == $c^{2}$ == when you scale by $c$ . If any of those four surprised you, reread that trap.

Connections

Measures of dispersion — variance, standard deviation — the parent this bank stress-tests.
Mean, Median, Mode — measures of central tendency — the centre every deviation is measured from.
Coefficient of Variation — needed for the relative-spread "why" question above.
Normal Distribution — where SD earns its interpretation via the 68-95-99.7 rule.
Covariance and Correlation — the two-variable generalisation of these same traps.