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Cambridge IGCSE and O Level

Most scientific calculators have a mode called STAT or STATISTICS. This can be used as a checking tool when finding the mean.

But first, it’s important to remember that if you don’t show your method and ONLY use your calculator, you may not be awarded full marks. You MUST make sure that you understand how to find the mean from a simple list or frequency table and, if your course requires it, an estimate of the mean from a grouped frequency table, before you think about using your calculator as a checking tool.

Okay, before you look at how to use the STAT mode, you need to understand the notation used.

Here’s a frequency table for values of x.
Let’s find the mean of x.
As a skill check, you may wish to pause while you do this.
The mean of x is 18 divided by 10 which is 1.8

When the data values you enter are x values, the mean is called ‘x bar’. It will look like an x with a bar over the top. So we say x bar is 1.8.

Here is a frequency table for values of y. Let’s find the mean of y.
Again, as a skill check, you may wish to pause while you do this.
The mean of y is 41 divided by 10 which is 4.1

When the data values you enter are y values, the mean is called ‘y bar’.
It will look like a y with a bar over the top. So we say y bar is 4.1

Bar notation is used over the top to show the value represents the mean of a set of values. Let’s look at an example question, but first it’s a good idea for you to have your own calculator in front of you so you can use it during this presentation. Some calculators have settings which allow you to have the frequency column or row switched off. It is more useful if the frequency is turned on as you can use this to find the mean of lists or frequency tables.

Let’s look at how to do this for this particular calculator.

First you need to press SHIFT and SETUP to enter the correct menu.
You then use the centre arrows to go down and look at all the options.
You need option 1: STATISTICS on the second screen.
When you press 1 you are given two more options.
You choose option 1 to make sure the frequency column is on and included in our data table.

If you then press MENU , option 2 for STATISTICS and then option 1 for 1- variable, you can see that you have a blank table with columns for x and frequency. Okay, here is a Cambridge Examination question. Notice that this question is worth 2 marks in the examination. A good tip is that when a question is worth more than 1 mark, you should always show your method as well as your answer. Let’s first work out the answer in the usual way and then look at how to use the calculator to check your answer.

You know that, to find the mean, you need to total the data values and then divide that by the number of data values.
You may wish to pause while you do this.
The total of all the data values is 284.
There are 8 data values in the list so the mean is 284 divided by 8 which is 35.5.
Let’s look at how to check this using a calculator. Your calculator may already be in STATISTICS mode. If it’s not you need to choose this mode.
For this calculator, you press MENU and then choose option 2, STATISTICS.
As you only want to find the mean of one set of data values, you choose option 1, 1-Variable.

You can then see the frequency table for the x values.
Your data values are going to be the x values.
You enter the values into the column on the left hand side.

So you press 34 and then the equals button to enter it.
The frequency column on the right hand side automatically becomes 1 when you do this.
For this list, all the frequencies will be 1 as the data values in the list are all different.
You enter all the other values in the same way.

When you’ve entered all the values in the list, you press the AC button to return to the main screen. You now need to look at the STATISTICS options for the data.
To do this you press the option key.
You then have 3 options.
The mean value is included in the various statistics for option 2,1- Variable Calc.
So you press 2 and find that the value of the mean, x bar, is 35.5.
So your original working seems to be correct.
If the mean value shown on your calculator does not match the mean value you found in your written work, you need to do some checks.
As well as finding the mean, you can find the total of all the data values and the number of data values using STATISTICS mode.
To check the number of data values, use the centre arrow keys to go down to the next screen.
The number of data values is called n.
You can see that it is correct at 8.

To check the total of all the values, use the centre arrow keys to go up to the previous screen.
The second option in the list has a symbol that looks like an E.
This is the Greek letter Sigma and mathematicians use it to represent a sum.
So, the total of all the data values is called sigma x and is 284.
If you didn’t have 284 in your written work then you need to check your addition. If that is correct, you need to check your data table to make sure your calculator entries are correct too.

If your calculator does not show n as 8 and sigma x as 284 then, to check your data table, press the OPTN key twice and then choose option 3, DATA.

This seems like a lot to remember, but if you practice, you’ll soon recall what to do! Make sure you know how to clear the data from the STATISTICS memory for your calculator before you try the next question. Your friend Lee has attempted to answer this Cambridge examination question. The answer cannot be correct. Can you explain why? You may wish to pause while you think about this.

Okay, 7.11 is much bigger than all the data values.
This cannot be correct as the mean always lies between the smallest and largest data values.
Here these are 2.7 and 4.1 kg.

Lee has correctly divided by 10, as there are 10 babies, so the total of all the values must be incorrect. Let’s practice using a calculator to correct Lee’s error and find the correct value of the mean. For this calculator, you press MODE and then choose option 1, STAT.
As you only want to find the mean of one set of data values, you choose the single data option which is 0, SD.
You can then see the frequency table for x values.
You enter the data values as before. You may wish to pause while you do this. Once you have entered all the data press the on/C button to return to the main screen.
You now press ALPHA and number 8 to look at the STAT options for the data.

There are 3 options on the screen and, as you want to look at the statistics for our data, you choose option 0, STATISTICS VAL.
You can see that the mean of the x values, x bar, is actually 3.42, which is between 2.7 and 4.1.

If you use the centre arrows to go down and look at all the options, you see that the sum of all the data, sigma x, was 34.2 not 71.1.

You can now correct Lee’s error. Here’s another Cambridge examination question.
You are given some information about goals scored in football matches.
This time you are asked to find the mean from a frequency table.
Before you look at how to use the calculator to check the answer, you will work out the answer in the usual way. You may wish to pause while you do this.

The total number of goals scored was 80.
There were 50 matches played and so the mean is 80 divided by 50 which is 1.6.

Now let’s look at how to enter this data into a calculator and find the mean. For this calculator you press the data key.
Three columns appear on screen.
You’ll use the left column for the x values and the middle column for the frequencies.
The x values in this example are the number of goals scored.

You enter the x values in the left hand column, called L1.
As before, you key in each value and then press ‘enter’.

You enter the frequencies in the second column, called L2.
To move across to this column you use the arrow keys.
Alongside each x value, you enter the correct frequency.

When the data has all been entered, you press clear to return to the main screen. You now press the 2nd and data keys to look at the STAT options for the data.
You’ve 2 options.
You only have one set of data and so you choose option 1, 1-Var Stats.

You then need to enter the names of the columns for the data values and frequencies.
The data is in column L1. The frequencies are in column L2.
These column names are already highlighted on the screen. Using the down arrow twice you can choose these columns.

When ‘calc’ is flashing, you press the enter key and the statistics you want are shown.

