Measures of Variation

— Probability & Statistics 1 · AS & A Level Maths (9709) · Chapter 3 —

★ Exam Tip: Standard deviation, variance from totals (Σx, Σx²), combined sets and coded data Σ(x−b) are favourites in Paper 6 — usually a 4–7 mark question. Practise the algebra until it is automatic.

0. Why measure variation?

A measure of central tendency alone does not describe a data set — it says where the middle is, not how spread out the values are. Two sets can have the same mean/median/mode and be totally different.

Best summary = centre (mean / median) + variation (also called spread or dispersion).

runs in 8 matches: A = 25 30 31 26 31 28 29 24, B = 2 70 1 0 43 1 104 3 (both total 224)

Both batters: mean = 224 ÷ 8 = 28. A's scores are consistent, B's are varied — a measure of variation shows this.

3 common measures: ① range ② interquartile range (IQR) ③ standard deviation (SD)

1. The Range

Range = largest value − smallest value

+ Easy to calculate.
− Uses only the two extreme values — ignores the middle values.
Often more informative to quote min and max than just the difference.

e.g. lowest mark 6, highest 19 ⇒ range = 19 − 6 = 13.

Grouped data: exact values unknown, so find a least and greatest possible range using the class boundaries.
Least = (lower bdy of the max) − (upper bdy of the min) | Greatest = (upper bdy of the max) − (lower bdy of the min).

Worked Example 3.1range · rounded data

To the nearest cm, tallest and shortest pupils are 169 cm and 150 cm. Find least & greatest possible range.

Rounded to nearest cm ⇒ 168.5 ≤ h < 169.5 and 149.5 ≤ h < 150.5.

Least168.5 − 150.5= 18 cm

Greatest169.5 − 149.5= 20 cm

2. Interquartile Range & Percentiles

Quartiles split the ordered data into 4 equal-sized parts:

Q₁ = lower quartile | Q₂ = median (middle quartile) | Q₃ = upper quartile

IQR = Q₃ − Q₁ (upper quartile − lower quartile)

each gap between markers holds ¼ of the data

Why IQR beats range: it measures the spread of the central half, is relatively unaffected by outliers (extreme values), and can be found even when the exact extreme values aren't known.

Ungrouped data — finding quartiles

Median Q₂ is at the n + 12 th value

Even n: split into lower half & upper half. Q₁ = median of lower half, Q₃ = median of upper half.
Odd n: split at the median and discard the median. Q₁, Q₃ = medians of the lower / upper halves.

Tip: find quartiles by inspection (split & take medians) — don't memorise position formulae.

Common mistake: forgetting to discard the median when n is odd, or forgetting to order the data first.

Worked Example 3.2IQR · even n = 8

IQR of the ordered values 2, 5, 9, 13, 29, 33, 49, 55.

Lower half: 2 5 9 13 | Upper half: 29 33 49 55.

Q₁ = 5 + 92 = 7 Q₃ = 33 + 492 = 41 IQR = Q₃ − Q₁ = 41 − 7 IQR = 34

Worked Example 3.3IQR · odd n = 7

IQR of 69, 17, 43, 6, 73, 77, 39.

Order: 6, 17, 39, 43, 69, 73, 77 ⇒ median = 4th = 43 (discard it).

Q₁ = 17 (median of 6 17 39) Q₃ = 73 (median of 69 73 77) IQR = 73 − 17 = 56

Worked Example 3.4IQR · stem-and-leaf, n = 13

Stem-and-leaf (key 14|2 = 142): 14| 2 2 4 8 9 15| 1 3 5 6 7 9 16| 5 8

Q₂ at 13 + 12 = 7th value = 153 (discard) Lower half 142 … 151 ⇒ Q₁ = 144 + 1482 = 146 Upper half 155 … 168 ⇒ Q₃ = 157 + 1592 = 158 IQR = 158 − 146 = 12

(each half has 6 values ⇒ its median is the 3.5th value, i.e. average of the 3rd & 4th)

