Coefficient of variation

The standard deviation $\sigma$ is in the same units as the data. So it tells you absolute spread. But it doesn't help compare spreads of different data sets if their means or units differ. The coefficient of variation fixes this.

Definition

$\boxed{\text{CV} = \frac{\sigma}{\bar{x}} \times 100\%.}$

This is the standard deviation expressed as a percentage of the mean. It is a pure number , no units. So you can compare CVs across data sets with different units.

Two data sets are equally consistent if their CVs are equal. The one with the smaller CV is "more consistent" (less variable relative to its mean).

When CV is appropriate

• Data should be on a ratio scale with a meaningful zero (length, weight, time, count).
• Mean should be positive and nonzero. (If $\bar{x} = 0$, CV is undefined.)
• Useful for comparing variability of measurements with very different magnitudes.

Comparing data sets

If data set A has $\bar{x}_A = 100$ and $\sigma_A = 20$, CV is $20\%$. If data set B has $\bar{x}_B = 50$ and $\sigma_B = 15$, CV is $30\%$.

Both data sets have similar absolute SDs, but B is more variable relative to its mean. So A is more consistent.

Worked examples

Example 1. A factory's monthly outputs (units): mean $500$, SD $25$. Find CV.

$\text{CV} = 25/500 \times 100 = 5\%$.

Example 2. Compare consistency:

• Worker A: mean wage $200$, SD $24$.
• Worker B: mean wage $300$, SD $30$.

CV_A $= 24/200 \times 100 = 12\%$. CV_B $= 30/300 \times 100 = 10\%$.

Worker B is more consistent (lower CV).

Example 3. A cricket batsman scored in $5$ matches: $40, 50, 70, 60, 55$.

$\bar{x} = 275/5 = 55$. Squared deviations: $225, 25, 225, 25, 0$. Sum $= 500$. $\sigma^2 = 100$, $\sigma = 10$.

CV $= 10/55 \times 100 \approx 18.18\%$.

Example 4. Two cricket batters' scores (in $10$ matches):

• Player X: mean $60$, SD $12$.
• Player Y: mean $50$, SD $10$.

CV_X $= 12/60 \times 100 = 20\%$. CV_Y $= 10/50 \times 100 = 20\%$.

Equally consistent.

Example 5. A class has $30$ students with mean marks $75$ and SD $9$. Another with $20$ students has mean $60$ and SD $8$. Which is more consistent?

CV1 $= 9/75 \times 100 = 12\%$. CV2 $= 8/60 \times 100 = 13.33\%$.

Class 1 is more consistent.

Try it yourself

1. SD of marks of $7$ students: $\sigma = 4$, mean $20$. Find CV.
2. A: mean income $30{,}000$, SD $3000$. B: mean income $50{,}000$, SD $4000$. Which is more consistent?
3. Heights of two groups: A , mean $160$ cm, SD $8$ cm; B , mean $170$ cm, SD $7$ cm. Compare consistency.
4. For data $5, 10, 15, 20, 25$, find mean, SD, and CV.
5. Two students score in $5$ tests with means $80, 75$ and SDs $4, 5$ respectively. Which is more consistent?
6. The CV of a data set is $40\%$ and mean is $60$. Find SD.
7. The CV of two distributions are $25\%$ and $30\%$. The SDs are $12$ and $15$. Find their means.
8. A's runs over $10$ innings have mean $55$ SD $11$. B's are mean $50$ SD $9$. Comment.
9. Mean and SD of $7$ items are $50, 6$. After adding $50$ to each, find new CV.
10. Mean and SD of marks for boys are $80, 12$; for girls $76, 10$. Combined mean is $78$ (assume equal numbers). Find combined SD and combined CV.
11. CV is more useful than SD for comparing because: explain in one sentence.
12. Data set has mean $0$. Why is CV undefined?

Pitfalls / Tricks

• Always multiply by $100$ to express CV as percentage.
• CV is useful only when $\bar{x}$ is positive and nonzero.
• Lower CV = more consistent.
• Insight. CV measures relative spread. It's the right number when comparing variability across different magnitudes.