You can see that the mean number of goals , x bar, is 1.6, so your original calculations seem correct.
If you use the arrow key to move down the screen, you can also see that the sum of all the goals, sigma x, is 80. Again this agrees with your values. Right, let’s try this Cambridge examination question. You are given some information about lambs and sheep.
You’re asked to calculate the mean number of lambs per sheep from a frequency table.
Showing as much method as you think necessary, find this mean.
You may wish to pause while you do this.

This question is worth 3 marks. For the first mark, you need to show that you’ve added the 4 part totals.
For the second mark you need to show that you’ve divided your grand total by 20.
For the last mark, you must have the correct answer and have made no errors.

Now enter this data into your calculator and check that you have found the mean correctly. Here are the calculations you should’ve shown to earn all 3 marks.
The total of all the lambs is 37.
37 divided by 20 is 1.85
And so the mean number of lambs per sheep is 1.85.

Now let’s look at how you should’ve entered this into the calculator to check your working. On this calculator, if you’re already in STATISTICS mode, you can enter the data by pressing the option key and then choosing option 3, DATA.
The x values are the number of lambs.
You enter this in the left hand column.
You then use the arrow keys to move across to the frequency column and up to the first x value.
You enter the correct frequency for each x value.
You then press AC to return to the main screen. To open the STATISTICS menu, press the Option key.
The mean value is included in the various statistics for option 2, 1-Variable Calc.
You press 2 and find that the value of the mean, x bar, is 1.85.

If the mean value shown on your calculator doesn’t match the mean value you found in your written work, you need to do some checks. First, check to see if the number of data values in the calculator is correct.
To do this use the centre arrow keys to go down to the next screen.
The number of data values, n, should be 20.

To check that you have the total of all the data values correct use the centre arrow keys to go up to the previous screen.
The total of the x values, sigma x, should be 37.
If you didn’t have sigma x equal to 37 in your written work then you need to check the part totals and addition. If that is correct, you need to check your data table to make sure your calculator entries are correct too.

If your calculator does not show n as 20 and sigma x as 37 then, to check your data table,
press the Option key twice and choose option 3, DATA.

Checking the total of all the data values, sigma x, and the number of data values, n, is useful if your mean value is not correct. You could of course check these values anyway, as this should give you confidence that your working values are correct too. So, for your calculator, you should know how to do the following;
Enter data values and frequencies.
Check the value of the mean, x bar, for your data set.
Find the total of the data values, sigma x.
Find the number of data values, n.

You should also;
only use your calculator as a checking tool and
always show enough method to be sure of earning full marks.

Remember, the mean will always be between the smallest and largest data values. If your course requires you to find an estimate of the mean from a grouped frequency table, you should also show enough method to make sure you earn all the marks. For the calculator, you find the mid-value of each group and these are entered as your values of x. So there is one extra step, which is finding the mid-values.

In this Cambridge examination question you are given some information about the distances 80 women threw a javelin. You are asked to calculate an estimate of the mean distance the javelin was thrown from a grouped frequency table. Using mid-values and showing as much method as you think necessary, find an estimate of this mean. You may wish to pause while you do this. This question is worth 3 marks. Here are the calculations you should’ve shown to earn all 3 marks.

The total of all the distances is 2710.
2710 divided by 80 is 33.875
So an estimate of the mean distance the javelin was thrown is 33.875 metres.

When there are a lot of calculations to work out, checking your answer is even more important, so let’s look at how you can enter this into the calculator to check your working. On this calculator, if you’re already in STAT mode, you can enter the data by pressing the CHANGE key.
The x values are mid values.
You enter these in the left hand column.
You then use the arrow keys to move across to the frequency column and up to the first x value.
You enter the correct frequency for each x value.
You then press the on/C button to return to the main screen. To open the STAT menu, press ALPHA and 8.
There are 3 options on the screen and, as you want to look at the statistics for the data, you choose option 0, STATISTICS VAL. You can see that the mean of the x values, x bar, is 33.875.

If the mean value shown on your calculator does not match the mean value you found in your written work, you need to do some checks. First, check to see if the total of the frequencies in the calculator is correct at 80. The first entry in the list of statistics is for n and it is 80, so it is likely that you have entered the frequencies correctly into the calculator.

To check that you have the sum of all the values correct, you use the centre arrow key to go down and check sigma x.
You can see that the value of sigma x is 2710.
If you didn’t have sigma x equal to 2710 in your written work then you need to check the part totals and addition. If that is correct, you need to check your data table to make sure your calculator entries are correct too.

If your calculator does not show n as 80 and sigma x as 2710, you can check your data table by pressing the CHANGE key when in STAT mode.

Remember, checking the total of all the data values, sigma x, and the number of data values, n, as well as the value of the mean, x bar, should give you confidence that your working values are correct too. Arit has attempted a solution for this Cambridge Examination question. The question is about the time it takes some runners to run 10 kilometres. This question is worth 4 marks. Arit has not shown the full method. If Arit is correct, she would earn 4 marks. However, you should be able to see that Arit cannot be correct as her answer is much more than 80 minutes.

First, you can check Arit’s mid-values. The groups are not all equally spaced and so you need to take care when checking. Has Arit found the mid-values correctly? You may wish to pause while you check this.

Yes, Arit has found the mid-values correctly. Now, using the STAT mode on your calculator, enter the data and check the total of all the times, sigma x. Is Arit’s answer of 9840 correct? You may wish to pause while you do this. To check Arit’s answer using this calculator you press the data key to show the data table.
You use the left column for the x values and the middle column for the frequencies.
The x values in this example are the mid-values.

As before, you key in each value and then press ‘enter’.
To move across to the middle column you use the arrow keys.
Alongside each x value, you enter the correct frequency.
When the data has all been entered, press the clear button to return to the main screen. You press the 2nd and data keys to look at the STAT options for our data.
You have 2 options.
You only have one set of data and so you choose option 1, 1-Var Stats.
The data is in column L1. The frequencies are in column L2.
Using the down arrow twice you can choose these columns.

When ‘calc’ is flashing, you press the enter key and the statistics you want are shown.
If you use the arrow key to move down the screen, you can see that an estimate of the sum of all the times, sigma x, is 9840.
Arit has correctly found the total of all the times.

However, you know that Arit’s final answer is incorrect.
Write down the extra step Arit should show to complete her method and correct her answer.
Then check the mean, x bar using the STAT mode of your calculator.
Finally, how many marks do you think Arit should be given for her solution to this question?
You may wish to pause while you do this. The step that Arit has forgotten to write down is the total of all the times divided by the number of times.
This is 9840 divided by 200 which is 49.2
You now need to check this in the list of statistics on the calculator. Using the arrow key to move back up the list of statistics on the screen, you can see that the mean, x bar, is 49.2.
Arit’s solution is now correct and would earn 4 marks.

How many marks would her original incorrect and incomplete answer be given? Arit’s original answer would have earned 1 mark for the correct use of the mid-values and 1 mark for finding the correct total for all the times.

