Measures of dispersion — variance, standard deviation: Exercises

Before we start, one shared reminder of the four symbols we use, so nothing is unearned:

Level 1 — Recognition

L1.1 A dataset has variance $σ^{2} = 49 cm^{2}$ . State the standard deviation and its units.

Recall Solution L1.1

SD is the square root of variance: $σ = 49 = 7$ . Variance was in $cm^{2}$ , so taking the root brings us back to the data's own units: $σ = 7 cm$ .

L1.2 For the data $5, 5, 5, 5$ , what is the variance? Answer without computing a sum.

Recall Solution L1.2

Every value is identical, so every deviation $d_{i} = 5 - 5 = 0$ . The average of zeros is zero: $σ^{2} = 0$ . (This is the only case where variance is exactly 0.)

L1.3 Which expression is the variance shortcut: (a) $(\overset{x}{ˉ})^{2} - \overline{x^{2}}$ , or (b) $\overline{x^{2}} - (\overset{x}{ˉ})^{2}$ ?

Recall Solution L1.3

(b) $\overline{x^{2}} - (\overset{x}{ˉ})^{2}$ . Read the name: mean of the squares minus the square of the mean. It must be $\geq 0$ ; option (a) would be negative, which a spread can never be.

Level 2 — Application

L2.1 Data (population): $3, 7, 7, 19$ . Find $\overset{x}{ˉ}$ , then $σ^{2}$ using the definition, then $σ$ .

Recall Solution L2.1

Mean: $\overset{x}{ˉ} = \frac{3 + 7 + 7 + 19}{4} = \frac{36}{4} = 9$ . Deviations: $3 - 9 = - 6$ , $7 - 9 = - 2$ , $7 - 9 = - 2$ , $19 - 9 = 10$ . Squares: $36, 4, 4, 100$ . Sum $= 144$ . Variance: $σ^{2} = \frac{144}{4} = 36$ . SD: $σ = 36 = 6$ .

L2.2 Same data $3, 7, 7, 19$ , now via the shortcut $σ^{2} = \overline{x^{2}} - (\overset{x}{ˉ})^{2}$ .

Recall Solution L2.2

$\sum x_{i}^{2} = 9 + 49 + 49 + 361 = 468$ , so $\overline{x^{2}} = \frac{468}{4} = 117$ . $(\overset{x}{ˉ})^{2} = 9^{2} = 81$ . $σ^{2} = 117 - 81 = 36$ ✓ — identical to L2.1, exactly as the algebra promised.

L2.3 Data (population): $1, 2, 3, 4, 5$ . Find $σ^{2}$ and $σ$ .

Recall Solution L2.3

$\overset{x}{ˉ} = \frac{15}{5} = 3$ . Deviations: $- 2, - 1, 0, 1, 2$ ; squares $4, 1, 0, 1, 4$ ; sum $= 10$ . $σ^{2} = \frac{10}{5} = 2$ . $σ = 2 \approx 1.414$ .

Level 3 — Analysis

L3.1 Reaction times (seconds) from 3 volunteers, treated as a sample of all people: $0.20, 0.30, 0.40$ . Find the sample variance $s^{2}$ and sample SD $s$ .

Recall Solution L3.1

$\overset{x}{ˉ} = \frac{0.20 + 0.30 + 0.40}{3} = 0.30$ . Deviations: $- 0.10, 0, 0.10$ ; squares $0.01, 0, 0.01$ ; sum $= 0.02$ . Sample → divide by $n - 1 = 2$ : $s^{2} = \frac{0.02}{2} = 0.01$ . $s = 0.01 = 0.10$ s.

L3.2 The same three numbers $0.20, 0.30, 0.40$ are now the entire population of a tiny experiment. Find $σ^{2}$ . Compare with L3.1 and explain the direction of the difference.

Recall Solution L3.2

Population → divide by $n = 3$ : $σ^{2} = \frac{0.02}{3} \approx 0.00667$ . This is smaller than the sample $s^{2} = 0.01$ . Dividing the same sum of squares ( $0.02$ ) by the larger number ( $3$ vs $2$ ) gives a smaller value. Bessel's $n - 1$ deliberately inflates the sample estimate to correct the downward bias caused by using $\overset{x}{ˉ}$ (chosen to minimise that very sum).

L3.3 A student reports " $s^{2} = 0.02/3$ " for a sample of 3. Is the divisor correct? Give the right value.

Recall Solution L3.3

No. For a sample, divide by $n - 1 = 2$ , not $n = 3$ . The correct sample variance is $s^{2} = 0.02/2 = 0.01$ . Dividing by $3$ would give the population variance $\approx 0.00667$ , which underestimates the spread of the wider population.

Level 4 — Synthesis

L4.1 (Shift invariance). Dataset $A = {10, 20, 30}$ has population variance $σ_{A}^{2}$ . A new set adds $c = 100$ to each: $B = {110, 120, 130}$ . Find $σ_{A}^{2}$ and $σ_{B}^{2}$ and confirm the property $Var (x + c) = Var (x)$ .

Recall Solution L4.1

A: $\overset{x}{ˉ}_{A} = 20$ ; deviations $- 10, 0, 10$ ; squares $100, 0, 100$ ; sum $200$ ; $σ_{A}^{2} = 200/3 \approx 66.67$ . B: $\overset{x}{ˉ}_{B} = 120$ ; deviations $- 10, 0, 10$ (the same, because both values and mean shifted by $+ 100$ ); squares identical; $σ_{B}^{2} = 200/3 \approx 66.67$ . Equal ✓. Adding a constant slides every point and the mean by the same amount, so the deviations — and hence the spread — are untouched.

