Standard deviation and coefficient of variation

Business Statistics · Unit III · Topic 8 of 15 · 5 min read

Standard deviation by the actual-mean, assumed-mean and step-deviation methods, variance and CV, and the set comparisons: the January 2026 grouped table, Sachin versus Sidhu, and the two factories' wages.

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On this page

1. Meaning and formulas
2. Worked: grouped data (January 2026)
3. Worked: Sachin and Sidhu (January 2025)
4. Worked: two factories (June 2025)
How this is examined

Cases checked to exist. Read the judgment before quoting a holding.

Every case cited here has been checked to exist — the citation resolves and names the right parties. The propositions attributed to each case have not been verified line by line against the judgment. Case names link through to the full judgment: read it before quoting a holding in an examination answer, and never carry a proposition from notes into court on trust.

The examinable core. The ten-mark Unit III problem is usually a comparison: “who is the better scorer, who is more consistent?” or “which factory pays more, which is more uniform?”. The higher mean answers the first part; the lower coefficient of variation answers the second. Never compare consistency by SD alone when the means differ.

1. Meaning and formulas

Standard deviation (Karl Pearson, 1893) is the square root of the mean of the squared deviations from the arithmetic mean. It is denoted σ; its square σ² is the variance. Squaring removes the signs legitimately, which is why SD is the best measure of dispersion.

Individual, actual mean: σ = √(Σ(x − x̄)² / n).
Individual, assumed mean: σ = √(Σd²/n − (Σd/n)²), d = x − A.
Grouped, step deviation: σ = h × √(Σfd′²/N − (Σfd′/N)²), d′ = (m − A)/h.
Coefficient of SD = σ / x̄. Coefficient of variation (CV) = σ / x̄ × 100 (Karl Pearson). The series with the lower CV is more consistent, uniform, stable or homogeneous.

Properties: SD is unaffected by change of origin (adding a constant) but affected by change of scale; it is least when taken from the mean; for a normal distribution x̄ ± 1σ covers 68.27% of items, ± 2σ 95.45%, ± 3σ 99.73%; combined SD of two groups can be found.

Merits: rigidly defined; based on all items; capable of algebraic treatment; least affected by sampling fluctuations; the basis of correlation, regression, skewness and sampling theory. Demerits: harder to compute; gives more weight to extreme items because deviations are squared; cannot compare series in different units (use CV).

2. Worked: grouped data (January 2026)

Classes 0–5 to 35–40 with f = 3, 9, 15, 22, 26, 19, 13, 8. Calculate CV. A = 22.5, h = 5.

Class	f	m	d′	fd′	fd′²
0–5	3	2.5	−4	−12	48
5–10	9	7.5	−3	−27	81
10–15	15	12.5	−2	−30	60
15–20	22	17.5	−1	−22	22
20–25	26	22.5	0	0	0
25–30	19	27.5	1	19	19
30–35	13	32.5	2	26	52
35–40	8	37.5	3	24	72
Total	115			−22	354

x̄ = 22.5 + (−22/115) × 5 = 22.5 − 0.957 = 21.54.
σ = 5 × √(354/115 − (22/115)²) = 5 × √(3.0783 − 0.0366) = 5 × √3.0417 = 5 × 1.7440 = 8.72.
CV = 8.72 / 21.54 × 100 = 40.48%.

The 100-mark January 2026 paper set the same on 0–5 to 25–30 (f = 3, 6, 9, 8, 5, 7): with A = 17.5, Σfd′ = −11, Σfd′² = 93, N = 38, giving x̄ = 16.05, σ = 7.69, CV = 47.89%.

3. Worked: Sachin and Sidhu (January 2025)

Sachin: 12, 115, 6, 73, 7, 19, 119, 36, 84, 29. Sidhu: 47, 12, 76, 42, 4, 51, 37, 48, 13, 0. Who is the better scorer? Who is more consistent?

Sachin: Σx = 500, x̄ = 50. Sidhu: Σx = 330, x̄ = 33. Deviations from the actual mean:

Sachin x	d	d²	Sidhu x	d	d²
12	−38	1,444	47	14	196
115	65	4,225	12	−21	441
6	−44	1,936	76	43	1,849
73	23	529	42	9	81
7	−43	1,849	4	−29	841
19	−31	961	51	18	324
119	69	4,761	37	4	16
36	−14	196	48	15	225
84	34	1,156	13	−20	400
29	−21	441	0	−33	1,089
500	0	17,498	330	0	5,462

Sachin: σ = √(17,498/10) = √1,749.8 = 41.83; CV = 41.83/50 × 100 = 83.66%.
Sidhu: σ = √(5,462/10) = √546.2 = 23.37; CV = 23.37/33 × 100 = 70.82%.

Conclusion: Sachin is the better scorer (higher average, 50 against 33); Sidhu is more consistent (lower CV). The August 2024 version (batsmen A and B) works the same way: A has x̄ = 46.1, σ = 25.50, CV = 55.32%; B has x̄ = 50, σ = 24.43, CV = 48.86%, so B is both the better scorer and more consistent.

4. Worked: two factories (June 2025)

Daily wages 120–130 to 180–190. Factory X: 15, 30, 44, 60, 30, 14, 7. Factory Y: 25, 40, 60, 35, 12, 15, 5. A = 155, h = 10, d′ = −3 to +3.

d′	f_X	fd′	fd′²	f_Y	fd′	fd′²
−3	15	−45	135	25	−75	225
−2	30	−60	120	40	−80	160
−1	44	−44	44	60	−60	60
0	60	0	0	35	0	0
1	30	30	30	12	12	12
2	14	28	56	15	30	60
3	7	21	63	5	15	45
Total	200	−70	448	192	−158	562

X: x̄ = 155 − 70/200 × 10 = Rs. 151.50; σ = 10 × √(2.24 − 0.1225) = 10 × √2.1175 = 14.55; CV = 9.61%.
Y: x̄ = 155 − 158/192 × 10 = Rs. 146.77; σ = 10 × √(2.9271 − 0.6772) = 10 × √2.2499 = 15.00; CV = 10.22%.

Conclusion: (i) Factory X pays higher wages, both on average (Rs. 151.50 against Rs. 146.77) and in total (200 × 151.50 = Rs. 30,300 against 192 × 146.77 = about Rs. 28,180). (ii) Factory X also has the more consistent wage structure (CV 9.61% against 10.22%).

Trap. In the step-deviation formula the correction term is (Σfd′/N)², squared after dividing. For Factory Y it is (158/192)² = 0.6772, not 158²/192.

How this is examined

CV from a grouped table: January 2026 (and SD with CV in the 100-mark January 2026 paper). Better scorer and more consistent: Sachin and Sidhu (January 2025), batsmen A and B (August 2024). Two factories: June 2025. Start the answer by stating the decision rule (higher mean, lower CV), then the table, then a two-line conclusion.

Now try the real questions: Business Statistics past question papers.