Mean, median and mode: Grade 12 Mathematics

By the end of this lesson you should be able to

Calculate the mean, median, mode and range of a set of data
Find averages from a frequency table
Estimate the mean and identify the modal class for grouped data
Choose the most suitable average for a situation

Three ways to describe the “middle”

An average is a single value that represents a set of data. There are three common averages, and each answers a slightly different question.

Average	What it is	How to find it
Mean	The “fair share” value	$\dfrac{\text{sum of all values}}{\text{number of values}}$
Median	The middle value	Put the data in order, then take the middle one (or the mean of the middle two)
Mode	The most common value	The value that appears most often

The range measures spread: $\text{range} = \text{largest value} - \text{smallest value}$ .

Worked examples with raw data

Example 1. Find the mean, median, mode and range of $5, 8, 6, 8, 9, 4, 8, 7$ .

Mean: $\dfrac{5 + 8 + 6 + 8 + 9 + 4 + 8 + 7}{8} = \dfrac{55}{8} = 6.875$

Median: first put the values in order: $4, 5, 6, 7, 8, 8, 8, 9$ . There are $8$ values, so the median is the mean of the 4th and 5th values: $\dfrac{7 + 8}{2} = 7.5$ .

Mode: $8$ , which appears three times.

Range: $9 - 4 = 5$ .

Frequency tables

When a value is repeated many times, data is often shown in a frequency table. To find the mean, multiply each value by its frequency, add these products, and divide by the total frequency.

Example 2. The table shows the scores of 20 learners in a short test.

Score $x$	1	2	3	4	5
Frequency $f$	2	5	8	3	2

Mean: $\sum fx = 1(2) + 2(5) + 3(8) + 4(3) + 5(2) = 2 + 10 + 24 + 12 + 10 = 58$ , and $\sum f = 20$ .

\text{mean} = \frac{58}{20} = 2.9

Median: with $20$ values, the median lies between the 10th and 11th values. The cumulative frequencies are $2$ , $7$ , $15$ , so both the 10th and 11th values are in the score $3$ group. The median is $3$ .

Mode: the highest frequency is $8$ , so the mode is a score of $3$ .

Grouped data

When data is grouped into class intervals, you do not know the exact values, so you can only estimate the mean. Use the midpoint of each class as its representative value.

Example 3. The marks of 30 learners are grouped in the table. Estimate the mean mark and state the modal class.

Marks	Frequency $f$	Midpoint $x$	$fx$
10 – 19	4	14.5	58
20 – 29	6	24.5	147
30 – 39	10	34.5	345
40 – 49	8	44.5	356
50 – 59	2	54.5	109
Total	30		1015

\text{estimated mean} = \frac{\sum fx}{\sum f} = \frac{1015}{30} = 33.8

The modal class is the class with the highest frequency: $30 - 39$ .

The median falls in the class that contains the 15th and 16th values. The cumulative frequencies are $4$ , $10$ , $20$ , so the median class is also $30 - 39$ .

Extension: estimating the median. Some courses also estimate the median by interpolation. With $L$ the lower class boundary of the median class, $F$ the cumulative frequency before it, $f$ its frequency and $w$ its width:

\text{median} \approx L + \left(\frac{n/2 - F}{f}\right) \times w

For Example 3: $29.5 + \dfrac{15 - 10}{10} \times 10 = 34.5$ .

Which average should you use?

Use the mean when values are fairly even and you want to use all the data.
Use the median when the data has extreme values (outliers) that would distort the mean.
Use the mode for categories or when you want the most popular choice, such as the most common shoe size.

Example 4. Five workers earn K2 000, K2 200, K2 100, K2 300 and K15 000 a month. Which average is more suitable?

The total is K23 600, so the mean is K4 720. But four of the five earn about K2 000 to K2 300, so K4 720 is misleading. The median, K2 200, describes a typical worker much better. The one very large salary pulls the mean up but does not affect the median.

Common mistakes

Not ordering the data before finding the median. The median is the middle of the ordered list, not the middle of the list as it was written.

Dividing by the wrong number. For a frequency table, divide $\sum fx$ by $\sum f$ , not by the number of columns or classes.

Using class limits instead of midpoints. For a class $20 - 29$ , use $24.5$ as the representative value.

Giving the frequency as the mode. The mode is the value (or class) that occurs most often, not the number of times it occurs.

Practice questions

Find the mean, median and mode of $3, 9, 4, 7, 4, 8, 4$ . Give the mean to 2 decimal places.
The mean of five numbers is $12$ . Four of them are $10, 15, 8, 14$ . Find the fifth number.
A frequency table shows $x = 0, 1, 2, 3, 4$ with frequencies $3, 7, 10, 6, 4$ . Find the mean (to 2 decimal places), the median and the mode.
Estimate the mean and state the modal class of this data.

Class 0 – 9 10 – 19 20 – 29 30 – 39 40 – 49
Frequency 5 12 18 10 5
The mean of six numbers is $8$ . When a seventh number is added, the mean becomes $9$ . Find the seventh number.
Find the range, mode and median of $12, 15, 11, 15, 18, 20, 15, 14$ .
A shop records these daily sales: K180, K200, K190, K210, K1 500. Say which average is more suitable and give a reason.

Class	0 – 9	10 – 19	20 – 29	30 – 39	40 – 49
Frequency	5	12	18	10	5

Show answers

Sum $= 39$ , so the mean is $\dfrac{39}{7} = 5.57$ . In order: $3, 4, 4, 4, 7, 8, 9$ , so the median is $4$ and the mode is $4$ .
The total must be $5 \times 12 = 60$ . The four known numbers add to $47$ , so the fifth is $60 - 47 = 13$ .
$n = 30$ and $\sum fx = 0 + 7 + 20 + 18 + 16 = 61$ , so the mean is $\dfrac{61}{30} = 2.03$ . The cumulative frequencies are $3, 10, 20, \ldots$ , so the 15th and 16th values are both $2$ , giving a median of $2$ . The mode is $2$ .
Midpoints: $4.5, 14.5, 24.5, 34.5, 44.5$ . $\sum fx = 22.5 + 174 + 441 + 345 + 222.5 = 1205$ and $\sum f = 50$ , so the estimated mean is $24.1$ . The modal class is $20 - 29$ .
The total of the seven numbers is $7 \times 9 = 63$ . The first six total $6 \times 8 = 48$ , so the seventh number is $63 - 48 = 15$ .
In order: $11, 12, 14, 15, 15, 15, 18, 20$ . Range $= 20 - 11 = 9$ , mode $= 15$ , median $= \dfrac{15 + 15}{2} = 15$ .
The median (K200) is more suitable. The K1 500 day is an outlier that would raise the mean to K456, which does not represent a typical day.

Summary

Mean: total divided by the number of values. Median: the middle of the ordered data. Mode: the most common value. Range: largest minus smallest.
For frequency tables, the mean is $\dfrac{\sum fx}{\sum f}$ .
For grouped data, use class midpoints to estimate the mean, and name the modal class by its highest frequency.
Choose the median when unusually large or small values would distort the mean.

Now practise this topic

Real questions show you what still needs work. Try a past paper under timed conditions, or take a short quiz.

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