GCSEStatisticsAQAHigherRevision NotesProcessing, Representing & Analysing DataMeasures of DispersionStandard Deviation

Standard Deviation (AQA GCSE Statistics: Higher): Revision Note

Exam code: 8382

Written by: Roger B

Reviewed by: Dan Finlay

Updated on 8 April 2024

Standard Deviation

What is the standard deviation of a data set?

The standard deviation of a data set is a measure of dispersion (i.e. a measure of spread)
- It measures how the data is spread out relative to the mean
  - If the standard deviation is small then most data values are close to the mean
  - If the standard deviation is large then many data values will be further away from the mean
- If the data has units (seconds, cm, etc.), then the standard deviation has the same units as the values in the data set.
The Greek letter $σ$ (lower case sigma) is often used for standard deviation

How do I calculate the standard deviation for a data set?

There are two different formulas you can use to calculate the standard deviation
- Usually the second formula will be the quickest one to use
  - But make sure you know how to use both of them
$Standard deviation = \sqrt{\frac{1}{n} \sum {(x - \bar{x})}^{2}}$
- In this formula:
  - $n$ is the number of values in the data set
  - $\bar{x}$ is the mean of the data set
  - $x$ is 'any value' in the data set
$Standard deviation = \sqrt{\frac{\sum x^{2}}{n} - {(\frac{\sum x}{n})}^{2}}$
- In this formula:
  - $n$ is the number of values in the data set
  - $\sum x$ is the sum of all the data values
  - $\sum x^{2}$ is the sum of the squares of all the data values
- Sometimes a question will give you the values of $\sum x$ and $\sum x^{2}$ for a data set
  - In that case definitely use this formula!
Both formulas are on the exam formula sheet
- So you don't need to remember them
- You just need to know how to use them

Examiner Tips and Tricks

Your calculator may be able to calculate the standard deviation for a list of data values

Worked Example

For the following set of data values

6 9 2 11 5

(a) Calculate the mean.

Add up the values and divide by the number of values (5)

$\frac{6 + 9 + 2 + 11 + 5}{5} = 6.6$

mean = 6.6

(b) Calculate the standard deviation using $\sqrt{\frac{1}{n} \sum {(x - \bar{x})}^{2}}$ .

It is easiest to set up a table to work out the different values

$x$	$x - \bar{x}$	${(x - \bar{x})}^{2}$
$6$	$6 - 6.6 = - 0.6$	${(- 0.6)}^{2} = 0.36$
$9$	$9 - 6.6 = 2.4$	${(2.4)}^{2} = 5.76$
$2$	$2 - 6.6 = - 4.6$	${(- 4.6)}^{2} = 21.16$
$11$	$11 - 6.6 = 4.4$	${(4.4)}^{2} = 19.36$
$5$	$5 - 6.6 = - 1.6$	${(- 1.6)}^{2} = 2.56$
total		$0.36 + 5.76 + 21.16 + 19.36 + 2.56 = 49.2$

Now we have all the values to put into the formula

$\sqrt{\frac{1}{5} (49.2)} = \sqrt{9.84} = 3.136877 . . .$

standard deviation = 3.14 (3 s.f.)

It is easiest to set up a table to work out the different values

	$x$	$x^{2}$
	$6$	$6^{2} = 36$
	$9$	$9^{2} = 81$
	$2$	$2^{2} = 4$
	$11$	$11^{2} = 121$
	$5$	$5^{2} = 25$
total	$6 + 9 + 2 + 11 + 5 = 33$	$36 + 81 + 4 + 121 + 25 = 267$

Now we have all the values to put into the formula

$\sqrt{\frac{267}{5} - {(\frac{33}{5})}^{2}} = \sqrt{9.84} = 3.136877 . . .$

standard deviation = 3.14 (3 s.f.)

Standard Deviation from a Table

How do I find the standard deviation for data in a table?

A data set may be presented in a table of data values and associated frequencies
- In this case the formulas to use are different
- These formulas are not on the exam formula sheet
  - So you need to remember them
  - But note that they are closely related to the basic formulas
- Usually the second formula will be the quickest one to use
  - But make sure you know how to use both of them
$Standard deviation = \sqrt{\frac{\sum f {(x - \bar{x})}^{2}}{\sum f}}$
- In this formula:
  - $\bar{x}$ is the mean of the data set
  - $x$ is 'any value' in the data set
  - $f$ is the frequency associated with a particular data value $x$
  - $\sum f$ is the sum of all the frequencies (this is the same as the total number of data values in the data set)
$Standard deviation = \sqrt{\frac{\sum f x^{2}}{\sum f} - {(\frac{\sum f x}{\sum f})}^{2}}$
- In this formula:
  - $x$ is 'any value' in the data set
  - $f$ is the frequency associated with a particular data value $x$
  - $\sum f x$ is the sum of $frequency \times data value$ for all the data values in the set
  - $\sum f x^{2}$ is the sum of $frequency \times {(data value)}^{2}$ for all the data values in the set
  - $\sum f$ is the sum of all the frequencies (this is the same as the total number of data values in the data set)
- Sometimes a question will give you the values of $\sum f x$ and $\sum f x^{2}$ for a data set
  - In that case definitely use this formula!