As Arit didn’t check her answer on her calculator and didn’t show her full method, she lost half the marks for this question! It is always worth checking your answer on your calculator. To use your calculator, to check an estimate of the mean from a grouped frequency table do the following.
Find the mid-values of your groups.
In the data table, enter the mid-values for x and enter the frequencies.
Check the value of the mean, x bar, for your data set.
Check the estimate of the total of the data values, sigma x.
Check the number of data values, n.

If your calculator values do not agree with your written values, check your written values first. If your written values are correct, check your calculator data table. And remember, that when there are a lot of calculations to work out, it is always a good idea to check your answer.

Good Luck!

Calculate the mean video transcript

Most graphic display calculators have a mode called STAT or STATISTICS. This can be used as a checking tool when finding the mean.

But first, it’s important to remember that if you don’t show your method and ONLY use your calculator, you may not be awarded full marks in an exam question. You MUST make sure that you understand how to find the mean from a simple list or frequency table and, if your course requires it, an estimate of the mean from a grouped frequency table, before you think about using your calculator as a checking tool.

Okay, before you look at how to use the STATISTICS mode to do this, you need to understand the notation used.

Here’s a frequency table for values of x.
Let’s find the mean of x.
As a skill check, you may wish to pause while you do this.
The mean of x is 18 divided by 10 which is 1.8

When the data values you enter are x values, the mean is called ‘x bar’. It will look like an x with a bar over the top. So we say x bar is 1.8.

Here’s a frequency table for values of y. Let’s find the mean of y.
Again, as a skill check, you may wish to pause while you do this.
The mean of y is 41 divided by 10 which is 4.1
When the data values you enter are y values, the mean is called ‘y bar’. It will look like a y with a bar over the top. So we say y bar is 4.1. Bar notation is used over the top to show the value represents the mean of a set of values.

Let’s look at an example question, but first it’s a good idea for you to have your own calculator in front of you so you can use it during this presentation. Okay, here is a Cambridge examination question. Notice that this question is worth 2 marks in the examination. A good tip is that when a question is worth more than 1 mark, you should always show your method as well as your answer. Let’s first work out the answer in the usual way, and then look at how to use the calculator to check our answer.

You know that, to find the mean, you need to total the data values and then divide that by the number of data values. You may wish to pause while you do this.

The total of all the data values is 284. There are 8 data values in the list, so the mean is 284 divided by 8 which is 35.5. Let’s look at how to check this using a calculator. Your calculator may already be in STATISTICS mode. If it’s not you need to choose this mode.
For this calculator, you press MENU and choose option 2, STATISTICS.
Four columns appear on the screen.
You enter the values into the column on the left hand side, List 1.
You press 34 and then EXE, which means execute, to enter it.
You enter all the other values in the same way. You then choose CALC from the menu options at the bottom of the screen by pressing F2.
Before you find the mean, you need to check the settings are correct.
The calculator needs to know that you only have 1 of each value.
You press F6 for SET.
As you only want to find the mean of one set of data values, you need to check the settings for 1 Var.
When you have a list, 1 Var Frequency needs to be equal to 1, as there is only one of each data value in the list.
If it is not already 1, choose it from the two options at the bottom of the screen by pressing F1.
Now you’ve done that you click EXE to go back to the list and then choose option 1, 1-Var by pressing F1. When you have one set of data values, the calculator calls them x values. You can see that the first entry in the list is the mean, x bar, and it is 35.5. Your original working seems to be correct. If the mean value shown on your calculator does not match the mean value you found in your written work, you need to do some checks.

As well as finding the mean, you can check the total of all the data values and the number of data values. The number of data values is called n. It is the last entry in the list, and you can see that it is correct at 8.

The total of all the data values is the second entry in the list. The symbol that looks like an E is the Greek letter Sigma and mathematicians use this to represent a sum. So, the total of all the data values is called sigma x. You can see that this is 284, as you originally calculated.

If you didn’t have the total of all the data values equal to 284 then you need to check the addition in your written work. If your written work is correct, but your calculator does not show n as 8 and sigma x as 284, then you need to check your data table. To do this press EXIT.

This seems like a lot to remember, but if you practice, you’ll soon recall what to do! Make sure you know how to clear the data from the STATISTICS memory for your calculator before you try the next question. Right, your friend Lee has attempted to answer this Cambridge examination question. The answer cannot be correct. Can you explain why? You may wish to pause while you think about this.

7.11 is much bigger than all the data values. This cannot be correct as the mean always lies between the smallest and largest data values. Here these are 2.7 and 4.1 kilograms.

Lee has correctly divided by 10, as there are 10 babies, so the total of all the values must be incorrect. Let’s practice using your calculator to correct Lee’s error and find the correct value of the mean. For this calculator, you press STAT and then choose option 1, EDIT.
Three columns appear on screen.
You will use the left column, L1, for the data values.
You enter the values as before. You may wish to pause while you do this. Once you have entered all the data press the 2nd and MODE keys to return to the main screen.
You now press the STAT key and use the arrow keys to move across to the CALC tab.
As you have one set of data values, you need to choose 1-Var Stats.
You press 1. The calculator needs to know where you have saved the data values.
List should be L1 as that is where you saved the data values. If it is not L1 press the 2nd and 1 keys and then enter.
The frequency of the list should be blank as you have a list not, a frequency table. If it’s not blank, press CLEAR to remove what is there and then enter.
Calculate is then highlighted and when you press enter again all the statistics you want are displayed in one table.
As before, when you have one set of data values, the calculator calls them x values.
You can see that the mean of the x values, x bar, is actually 3.42, which is between 2.7 and 4.1.
You can also see that the total of all the data values, sigma x, is 34.2 not 71.1
You can now correct Lee’s error. Here is another Cambridge examination question. You are given some information about goals scored in football matches. This time you are asked to find the mean from a frequency table. Before you look at how to use the calculator to check your answer, you will work out the answer in the usual way. You may wish to pause while you do this.

Okay, the total number of goals scored was 80. There were 50 matches played and so the mean is 80 divided by 50 which is 1.6. Now let’s look at how to enter this data into a calculator and find the mean. For this calculator, you press MENU and choose option 2, STATISTICS.
Four columns appear on the screen.
You enter the number of goals into the column on the left hand side, List 1.
The x values in this example are the number of goals scored.
As before, you key in each value and then press EXE.
You enter the frequencies into the second column, List 2.
To move across to this column you use the arrow keys.
Alongside each x value, you enter the correct frequency. You then choose CALC from the menu options at the bottom of the screen by pressing F2.
Before you find the mean, you need to check the settings are correct.
The calculator needs to know that you are using a frequency table.
You press F6 for SET.
As before, you need to check the settings for 1 Var.
This time, 1 Var Frequency needs to be List 2, as your frequencies are written in this list.
If it’s not already List 2, choose List from the two options at the bottom of the screen by pressing F2 and then enter 2 and EXE.
Now you have done that press EXE to go back to the list and then choose option 1, 1-Var by pressing F1. You can see that the mean number of goals , x bar, is 1.6, so your original calculations seem correct. We can also see that the total of all the goals, sigma x, is 80. Again this agrees with your values. The number of data values, n, is also correct at 50. In this Cambridge examination question you are given some information about lambs and sheep. You are asked to calculate the mean number of lambs per sheep from a frequency table. Showing as much method as you think necessary, find this mean. You may wish to pause while you do this.