Explore 3.1five-number summaries

Set	min	Q₁	Q₂	Q₃	max
A: 2 2 3 11 11 21 22	2	2	11	21	22
B: 6 6 6 11 13 17 19 20	6	6	12	18	20
C: 9 15 28 32 35 49	9	15	30	35	49
D: 5 7 9 10 11 12 12 16 17	5	8	11	14	17

① median mid-way between min & max — sometimes (true only for D)
② median mid-way between Q₁ & Q₃ — sometimes (B and D)
③ IQR = exactly ½ range — sometimes (D: 6 and 12)
④ Q₃ − Q₂ > Q₂ − Q₁ — sometimes (A: 10 > 9 ✓; B: 6 = 6 ✗; C: 5 < 15 ✗)

Grouped data — use a cumulative frequency graph

Read values at any position of the ordered data off a cumulative frequency (CF) graph.

Quartile	Lower Q₁	Median Q₂	Upper Q₃
Position	n4 or ¼Σf	n2 or ½Σf	3n4 or ¾Σf

Note: for graphs use n/4, n/2, 3n/4 (with n = Σf) — not (n+1)/4 etc.

Percentile: the nth percentile is the value that is n% of the way through the data.
Q₁, Q₂, Q₃ = 25th, 50th, 75th percentiles.
e.g. 320 values: Q₁, Q₂, Q₃ at 80th, 160th, 240th; 90th percentile at (0.90 × 320) = 288th value.
Range of the middle 80% = 90th percentile − 10th percentile.

Worked Example 3.5CF graph · 500 people, task time

From the CF polygon (min 2 min, max 30 min) estimate: a greatest possible range, b IQR, c 95th percentile.

locate the position on the CF axis → across to the curve → down to read the value

a greatest range = width of polygon = 30 − 2 = 28 min b Q₁ at 5004 = 125th ⇒ Q₁ ≈ 8.0 Q₃ at 3 × 5004 = 375th ⇒ Q₃ ≈ 14.5 IQR ≈ 14.5 − 8.0 = 6.5 min c 95th percentile at 0.95 × 500 = 475th ⇒ ≈ 24.0 min

3. Box-and-Whisker Diagrams

A box plot shows 5 key values: smallest, Q₁, median Q₂, Q₃, largest.

Draw on graph paper with a scale; label the variable and units (e.g. "Length (cm)").
Box = IQR (Q₁ to Q₃) with a line at the median. Whiskers go from the box out to min and max.
The whisker (range) is NOT drawn through the box.

x: min 1, Q₁ 4, Q₂ 6, Q₃ 11, max 14 | y: min 0, Q₁ 3, Q₂ 9, Q₃ 12, max 15

	Median	Range	IQR
Dataset x	6	14 − 1 = 13	11 − 4 = 7
Dataset y	9	15 − 0 = 15	12 − 3 = 9

Compare medians (average): x-values are on average less than y-values.

Compare range & IQR (spread): y is more varied than x.

Skewness from a box plot

positive skew (tail to larger values)

symmetric

negative skew (tail to smaller values)

Check: dataset x has Q₃−Q₂ = 5 > Q₂−Q₁ = 2 ⇒ positive; y has Q₂−Q₁ = 6 > Q₃−Q₂ = 3 ⇒ negative.

The longer tail of the data's curve is on the side of the larger values for positive skew, smaller values for negative skew.

4. Variance & Standard Deviation

Explore 3.2why not mean deviation?

Deviation of a value = x − x̄ (positive above the mean, negative below). Take 5 numbers with mean 10, e.g. 4, 8, 10, 12, 16:

deviations −6, −2, 0, 2, 6 add up to 0

Sum of deviations is always 0 for any data set (positives cancel negatives) ⇒ mean deviation is useless. We need every deviation to be ≥ 0.

Option 1 – mean absolute deviation: Σ|x − x̄|n (|x − x̄| = drop the minus sign). Awkward for big data & algebra ⇒ not used.

Option 2 – square the deviations (our choice): mean of (x − x̄)² is the variance.

Var(x) = Σ(x − x̄)²n = Σx²n − x̄² = Σx²n − (Σxn)²

Variance has squared units (m² if data in m) ⇒ take the square root to get the standard deviation, in the original units. Written SD(X).