L4.2 (Scaling). Take $A = {10, 20, 30}$ with $σ_{A}^{2} = 200/3$ . Multiply each value by $c = 3$ : $C = {30, 60, 90}$ . Predict $σ_{C}^{2}$ and $σ_{C}$ using the scaling rule, then verify.

Recall Solution L4.2

Predict: $Var (c x) = c^{2} Var (x)$ , so $σ_{C}^{2} = 3^{2} \cdot \frac{200}{3} = 9 \cdot \frac{200}{3} = 600$ . And $σ_{C} = ∣ c ∣ σ_{A} = 3 200/3 = 3 \cdot 66.67 \approx 24.49$ . Verify directly: $\overset{x}{ˉ}_{C} = 60$ ; deviations $- 30, 0, 30$ ; squares $900, 0, 900$ ; sum $1800$ ; $σ_{C}^{2} = 1800/3 = 600$ ✓. Stretching by $3$ tripled every distance; squared distances grew by $3^{2} = 9$ .

L4.3 (Combining both). From $A$ , build $D = 3 x + 100 = {130, 160, 190}$ . Without recomputing deviations from scratch, find $σ_{D}^{2}$ using both properties.

Recall Solution L4.3

The $+ 100$ shift does nothing to variance (shift invariance). The $\times 3$ multiplies variance by $3^{2} = 9$ . $σ_{D}^{2} = 9 \cdot σ_{A}^{2} = 9 \cdot \frac{200}{3} = 600$ . (Same as $C$ , because the constant $+ 100$ is irrelevant to spread.) SD: $σ_{D} = 3 200/3 \approx 24.49$ .

Figure — Measures of dispersion — variance, standard deviation

Look at the red band above: it's the same width for $A$ and $A + c$ (pure slide) but stretches by factor $3$ for $3 A$ — the picture behind L4.1–L4.3.

Level 5 — Mastery

L5.1 (Prove $\overline{x^{2}} \geq (\overset{x}{ˉ})^{2}$ ). Show the mean of squares is never below the square of the mean, using only the shortcut and the definition of variance.

Recall Solution L5.1

By definition, $σ^{2} = \frac{1}{n} \sum (x_{i} - \overset{x}{ˉ})^{2}$ . This is an average of squares, and each square $(x_{i} - \overset{x}{ˉ})^{2} \geq 0$ , so their average is $\geq 0$ ; hence $σ^{2} \geq 0$ . By the shortcut, $σ^{2} = \overline{x^{2}} - (\overset{x}{ˉ})^{2}$ . Combining: $\overline{x^{2}} - (\overset{x}{ˉ})^{2} \geq 0 \Rightarrow \overline{x^{2}} \geq (\overset{x}{ˉ})^{2}$ . Equality holds iff every $(x_{i} - \overset{x}{ˉ})^{2} = 0$ , i.e. all values identical. $■$

L5.2 (Reverse-engineer the data). A two-point population ${a, b}$ with $a < b$ has mean $\overset{x}{ˉ} = 5$ and SD $σ = 3$ . Find $a$ and $b$ .

Recall Solution L5.2

For two points, both deviations have equal magnitude $\frac{b - a}{2}$ (each is exactly half the gap from the midpoint mean). $σ^{2} = \frac{1}{2} [(\frac{b - a}{2})^{2} + (\frac{b - a}{2})^{2}] = (\frac{b - a}{2})^{2}$ . So $σ = \frac{b - a}{2} = 3 \Rightarrow b - a = 6$ . With midpoint $\overset{x}{ˉ} = 5$ : $a = 5 - 3 = 2$ , $b = 5 + 3 = 8$ . Check: ${2, 8}$ , mean $5$ , deviations $\pm 3$ , squares $9, 9$ , $σ^{2} = 18/2 = 9$ , $σ = 3$ ✓.

L5.3 (Which is more spread?). Two classes: Class P scores ${40, 50, 60}$ ; Class Q scores ${20, 50, 80}$ . Both share the same mean. Without full computation, argue which has larger variance, then confirm numerically.

Recall Solution L5.3

Both have $\overset{x}{ˉ} = 50$ . Q's values sit farther from $50$ ( $\pm 30$ ) than P's ( $\pm 10$ ), so Q must have larger variance. P: deviations $- 10, 0, 10$ ; squares $100, 0, 100$ ; sum $200$ ; $σ_{P}^{2} = 200/3 \approx 66.67$ . Q: deviations $- 30, 0, 30$ ; squares $900, 0, 900$ ; sum $1800$ ; $σ_{Q}^{2} = 1800/3 = 600$ . $σ_{Q}^{2} = 600 > 66.67 = σ_{P}^{2}$ ✓. In fact Q's spread is $3 \times$ P's, so its variance is $3^{2} = 9 \times$ larger — the scaling rule in disguise.

Connections

Mean, Median, Mode — measures of central tendency — every exercise starts by locating the centre.
Coefficient of Variation — for L5.3, dividing $σ$ by $\overset{x}{ˉ}$ compares spread fairly when means differ.
Normal Distribution — the SD you compute here is the ruler for the 68-95-99.7 rule.
Covariance and Correlation — L4's scaling generalises to how paired variables co-vary.
Least Squares Regression — L5.1's " $\overset{x}{ˉ}$ minimises sum of squares" is the seed of regression.

Exercises — Measures of dispersion — variance, standard deviation