This question is worth 3 marks. For the first mark, you need to show that you’ve added the 4 part totals. For the second mark you need to show that you’ve divided your grand total by 20. For the last mark, you must have the correct answer and have made no errors. Now enter this data into your calculator and check that you have found the mean correctly. Here are the calculations you should’ve shown to earn all 3 marks.
The total of all the lambs is 37.
37 divided by 20 is 1.85
And so the mean number of lambs per sheep is 1.85.
Now let’s look at how you should’ve entered this into the calculator to check your working. For this calculator, you press STAT and then choose option 1, EDIT.
Three columns appear on screen.
The x values are the number of lambs.
You enter the x values in the left hand column, L1, as before.
You then use the arrow keys to move across to column L2.

You enter the correct frequency for each number of lambs in this column.
You may wish to pause while you do this. Once you have entered all the data, press the 2nd and MODE keys to return to the main screen. You now press the STAT key and use the arrow keys to move across to the CALC tab.
As you have one set of data values, you need to choose 1-Var Stats.
You press 1. The calculator needs to know where you have saved the data values. List should be L1 as that is where we saved the number of lambs. If it is not L1, press 2nd and 1 then enter.
The frequency of the list should be L2 as that is where you saved the frequencies. If it is not L2, press 2nd and 2 then enter.
Calculate is then highlighted and when you press enter again all the statistics you want are displayed in one table.
You can see that the mean of the x values, x bar, is 1.85.
You can also see that the total of all the data values, sigma x, is 37 and the number of data values, n, is 20.
If you didn’t have sigma x equal to 37 in your written work, then you need to check the part totals and addition. If that is correct, you need to check your data table to make sure your calculator entries are correct as well.

If your calculator does not show n as 20 and sigma x as 37 then, to check your data table,
press STAT and 1.

Checking the total of all the data values, sigma x, and the number of data values, n, is useful if your mean value is not correct. You could of course check these values anyway, as this should give you confidence that your working values are correct as well. So, for your calculator, you should know how to do the following;
Enter data values and frequencies.
Check the value of the mean, x bar, for your data set.
Find the total of the data values, sigma x.
Find the number of data values, n.

You should also;
only use your calculator as a checking tool and
always show enough method to be sure of earning full marks.

Remember, the mean will always be between the smallest and largest data values. Right, let’s consider another scenario. If your course requires you to find an estimate of the mean from a grouped frequency table, you should also show enough method to make sure you earn all the marks.

For the calculator, you find the mid-value of each group and these are entered as your values of x.
So there is one extra step you have to do, which is finding the mid-values.

In this Cambridge International examination question you are given some information about the distances 80 women threw a javelin. You are asked to calculate an estimate of the mean distance the javelin was thrown from a grouped frequency table. Using mid-values, and showing as much method as you think necessary, find an estimate of this mean. You may wish to pause while you do this. This question is worth 3 marks, and here are the calculations you should have shown to earn all 3 marks.
The total of all the distances is 2710.
2710 divided by 80 is 33.875

So, an estimate of the mean distance the javelin was thrown is 33.875 metres. When there are a lot of calculations to work out, checking your answer is even more important, so let’s look at how you can enter this into the calculator to check your working. On this calculator, from the MENU screen, choose option 2 STATISTICS.
The x values are mid values.
As before, you enter these in the left hand column, List 1.
You then use the arrow keys to move across to List 2, which will be the frequency column.
You then enter the correct frequency for each x value. You then choose CALC from the menu options at the bottom of the screen by pressing F2.
Before you find the mean, you need to check the settings are correct.
The calculator needs to know that you are using a frequency table.
You press F6 for SET.
As before, you need to check the settings for 1 Var.
This time, 1 Var Frequency needs to be List 2, as the frequencies are written in this list.
If it is not already List 2, choose List from the two options at the bottom of the screen by pressing F2 and then enter 2 and EXE.
Now you have done that press EXE to go back to the list and then choose option 1, 1-Var by pressing F1.
We can see that the mean of the x values, x bar, is 33.875.

If the mean value shown on your calculator does not match the mean value you found in your written work, you need to do some checks. First, check to see if the total of the frequencies in the calculator is correct at 80. You can see that n is 80 so it is likely that you have entered the frequencies correctly into the calculator.

To check that you have the sum of all the values correct, you check sigma x.
You can see that the value of sigma x is 2710.
If you didn’t have sigma x equal to 2710 in your written work then you need to check the part totals and addition. If that is correct, you need to check your data table to make sure your calculator entries are correct as well.

If your calculator doesn’t show n as 80 and sigma x as 2710, you can check your data table by pressing EXIT.
Remember, checking the total of all the data values, sigma x, and the number of data values, n, as well as the value of the mean, x bar, should give you confidence that your working values are correct. Okay, let’s look at a final example. Here Arit has attempted a solution for this Cambridge examination question. The question is about the time it takes some runners to run 10 kilometres. This question is worth 4 marks, but Arit has not shown the full method, and even If Arit is correct, she wouldn’t earn all 4 marks. However, you should also be able to see that Arit cannot be correct as her answer is much more than 80 minutes.

So let’s review Arit’s answer. First, you can check Arit’s mid-values.
The groups are not all equally spaced and so you need to take care when checking. Has Arit found the mid-values correctly? You may wish to pause while you check this.

Yes, Arit has found the mid-values correctly. Now, using the STAT mode on your calculator, enter the data and check the total of all the times, sigma x. Is Arit’s answer of 9840 correct? You may wish to pause while you do this. To check Arit’s answer using this calculator you press the STAT and 1 keys to show the data table. You use the left column, L1, for the x values and the middle column, L2, for the frequencies.
The x values in this example are the mid-values.
As before, you key in each value and then press ‘enter’.
To move across to the middle column you use the arrow keys.
Alongside each x value, you enter the correct frequency.
Once you have entered all the data press the 2nd and MODE keys to return to the main screen. You now press the STAT key and use the arrow keys to move across to the CALC tab.
As you have one set of data values, you need to choose 1-Var Stats. So press 1.

The calculator needs to know where you have saved the data values. List should be L1 as that is where you saved the mid-values . If it is not L1, press 2nd and 1 then enter.

The frequency of the list should be L2 as that is where you saved the frequencies. If it is not L2, press 2nd and 2 then enter.

Calculate is then highlighted and when you press enter again all the statistics you want are displayed in one table.

