Mean (M) and Standard Deviation (S) of different datasets are given below.…
2025
Mean (M) and Standard Deviation (S) of different datasets are given below. Compute the coefficient of variation of each dataset and arrange in ascending order.
A. M = 60, S = 14
B. M = 70, S = 16
C. M = 80, S = 5
D. M = 90, S = 4
- A.
D, B, C, A
- B.
A, C, B, D
- C.
C, A, D, B
- D.
D, C, B, A
Show answer & explanation
Correct answer: D
The coefficient of variation (CV) is a relative measure of dispersion, defined as CV = (Standard Deviation / Mean) × 100. Because it expresses the standard deviation as a percentage of the mean, CV lets datasets with different means and different units be compared on the same relative scale — the raw standard deviation alone cannot do this.
Apply this formula to each of the four datasets:
Dataset | Mean (M) | S.D. (S) | CV = (S/M) × 100 |
|---|---|---|---|
A | 60 | 14 | ≈ 23.33% |
B | 70 | 16 | ≈ 22.86% |
C | 80 | 5 | 6.25% |
D | 90 | 4 | ≈ 4.44% |
Arranging these four percentages from smallest to largest gives: 4.44% (D) < 6.25% (C) < 22.86% (B) < 23.33% (A) — that is, the ascending sequence D, C, B, A.
Cross-check by comparing the CVs pairwise as fractions, using cross-multiplication instead of decimals (this avoids any rounding error):
D vs C: 4 × 80 = 320 and 5 × 90 = 450. Since 320 < 450, 4/90 < 5/80 — dataset D's CV is indeed lower than dataset C's.
C vs B: 5 × 70 = 350 and 16 × 80 = 1280. Since 350 < 1280, 5/80 < 16/70 — dataset C's CV is indeed lower than dataset B's.
B vs A: 16 × 60 = 960 and 14 × 70 = 980. Since 960 < 980, 16/70 < 14/60 — dataset B's CV is indeed lower than dataset A's.
Every pairwise check agrees with the direct calculation, confirming the ascending order D, C, B, A.