Following are the marks obtained, out of 100, by two students Ravi and Hashina in 10 tests. | Ravi | 25 | 50 | 45 | 30 | 70 | 42 | 36 | 48 | 35 | 60 | |---|---|---|---|---|---|---|---|---|---|---| | Hashina | 10 | 70 | 50 | 20 | 95 | 55 | 42 | 60 | 48 | 80 | Who is more intelligent and who is more consistent?

> [!TLDR] > Hashina has the higher mean (53 vs 44.1), so she is more intelligent; Ravi has the lower coefficient of variation (29.49% vs 45.94%), so he is more consistent. ### The idea - **Intelligence** → overall performance → compare the **mean** score (higher is better). - **Consistency** → how little the scores vary → compare the **coefficient of variation** $\text{C.V.}=\dfrac{\sigma}{\bar{x}}\times100$ (lower is more consistent). C.V. is the correct measure here because the two students have different means. ### Step 1 — Means $$ \bar{x}_R=\frac{25+50+45+30+70+42+36+48+35+60}{10}=\frac{441}{10}=44.1 $$ $$ \bar{x}_H=\frac{10+70+50+20+95+55+42+60+48+80}{10}=\frac{530}{10}=53 $$ Hashina's mean is higher, so **Hashina is more intelligent**. ### Step 2 — Ravi's standard deviation | $x_i$ | $x_i-\bar{x}_R$ | $(x_i-\bar{x}_R)^{2}$ | |---|---|---| | 25 | -19.1 | 364.81 | | 50 | 5.9 | 34.81 | | 45 | 0.9 | 0.81 | | 30 | -14.1 | 198.81 | | 70 | 25.9 | 670.81 | | 42 | -2.1 | 4.41 | | 36 | -8.1 | 65.61 | | 48 | 3.9 | 15.21 | | 35 | -9.1 | 82.81 | | 60 | 15.9 | 252.81 | | **Sum** | | **1690.90** | $$ \sigma_R=\sqrt{\frac{1690.90}{10}}=\sqrt{169.09}\approx 13.00. $$ ### Step 3 — Hashina's standard deviation | $x_i$ | $x_i-\bar{x}_H$ | $(x_i-\bar{x}_H)^{2}$ | |---|---|---| | 10 | -43 | 1849 | | 70 | 17 | 289 | | 50 | -3 | 9 | …

Following are the marks obtained, out of 100, by t – Class 11 Mathematics

NCERT Exemplar · Q21

Following are the marks obtained, out of 100, by two students Ravi and Hashina in 10 tests.

Ravi	25	50	45	30	70	42	36	48	35	60
Hashina	10	70	50	20	95	55	42	60	48	80

Who is more intelligent and who is more consistent?

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Hashina has the higher mean (53 vs 44.1), so she is more intelligent; Ravi has the lower coefficient of variation (29.49% vs 45.94%), so he is more consistent.

The idea

Intelligence → overall performance → compare the mean score (higher is better).
Consistency → how little the scores vary → compare the coefficient of variation $\text{C.V.}=\dfrac{\sigma}{\bar{x}}\times100$ (lower is more consistent). C.V. is the correct measure here because the two students have different means.

Step 1 — Means

\bar{x}_R=\frac{25+50+45+30+70+42+36+48+35+60}{10}=\frac{441}{10}=44.1

\bar{x}_H=\frac{10+70+50+20+95+55+42+60+48+80}{10}=\frac{530}{10}=53

Hashina's mean is higher, so Hashina is more intelligent.

Step 2 — Ravi's standard deviation

$x_i$	$x_i-\bar{x}_R$	$(x_i-\bar{x}_R)^{2}$
25	-19.1	364.81
50	5.9	34.81
45	0.9	0.81
30	-14.1	198.81
70	25.9	670.81
42	-2.1	4.41
36	-8.1	65.61
48	3.9	15.21
35	-9.1	82.81
60	15.9	252.81
Sum		1690.90

\sigma_R=\sqrt{\frac{1690.90}{10}}=\sqrt{169.09}\approx 13.00.