You can see that an estimate of the sum of all the times, sigma x, is 9840. So Arit has correctly found the total of all the times. However, you know that Arit’s final answer is incorrect. You can see that the mean, x bar, should be 49.2.

Write down the extra step Arit should show to complete her method and correct her answer.

Finally, how many marks do you think Arit should be given for her solution to this question? You may wish to pause while you do this. The step that Arit has forgotten to write down is the total of all the times divided by the number of times.
This is 9840 divided by 200 which is 49.2

Arit’s solution is now correct and would earn 4 marks, but how many marks would her original incorrect and incomplete answer be given? Arit’s original answer would have earned 1 mark for the correct use of the mid-values and 1 mark for finding the correct total for all the times. As Arit did not check her answer on her calculator and did not show her full method, she lost half the marks for this question. You can see that it is always worth checking your answer on your calculator! So, to use your calculator to check an estimate of the mean from a grouped frequency table do the following.
Find the mid-values of your groups.
In the data table, enter the mid-values for x and enter the frequencies.
Check the value of the mean, x bar, for your data set.
Check the estimate of the total of the data values, sigma x.
Check the number of data values, n.
If your calculator values do not agree with your written values, check your written values first.
If your written values are correct, check your calculator data table.
And remember, that when there are a lot of calculations to work out, it is always a good idea to check your answer.

Good luck!

Drawing and sketching graphs video transcript

Sometimes you will be asked to draw a graph of a line or curve accurately, and sometimes you will be asked to sketch a graph. So what’s the difference?

Well, when you need to draw a graph accurately, often you’ll be asked to draw it on graph paper. You’re usually asked to find and plot pairs of coordinates and join them together correctly with a straight line or curve. The instruction you’ll be given will be ‘plot and draw’ or just ‘draw’ and you are often given the set of x values that you should use. If you’re not given a set of x values then you should make full use of the graph paper you are given.

You may also need to draw a graph accurately because you have to use it in some way. For example, you may have to find an approximate solution for an equation.

When the instruction in the question is ‘Sketch the graph’, then you are usually only given the x-axis and the y-axis. You don’t plot points and you’re not expected to measure and mark an accurate scale. Occasionally, a scale is marked on either the x-axis or the y-axis or both, but this is not common.

For a curve, the instruction ‘Sketch’ means ‘Make a simple freehand drawing showing the key features’. When you need to make a sketch, the instruction ‘sketch’ will be used in the question and you are not always given the set of x values that you should use. In this case, you should draw the full shape of the graph so that all the key features can be seen and, usually, mark the coordinates of any points where the graph meets or crosses the x-axis and the y-axis. You may also be asked to mark the position of other key points, such as turning points, if there are any.

Remember that a straight line graph should always be drawn with a ruler, whether it is a sketch or an accurate drawing, and like a curve, your drawing should show the key features.

Let’s begin by looking at how to draw straight line graphs. When the equation of the graph follows the pattern y = mx + c, for example y = 2x + 1, the graph will be a straight line. In this Cambridge examination question, the equation of the line is y = 4 – x.

If you rewrite this as y = -x + 4, you can see it fits the pattern for the equation of a straight line. First, you have to find the coordinates of two points. These points are the points of intersection with each of the coordinate axes. How can you find where the line crosses the x axis and the y axis?

You may wish to pause while you think about this.

Okay, on the x-axis, the value of y is always 0. This means you can find the answer to part one by putting y = 0 in the equation of the line. You can then solve the equation for the value of x.

So let’s consider what happens on the y-axis? On the y-axis, the value of x is always 0. This means you can find the answer to part two by putting x = 0 in the equation of the line. We can then solve the equation for the value of y.

You may also know that the value of c in the equation y = mx + c is actually the y coordinate of this point. These two points are very important and are really useful whether you need to draw or sketch a line, so a good tip is to make sure you are confident with finding these two pairs of coordinates as they will help you a lot. In this part of the question you need to draw the line. You’ve been given graph paper and you have two points.
You don’t have a set of values for x, but if you make sure that you use the graph paper fully, you will have an acceptable answer.

You should plot each point and then draw, as accurately as you can with a ruler, the straight line that passes through the two points.

A good tip is to check a third point to try to make sure that you’ve not made an error. Reading from the graph, you can see that the point (1, 3) is on the line you have drawn. You can check this in the equation of the line. The equation is correct for this point and so your answer seems to be correct.

Notice that, in this question, you only found and plotted 2 points. More often, when you are asked to draw a graph, you are asked to complete a table of values and plot several points. You will see examples of this later. Also, the x term in the equation of this line is negative, and this results in a line which slopes down. Let’s now look at a Cambridge examination question that requires a sketch of a graph. Notice that, in this question, you’re not told the graph will be a line. However, if you check the equation of the function we have been given, you can see that it fits the pattern for y = mx + c.

The graph you sketch will need to be a ruled line. Just like the previous question, it is helpful if you know where the line cuts the x-axis and the y-axis. There are no marks for doing this in this question, but it is a useful task to complete.

• When x is 0, the value of y is -3.
• When y is 0, the value of x is 3.

There is no scale marked on the x axis and the y axis, so you don’t have to measure and mark a scale, and an approximate position is good enough.

You mark 3 on the x axis and -3 on the y axis and then draw, as accurately as you can with a ruler, the straight line through these points. You make the line long enough to make full use of the diagram we have been given. Notice that the x term in the equation of this line is positive. This results in a line which slopes up. Okay, let’s try another example. Your friend Lee has attempted to answer this Cambridge examination question. Let’s check Lee’s working and answer.

When you substitute -1 for x in the equation of the function, you get 9 not 7 like Lee. Lee has calculated that minus 1 squared is minus 1. This is a common error when squaring negative numbers using a calculator. You can avoid this error by always using brackets when squaring negative numbers.

• When we substitute 2 for x, we do get -3 and so Lee’s second point is correct.
• When we substitute 3 for x we also get -3 and so Lee’s third point is correct as well.

If you are allowed to use a calculator to find the values, a good tip is to check to see if your calculator has a TABLE mode. You can use this mode to find and check coordinates.

Now you have corrected Lee’s coordinates, let’s look at his graph. Is it correct? You may wish to pause while you think about this.

The instruction in the question is ‘draw’ and we have graph paper, so we should plot points and join them together, which Lee has done. We know Lee’s first point is not correct. However, he has plotted all his points correctly using the values in his original table. This means he has tried to draw the graph from x equals minus 1 to x equals 5, which is correct.

You can see that Lee has joined his plotted points with straight lines. Is this correct?

No, the function is a quadratic function. It has the pattern y equals ax squared plus bx plus c. If you could plot all the points in between those that have already been plotted, you would have a smooth curve, so the plots should not be joined with straight lines. You should also be careful to ensure that the curve should not have a flat bottom. The lowest point on the graph is a single point, not a line. So the graph Lee should have drawn should be more like this one.

