Coefficient of variation and changing every value

What you will be able to do

• compare the steadiness of two data sets with the coefficient of variation;
• say when the coefficient of variation should not be used;
• find the new mean and spread when every value is shifted, scaled or both.

Which one is more steady?

A bakery bakes small cakes and big cakes. The small cakes have a mean mass of $250\ \text{g}$ with a standard deviation of $15\ \text{g}$. The big cakes have a mean mass of $1000\ \text{g}$ with a standard deviation of $40\ \text{g}$. The big cakes have the larger standard deviation. But a $40\ \text{g}$ miss on a cake of $1000\ \text{g}$ is a small share, and a $15\ \text{g}$ miss on a cake of $250\ \text{g}$ is a bigger share. Which batch is more consistent? To decide, we compare the spread relative to the mean.

The coefficient of variation

The coefficient of variation is

It has no unit, so it can compare data in different units. The data set with the smaller CV is more consistent. Here:

• small cakes: $\dfrac{15}{250} \times 100 = 6\%$;
• big cakes: $\dfrac{40}{1000} \times 100 = 4\%$.

The big cakes are more consistent, even though their standard deviation is larger.

When not to use it. The coefficient of variation makes sense only when a value of zero means "none of it", as with mass, length or time. It is meaningless for a temperature in degrees Celsius, because $0\ ^\circ\text{C}$ is not "no temperature". The same data would give a different CV in kelvin. It is also useless when the mean is close to zero. A-Level, IB and AP courses seldom use the CV. Some call it the relative standard deviation.

Changing every value

A teacher adds $5$ bonus marks to every paper. A shop converts every price from one currency to another. A thermometer reads $2\ ^\circ\text{C}$ too high. In each case, every value $x$ becomes a new value $y = a x + b$. We can find the new mean and spread from the old ones, without redoing the data.

Adding $b$ to every value. Everything moves along the number line together, so the mean moves by $b$, and so does the median. The distances between values do not change. So the range, the mean deviation, the variance and the standard deviation stay the same.

Multiplying every value by $a$. All the distances are multiplied by $\lvert a \rvert$. The mean is multiplied by $a$. The standard deviation and the mean deviation are multiplied by $\lvert a \rvert$, and the variance is multiplied by $a^2$.

Both. For $y = a x + b$,

The absolute value matters because a spread is never negative. If $a = -2$, the data are reflected and stretched by $2$, and the standard deviation is multiplied by $2$, not $-2$.

The coefficient of variation behaves in its own way. Multiplying by $a$ multiplies the mean and the standard deviation by the same factor, so the CV does not change. Adding $b$ changes the mean but not the standard deviation, so the CV changes.

A worked example

A class has marks out of $50$ with mean $36$ and standard deviation $5$. The marks are scaled to be out of $100$ by multiplying by $2$. Then $4$ bonus marks are added.

1. After scaling. The mean is $72$ and the standard deviation is $10$. The CV is $\dfrac{10}{72} \times 100 = 13.89\%$, the same as before, since $\dfrac{5}{36} = 0.1389$.
2. After the bonus. The mean is $72 + 4 = 76$, and the standard deviation is still $10$.
3. The new CV. $\dfrac{10}{76} \times 100 = 13.16\%$. It is smaller, because the same spread now sits on a larger mean.

Practise with new numbers

Try it in Python

The first program finds the coefficient of variation of the two batches of cakes. The second multiplies and shifts a small data set and compares the new mean and standard deviation with the predictions.

Common mistakes

• Adding $b$ to the standard deviation. Adding the same number to every value does not change the spread.
• Forgetting the absolute value. A standard deviation is never negative, so multiply by $\lvert a \rvert$.
• Multiplying the variance by $a$. The variance is multiplied by $a^2$.
• Using the CV for temperature in degrees Celsius. It is not meaningful there.
• Choosing the larger standard deviation as the less steady one. Compare the CV, not the standard deviation, when the means differ.

Remember these

• $\text{CV} = \dfrac{\sigma}{\bar{x}} \times 100\%$. The smaller CV means the more consistent data.
• For $y = a x + b$: $\bar{y} = a\bar{x} + b$, $\sigma_y = \lvert a \rvert\,\sigma_x$, $\sigma_y^2 = a^2\sigma_x^2$.
• Adding a number leaves every measure of spread unchanged.
• Multiplying leaves the CV unchanged. Adding does not.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A data set has mean $80$ and standard deviation $12$. Find the coefficient of variation.

2. The values have mean $12$ and standard deviation $3$. Each value is increased by $6$. Find the new mean and standard deviation.

3. The values have mean $12$ and standard deviation $3$. Each value is multiplied by $-2$. Find the new mean, standard deviation and variance.

4. Why can we not decide, from the standard deviations alone, whether the heights in centimetres or the masses in kilograms of a group of students are more variable?

5. Temperatures in a town have mean $25\ ^\circ\text{C}$ and standard deviation $4\ ^\circ\text{C}$. Find the mean and the standard deviation in degrees Fahrenheit, using $F = 1.8C + 32$.

References

• OpenStax, Introductory Statistics, Chapter 2, section on measures of the spread of the data.
• NCERT, Mathematics for Class XI, Chapter 13 (Statistics), section on analysis of frequency distributions (the coefficient of variation).