Frequency Tables Worksheets | Questions and Revision | MME

# Frequency Tables Worksheets, Questions and Revision

Level 4 Level 5

## What you need to know

### Frequency Tables

If we have collected a lot of data, we might display it with a frequency table. 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.

This works a little differently with grouped frequency table – where you might be asked to estimate the mean. To get to grips with frequency tables it would be good to first revise how to calculate the mean, median, mode and range

### Interpreting Frequency Tables

Below is a frequency table of data based on survey wherein 89 women were asked what their shoe size was. The 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

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

## Example 1: Constructing Frequency Tables

66 People were asked about how many bathrooms they had in their house. 30 people had 1 bathroom. 21 people had 2, 5 people had 3 and 7 people had bathrooms, the rest had 5 or more. Use this information to construct a frequency table.

We have number of bathrooms in our left column, then frequency in the right. The only calculation that has to be done is

$30+21+5+7 = 63$

As the question tells us there are 66 people in total, this must mean there are 3 people who have 5 or more bathrooms. The table therefore looks like this.

## Example 2: Using Frequency Tables

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.

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

Median: 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:

\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}

### Example Questions

a) The number of bathrooms with the highest frequency is the 1 bathroom category, so the mode is 1.

To find the median, we need to find the middle value(s). In order to find the middle value(s), we need to find how many values there are in total. The number of values in total is the sum of all the frequencies:

There are $30+21+5+7+3=66$ values in total.

Since the number of values is an even number, this means that there is no single middle value, so we will need to locate the two middle values. To find the middle values we need to use the formula $\dfrac{n + 1}{2}$ where $n$ represents the total number of values:

$\dfrac{66 + 1}{2} = 33.5$

This means that the median is halfway between the 33rd and the 34th value.

If we go back to the frequency table, we can see that the first 30 values are in the 1 bathroom category, and the following 23 values are in the 2 bathroom category. Therefore values 33 and 34 are in the 2 bathroom category. Since the 33rd and the 34th values are identical, then the median is simply 2 bathrooms.

b) It is not possible to calculate the mean due to the fact that there is a category of ‘5 bathrooms or more’. We do not know exactly how many bathrooms people have who are in this category (they could have 5, they could have 500!).

a) Working out the mode is the easy part. Which category was the most common (has the highest frequency)?

1 goal per game is therefore the mode.

b) To find the median, we need to find the middle value(s). In order to find the middle value(s), we need to find how many values there are in total. The number of values in total is the sum of all the frequencies:

There are $7+14+13+8+3+4+1=50$ values in total.

Since the number of values is an even number, this means that there is no single middle value, so we will need to locate the two middle values. To find the middle values we need to use the formula $\dfrac{n + 1}{2}$ where $n$ represents the total number of values:

$\dfrac{50 + 1}{2} = 25.5$

This means that the median is halfway between the 25th and the 26th value.

If we go back to the frequency table, we can see that the first 7 values are in the 0 goals category, and the following 14 values are in the 1 goal category. This means that the first 21 values fall in the 0 goal or the 1 goal category. The following 13 values fall into the 2 goal category, so values 25 and 26 must be in this category. Since the 25th and the 26th values are identical, then the median is simply 2 goals.

c) The mean is the total number of goals divided by the total number of games. In this question, the frequency represents the total number of games which is 50 (which we had already calculated from the previous question).

To work out the total number of goals, we need to multiply the number of goals by the frequency (if the team scored 5 goals on 4 occasions, then the team scored 20 goals in these 4 matches combined):

$7\times 0\text{ goals} = 0\text{ goals}$

$14\times 1\text{ goal} = 14\text{ goals}$

$13\times 2\text{ goals} = 26\text{ goals}$

$8\times 3\text{ goals} = 24\text{ goals}$

$3\times 4\text{ goals} = 12\text{ goals}$

$4\times 5\text{ goals} = 20\text{ goals}$

$1\times 6\text{ goals} = 6\text{ goals}$

Now that we know how many goals were scored in each category, we can work out the total number of goals scored:

$\text{ Total number of goals scored} = 0+14+26+24+12+20+6=102\text{ goals scored}$

If the team scored 102 goals in 50 games, then the mean number of goals scored can be calculated as follows:

$102\text{ goals} \div 50\text{ games}= 2 \text{ goals (to the nearest goal)}$

a) The modal salary band is the salary band which is the most common (has the highest frequency). This is clearly the £30,000 – £34,999 category since 104 staff members fall into this category, more than any other.

b) To find the median, we need to find the middle value(s). In order to find the middle value(s), we need to find how many values there are in total. The number of values in total is the sum of all the frequencies:

There are $63+86+96+104+65+25=439$ values in total.

Since the number of values is an odd number, this means that there is a middle value (this is easier than when there is an even number of values resulting in there being two middle values). To find the middle value we need to use the formula $\dfrac{n + 1}{2}$ where $n$ represents the total number of values:

$\dfrac{439 + 1}{2} = 220$

This means that the median is 220th value.

If we go back to the frequency table, we can see that the first 63 values are in the £15,000 – £19,999 category, and the following 86 values are in the £20,000 – £24,999 category. This means that the first 149 values fall into either the £15,000 – £19,999 category or the £20,000 – £24,999 category. The following 96 values fall into the £25,000 – £29,999 category, so the 220th value must be in this category, so the median salary band is £25,000 – £29,999.

c)  Since we are dealing with grouped data, it is not possible to calculate a mean. For example, the people in the £40,000 + category could be earning £40,001 or £40 million.

a) There is a lot to do for this question, but provided we tackle each part in order, and logically, it should not pose too much difficulty.

