Standard deviation tells you the spread of one data set in its own units. But suppose you want to compare the consistency of two sets that are measured differently, or that simply have very different averages — say monthly rainfall against daily temperature, or a batsman who averages $80$ against one who averages $25$. A raw standard deviation cannot settle that comparison fairly, because a larger mean naturally tends to come with a larger spread.
Begin with the easiest case. When two distributions have equal means, the means cancel out of any fair comparison, so you simply compare standard deviations directly: the set with the smaller $\sigma$ is the more consistent. This is the situation in many textbook problems where, for example, two sections sit the same test and score the same average.
The general fix, when the means differ, is to express the spread as a percentage of the mean. This is the coefficient of variation (C.V.), a unit-free, relative measure of dispersion:
$$\text{C.V.}=\dfrac{\sigma}{\bar{x}}\times 100\quad(\bar{x}\neq 0)$$
Because the units cancel in the ratio $\dfrac{\sigma}{\bar{x}}$, the C.V. is a pure number (a percentage) and can compare any two distributions directly. The rule is simple:
$$\text{smaller C.V.}\ \Rightarrow\ \text{more consistent (less variable)}$$
The data set with the greater C.V. is the more variable (less consistent); the one with the smaller C.V. is the more uniform or reliable. The comparison below shows two players with the same average but different consistency:
Correcting wrong observations. A frequent exam task gives a mean and standard deviation that were computed with one or more misread values, and asks for the corrected figures. The method is mechanical and rests on two running totals, $\sum x_i$ and $\sum x_i^2$, both of which you can recover:
$$\sum x_i=n\bar{x},\qquad \sum x_i^2=n\big(\sigma^2+\bar{x}^2\big)$$
Then subtract each wrong value and add the right one to both totals — $\sum x_i^{\text{corr}}=\sum x_i-(\text{wrong})+(\text{correct})$ and likewise for the squares — and recompute the corrected mean and standard deviation from
$$\bar{x}_{\text{corr}}=\dfrac{\sum x_i^{\text{corr}}}{n},\qquad \sigma_{\text{corr}}=\sqrt{\dfrac{\sum x_i^{2,\text{corr}}}{n}-\bar{x}_{\text{corr}}^{\,2}}$$
The number of observations $n$ does not change (you are correcting values, not adding or removing them), which is the key to why both totals can simply be patched.
Deeper Insight — why "relative" spread beats "absolute" spread: The coefficient of variation exists because dispersion only means something in proportion to the scale you are working at. A standard deviation of two kilograms is enormous for the weights of medicines but trivial for the weights of adults; the same number tells opposite stories depending on the average it is set against. By dividing by the mean we strip out that scale, leaving a clean statement of "how much does the data wobble relative to its own typical size?" This is exactly why C.V. is the right tool for comparing consistency across players, factories, crops or share prices that operate at different levels. Two cautions worth carrying forward: the C.V. is undefined when $\bar{x}=0$ and becomes unstable when the mean is close to zero, and it is only meaningful for data on a ratio scale with a true zero — comparing the C.V. of temperatures in Celsius would be nonsense, since the choice of zero is arbitrary. Used within those limits, the C.V. turns the vague word "consistent" into a number you can defend.
A data set has mean $\bar{x}=40$ and standard deviation $\sigma=8$. Find its coefficient of variation.
Solution- Apply $\text{C.V.}=\dfrac{\sigma}{\bar{x}}\times 100$.
- $\text{C.V.}=\dfrac{8}{40}\times 100=20\%$.
Answer: C.V. $= 20\%$.
Two firms A and B pay mean wages of $\bar{x}_A=5250$ and $\bar{x}_B=5200$, with standard deviations $\sigma_A=900$ and $\sigma_B=816$. Which firm has more consistent wages?
Solution- Firm A: $\text{C.V.}_A=\dfrac{900}{5250}\times 100\approx 17.14\%$.
- Firm B: $\text{C.V.}_B=\dfrac{816}{5200}\times 100\approx 15.69\%$.
- Firm B has the smaller coefficient of variation.
Answer: Firm B has more consistent wages (lower C.V. $\approx 15.69\%$).
Batsman X averages $50$ runs with $\sigma=5$; batsman Y averages $40$ runs with $\sigma=5$. Both have the same standard deviation — who is the more consistent scorer?
Solution- Equal $\sigma$ does not mean equal consistency, because the means differ — compare C.V.
- X: $\text{C.V.}=\dfrac{5}{50}\times 100=10\%$.
- Y: $\text{C.V.}=\dfrac{5}{40}\times 100=12.5\%$.
- X has the smaller C.V.
Answer: Batsman X is more consistent (C.V. $=10\%$ vs $12.5\%$).
For the data $5, 7, 9, 11, 13$, find the mean, standard deviation, and coefficient of variation.
Solution- Mean $\bar{x}=\dfrac{5+7+9+11+13}{5}=\dfrac{45}{5}=9$.
- Deviations $(x_i-9)$: $-4, -2, 0, 2, 4$; squares $16, 4, 0, 4, 16$ summing to $40$.
- $\sigma=\sqrt{\dfrac{40}{5}}=\sqrt{8}=2\sqrt{2}\approx 2.83$.
