Coefficient of variation and interpreting spread – Measures of Dispersion (Senior 4 Mathematics) | Ganzaa

Measures of Dispersion

Lessons

Senior 4 · Mathematics

Measures of Dispersion

3 lessons

Lesson 3 of 3

Coefficient of variation and interpreting spread

25 min

Is 60 000 Frw a lot of spread?

At a livestock market, the prices of cows have a standard deviation of about 60 000 Frw, and the prices of chickens a standard deviation of about 1 500 Frw (illustrative figures). It is tempting to say cow prices vary much more. But a cow costs around 450 000 Frw and a chicken around 6 000 Frw. A difference of 1 500 Frw on a chicken is a quarter of its price; a difference of 60 000 Frw on a cow is less than a seventh. To compare spreads fairly when the means are very different, or when the data are in different units, we measure the spread relative to the mean.

Definition

Coefficient of variation (CV): the standard deviation expressed as a percentage of the mean,

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

It has no units, so it can compare data measured in francs with data measured in kilograms. A smaller CV means the data are more consistent (less variable relative to their size).

Condition: use it only for data that cannot be negative and whose mean is clearly above zero (prices, masses, times, counts).

Example: cows: $\frac{60\,000}{450\,000} \times 100\% \approx 13.3\%$ ; chickens: $\frac{1500}{6000} \times 100\% = 25\%$ . Relative to their price, chicken prices vary almost twice as much.

Worked example 1: which student is more consistent?

Over a term, Keza's weekly quiz marks (out of 20) had mean 15 and standard deviation 2.4. Gatete's marks in a longer test (out of 100) had mean 68 and standard deviation 8.5. Whose marks were more consistent?

Solution

\text{CV}_{\text{Keza}} = \frac{2.4}{15} \times 100\% = 16\%, \qquad \text{CV}_{\text{Gatete}} = \frac{8.5}{68} \times 100\% = 12.5\% .

Gatete's marks have the smaller CV, so relative to their level they were more consistent, even though his standard deviation (8.5) is larger than Keza's (2.4). Comparing 8.5 with 2.4 directly would be comparing marks out of 100 with marks out of 20.

Warning

When the CV misleads. Temperatures in °C can be 0 or negative, so their mean can be near zero and the CV meaningless (a mean of $0.5^\circ\mathrm{C}$ with $\sigma = 3^\circ\mathrm{C}$ gives a "CV" of 600%). Also notice what a change of data does: multiplying every value by $k \gt 0$ multiplies both $\sigma$ and $\bar{x}$ by $k$ , so the CV does not change (converting francs to dollars leaves it the same); but adding a constant changes $\bar{x}$ and not $\sigma$ , so it does change the CV.

Interpreting real data: district population density in 2022

The 2022 Population and Housing Census, run by the National Institute of Statistics of Rwanda (NISR), gives the population density of every district: the number of residents per square kilometre. Here are the districts of two provinces:

Southern Province	Density (per km²)	Western Province	Density (per km²)
Nyanza	545	Karongi	482
Gisagara	584	Rutsiro	565
Nyaruguru	409	Rubavu	1614
Huye	657	Nyabihu	642
Nyamagabe	441	Ngororero	551
Ruhango	573	Rusizi	871
Muhanga	555	Nyamasheke	632
Kamonyi	683

Source: NISR, Fifth Population and Housing Census, Rwanda 2022: Main Indicators, Table 1.

Worked example 2: summarising and comparing the two provinces

Southern ( $n = 8$ ): $\sum x = 4447$ , $\sum x^2 = 2\,534\,335$ .

\bar{x} = \frac{4447}{8} \approx 556, \qquad \sigma^2 = \frac{2\,534\,335}{8} - 555.875^2 \approx 7795, \qquad \sigma \approx 88.3, \qquad \text{CV} \approx 15.9\% .

Western ( $n = 7$ ): $\sum x = 5357$ , $\sum x^2 = 5\,030\,375$ .

\bar{x} = \frac{5357}{7} \approx 765, \qquad \sigma^2 = \frac{5\,030\,375}{7} - 765.29^2 \approx 132\,963, \qquad \sigma \approx 365, \qquad \text{CV} \approx 47.6\% .

Quartiles. Southern, ordered: 409, 441, 545, 555, 573, 584, 657, 683, so $Q_{1} = 441 + 0.25 (104) = 467$ , median $= 564$ , $Q_{3} = 584 + 0.75 (73) = 638.75$ , IQR $\approx 172$ . Western, ordered: 482, 551, 565, 632, 642, 871, 1614, so $Q_{1} = 551$ , median $= 632$ , $Q_{3} = 871$ , IQR $= 320$ .

Outlier test for Western: upper fence $= 871 + 1.5 (320) = 1351$ . Rubavu (1614) is above it, so it is an outlier.

