What is S2.3: Averages and standard deviation?

The mean uses every value, while the median is resistant to extremes; choose a measure that suits the distribution and context.

How many marks are S2.3 questions worth?

S2.3 practice drills on this page are worth 2–7 marks each; marks in a real exam vary by question.

Which past exam papers has S2.3 appeared in?

We have not yet indexed a verified past-paper appearance for S2.3 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

S2.3

Interpret measures of central tendency and variation, extending to standard deviation; be able to calculate standard deviation, including from summary statistics.

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Averages and standard deviation

Worked answers and methods for S2.3 on Edexcel A-level Maths 9MA0.

Explanation

The mean uses every value, while the median is resistant to extremes; choose a measure that suits the distribution and context.
Range and interquartile range measure spread using endpoints or quartiles; standard deviation measures typical spread about the mean using all observations.
For $n$ data values, use $\sigma=\sqrt{\frac{\sum x^2}{n}-\left(\frac{\sum x}{n}\right)^2}$ unless a different convention is stated.
When groups are combined, add $n$ , $\sum x$ and $\sum x^2$ before recalculating; averaging separate standard deviations is not valid.
Coding data by $y=(x-a)/b$ simplifies calculation, after which transform the statistics back; grouped-data percentiles require linear interpolation within the relevant class.

Worked example

For $20$ observations, $\sum x=310$ and $\sum x^2=5020$ . Calculate the population standard deviation to $3$ significant figures.

1.The mean is $310/20=15.5$ .
2.Hence $\sigma=\sqrt{5020/20-15.5^2}=\sqrt{251-240.25}=\sqrt{10.75}=3.278\ldots$ , so the standard deviation is $3.28$ .

Answer: $3.28$

Common mistakes

Don't average two group means without weighting them by their different sample sizes.
Don't use $S_{xx}/n$ as the standard deviation without taking the square root.

Exam tip

Substitute the stated divisor into the variance formula, keep the square root until the end and check the result is non-negative.

Worked practice

Tier 1 · Easy

Calculate the mean and population standard deviation of

4,7,7,8,9

. Give the standard deviation to

3

significant figures.

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	Mean $=7$ . Standard deviation $=1.67$ to $3$ significant figures.	3
Notes
$\sum x=35$ and $\sum x^2=259$ , so $\bar{x}=35/5=7$ . Then $\sigma=\sqrt{259/5-7^2}=\sqrt{2.8}=1.673\ldots$ , giving $1.67$ .

(3 marks)

Tier 2 · Standard

A data set

x

has mean

12

and standard deviation

3

. A new variable is defined by

y=5+2x

. Find the mean and standard deviation of

y

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	Mean $29$ Standard deviation $6$	3
Notes
Adding $5$ shifts every value and hence adds $5$ to the mean but does not change the spread. Multiplying by $2$ doubles both the mean contribution and the standard deviation. Thus the new mean is $5+2(12)=29$ and the new standard deviation is $\|2\|(3)=6$ .

(3 marks)

Tier 3 · Hard

Group A has

12

values with mean

18

and

\sum x^2=3996

. Group B has

8

values with mean

24

and

\sum x^2=4736

. Find the mean and population standard deviation of all

20

values.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	Combined mean $=20.4$ . Combined standard deviation $=4.52$ to $3$ significant figures.	5
Notes
The combined sum is $12(18)+8(24)=408$ , so the combined mean is $408/20=20.4$ . The combined sum of squares is $3996+4736=8732$ . Therefore $\sigma=\sqrt{8732/20-20.4^2}=\sqrt{436.6-416.16}=\sqrt{20.44}=4.521\ldots$ , giving $4.52$ .

(5 marks)

Tier 1 · Easy

Eight readings have mean

14

. A ninth reading,

16

, is added. Find the mean of all nine readings, giving your answer to

3

significant figures.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$14.2$ to $3$ significant figures	2
Notes
The original total is $8(14)=112$ . After adding $16$ , the total is $128$ , so the new mean is $128/9=14.222\ldots=14.2$ to $3$ significant figures.

(2 marks)

Tier 2 · Standard

Machine A fills

10

containers with

\sum x=200

and

\sum x^2=4040

. Machine B fills

8

containers with

\sum x=164

and

\sum x^2=3400

. Calculate the mean and population standard deviation for each machine, and compare their consistency. Give standard deviations to

3

significant figures.

(5)

(Total for Question 5 is 5 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	Machine A: mean $20$ , standard deviation $2.00$ . Machine B: mean $20.5$ , standard deviation $2.18$ . Machine A is slightly more consistent because it has the smaller standard deviation.	5
Notes
For A, $\bar{x}=200/10=20$ and $\sigma_A=\sqrt{4040/10-20^2}=2$ . For B, $\bar{x}=164/8=20.5$ and $\sigma_B=\sqrt{3400/8-20.5^2}=\sqrt{4.75}=2.179\ldots$ . Smaller standard deviation means less spread, so A is more consistent.