Key Point 3.3 — ungroupedSD = √Var = √Σ(x − x̄)²n = √Σx²n − x̄² , x̄ = Σxn
grouped (x = class mid-value)SD = √Σ(x − x̄)²fΣf = √Σx²fΣf − x̄² , x̄ = ΣxfΣf

Memory: variance = mean of the squares − square of the mean.

Meaning: low SD ⇒ values close to the mean; high SD ⇒ widely spread. (e.g. coffee machine set to 400 ml: high SD ⇒ some customers cheated, some cups overflow!)

Common mistakes: ① Σx² ≠ (Σx)². Σx² = add the squares; (Σx)² = add, then square. ② x²f = x × xf (or x² × f) — never (xf)².

Worked Example 3.6SD · ungrouped

For 3, 9, 15, 24, 29 find a the SD, b which numbers are more than 1 SD from the mean.

Var = Σx²n − (Σxn)² 3²+9²+15²+24²+29²5 − (3+9+15+24+295)² = 17325 − 805² = 346.4 − 16² = 90.4 SD = √90.4 = 9.51 (3 s.f.) b mean ± SD: 16 − 9.51 = 6.49 and 16 + 9.51 = 25.51 outside 6.49 ≤ x ≤ 25.51 ⇒ 3 and 29

Worked Example 3.7SD · frequency table

x = 12, 14, 16 with f = 13, 28, 10. Find SD to 3 s.f. — extend the table:

x	f	xf	x²f = x × xf
12	13	156	12×156 = 1872
14	28	392	14×392 = 5488
16	10	160	16×160 = 2560
Σf = 51		Σxf = 708	Σx²f = 9920

SD = √( Σx²fΣf − x̄² ) √( 992051 − (70851)² ) SD = 1.34

Never use a rounded mean! Here x̄ = 708/51 = 13.88… If you use 13.9: SD = √(9920/51 − 13.9²) = 1.14 instead of 1.34 — an error of 0.2 (≈ 15%). Always use Σx/n or Σxf/Σf (keep the exact fraction in the calculator).

Worked Example 3.8estimate of SD · grouped (class intervals)

Heights of 20 children: 1.2– (f = 2), 1.4– (f = 12), 1.5–1.7 (f = 6). Grouped ⇒ use class mid-values x = 1.3, 1.45, 1.6 ⇒ an estimate.

x	1.3	1.45	1.6	totals
f	2	12	6	Σf = 20
xf	2.6	17.4	9.6	Σxf = 29.6
x²f	3.38	25.23	15.36	Σx²f = 43.97

SD = √( 43.9720 − (29.620)² ) = √0.0081 = 0.09 m

method for SD from a frequency table (use mid-values for grouped data)

Explore 3.3interpreting SD / variance

High SD of property prices: mixed area — cheap flats beside luxury houses; buyers of very different wealth.
High variance of monthly sales: seasonal product (e.g. ice cream, heaters, festival goods).
SD of exam marks ≈ 0: nearly everyone scored the same ⇒ exam does not separate students (too easy / too hard) ⇒ not very useful.
Low variance of procedure times: teams are consistent, efficient and well-trained (standard procedure).

Explore 3.4Andrea v Billie · laps data

Laps (4–8: 6 adults, 9–13: 10, 14–18: 4). Andrea used mid-values 6, 11, 16 (mean 10.5); Billie used 6.5, 11.5, 16.5 (mean 11). Both got SD = 3.5.

Billie's values are all 0.5 more ⇒ the whole set is translated ⇒ same SD (spread unchanged), but a different mean.

Laps completed are whole numbers ⇒ class boundaries 3.5–8.5, 8.5–13.5, 13.5–18.5 ⇒ mid-values 6, 11, 16. Andrea's mean (10.5) is the better one; for SD either works.

SD vs IQR

Advantages of SD	Disadvantages of SD
much simpler to calculate than IQR	far more affected by extreme values than IQR
data need not be ordered	gives more weight to large deviations than small ones
easier to use algebraically (advanced work)
uses all data values

With outliers / skewed data the IQR (and a box plot) may describe the data better than mean & SD.