Without Rubavu the six Western districts have $\bar{x} \approx 624$ , $\sigma \approx 123$ and $\text{CV} \approx 19.7\%$ : only a little more variable than the South.

Writing the conclusion

A good statistical conclusion compares centre, spread and shape, names any outliers, and explains them in context. For the census data:

In 2022 the typical Western Province district was a little more densely populated than the typical Southern district (medians 632 and 564 people per km²). Southern districts were very consistent (CV about 16%). The Western figures were far more spread out (CV about 48%), but almost all of this comes from one outlier, Rubavu (1614 per km²), a district with a large town. Without Rubavu the Western CV falls to about 20%, close to the South's.

Notice which measures were chosen. Because of the outlier, the Western mean (765) is higher than six of its seven districts, so the median and IQR describe a typical Western district better than the mean and standard deviation.

The data are…	Describe the centre with	Describe the spread with
roughly symmetric, no outliers	the mean	the standard deviation (and CV to compare)
skewed, or with outliers	the median	the IQR
very small groups, or categories	the mode or a table	the range, with care

One more caution: the province's own density (Western: 693 per km²) is not the mean of the district densities (765), because the districts have different areas. A mean of rates is not the overall rate.

Tip

For data that are roughly bell-shaped, about two thirds of the values lie within one standard deviation of the mean, and about 95% within two. So "mean 150 minutes, standard deviation 4 minutes" tells a passenger that almost every trip takes between about 142 and 158 minutes. Use this as a quick sense check of any standard deviation you calculate.

zedstatistics: What is the Coefficient of Variation? Watch the examples that compare data on very different scales, and the warning about data whose mean is near zero.

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Practice

Try each one before opening the answers. They get harder as you go.

A data set has mean 80 and standard deviation 6. Find its CV.
Bags of maize flour from machine P have mean 2.02 kg and standard deviation 0.04 kg; bags of sugar from machine Q have mean 1.01 kg and standard deviation 0.03 kg. Which machine fills more consistently?
A set of prices has mean 2000 Frw and CV 15%. Find the standard deviation. What happens to the mean, the standard deviation and the CV if every price rises by 10%?
The same prices all rise by 300 Frw instead. Find the new CV.
The marks of two classes are: class X mean 54, standard deviation 9; class Y mean 61, standard deviation 15. Write two sentences comparing the classes.
From the census table above, find the mean and standard deviation of the three City of Kigali district densities in 2022: Nyarugenge 2830, Gasabo 2056, Kicukiro 2944. Compare their CV with the Southern Province's.
A data set of household incomes has mean 180 000 Frw and median 95 000 Frw. What does this suggest about its shape, and which measure of spread would you report?

Show answersHide answers

$\frac{6}{80} \times 100\% = 7.5\%$ .
P: $\frac{0.04}{2.02} \approx 1.98\%$ ; Q: $\frac{0.03}{1.01} \approx 2.97\%$ . Machine P is more consistent.
$\sigma = 0.15 \times 2000 = 300$ Frw. A 10% rise multiplies every price by 1.1: mean 2200 Frw, standard deviation 330 Frw, CV still 15%.
Mean 2300 Frw, standard deviation still 300 Frw, so CV $= \frac{300}{2300} \times 100\% \approx 13.0\%$ .
Class Y did better on average (mean 61 against 54). Class X was more consistent: CV $\frac{9}{54} \approx 16.7\%$ against $\frac{15}{61} \approx 24.6\%$ for Y.
$\bar{x} = \frac{7830}{3} = 2610$ ; $\sum x^2 = 20\,903\,172$ , $\sigma^2 = \frac{20\,903\,172}{3} - 2610^2 = 155\,624$ , $\sigma \approx 394$ ; CV $\approx 15.1\%$ , almost the same relative spread as the South (15.9%), although Kigali's densities are about five times larger.
The mean is far above the median, so the data are positively skewed (a few very high incomes). Report the median and the IQR.

Quick check

Data set P has mean 50 and standard deviation 5. Data set Q has mean 400 and standard deviation 20. Which statement is correct?

Key points

$\text{CV} = \frac{\sigma}{\bar{x}} \times 100\%$ compares spreads of data with different means or units; smaller CV = more consistent.
Use the CV only for positive data with a mean well above zero.
Report centre, spread, shape and outliers, in context; use the median and IQR when there are outliers or skew.
In 2022, Southern district densities had CV ≈ 16%; Western ≈ 48%, almost all due to Rubavu.

1Is 60 000 Frw a lot of spread?

2Interpreting real data: district population density in 2022

3Writing the conclusion

Is 60 000 Frw a lot of spread?

Interpreting real data: district population density in 2022

Writing the conclusion