# Grouped Frequency Tables *Revision and Worksheets*

## What you need to know

Sometimes when we’re dealing with a lot of data at once it’s easier to separate the values into groups. Then, we can display it in what is called a grouped frequency table, which looks something like this:

This particular example is a set of data on the heights of a selection of adults. To understand what this table is saying, let’s look at the first group (also known as a class):

150<h\leq 160This notation means that all people who fall within this group must’ve been

“taller than 150cm but no taller than 160cm.”

It’s important to understand that this group doesn’t include people who are exactly 150cm tall, but it does include people who are exactly 160cm tall. This distinction matters, because we have to be clear about which group people on the edges of their class will fit into.

So, we know there are 14 people who are taller than 150cm but no taller than 160cm. Those 14 people could include heights 151cm, 160cm, 159cm, 152cm, but the important thing to note is that we don’t know. We simply don’t know how tall each of those 14 people were, only that they fall in that group. This is the disadvantage of grouped frequency tables.

NOTE: the groups in a grouped frequency table do not also have the same “width”, i.e. the first group goes from 150 to 160, meaning it has a width of 10, whilst the third group goes from 170 to 175, meaning it only has a width of 5.

Here are links to two of the main things you’ll be using grouped frequency tables for:

## Example Questions

There are several good ways to do this question, one of which is to cross of each number as you go and tally them in the appropriate groups. For clarity, here is a full list of which heights are in each group.

• The 0<h\leq 20 group:

7, 9, 15, 19, 19

• The 20<h\leq 30 group:

21, 22, 25, 25, 27, 28, 30

• The 30<h\leq 40 group:

31, 32, 32, 33, 35, 37, 38, 39

• The 40<h\leq 70 group:

46, 51, 55, 61

So, the completed table looks like:

2) 100 people were given a puzzle to solve. Their times, in minutes, are recorded in the grouped frequency table below.

a) State the number of people who took over 2 minutes to complete the puzzle.

b) If you complete the puzzle in 90 seconds or less, you won a prize. Work out the proportion of the people taking the test who won the prize. Give your answer as a fraction in its simplest form.

a) The bottom two groups in the table amount to the total number of people who took over 2 minutes. Therefore, the total is

19+19=38\text{ people took over 2 minutes to solve the puzzle}

b) First of all, we need to convert 90 seconds to minutes.

90=60+30\equiv 1\text{ minute }+\dfrac{1}{2} \text{ a minute }=1.5\text{ minutes}

So, we’re looking for the number of people who took up to 1.5 minutes to solve the test. This amounts to the first two groups in the table:

8+22=30

There are 100 people in total, so to find this value as a proportion, we write it as a fraction over 100:

\dfrac{30}{100}=\dfrac{3}{10}

This fraction cannot be simplified further, so we are done.