- $\text{C.V.}=\dfrac{2.83}{9}\times 100\approx 31.4\%$.
Answer: $\bar{x}=9$, $\sigma\approx 2.83$, C.V. $\approx 31.4\%$.
The C.V. of a distribution is $25\%$ and its standard deviation is $12.5$. Find the mean.
Solution- Rearrange $\text{C.V.}=\dfrac{\sigma}{\bar{x}}\times 100$ to give $\bar{x}=\dfrac{\sigma}{\text{C.V.}}\times 100$.
- $\bar{x}=\dfrac{12.5}{25}\times 100=50$.
Answer: Mean $\bar{x}=50$.
Two crops give yields with $\bar{x}_1=60$, $\sigma_1=9$ and $\bar{x}_2=75$, $\sigma_2=12$. Which crop's yield is more stable?
Solution- Crop 1: $\text{C.V.}_1=\dfrac{9}{60}\times 100=15\%$.
- Crop 2: $\text{C.V.}_2=\dfrac{12}{75}\times 100=16\%$.
- Crop 1 has the smaller C.V., so its yield varies less relative to its mean.
Answer: Crop 1 is more stable (C.V. $=15\%$ vs $16\%$).
The scores of two batsmen A and B in $10$ innings each give: A — mean $50$, S.D. $5.1$; B — mean $44$, S.D. $6.16$. Decide who is the better run-getter and who is the more consistent.
Solution- "Better run-getter" is decided by the higher mean: A's $50$ beats B's $44$.
- "More consistent" is decided by the smaller C.V.
- A: $\text{C.V.}_A=\dfrac{5.1}{50}\times 100=10.2\%$.
- B: $\text{C.V.}_B=\dfrac{6.16}{44}\times 100=14\%$.
- A has both the higher mean and the lower C.V.
Answer: Batsman A is the better and the more consistent run-getter (C.V. $10.2\%$ vs $14\%$).
Two distributions have the same mean $\bar{x}=30$. The first has variance $25$ and the second variance $49$. Which is more consistent, and what are the two C.V.s?
Solution- Equal means, so direct comparison of $\sigma$ already decides it: $\sigma_1=\sqrt{25}=5$, $\sigma_2=\sqrt{49}=7$.
- The first, with smaller $\sigma$, is more consistent.
- $\text{C.V.}_1=\dfrac{5}{30}\times 100\approx 16.67\%$; $\text{C.V.}_2=\dfrac{7}{30}\times 100\approx 23.33\%$.
Answer: The first distribution is more consistent (C.V. $\approx 16.67\%$ vs $\approx 23.33\%$).
The mean of $100$ observations is $50$ and their standard deviation is $5$. If $\sum x_i^2$ is required, find it; then find the corrected mean and standard deviation if it is later found that an observation $50$ was misread and should have been $80$.
Solution- $\sum x_i=100\times 50=5000$ and $\sum x_i^2=n(\sigma^2+\bar{x}^2)=100(25+2500)=252500$.
- Correct the totals: $\sum x_i^{\text{corr}}=5000-50+80=5030$, so $\bar{x}_{\text{corr}}=\dfrac{5030}{100}=50.3$.
- $\sum x_i^{2,\text{corr}}=252500-50^2+80^2=252500-2500+6400=256400$.
- $\sigma_{\text{corr}}^2=\dfrac{256400}{100}-(50.3)^2=2564-2530.09=33.91$.
- $\sigma_{\text{corr}}=\sqrt{33.91}\approx 5.82$.
Answer: Corrected mean $=50.3$ and corrected standard deviation $\approx 5.82$.
The mean and standard deviation of $20$ observations are found to be $10$ and $2$. On rechecking, two observations were wrong: $8$ should have been $12$ and $6$ should have been $4$. Find the corrected mean and standard deviation.
Solution- $\sum x_i=20\times 10=200$; $\sum x_i^2=n(\sigma^2+\bar{x}^2)=20(4+100)=2080$.
- Net change to $\sum x_i$: $-8+12-6+4=+2$, so $\sum x_i^{\text{corr}}=202$ and $\bar{x}_{\text{corr}}=\dfrac{202}{20}=10.1$.
- Change to $\sum x_i^2$: $-8^2+12^2-6^2+4^2=-64+144-36+16=60$, so $\sum x_i^{2,\text{corr}}=2140$.
- $\sigma_{\text{corr}}^2=\dfrac{2140}{20}-(10.1)^2=107-102.01=4.99$.
- $\sigma_{\text{corr}}=\sqrt{4.99}\approx 2.23$.
Answer: Corrected mean $=10.1$ and corrected standard deviation $\approx 2.23$.
The coefficient of variation of two distributions are $60\%$ and $70\%$, and their standard deviations are $21$ and $16$ respectively. Find their means.
Solution- Rearrange C.V. $=\dfrac{\sigma}{\bar{x}}\times 100$ to $\bar{x}=\dfrac{\sigma}{\text{C.V.}}\times 100$.
- First: $\bar{x}_1=\dfrac{21}{60}\times 100=35$.
- Second: $\bar{x}_2=\dfrac{16}{70}\times 100\approx 22.86$.
Answer: The means are $\bar{x}_1=35$ and $\bar{x}_2\approx 22.86$.