If you look at pairs of points with matching y coordinates, you can see that the graph has a line of symmetry which passes through the lowest point. Notice also that the x squared term in the equation is positive. This means that the curve opens upward. Sometimes we say it is a ‘happy’ curve or a ‘smile’ shape.

When the function whose graph we are trying to draw has terms that are powers of x, such as x squared or x cubed, for example, the graph we draw should be a smooth curve. Now let’s look at another Cambridge examination question that requires a sketch of a graph. Notice that, in this question, you’re not told that the graph will be a curve. However, if you multiply out and simplify the right hand side of the equation you have been given, you can see that it fits the pattern for y = ax2 + bx + c, so this is a quadratic function just like the previous question. So the graph you sketch will need to be a smooth curve.

As the x squared term is positive, the curve will open upward, so it will be a happy curve or a smile shape. It will have a lowest point and be symmetrical. The sketch should show the symmetry of the curve and not be pointed or flat at the bottom.

It is helpful if you know where the curve cuts the x-axis and the y-axis. There are no marks for doing this in this question, but it is still a useful task to complete.

• When x is 0, the value of y is 1.
• When y is 0, the value of x is 1.

Notice that this curve doesn’t cut the x-axis in two places as, when y = 0, there is only one value of x.
What does this mean? Well the graph must touch the x-axis. In other words, the x-axis is a tangent to this curve.
You may wish to pause while you try to sketch this graph.

There is no scale marked on the x-axis and the y-axis. You do not have to measure and mark a scale, an approximate position is good enough. So you mark 1 on the x-axis and 1 on the y-axis and then draw, a smooth curve through these points. If you find it difficult to sketch the curve through the points, a good tip is to sketch the curve in the correct position first and then mark the points afterwards. In this question you are again asked to complete a table of values and then draw a graph for a particular set of x values. What should this graph look like? Is it a straight line, the graph of a quadratic function or is it something else?

If you rearrange the right hand side of the equation, youSame for this oen as well can see that it fits the pattern for y = ax2 + bx + c , so it will be the graph of a quadratic function. Now complete the table and draw the graph. You may wish to pause while you do this.

You should have the y coordinates -5, -2 and -5. When you plot them and draw a smooth curve through them, you should have a diagram like this. Notice that the curve is symmetrical, but this time, it does not have a lowest point, it has a highest point.

The line of symmetry still passes through this point. The graph does not open upward. This time, it opens downward. Sometimes we say this is a ‘sad’ curve or a ‘frown’ shape. Why is this?

In this equation the x squared term is negative. When you have a negative x squared term in your quadratic function, the graph will always open downwards. Notice that this curve does not cut the x-axis at all. We have now seen graphs of quadratic functions that cut the x-axis twice, touch the x-axis once and that do not cut the x-axis at all. This is possible for the curves that open upwards or downwards. In this question you are again asked to sketch a graph. What should this graph look like? Does it fit the pattern for a quadratic function?

Yes, we can see that it fits the pattern for y = ax2 + bx + c
The graph you sketch will need to be a smooth curve. You may wish to pause while you try to sketch this graph.

As the x squared term is negative, the curve will open downward, so it will be a sad curve or a frown shape.
It will have a highest point and be symmetrical. The sketch should show the symmetry of the curve and not be pointed or flat at the top.

Where does the curve cut the x-axis and the y-axis?
When x is 0, the value of y is 3.
When y is 0, the values of x are 3 or -1.
So when y = 0, there are two x values so the curve cuts the x-axis in two places.

As usual, there is no scale marked on the x-axis and the y-axis and an approximate position is good enough when sketching the graph.

You mark -1 and 3 on the x axis and 3 on the y-axis and then draw, a smooth curve through these points.
Again, if you find it difficult to sketch the curve through the points, sketch the curve in the correct position first and then mark the points afterwards.

Notice that the highest point is to the right of the y-axis. The symmetry of the curve means that the x-coordinate of this point is the mid-value of -1 and 3, which is 1. You do not need to mark this unless you are asked to, but it is useful information when making a sketch. In its simplest form, a reciprocal function is a fraction with a number for its numerator and x for its denominator. There are more complex reciprocal functions, but these are not studied at this level. Let’s have a look at an example.

Jay has attempted to answer this Cambridge examination question about drawing the graph of the reciprocal function, y = 8 over x. Let’s check Jay’s working and answer.

The y coordinates Jay has found seem to be correct, and Jay has also plotted all 10 points correctly. However, before you look at Jay’s graph, you need to understand simple reciprocal functions. Let’ss look at the table. Notice that there’s a grey section between -1 and 1. Why is this?

It’s not possible for x to have the value 0 as 8 divided by 0 has no mathematical meaning. This means that there should be a break in the graph and it should be in 2 sections and not be one continuous graph.

Clearly Jay’s graph is not correct. You can see that there are two sets of values for x for which we need to draw the graph. This is another clue that you need to draw two sections, not one. You must only draw the graph from -5 to -1 and from 1 to 5. If you draw parts of the graph for other values of x, even though it might be possible to do that, you will not earn full marks.

Jay should’ve joined the first five points she plotted with a smooth curve to form one section of the graph.
Then the other five points should’ve been joined with a smooth curve to form the second section of the graph.
Her graph should have looked like this.

Notice that there should be nothing drawn between x = -1 and x = 1, which is correct for this question. If your course requires it, you may need to sketch the graph of a simple reciprocal function as well.

y = 1 over x is a fraction with a number for its numerator and x as its denominator. You now know that for this type of function you must draw two sections, as x cannot be 0. In this question you are not given two particular sets of x values to use, so there are no restrictions on the sections that you draw.

You need to look at what happens to the curve as x is very large or very small, and you need to do this for values of x that are greater than and less than 0.

First, if you rearrange the equation of the curve you see that x is equal to 1 over y. This means that, as well as x not being 0, y cannot be 0 either. So the graph cannot cut the y-axis and it cannot cut the x-axis either. When x is very large what happens to y?

Let’s say x is 1000. Then y would be 1 over 1000 or 0.001. As x gets larger and larger, y gets smaller and smaller, but it never reaches 0. The graph gets closer and closer to the x-axis but it will never reach it as y cannot be 0. The x-axis is called an asymptote of this section of the graph.

When x is positive and very close to 0 what happens to y? Let’s say x is 0.0001. Then y would be 10000. As positive values of x get closer and closer to 0, y gets larger and larger. The graph gets closer and closer to the y-axis, but it will never reach it as x cannot be 0. The y-axis is also an asymptote of this section of the graph.

An asymptote often causes a break in the graph, so there will be more than one section. When x is negative what happens to y? Let’s say x is minus 1000. Then y would be 1 over minus 1000 or 0.001. As x gets larger and larger in the negative direction, y gets closer and closer to 0 but it never reaches 0. The graph gets closer and closer to the x-axis but it will never reach it as y cannot be 0. The x-axis is an asymptote of this section of the graph too.

