Range and mean deviation

What you will be able to do

• find the range and the coefficient of range;
• find the mean deviation about the mean or about the median, for a list, a frequency table and grouped data;
• explain why the deviations are taken without their sign, and why the median gives the smallest mean deviation.

The same average, a different story

Two machines fill bags of rice, each meant to hold $500\ \text{g}$. Machine A fills five bags with $498, 499, 500, 501$ and $502\ \text{g}$. Machine B fills five bags with $480, 490, 500, 510$ and $520\ \text{g}$. Both machines have a mean of $500\ \text{g}$. Yet nobody would buy rice from machine B with the same trust. The mean cannot see this difference. We need a number for the spread, also called the dispersion, of the data.

The range

The simplest measure is the range:

Machine A has range $502 - 498 = 4\ \text{g}$. Machine B has range $520 - 480 = 40\ \text{g}$. The range is quick, and it has the same unit as the data. But it uses only two values, so one odd reading changes it a lot.

To compare two data sets in different units, use the coefficient of range:

It has no unit. For machine A it is $\dfrac{4}{1000} = 0.004$ and for machine B it is $\dfrac{40}{1000} = 0.04$.

Deviations from a centre

A deviation is the distance of a value from a centre $a$, written $x - a$. It is positive for a value above the centre and negative below it. Suppose we take the mean as the centre. Then the deviations always add up to zero, because

So the sum of the deviations says nothing. We remove the sign by taking the absolute value, $\lvert x - a \rvert$, which is the distance without direction.

The mean deviation

The mean deviation about a centre $a$ is the mean of the distances:

For a frequency table, and for grouped data with the class marks as $x$,

The centre is usually the mean or the median. Say which one: the mean deviation about the mean and the mean deviation about the median are different numbers.

For machine A the distances from the mean $500$ are $2, 1, 0, 1, 2$, so $\text{MD} = \dfrac{6}{5} = 1.2\ \text{g}$. For machine B they are $20, 10, 0, 10, 20$, so $\text{MD} = \dfrac{60}{5} = 12\ \text{g}$. Now the difference is clear: machine B misses by ten times as much.

A-Level, IB and AP courses do not use the mean deviation. They use the standard deviation and the interquartile range, which come next. The Indian syllabus asks for all of them.

The curve in the simulation shows a fact worth remembering: the mean deviation is smallest when the centre is the median. The mean gives a slightly larger value. This is why a question that says "about the median" has an answer no larger than "about the mean", for a list of values.

Coefficient of mean deviation

To compare spreads in different units, divide by the centre:

It is a pure number. Machine B has $\dfrac{12}{500} = 0.024$.

A worked example with grouped data

Use the marks of $50$ students from the last lesson, with class marks $5, 15, 25, 35, 45, 55$ and frequencies $4, 8, 15, 10, 7, 6$. The mean is $30.2$.

1. The distances. From $30.2$ the class marks are $25.2, 15.2, 5.2, 4.8, 14.8, 24.8$ away.
2. Multiply by the frequencies. $4 \times 25.2 = 100.8$, then $121.6$, $78$, $48$, $103.6$ and $148.8$.
3. Add and divide. The total is $600.8$. So $\text{MD} = \dfrac{600.8}{50} = 12.016$ marks.
4. The coefficient. $\dfrac{12.016}{30.2} = 0.398$.

Practise with new numbers

Try it in Python

The program finds the mean deviation of the five readings in the simulation about the mean, about the median, and about several other centres.

Common mistakes

• Dropping the absolute value. Without it the deviations from the mean add up to zero.
• Using the wrong centre. Read whether the question says about the mean or about the median.
• Dividing by the number of classes. For a frequency table divide by $N = \sum f$.
• Forgetting the frequencies. In a frequency table each distance is multiplied by its frequency.

Remember these

• Range $=$ largest $-$ smallest. Coefficient of range $= \dfrac{\text{largest} - \text{smallest}}{\text{largest} + \text{smallest}}$.
• $\text{MD}(a) = \dfrac{\sum f \lvert x - a \rvert}{N}$, about the mean or the median.
• The sum of the deviations from the mean is zero.
• The mean deviation is smallest about the median.

Check yourself

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

1. Find the range and the coefficient of range of $12, 7, 20, 15, 9$.

2. Find the mean deviation about the mean of $2, 5, 9, 12, 17$.

3. Show that the sum of the deviations of $n$ values from their mean is zero.

4. The value $2$ occurs three times, $4$ four times and $6$ three times. Find the mean deviation about the mean.

5. A data set has median $40$ and mean deviation about the median $6$. Find the coefficient of mean deviation.

References

• OpenStax, Introductory Statistics, Chapter 2, section on measures of the spread of the data.
• NCERT, Mathematics for Class XI, Chapter 13 (Statistics), sections on the range and the mean deviation.