5. Calculating from Totals & Combined Sets

Ungrouped: use totals n, Σx, Σx². Grouped: use Σf, Σxf, Σx²f. Rearrange the variance formula to find a missing total.

Worked Example 3.9find Σx²

n = 25, Σx = 275, Var(x) = 7. Find Σx².

Σx²25 − (27525)² = 7 Σx² = 25 × [ 7 + (27525)² ] Σx² = 3200

Combined sets

Treat all values together: add the totals of the two sets, then use the formulae.

Key Point 3.4Mean = Σx + Σyn_x + n_y Var = Σx² + Σy²n_x + n_y − (Σx + Σyn_x + n_y)²

Common mistake: Var(combined) is NOT Var(x)+Var(y)2. Check: {1,2,3,4} var 1¼, {4,5,6} var ⅔, combined {1,2,3,4,4,5,6} var ≈ 2.53 (not 0.96).

Worked Example 3.10combined SD

Boys (10): Σx = 1650, Σx² = 275 490. Girls (15): Σy = 2370, Σy² = 377 835. SD of all 25 children (3 s.f.)?

Σx² + Σy² = 275490 + 377835 = 653 325 Σx + Σy = 1650 + 2370 = 4020 SD = √( 65332525 − (402025)² ) SD = 16.6 cm (Var boys = 324, Var girls = 225, but Var combined ≠ (324+225)/2)

Worked Example 3.11find a missing total from combined SD

120 boys (x): Σx = 7020, Σx² = 424 320. 80 girls (y): Σy² = 352 130. SD of all = 10. Find the girls' mean.

424320 + 352130120 + 80 − (7020 + Σy200)² = 10² 3882.25 − (7020+Σy)²40000 = 100 (7020 + Σy)² = 40000 × 3782.25 = 151 290 000 7020 + Σy = √151290000 = 12 300 (positive root: marks ≥ 0) Σy = 5280 ⇒ mean = 528080 = 66

6. Coded Data — adding / subtracting

Explore 3.5Amber, Buti, Chen

Amber (x): 12 17 11 9 16 | Buti = x − 1: 11 16 10 8 15 | Chen = x + 3: 15 20 14 12 19.

same pattern of spread, just shifted left / right

All three: mean 13 / 12 / 16, but Var = 9.2, SD ≈ 3.03, range = 8, IQR = 6.5 — identical.

Adding/subtracting a constant translates the data: mean shifts, all measures of spread (variance, SD, range, IQR) are unchanged.

Key Point 3.5Var(x) = Var(x − b)
ungrouped: Σx²n − x̄² = Σ(x − b)²n − (Σ(x − b)n)²
grouped: Σx²fΣf − x̄² = Σ(x − b)²fΣf − (Σ(x − b)fΣf)²

Careful: the variance of coded data is the same as for x, but the mean is not! Recover it: x̄ = mean(x − b) + b. Then get Σx = n x̄ and Σx² from the variance formula.

Worked Example 3.12two coded sets combined

8 values of x: Σ(x−10)² = 1490, Σ(x−10) = 100. 12 values of y: Σ(y+5)² = 5139, Σ(y+5) = 234. Find Var of all 20.

x x̄ = 1008 + 10 = 22.5 Σx = 8 × 22.5 = 180 Var(x) = 14908 − (1008)² = 30 Σx²8 − 22.5² = 30 Σx² = 4290

y ȳ = 23412 − 5 = 14.5 Σy = 12 × 14.5 = 174 Var(y) = 513912 − (23412)² = 48 Σy²12 − 14.5² = 48 Σy² = 3099

Var = 4290 + 30998 + 12 − (180 + 17420)² Var = 56.16

Worked Example 3.13SD from coded frequency table

Number of brothers x for 20 girls; table gives x − 1.

x − 1	0	1	2	3	4	total
f	2	4	8	5	1	Σf = 20
(x−1)f	0	4	16	15	4	39
(x−1)²f	0	4	32	45	16	97

SD(x) = SD(x − 1) = √( 9720 − (3920)² ) = 1.023

7. Coded Data — multiplying (ax − b)

Multiplying stretches the set (changes spread); adding just translates it (no effect on spread).

