# Frequency Tables Worksheets, Questions and Revision

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## What you need to know

If we have collected a lot of data, we might display it with a frequency table. These contain two rows/columns: one with all the values that appeared when we collected the data, and one containing the number of times each value showed up. You should know how to read a frequency table and know how to use a frequency table to calculate the mean, median, and mode of the data.

Example: Below is a frequency table of data based on survey wherein 89 women were asked what their shoe size was. Calculate the mean, median, and mode of the data.

This frequency table tells us: 5 of the people asked had size 4 feet, 12 had size 4.5 feet, 18 had size 5 feet, and so on.  If this data were written as a list, it would begin like

$4, 4, 4, 4, 4, 4.5, 4.5, 4.5, 4.5, 4.5, 4.5, 4.5, 4.5, 4.5, 4.5, 4.5, 4.5, 5, 5, 5, 5, ...$

89 people is a lot of data points, so you can see why we put it into a table. That said, keeping in your head that it can be written as a list like this is useful. Now let’s find those averages.

– Mode: simply identify the shoe size with the highest frequency: 5.

– Median: we’re still looking for the middle data point. In this case, the number of people is 89, so the median is the $\frac{89 + 1}{2} = 45\text{th}$ term. Frequency tables present all the data to us in order, so to find the 45th term we have to add the frequencies as we go along as such:

$5 + 12 + 18 = 35$, so the 35th person is the last one with size 5 feet.

$5 + 12 + 18 + 19 = 54$, so the 54th person is the last one with size 5.5 feet.

Clearly, the 45th person is somewhere between these two, and all the people between these two data points fall into the size 5.5 category, thus the median is 5.5.

– Mean: to calculate the mean, we add up all the data points and divide by 89. Our frequency table tells us that there are 5 people who are size 4 and 12 who were size 4.5 etc, so rather than adding them up 1-by-1, we can do $(5 \times 4) + (12 \times 4.5)$ etc. So:

\begin{aligned}\text{Mean }=&[(5\times 4)+(12\times 4.5)+(18\times 5)+(5.5\times 19)+(6\times 11)+(6.5\times 4)\\&+(7\times 8)+(7.5\times 5)+(8\times 5)+(8.5\times 0)+(9\times 2)]\div 89 \\ &= 5.8 \text{ (1dp)}\end{aligned}

Now, we must also see how to work with a grouped frequency table – a method often used for displaying continuous data, e.g. length or weight. In a grouped frequency table, values aren’t all recorded/displayed individually. Instead, the range of all possible values is split into groups or classes, and we write down how many data points appeared in each class.

Example: The grouped frequency table below shows data on the weights of 117 cats. Find: the modal class, the class containing the median, and an estimate for the mean. Additionally, construct a frequency polygon from this table.

You can see that we aren’t explicitly asked for the mean, median, or mode. This is because we don’t know the individual weights of the cats, we only know how many fell in between certain values. As a result, we can only find estimates of these measures/say whereabouts they are.

– The modal class: we have no idea what number was most common, but we can see which class has the highest frequency – this is the modal class. Here, this is $3.5 < w \leq 4$.

– The class containing the median: there are 117 cats in total, so the median is the $\frac{117 + 1}{2} = 59\text{th}$ cat. We don’t know what this is, but we can say which class it is in. We know that $22+14=36$ cats weighed less than or equal to 3.5kg, and we also know that $22+14+39=75$ cats weighed less than or equal to 4kg, so the 59th cat (the median) must be somewhere in the $3.5 < w \leq 4$ class.

– The estimate for the mean: we can’t find the exact mean, so we must estimate. To do this, we pretend that every data point in each class is situated right at its midpoint (the easiest way to find the midpoint is to add up the lower bound and the upper bound and divide by 2).

For example, the midpoint of the first class is 2.5kg, so we pretend that the first 9 cats all weighed 2.5kg. At this point, it’s a good idea to add another column to your table containing the midpoint values. Then, treat these values like the actual values, and calculate the mean like in the first example.

$\text{Mean } = \dfrac{(22\times 2.5)+(14\times3.25)+(39\times 3.75)+(12\times 4.25)+(13\times 5.25)}{117} = 3.7 \text{ kg (1dp)}$

The final part of the question asks us to construct a frequency polygon.

This is nothing to worry about, all one must do is this: plot the frequency on the y-axis with the midpoint of each class on the x-axis, plot the smallest/largest values of weight on the x-axis at zero on the y-axis, and then join the points with straight lines. In this case, the result looks like the graph on the right.

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### Example Questions

The most common number of bathrooms is 1, so the mode is 1.

There are $30+21+5+7+3=66$ data points in total, so the median is the $\dfrac{66+1}{2} = 33.5\text{th}$ term. The first 30 terms are 1, and the next 21 terms are 2, so clearly both the 33rd and 34th terms are 2, therefore the median is 2.

It is impossible to find the mean because we don’t how many bathrooms the three people in the “5 or more” category actually have in their home.

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Then we add together the results of multiply each frequency by its associated midpoint and divide the answer by $12+18+34+33+19=116$.

$\text{Mean } = \dfrac{(12\times 5)+(18\times 12)+(34\times 17)+(33\times 26)+(19\times 38.5)}{116} = 21.1 \text{mins (3sf)}$

Now, construct a frequency polygon by plotting the midpoints on the x-axis with their associated frequencies on the y-axis, and joining the points with straight lines.

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GCSE MATHS

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