Mean & Standard Deviation (SQA National 5 Maths): Revision Note

Exam code: X847 75

Roger B

Written by: Roger B

Reviewed by: Dan Finlay

Updated on

Finding the mean & standard deviation

What are the mean and standard deviation of a data set?

  • The mean of a data set is a type of average

    • Its value gives a 'typical value' for a number in the data set

    • The median is another type of average

  • The standard deviation is a measure of spread

    • It tells you how spread out the data set is around the mean

      • The larger the standard deviation, the more spread out the data is

      • The smaller the standard deviation, the less spread out the data is

        • If the standard deviation is zero, then all the data values are equal to the mean!

    • The interquartile range is another measure of spread

  • The mean and standard deviation are used together to analyse a set of data

    • The median and interquartile range may also be used together to analyse a data set

    • But you should not use the mean with the interquartile range, or the median with the standard deviation

Examiner Tips and Tricks

An exam question will tell you which average, and which measure of spread, to calculate for a given data set.

How do I find the mean of a data set?

  • The mean is the sum of the data values divided by the number of values

    • The mean of 1, 2, 6 is (1 + 2 + 6) ÷ 3 = 3

  • The mean can be fraction or a decimal

    • It may need rounding

    • You do not need to force it to be a whole number

      • You can have a mean of 7.5 people, for example!

  • The mean of a data set is sometimes represented by x¯

    • Then the mean can be written as a formula

      • x¯=Σxn

    • where

      • Σx is the sum of all the data values

      • n is the number of data values

    • This formula is not given to you in the exam

How do I find the standard deviation of a data set?

  • The Formulae List in the exam paper gives you two different ways to calculate the standard deviation, s

    • s=Σ(xx¯)2n1

    • s=Σx2(Σx)2nn1

  • In those formulae:

    • n is the sample size (the number of values in the data set)

    • x¯ is the mean of the data set

    • x stands for 'a value in the data set'

    • Σx2 is the sum of the squares of the data values

      • Square each data value, then add those squared values together

    • (Σx)2 is the square of the sum of the data values

      • Find the sum of all the data values, then square the sum

    • Σ(xx¯)2 is the sum of the squares of the differences between the data values and the mean

      • Find the difference between each data value and the mean

      • Square each one of those differences

      • Then add those squared differences together

  • It can be useful to set up a table to work out these values and sums

    • See the Worked Example

Examiner Tips and Tricks

If you have already calculated the mean, then the  s=Σ(xx¯)2n1 form of the standard deviation formula will usually be easier to use.

But either formula will give you the same answer if you use it correctly.

Worked Example

A teacher recorded the number of correct answers achieved by a sample of seven students in a short mathematics test. The results for School A were:

19, 21, 16, 22, 17, 19, 26

Calculate the mean and standard deviation of the number of correct answers achieved by the students in School A.

Answer:

To calculate the mean

  • Find the sum of the data values

  • and divide it by the number of data values (7)

mean=19+21+16+22+17+19+267=1407

mean = 20

To calculate the standard deviation, there are two different formulae you can use

Method 1

Using the formula s=Σ(xx¯)2n1

  • The mean, x¯, is 20, as calculated above

  • n=7

Start by finding the value of Σ(xx¯)2

x

xx¯

(xx¯)2

19

19-20=-1

(-1)2=1

21

21-20=1

12=1

16

16-20=-4

(-4)2=16

22

22-20=2

22=4

17

17-20=-3

(-3)2=9

19

19-20=-1

(-1)2=1

26

26-20=6

62=36

sum:  68

Substitute the values into the formula

s=6871=3.366501...

Round to a sensible degree of accuracy

  • If a question doesn't say otherwise, 3 significant figures is usually a good choice

standard deviation = 3.37  (3 s.f.)

Method 2

Using the formula s=Σx2(Σx)2nn1

  • n=7

Find the values of Σx and Σx2

x

x2

19

192=361

21

212=441

16

162=256

22

222=484

17

172=289

19

192=361

26

262=676

sum:  140

sum:  2868

Substitute the values into the formula

s=2868(140)2771=3.366501...

Round to a sensible degree of accuracy

  • If a question doesn't say otherwise, 3 significant figures is usually a good choice

standard deviation = 3.37  (3 s.f.)

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Roger B

Author: Roger B

Expertise: Development Editor

Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

Dan Finlay

Reviewer: Dan Finlay

Expertise: Portfolio Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.