Let’s now say x is 0.0001. Then y would be 10000. So, as negative values of x get closer and closer to 0, y gets larger and larger in the negative direction. The graph gets closer and closer to the y-axis but will never reach it as x cannot be 0. The y-axis is also an asymptote of this section of the graph.

Notice that, for this graph, when the x coordinates are positive the y coordinates are positive and when the x coordinates are negative, the y-coordinates are also negative. You now need to sketch the graph of this function. What type of function is this?

If we rewrite it, we can see that it is a fraction with a number for its numerator and x for its denominator, so this is a reciprocal function. In this case:

• When the x coordinates are positive, will the y coordinates be positive or negative?
• When the x coordinates are negative, will the y coordinates be positive or negative?

Think about this and use the logic from the previous example to help you sketch this graph. You may wish to pause while you do this.

When the x coordinates are positive, the y coordinates will be negative.
When the x coordinates are negative, the y coordinates will be positive.
This means that the sections of this graph will be in the top left and bottom right corners of the diagram. The x-axis and the y-axis will be asymptotes, as they were before. This is what you should have drawn.

Notice that, when sketching, the only important thing about the number on top of the reciprocal function is whether it is positive or negative. Apart from that, the size of the number does not really change the shape of our graph, as you do not mark any values on the diagram. The shape is basically the same whether we have -2 over x or -10 over x, for example. So, when asked to draw the graph of a function, you need to do so accurately. This means you will find and plot points and then join them together.

When asked to sketch the graph of a function, the only points you are likely to need are any points where the graph cuts or touches the x-axis or the y-axis.

For a curve, making a sketch means making a simple freehand drawing showing the key features.

For a straight line, always use a ruler whether you are sketching or drawing.

The pattern y = mx + c is the equation of a straight line,
The pattern y = ax2 + bx + c is the equation of a quadratic function and
The pattern y = "a" /"x" is the equation of a reciprocal function.

Remember, read each question carefully before you decide what you need to do, and use your calculator carefully when finding values.

Good luck!

Cambridge International AS & A Level

Graph sketching video transcript

Graph sketching is an essential skill for many of the topics in AS&A Level Maths. Generally, you sketch a graph to help understand its function and behaviour. Sometimes you use the graph to check values you’ve obtained algebraically, for example when solving an equation. You can also use graphs to compare two different functions, when you’re looking for points of intersection, for solving geometrical problems or in understanding transformations.

What does it mean to sketch a graph? It means showing the shape of the graph and its essential features such as intercepts with the axes or coordinates of a vertex or turning point.

Note that sketching a graph is not the same as plotting. To plot a graph, you make a table of values then plot the individual points on a graph and join them up. When you sketch a graph, you just show its important features and its shape. It only needs to be roughly to scale.

Okay, let’s look at a variety of different types of graph and the key features needed in order to sketch them.

First, a straight line graph. Here’s an example. From the equation you can deduce its gradient and y-intercept. This is all you need for the sketch.

The y-intercept is -5. You can either mark -5 on the axis or the coordinates (0, -5), And the gradient is 3: it’s positive and fairly steep, so you can draw the line like this.

Now let’s consider the information you need to sketch a quadratic function. Here’s a general quadratic function.
If a is positive, the graph is shaped like this, like a smile.
If a is negative, the graph is this way up, more like a frown.
If a is 0, it’s not a quadratic function at all but a straight line.

Now let’s look at this example. Suppose you want to sketch the graph of y=f(x).

You can see that the y-intercept is 4, when x is 0. You can factorise the function and solve an equation to find the roots, or x-intercepts.

To find the vertex or turning point, you can complete the square. Can you think of another method of finding this point?

Well, the vertex should be half way between the roots with the line of symmetry through it, in this case x = 5/2. We also know the shape of the graph since the coefficient of x^2 is positive. To sketch the graph you just need to join the intercepts with a curve that’s the correct shape.

You’ve seen how to sketch a quadratic graph and now you’ll extend the reasoning to a cubic graph. Here’s the function: it’s given in factorised form.

You can put x equal to 0 and deduce that the y-intercept is 6. Solving the equation gives roots of -1, 2 and 3.
The coefficient of x^3 is positive so the graph is this shape. So to sketch the cubic curve you can join the points you’ve found.

You could also find the turning points, the maximum and minimum, by differentiation. Not all questions will require this level of detail, so think carefully about which features you need to show.

Sketch graphs also help you to find the domain and range of a function. Using the same quadratic function again, let’s look at this graph.

The domain of the function, the x-values or input, is any real number which you can write like this. The range of the function, the y-values or output, is all values from the vertex, y = -9/4, upwards. You write it like this because you’re including the value -9/4.

If you restrict the domain, the range may be affected; and the graph helps you to work out the new range.

For the domain x > 0, the range is still the same.
For the domain x > 3/2, you’re not including the vertex so the range has changed.
For the domain x ≥ 4, the range is all values from 0 upwards which you write like this.

So the sketch helps you to identify which part of the graph you’re interested in.

You can also think of a circle as a graph. The equation of a circle at the origin looks like this. For example if the radius is 5 then r^2 is 25

Here’s the graph, and a circle translated away from the origin looks like this. From the equation in this form, you can deduce the centre. Since the radius is still 5, you know the circle must overlap both the x- and y-axes. It goes up to 2 on the y-axis and across to -3 on the x-axis.

When you solve geometrical problems, sketching a graph helps you understand geometrical relationships and work out a strategy. Here’s an example:

We’ll start by making a sketch using the information in the question. You can find the centre of the circle and its radius is 2. Drawing in the tangents to the origin, you can see there are two. Let’s call them OS and OT. These are perpendicular to the radius. You can find the distance from O to the centre. Now you can use Pythagoras to calculate the length OT.

It’s important to be able to sketch trigonometric graphs in degrees. Check that you know the correct shape and the important features, which are:
the intercepts
the position of the maximum and minimum for the sin and cos graphs
and asymptotes for the tan graph.
You can use these graphs to find solutions when solving trigonometric equations, or to help with problem-solving.

You can use sketch graphs to help you understand some aspects of transformations.

Let’s start with the graph of y = x.

Suppose you translate it by 2 units parallel to the x-axis. What is the equation of the new line? Well, the new y-intercept is -2 but the gradient hasn’t changed. The line is y = x – 2.

But you could also describe this line as a translation of y=x by -2 units parallel to the y-axis, or 2 units in the negative y direction.

The equation is still y = x - 2 so what is different?

As you can see, under the first translation, each point on the line y = x was mapped horizontally to a point on the new line. But under the second translation, each point on y = x was mapped vertically to a different point on the new line. In each case, the result is a set of points that makes up the line y = x – 2.

