Mean, median and mode

What you will be able to do

• find the mean, median and mode of a list, a frequency table and grouped data;
• use an assumed mean and a step deviation to shorten the work;
• choose the average that suits the data, and explain why an extreme value moves the mean but not the median.

One number for the whole list

A school wants to tell new students how long they usually wait for the school bus. Eight students write down their waiting time in minutes: $14, 9, 12, 17, 9, 11, 9, 13$. Eight numbers are too many to remember. One number that stands for the list is easier. Such a number is called an average, or a measure of central tendency, because it shows where the data centres.

There are three common averages. Each answers a slightly different question.

The mean

Add all the values, then divide by how many there are. The mean of $x_1, x_2, \dots, x_n$ is

The symbol $\bar{x}$ is read "x bar". The symbol $\sum$ is the Greek capital sigma and means "add up". For the waiting times, the total is $94$ minutes and $n = 8$, so $\bar{x} = \dfrac{94}{8} = 11.75$ minutes.

When a value repeats, a frequency table saves writing. If the value $x_i$ occurs $f_i$ times, and $N = \sum f$ is the total number of values, then

The median

Put the values in order. The median is the middle one. With an odd number of values it is the value in position $\dfrac{n + 1}{2}$. With an even number it is the mean of the two middle values.

The ordered waiting times are $9, 9, 9, 11, 12, 13, 14, 17$. There are eight values, so the median is the mean of the fourth and fifth: $\dfrac{11 + 12}{2} = 11.5$ minutes. Half of the students waited no more than $11.5$ minutes.

The mode

The mode is the value that occurs most often. Here $9$ occurs three times, so the mode is $9$ minutes. A data set can have two modes (bimodal), or none, when every value occurs equally often.

Which average to use

The mean uses every value, so one very large or very small value pulls it. The median depends only on the order, so it stays steady. The mode is the only average that works for data that are not numbers, such as the most common shirt size.

For incomes, house prices and waiting times, a few very large values pull the mean up. A report that says "typical" then often uses the median.

Grouped data

When a table groups the data into classes, we no longer know the exact values. We take every value in a class to be the class mark, which is the middle of the class, also called the mid-point or the mid-interval value. Suppose 50 students scored these marks.

The mean. The class marks are $5, 15, 25, 35, 45, 55$. Then $\sum f x = 1510$ and $N = 50$, so $\bar{x} = 30.2$.

The median. Find the class where the running total of frequencies first reaches $\dfrac{N}{2} = 25$. The running totals are $4, 12, 27, \dots$, so the median class is $20$ to $30$. Then

Here $l$ is the lower boundary of the median class, $C$ is the running total before it, $f$ is its frequency and $h$ is its width. This gives $20 + \dfrac{25 - 12}{15} \times 10 = 28.67$. The formula assumes that the values spread evenly through the class. A-Level and IB texts call this linear interpolation.

The mode. The modal class is the class with the largest frequency, here $20$ to $30$. Then

where $f_1$ is the frequency of the modal class, $f_0$ the frequency of the class before it and $f_2$ the frequency of the class after it. This gives $20 + \dfrac{15 - 8}{30 - 8 - 10} \times 10 = 25.83$.

For data that is a little skewed, the three averages are linked by the rule of thumb $\text{mode} \approx 3\,\text{median} - 2\,\text{mean}$. Here it gives $3 \times 28.67 - 2 \times 30.2 = 25.6$, close to $25.83$.

A shorter way to the mean

With big numbers, take an assumed mean $A$, usually a class mark near the middle, and write $u = \dfrac{x - A}{h}$, where $h$ is the class width. Then

For the marks above, take $A = 25$ and $h = 10$. The values of $u$ are $-2, -1, 0, 1, 2, 3$, and $\sum f u = 26$. So $\bar{x} = 25 + 10 \times \dfrac{26}{50} = 30.2$, the same as before, with smaller numbers.

A frequency table that is not grouped

Thirty learners took a quiz and scored $10, 20, 30, 40$ or $50$ marks. The frequencies are $5, 8, 12, 4$ and $1$, so $N = 30$.

The mean is $\dfrac{780}{30} = 26$. The $14$th to the $25$th values are all $30$, so the $15$th and $16$th values, the two middle ones, are both $30$. The median is $30$. The mode is also $30$, the value with the largest frequency.

Practise with new numbers

Try it in Python

The first program finds the three averages of the waiting times with plain Python. The second finds all three for the grouped marks, and also finds the mean by the step-deviation method.

Common mistakes

• Dividing by the number of classes. For grouped data divide by $N$, the total of the frequencies, not by the number of classes.
• Taking the median before ordering. Sort the values first.
• Using the wrong class for the median. The median class is where the running total first reaches $\dfrac{N}{2}$, not the class with the largest frequency.
• Mixing up the three classes of the mode formula. $f_0$ is the frequency before the modal class and $f_2$ the frequency after it.
• Forgetting the width. Both the median and the mode formulas end with the class width $h$.

Remember these

• Mean: $\bar{x} = \dfrac{\sum f x}{N}$. Median: the middle value. Mode: the most common value.
• For grouped data use class marks for the mean, and the interpolation formulas for the median and the mode.
• An extreme value pulls the mean. The median is steady.
• With an assumed mean $A$ and step $u = \dfrac{x - A}{h}$: $\bar{x} = A + h\,\bar{u}$.

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 mean and the median of $7, 11, 4, 15, 8$.

2. A frequency table gives the value $1$ six times, $2$ nine times, $3$ three times and $4$ two times. Find the mean and the mode.

3. The numbers $3, 7, 8, x, 12, 15$ are in order, and the median is $9.5$. Find $x$.

4. A village has ten families. Nine earn about the same, and one earns a hundred times as much. Which average describes a typical family better, and why?

5. Find the mode of this grouped data: classes $0$ to $20$, $20$ to $40$, $40$ to $60$ and $60$ to $80$, with frequencies $4, 10, 14, 2$.

References

• OpenStax, Introductory Statistics, Chapter 2, section on measures of the centre of the data.
• NCERT, Mathematics for Class XI, Chapter 13 (Statistics), introduction on measures of central tendency. The mean, median and mode of grouped data are taught in the Class 10 book, Chapter 14.