Taxi: cost y = 4x + 3 ($3 fixed + $4/km). Distances {1, 2, 3} ⇒ costs {7, 11, 15}.

spacing in y is 4 × spacing in x: range(y) = 4·range(x)

Key Point 3.6Var(ax − b) = a² × Var(x) ⇔ Var(x) = 1a² × Var(ax − b)
SD(ax − b) = |a| × SD(x) Range, IQR also × |a|

Check with the taxi: Var(x) = ⅔, Var(4x+3) = 32/3 = 16 × ⅔ ✓; SD(x) = ¼ × SD(4x + 3).

Grouped / ungrouped coded totals: Var(x) = 1a² [ Σ(ax − b)²n − (Σ(ax − b)n)² ] (use Σf in place of n and add f for grouped).

Common mistake: The +/− b never changes variance; only the multiplier a does — and it is squared (a²). For SD it is just a.

Worked Example 3.14scaling

SD of brand-name prices is $24. Imitations cost 25% of brand price. Variance of imitation prices?

Var(0.25x) = 0.25² × Var(x) = 0.25² × 24² = 36 (dollars²)

Worked Example 3.15find Σx² from Σ(3x−1) totals

Σ(3x − 1)² = 9136, Σ(3x − 1) = 53, n = 10. Find Σx².

3x̄ − 1 = 5310 ⇒ x̄ = ⅓(5.3 + 1) = 2.1 Var(x) = 13² × Var(3x − 1) Σx²10 − 2.1² = 19 × ( 913610 − 5.3² ) Σx²10 − 4.41 = 98.39 Σx² = 1028

Appendix to 3.3why the two variance formulae agree (n = 3)

Var = ⅓[(x₁ − x̄)² + (x₂ − x̄)² + (x₃ − x̄)²] = ⅓[ x₁² + x₂² + x₃² − 2x̄(x₁ + x₂ + x₃) + 3x̄² ] = Σx²3 − 2x̄·x̄ + x̄² (since Σx/3 = x̄) = Σx²3 − x̄²

Challenge from the book: repeat for n = 2, n = 4, then any n (Σ(x − x̄)² = Σx² − 2x̄Σx + n x̄² = Σx² − n x̄²).

8. Full Chapter Cheat-Sheet

✓ Quick Recap — Measures of variation

Range = max − min (grouped: least/greatest from class boundaries). IQR = Q₃ − Q₁.
Ungrouped: median at (n+1)/2 th value; split halves (discard median if n odd) ⇒ Q₁, Q₃ = medians of halves.
Grouped / CF graph: Q₁ at n/4 (¼Σf), Q₂ at n/2, Q₃ at 3n/4; kth percentile at k% × n; middle 80% = P₉₀ − P₁₀.
Box plot: min, Q₁, Q₂, Q₃, max on a scale. Positive skew: Q₃−Q₂ > Q₂−Q₁; negative: Q₂−Q₁ > Q₃−Q₂; symmetric ≈ equal.
Var = Σx²/n − x̄² = Σ(x−x̄)²/n; grouped: Σx²f/Σf − x̄². SD = √Var. Use the exact mean.
Combined sets: mean = (Σx+Σy)/(nₓ+n_y); Var = (Σx²+Σy²)/(nₓ+n_y) − [(Σx+Σy)/(nₓ+n_y)]².
Coding: Var(x) = Var(x − b); Var(ax − b) = a² Var(x); Var(x) = (1/a²)Var(ax − b). Mean: x̄ = mean(x−b) + b.
SD: uses all data, easy algebra, but affected by outliers. IQR: robust to outliers, uses only the middle 50%.

Before you hand in — checklist: ① ordered the data? ② Σx² vs (Σx)²? ③ exact mean in the SD formula? ④ mid-values for grouped data? ⑤ units (variance = units²)? ⑥ for coded data, did I square a? ⑦ for combined sets, did I add totals rather than average variances?