(5 marks)

Tier 3 · Hard

For

25

observations of

x

, the coded variable

y=(x-50)/4

satisfies

\sum y=30

and

\sum y^2=100

. Find the mean and population standard deviation of

x

. One further observation,

x=62

, is then added. Find the new mean and population standard deviation. Give all non-integer answers to

3

significant figures.

(7)

(Total for Question 6 is 7 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	Original mean $=54.8$ and original standard deviation $=6.40$ . New mean $=55.1$ and new standard deviation $=6.43$ .	7
Notes
For $y$ , the mean is $30/25=1.2$ and the variance is $100/25-1.2^2=2.56$ , so $\sigma_y=1.6$ . Since $x=4y+50$ , $\bar{x}=4(1.2)+50=54.8$ and $\sigma_x=4(1.6)=6.4$ . Also $\sum x=1370$ and $\sum x^2=76100$ . After adding $62$ , these become $1432$ and $79944$ for $26$ values. Thus the new mean is $1432/26=55.0769\ldots$ and the new standard deviation is $\sqrt{79944/26-(1432/26)^2}=6.4266\ldots$ , giving $55.1$ and $6.43$ .

(7 marks)

Tier 2 · Standard

A set of

16

observations has mean

7.5

and population standard deviation

2.25

. Find

\sum x

and

\sum x^2

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$\sum x=120$ $\sum x^2=981$	4
Notes
The sum is $n\bar{x}=16(7.5)=120$ . Rearranging $\sigma^2=\sum x^2/n-\bar{x}^2$ gives $\sum x^2=n(\sigma^2+\bar{x}^2)=16(2.25^2+7.5^2)=16(61.3125)=981$ .

(4 marks)

Tier 3 · Hard

Eight observations have mean

10

and population standard deviation

3

. Six of the observations are

5,10,11,12,12,12

; the other two are

u

and

v

, where

u\leq v

. Find

u

and

v

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$u=5$ and $v=13$	6
Notes
The total sum is $8(10)=80$ , while the six known values sum to $62$ , so $u+v=18$ . The total sum of squares is $8(3^2+10^2)=872$ . The known squares sum to $678$ , so $u^2+v^2=194$ . Hence $2uv=(u+v)^2-(u^2+v^2)=324-194=130$ , giving $uv=65$ . Thus $u$ and $v$ are the roots of $t^2-18t+65=0=(t-5)(t-13)$ ; the stated order gives $u=5$ and $v=13$ .

(6 marks)

Tier 3 · Hard

A grouped frequency table has classes

0\leq x<10

10\leq x<20

20\leq x<30

and

30\leq x<50

, with respective frequencies

3

5

8

and

4

. Using class midpoints, estimate the mean and population standard deviation. Give the standard deviation to

3

significant figures and explain why both answers are estimates rather than exact statistics for the raw data.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	Estimated mean $=22.5$ Estimated population standard deviation $=11.1$ to $3$ significant figures The calculations replace every observation in a class by its midpoint, so the within-class positions of the raw observations are unknown.	6
Notes
Use midpoints $5,15,25,40$ . Then $\sum fx=3(5)+5(15)+8(25)+4(40)=450$ , so $\bar{x}=450/20=22.5$ . Also $\sum fx^2=3(25)+5(225)+8(625)+4(1600)=12600$ . Hence $\sigma=\sqrt{12600/20-22.5^2}=\sqrt{123.75}=11.124\ldots$ , giving $11.1$ . Midpoint substitution loses the unknown variation within each class.

(6 marks)

Q10

Tier 3 · Hard

10.

Two groups together contain

10

observations with mean

16.4

and population variance

28.44

. Group A contains

4

observations with mean

11

and population variance

5

. Find the mean and population standard deviation of the

6

observations in group B. Use unrounded values in your working and give the standard deviation to

3

significant figures.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	Mean of group B $=20$ Population standard deviation of group B $=3.42$ to $3$ significant figures	6
Notes
The combined sum is $10(16.4)=164$ and group A contributes $4(11)=44$ , so group B has sum $120$ and mean $20$ . From $\sigma^2=\sum x^2/n-\bar{x}^2$ , the combined sum of squares is $10(28.44+16.4^2)=2974$ , while group A contributes $4(5+11^2)=504$ . Thus group B has sum of squares $2470$ , variance $2470/6-20^2=35/3$ , and standard deviation $\sqrt{35/3}=3.41565\ldots=3.42$ .

(6 marks)

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Worked practice

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Other points in S2 Data presentation and interpretation

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