Examiner Tips and Tricks

Your calculator may be able to calculate the standard deviation for a list of data values and their associated frequencies

Worked Example

Kira collected data about the numbers of pet rabbits owned by the members of her local house rabbits association. This data is shown in the following table:

Number of rabbits	1	2	3	4	5
Frequency	2	6	4	6	2

Work out the standard deviation of this data set.

Method 1: using $\sqrt{\frac{\sum f x^{2}}{\sum f} - {(\frac{\sum f x}{\sum f})}^{2}}$

It is easiest to set up a table to work out the different values

number, $x$	$f$	$f x$	$f x^{2}$
$1$	$2$	$2 \times 1 = 2$	$2 \times 1^{2} = 2$
$2$	$6$	$6 \times 2 = 12$	$6 \times 2^{2} = 24$
$3$	$4$	$4 \times 3 = 12$	$4 \times 3^{2} = 36$
$4$	$6$	$6 \times 4 = 24$	$6 \times 4^{2} = 96$
$5$	$2$	$2 \times 5 = 10$	$2 \times 5^{2} = 50$
total	$2 + 6 + 4 + 6 + 2 = 20$	$2 + 12 + 12 + 24 + 10 = 60$	$2 + 24 + 36 + 96 + 50 = 208$

So $\sum f = 20$ , $\sum f x = 60$ and $\sum f x^{2} = 208$
That gives us everything we need to put into the formula

$\sqrt{\frac{208}{20} - {(\frac{60}{20})}^{2}} = \sqrt{1.4} = 1.183215 . . .$

standard deviation = 1.18 (3 s.f.)

Method 2: using $\sqrt{\frac{\sum f {(x - \bar{x})}^{2}}{\sum f}}$

It is easiest to set up a table to work out the different values

number, $x$	$f$	$f x$
$1$	$2$	$2 \times 1 = 2$
$2$	$6$	$6 \times 2 = 12$
$3$	$4$	$4 \times 3 = 12$
$4$	$6$	$6 \times 4 = 24$
$5$	$2$	$2 \times 5 = 10$
total	$2 + 6 + 4 + 6 + 2 = 20$	$2 + 12 + 12 + 24 + 10 = 60$

Now that we have the sum of the f and fx columns we can work out the mean
The sum of the fx column is the sum of all the data values
And the sum of the f column is the total number of data values

$\bar{x} = \frac{60}{20} = 3$

Now we can complete the rest of the table

number, $x$	$f$	$f x$	$x - \bar{x}$	${(x - \bar{x})}^{2}$	$f {(x - \bar{x})}^{2}$
$1$	$2$	$2$	$1 - 3 = - 2$	${(- 2)}^{2} = 4$	$2 \times 4 = 8$
$2$	$6$	$12$	$2 - 3 = - 1$	${(- 1)}^{2} = 1$	$6 \times 1 = 6$
$3$	$4$	$12$	$3 - 3 = 0$	${(0)}^{2} = 0$	$4 \times 0 = 0$
$4$	$6$	$24$	$4 - 3 = 1$	${(1)}^{2} = 1$	$6 \times 1 = 6$
$5$	$2$	$10$	$5 - 3 = 2$	${(2)}^{2} = 4$	$2 \times 4 = 8$
total	$20$	$60$			$8 + 6 + 0 + 6 + 8 = 28$

So $\sum f = 20$ and $\sum f {(x - \bar{x})}^{2} = 28$
That gives us everything we need to put into the formula

$\sqrt{\frac{28}{20}} = \sqrt{1.4} = 1.183215 . . .$

standard deviation = 1.18 (3 s.f.)

Standard Deviation for Grouped Data

How do I find the standard deviation for grouped data?

For grouped data we no longer have access to the original data values
- Therefore we can only find an estimate for the standard deviation
To calculate an estimate for the standard deviation for a set of grouped data:
- Use the same formulas as used for data in a table
  - See the 'Standard Deviation from a Table' spec point
- But use the midpoints of the class intervals as the data values
  - i.e. as the values for $x$ in the formulas
  - The mean $\bar{x}$ will also be an estimate where it appears in a formula

Examiner Tips and Tricks

Your calculator may be able to calculate an estimate for the standard deviation from a list of midpoints and their associated frequencies

Worked Example

Kira collected data about how long the pet rabbits, owned by the members of her local house rabbits association, took to eat their lunch. This data is shown in the following table:

Time, t (minutes)	0 ≤ t < 3	3 ≤ t < 6	6 ≤ t < 9	9 ≤ t < 12
Frequency	1	5	8	6

Work out an estimate for the standard deviation of this data set.