Okay, let’s make one final sketch to combine the ideas you’ve looked at. Suppose you want to sketch this graph. What does it remind you of? It’s clearly a quadratic function, so think of y = x^2

The bracket means it is translated 2 units along the x-axis. Think what difference the ½ outside the bracket makes. It’s a stretch scale factor ½ parallel to y, so the graph will be flatter.

Finally you add the 3 to all the y-coordinates so the graph is translated 3 units parallel to the y-axis.

So, how can you get really good at sketching graphs?

Well, practise sketching as often as you can. Think first, try sketching a graph, then check it using a graphing program such as Desmos or GeoGebra. Try to relate the algebra you’ve done to the graph and its features. That will help you to understand the graph and make sure it’s consistent with your working.

Good luck!

Argand diagrams and loci video transcript

Argand diagrams help us to visualise complex numbers. You can use them to understand relationships between numbers, to sketch loci and to solve equations and inequalities. Let’s start by thinking about what a complex number is.

You can write a complex number in several different ways.
If we write it in the form x + iy, it consists of two parts:
a real part, x, and
an imaginary part, y, which is the coefficient of i (the square root of -1)

An Argand diagram consists of real and imaginary axes, with equal scales on both axes.

Let’s plot the complex number z = 1+√3i . It’s just like plotting the point with coordinates (1, √3) on the Cartesian grid.
You could also plot the number as the position vector (■(1@√3))

There are other ways you can represent complex numbers using the modulus and the argument. Looking at the triangle in this Argand diagram, you can see that tanθ is √3 and so the argument, θ, is π/3.

By Pythagoras, you can work out the modulus (or magnitude) of z which is 2. So instead of plotting z using coordinates, you could use the modulus and argument to plot it in terms of a distance and an angle.

By convention, you generally give an argument between –π and π.

It’s helpful to plot complex numbers on an Argand diagram because this helps you to understand geometrical relationships between them.

For example, if you plot z and w, you can immediately see that complex conjugates are reflections of each other in the Real axis.

Adding and subtracting complex numbers works just like adding and subtracting vectors.
Suppose you want to add z1 and z2

The sum of these two complex numbers is at the 4th vertex of this parallelogram. What difference does it make if you subtract instead? Similarly, when subtracting, you can visualise the result by drawing a different parallelogram.

Let’s look at the geometrical implications of multiplying a complex number by a scalar.
Let’s start with this complex number, z, and plot it on the Argand diagram.

Multiplying z by 2, you can see that the real and imaginary components have doubled to give us 2z. On the Argand diagram, the distance from the origin has doubled so the modulus has doubled
but the argument has stayed the same.

Now if you divide z by 2 instead, the modulus has halved but the argument is still unchanged.

Consider a complex number, w, whose modulus is 3.

This statement means w can lie at any point in the complex plane that is 3 units from the origin.
You can describe this as a locus of points that are equidistant from the origin, in this case a circle of radius 3 centred at the origin.

So this equation describes a circle of radius 3 at the origin.

Suppose you want to show the region where the modulus of z is less than 3 rather than exactly 3. What difference would that make to the diagram?

You can imagine z in any position where its distance from the origin is less than 3. So in this case the locus of points is the shaded area inside the circle. Because the modulus of z cannot be equal to 3, you’re not including the circle itself. You show it as a dotted line.

You’ll extend the idea of the locus as a circle by considering this equation.

This tells you that the distance between z and 1 must be equal to 2.
1 is the point (1, 0) on the Real axis.
The distance is always 2 so your circle has a radius of 2. This time, the circle is centred on (1, 0), not the origin. You know that it will intersect the Real axis at -1 and 3 because these are both 2 units away from (1, 0). Let’s imagine that a circle is translated from the origin in two dimensions, not just along one of the axes. How can you find its centre?

Consider this inequality. It states that the distance between z and some point is less than or equal to 3, so that point must be the centre of the circle. How can you find the point?
If you rewrite the left hand side as the difference between z and another complex number, you can see that the centre must be 1 – 2i.

So on the Argand diagram, you can plot the point (1, -2). You’re interested in all points within a circle of radius 3, this time including the circle itself. When you sketch this locus, you need to think carefully about whether the circle will overlap with the Real or Imaginary axes.

What happens if there is a modulus statement on both sides of the equation? Is the locus still a circle?

Let’s start with this equation. As before, you can rewrite the statement to help work out what to plot on the Argand diagram. So you know that the two points you’re interested in are (3, 0) and (1, 1)

The equation you started with tells you that the distance between z and 3 is the same as the distance between z and 1 + i.

This means the set of points z lies on a line that is equidistant from the two points (3, 0) and (1, 1).
It’s the perpendicular bisector of a line joining those two points since the bisector will always be equidistant from those two points.

So far, you’ve looked at loci that are defined in terms of a distance, or modulus. But loci can be defined in terms of an angle, the argument, instead.

Suppose that the argument of z is π by 3. This tells us that any complex number z must lie on a line that starts at the origin and makes an angle of π by 3 with the Real axis. This is called a half-line because it doesn’t extend beyond the origin. If it did, the argument would be -2π by 3 instead.

Let’s consider this statement. As with the circles you looked at, you need to rewrite it as the difference between z and another complex number, in this case –3 + 2i

So you can plot the point (–3, 2) on the Argand diagram and add a horizontal dotted line to the right of the point. The argument is measured as π by 4 anticlockwise. This illustrates that the locus is a half line starting at the point (–3, 2) at an angle of π/4 to the horizontal.

Now let’s think about how you could sketch inequalities for loci defined in terms of an argument. Again, you start by finding a point so that you can plot it on the Argand diagram. The point is (-2, -1). From the horizontal line starting at that point, you measure an angle of π by 6.

The inequality states that the argument must be less than π by 6, so you draw the half-line as a dotted line then shade the region between the two dotted lines.

Now you’re going to combine two different loci. Let’s consider this question. You need to find the region that satisfies two inequalities at once. Let’s take it step by step.

As before, you can rewrite the statements as the difference between z and another complex number
in this case 2+ i so let’s plot it on the Argand diagram.

Looking at the format of the inequalities, you can see that the locus on the left is a circle, centre (2, 1)
while the locus on the right is a half line starting at the point (2, 1)

For the circle, the inequality means you’re interested in all points that are at least one unit from its centre,
so this time you’ll need the circle itself and the region outside it.

For the half line, you can add a horizontal dotted line to the right of the point (2, 1) to show the argument.
The inequality tells us we need the region where the argument is more than π by 6, but less than π by 4. These should be dotted lines because they’re not included in the region we need.

Finally, you can shade the region that satisfies both statements: it’s the region between these dotted lines and outside the circle.

So you’ve looked at how Argand diagrams can be used to visualise complex numbers. Now you can use them to understand relationships between numbers, to sketch loci and to solve equations and inequalities.

Good luck!

Topic outline