Step 3 — Hashina's standard deviation

$x_i$	$x_i-\bar{x}_H$	$(x_i-\bar{x}_H)^{2}$
10	-43	1849
70	17	289
50	-3	9

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CA Foundation 2025Set jan-20251 markMCQ
$Q. For a distribution the mean is 30. The standard deviation is 2, then coefficient of variation is (A) 6.67% (B) 9.45% (C) 7.5% (D) 2.5%$
›Reveal solutionSolution

$CV=\dfrac{\sigma}{\bar{x}}\times100=\dfrac{2}{30}\times100=6.67\%$ .

Step 1 — Formula for coefficient of variation
$CV = \dfrac{\sigma}{\bar{x}}\times 100\%$
Step 2 — Substitute the given values

With standard deviation $\sigma=2$ and mean $\bar{x}=30$ :
$CV = \dfrac{2}{30}\times100 = \dfrac{200}{30} = 6.67\%.$
Why the other options are wrong: $7.5\%$ (C) comes from $\tfrac{30}{2}$ or a mis-division; $2.5\%$ (D) forgets to divide by the mean correctly; $9.45\%$ (B) does not follow from the data. …
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CA Foundation 2025Set may-20251 markMCQ
$N = 14$
›Reveal solutionSolution
Mean $=280/14=20$ ; C.V. $=(\sigma/\bar x)\times100=(3/20)\times100=\mathbf{15}$ .

Step 1 — Find the mean
$\bar x = \dfrac{\Sigma fx}{N} = \dfrac{280}{14} = 20.$
Step 2 — Apply the coefficient-of-variation formula
$\text{C.V.} = \dfrac{\sigma}{\bar x}\times 100 = \dfrac{3}{20}\times 100 = 15.$
Step 3 — Check the traps

(C) 4.67 comes from wrongly dividing $\Sigma fx$ -type numbers or mixing up terms.

(A) 20 is just the mean, not the C.V.

(D) Zero would need $\sigma = 0$ (no variation), which is false here. …
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CA Foundation 2025Set sep-20251 markMCQ
$Q. Find the Coefficient of variation for the following numbers : 7, 5, 9, 3, 6 (A) 33.33 (B) 66.66 (C) 3 (D) 300$
›Reveal solutionSolution

CV = (SD ÷ Mean) × 100 = (2 ÷ 6) × 100 = 33.33%.

Step 1 — Mean

Data: 7, 5, 9, 3, 6.
$\bar x = \frac{7+5+9+3+6}{5} = \frac{30}{5} = 6$
Step 2 — Standard deviation

Deviations from 6: $1, -1, 3, -3, 0$ ; squared: $1, 1, 9, 9, 0$ , sum $= 20$ .
$\sigma = \sqrt{\frac{\sum (x-\bar x)^2}{n}} = \sqrt{\frac{20}{5}} = \sqrt{4} = 2$
Step 3 — Coefficient of variation

$CV = \frac{\sigma}{\bar x}\times 100 = \frac{2}{6}\times 100 = 33.33\%$ …
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CA Foundation 2022Set dec-20221 markMCQ
$Q. If the coefficient of variation and standard deviation are 30 and 12 respectively, then the arithmetic mean of the distribution is (A) 40 (B) 36 (C) 25 (D) 19$
›Reveal solutionSolution

Mean = SD × 100 / CV = 12 × 100 / 30 = 40.

Step 1 — Write the CV formula
$CV = \frac{\sigma}{\bar{X}}\times 100$
Step 2 — Solve for the mean
$30 = \frac{12}{\bar{X}}\times 100 \;\Rightarrow\; \bar{X} = \frac{12\times 100}{30} = 40$
Watch out

CV is expressed as a percentage, so keep the ×100. Dropping it (mean = 12/30 = 0.4) gives a nonsensical answer. …

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Concept understanding — Coefficient of Variation

Coefficient of Variation

The idea

Step 1 — Means

Step 2 — Ravi's standard deviation

Step 3 — Hashina's standard deviation

Step 1 — Formula for coefficient of variation

Step 2 — Substitute the given values

Step 1 — Find the mean

Step 2 — Apply the coefficient-of-variation formula

Step 3 — Check the traps

Step 1 — Mean

Step 2 — Standard deviation

Step 3 — Coefficient of variation

Step 1 — Write the CV formula

Step 2 — Solve for the mean

The idea

Step 1 — Means

Step 2 — Ravi's standard deviation

Step 3 — Hashina's standard deviation

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