Method 1: using $\sqrt{\frac{\sum f x^{2}}{\sum f} - {(\frac{\sum f x}{\sum f})}^{2}}$

It is easiest to set up a table to work out the different values
Remember to use the class interval midpoints as the x values

midpoint, $x$	$f$	$f x$	$f x^{2}$
$1.5$	$1$	$1 \times 1.5 = 1.5$	$1 \times 1 . 5^{2} = 2.25$
$4.5$	$5$	$5 \times 4.5 = 22.5$	$5 \times 4 . 5^{2} = 101.25$
$7.5$	$8$	$8 \times 7.5 = 60$	$8 \times 7 . 5^{2} = 450$
$10.5$	$6$	$6 \times 10.5 = 63$	$6 \times 10 . 5^{2} = 661.5$
total	$1 + 5 + 8 + 6 = 20$	$1.5 + 22.5 + 60 + 63 = 147$	$2.25 + 101.25 + 450 + 661.5 = 1215$

So $\sum f = 20$ , $\sum f x = 147$ and $\sum f x^{2} = 1215$
That gives us everything we need to put into the formula

$\sqrt{\frac{1215}{20} - {(\frac{147}{20})}^{2}} = \sqrt{6.7275} = 2.593742 . . .$

standard deviation = 2.59 (3 s.f.)

Method 2: using $\sqrt{\frac{\sum f {(x - \bar{x})}^{2}}{\sum f}}$

It is easiest to set up a table to work out the different values
Remember to use the class interval midpoints as the x values

midpoint, $x$	$f$	$f x$
$1.5$	$1$	$1 \times 1.5 = 1.5$
$4.5$	$5$	$5 \times 4.5 = 22.5$
$7.5$	$8$	$8 \times 7.5 = 60$
$10.5$	$6$	$6 \times 10.5 = 63$
total	$1 + 5 + 8 + 6 = 20$	$1.5 + 22.5 + 60 + 63 = 147$

Now that we have the sum of the f and fx columns we can work out the estimated mean
The sum of the fx column is the estimated sum of all the data values
And the sum of the f column is the total number of data values

$\bar{x} = \frac{147}{20} = 7.35$

Now we can complete the rest of the table

midpoint, $x$	$f$	$f x$	$x - \bar{x}$	${(x - \bar{x})}^{2}$	$f {(x - \bar{x})}^{2}$
$1.5$	$1$	$1.5$	$1.5 - 7.35 = - 5.85$	${(- 5.85)}^{2} = 34.2225$	$1 \times 34.2225 = 34.2225$
$4.5$	$5$	$22.5$	$4.5 - 7.35 = - 2.85$	${(- 2.85)}^{2} = 8.1225$	$5 \times 8.1225 = 40.6125$
$7.5$	$8$	$60$	$7.5 - 7.35 = 0.15$	${(0.15)}^{2} = 0.0225$	$8 \times 0.0225 = 0.18$
$10.5$	$6$	$63$	$10.5 - 7.35 = 3.15$	${(3.15)}^{2} = 9.9225$	$6 \times 9.9225 = 59.535$
total	$20$	$147$			$34.2225 + 40.6125 + 0.18 + 59.535$ $= 134.55$

So $\sum f = 20$ and $\sum f {(x - \bar{x})}^{2} = 134.55$
That gives us everything we need to put into the formula

$\sqrt{\frac{134.55}{20}} = \sqrt{6.7275} = 2.593742 . . .$

standard deviation = 2.59 (3 s.f.)

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

I would just like to say a massive thank you for putting together such a brilliant, easy to use website.I really think using this site helped me secure my top gradesin science and maths. You really did save my exams! Thank you.

Beth
IGCSE Student

This website is soooo useful and I can’t ever thank you enough for organising questions by topic like this. Furthermore, the name of the website could not have been more appropriate as it literally did SAVE MY EXAMS!

Fathima
A Level Student

Incredible! SO worth my money, the revision notes have everything I need to know and are so easy to understand. I actually enjoy revising! It makes me feel a lot more confident for my GCSEs in a few months.

Kate
GCSE Student

Absolutely brilliant, both my girls used it for A levels and GCSE. It's saves on paper copies, also beneficial exam questions ranked from easy to hard. It's removed a lot of stress from the exams.

Sameera
Parent

Just to say that your resources are the best I have seen and I have been teaching chemistry at different levels for about 40 years

Mark
Chemistry Teacher

Excellent

Standard Deviation (AQA GCSE Statistics: Higher): Revision Note

Standard Deviation

What is the standard deviation of a data set?

How do I calculate the standard deviation for a data set?

Examiner Tips and Tricks

Worked Example

Standard Deviation from a Table

How do I find the standard deviation for data in a table?

Examiner Tips and Tricks

Worked Example

Standard Deviation for Grouped Data

How do I find the standard deviation for grouped data?

Examiner Tips and Tricks

Worked Example

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Roger B

Reviewer: Dan Finlay