We know that the ratio of crocodiles that are less than a metre in length to crocodiles between 1 – 2 metres is $4 : 5$. We also know that 16 crocodiles are less than 1 metre in length. This figure of 16 is 4 times greater than the number given in the ratio, so the number of crocodiles that are between 1 – 2 metres must also be 4 times greater than its corresponding figure in the ratio (5):

$5\times4 = 20\text{ crocodiles between 1 – 2 metres}$

We are told that the twice as many crocodiles are between 2 – 3 metres in length than between 1 – 2 metres in length, so that means that he sees 40 crocodiles between 2 – 3 metres.

We are also told that $\frac{3}{8}$ of the number of crocodiles between 2 – 3 metres in length have a length of between 3 – 4 metres. We know that there are 40 crocodiles with a length of between 2 – 3 metres, so we can calculate this as follows:

$\dfrac{3}{8}\times40=15\text{ crocodiles between 3 – 4 metres}$

He sees 8 more crocodiles with length of 4 – 5 metres than with length of 3 – 4 metres. Therefore, he see $15+8=24$ crocodiles with length of 4 – 5 metres.

He sees 25% fewer crocodiles with a length of 5 -6 metres than with length of 4 – 5 metres. If he sees 25% fewer, this means that he sees 75% of the number of crocodiles of length 4 – 5 metres. This can be calculated as follows:

$24\times0.75 = 18\text{ crocodiles between 5 - 6 metres}$

As a result, the completed frequency table should look like this:

b) The modal length of crocodile is the length which is the most common (has the highest frequency). This is clearly the 2 – 3 metre category since 40 crocodiles are in this category, more than any other.

c) To find the median, we need to find the middle value(s). In order to find the middle value(s), we need to find how many values there are in total. The number of values in total is the sum of all the frequencies:

There are $16+20+40+15+24+18=133$ values in total.

Since the number of values is an odd number, this means that there is a middle value (this is easier than when there is an even number of values resulting in there being two middle values). To find the middle value we need to use the formula $\dfrac{n + 1}{2}$ where $n$ represents the total number of values:

$\dfrac{133 + 1}{2} = 67$

This means that the median is the 67th value.

If we go back to the frequency table, we can see that the first 16 values are in the 0 – 1 metre category, and the following 20 values are in the 1 – 2 metre category. This means that the first 36 values fall into either the 0 – 1 metre category or the 1 – 2 metre category. The following 40 values fall into the 2 – 3 metre category, so value 67 must also fall in this category, so the median length of crocodile is 2 – 3 metres.

a) We know that a total of 200 divers were surveyed. This means that the total of frequency column is 200. Therefore, if we subtract all the known values from 200, we can work out the value of $x$ and $y$ combined:

$200-15-76-32-9=108\text{ divers}$

Therefore $x + y = 108\text{ divers}$

We have been told that the ratio of $x$ to $y$ is $7 : 5$. This means that $x$ is $\frac{7}{12}$ of the total and $y$ is $\frac{5}{12}$ of the total. (We are dealing in twelfths here because the sum of the ratio is 12.)

We can calculate the value of $x$ as follows:

$\dfrac{7}{12}\times 108\text{ divers} = 63\text{ divers}$

We can calculate the value of $y$ as follows:

$\dfrac{5}{12}\times 108\text{ divers} = 45\text{ divers}$

b) The modal number of shark encounters is most common number of shark encounters (the category with the highest frequency). This is clearly the 2 shark encounters category since 76 divers fall into this category, more than any other.

c) To find the median, we need to find the middle value(s). In order to find the middle value(s), we need to find how many values there are in total. Since we have been told that there are 200 divers, we do not need to calculate this.

Since the number of values is an even number, this means that there is no single middle value, so we will need to locate the two middle values. To find the middle values we need to use the formula $\dfrac{n + 1}{2}$ where $n$ represents the total number of values:

$\dfrac{200 + 1}{2} = 101.5$

This means that the median is halfway between the 100th and the 101st value.

If we go back to the frequency table, we can see that the first 9 values are in the 0 shark encounters category, and the following 32 values are in the 1 shark encounter category. This means that the first 41 values fall in the 0 or the 1 shark encounter categories. The following 76 values fall into the 2 shark encounters category, so values 100 and 101 must be in this category. Since the 100th and the 101st values are identical, then the median is simply 2 shark encounters.

d) The mean is the total number of shark encounters divided by the total number of divers (200).

To work out the total number of shark encounters, we need to multiply the number of shark encounters by the frequency:

$9\times 0\text{ shark encounters} = 0\text{ shark encounters}$

$1\times 32\text{ shark encounters} = 32\text{ shark encounters}$

$2\times 76\text{ shark encounters} = 152\text{ shark encounters}$

$3\times 63\text{ shark encounters} = 189\text{ shark encounters}$

$4\times 45\text{ shark encounters} = 180\text{ shark encounters}$

$5\times 15\text{ shark encounters} = 75\text{ shark encounters}$

Now that we know how many shark encounters there are in each category, we can work out the total number of shark encounters:

$\text{Total number of shark encounters} = 0+32+152+189+180+75=628\text{ shark encounters}$

If 200 divers had a total of 628 shark encounters, then the mean number of shark encounters can be calculated as follows:

$6282\text{ shark encounters} \div 200\text{ divers}= 3 \text{ shark encounters (to the nearest whole number)}$

Level 4-5

Level 4-5

GCSE MATHS

GCSE MATHS

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