Frequency tables: finding averages

In a nutshell

Just like for sets of values, the mean, median, mode and range can be found from frequency tables too. The only difference is that these averages might refer to a category, rather than an individual number.

PROCEDURE

Mean	Create a new column: 'number $\times$ frequency' and calculate these values. Next, find the sum of all these values, then divide this number by the total frequency.
Median	Identify the category containing the middle value: use $\frac{n+1}{2}$ where n is the total frequency to find the middle position.
Mode	Identify the category with the highest frequency.
Range	Calculate the difference between the highest and lowest in the 'number of' column.

Example

This frequency table shows the number of siblings pupils in a class of $23$ have:

Number of siblings	Frequency
$0$	$6$
$1$	$12$
$2$	$3$
$3$	$2$
$4$	$0$

Find the mean, median, range and mode.

1. Mean:

Add the third column to the table.

Number of siblings

\times

frequency

0\times6=0

1\times12=12

2\times3=6

3\times2=6

4\times0=0

Use the formula:

$\frac{\text{sum of 'number}\times \text{frequency' column}}{\text{total frequency}}=\frac{12+6+6}{6+12+3+2}=\frac{24}{23}=\underline{1.043} \text{ (to 3 d.p})$

Note: The total number of students was given in the question here, but often you will need to calculate it yourself.

2. Median:

Identify the middle position.

$\frac{(n+1)}{2}=\frac{(23+1)}{2}=12$

The median is the category containing the $12th$ value cumulatively: $\underline1$

3. Mode:

Identify the category with the highest frequency.

$\underline1$

4. Range:

Work out the difference between the highest and lowest number of siblings.

$3-0 = \underline3$

Note: $4$ is not used here as there is no one with $4$ siblings.