© boardworks ltd 2005 1 of 49 d2 averages and range ks4 mathematics

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© Boardworks Ltd 2005 1 of 49 D2 Averages and range KS4 Mathematics

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Page 1: © Boardworks Ltd 2005 1 of 49 D2 Averages and range KS4 Mathematics

© Boardworks Ltd 20051 of 49

D2 Averages and range

KS4 Mathematics

Page 2: © Boardworks Ltd 2005 1 of 49 D2 Averages and range KS4 Mathematics

© Boardworks Ltd 20052 of 49

A

A

A

A

AD2.1 The mode

D2 Averages and range

Contents

D2.5 Comparing data

D2.2 The mean

D2.3 Calculating the mean from frequency tables

D2.4 The median

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The three averages and range

There are three different types of average:

MEDIAN

middle value

The range is not an average, but tells you how the data is spread out:

RANGE

largest value – smallest value

MODE

most common

MEAN

sum of valuesnumber of values

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This graph shows pupils’ favourite athletics events.

Favourite athletics event

0

5

10

15

20

Sprint

Long

dist

ance

runn

ing

Hurdle

s

High ju

mp

Long

jum

p

Triple

jump

Shot

Discus

Jave

lin

Fre

quen

cy

Which is the most popular event? How do you know?

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The mode is the item that occurs the most often in a data set.

The mode

The most common item is called the mode.The most common item is called the mode.

In the graph the mode is sprint because it is represented by the highest bar.

We could also say “The modal athletic event is sprint.”

Is it possible to have more than one modal value?

Is it possible to have no modal value?

Yes

Yes

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The mode

We could write out all the results in a list. The list would begin:

How many words (items) would there be in the list altogether?

How could we work out the mode from the list if we didn’t have the graph?

Can we tell how many pupils took part in the survey?

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14 15 15 13 12 14 15 0 1113 14 11 16 14 15 9 10 12

The mode

Here are the attendance figures at a weekly school athletics club for Year 11.

What is the modal number of pupils attending?

Are there any unusual results in the data set?

This result is called an outlier. Can you think of any possible reasons for the outlier?

If the data set were very long, what would be the best way to find the mode?

Discuss : Over how many weeks were the results collected?

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02468

101214161820

Sprint

Long

dist

ance

runn

ing

Hurdle

s

High ju

mp

Long

jum

p

Triple

jump

Shot

Discus

Jave

lin

Fre

quen

cy

Compare this graph to the previous one.

Favourite athletics event

What conclusions can you draw? Which two groups of pupils could be represented by the two graphs?

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How many sports do you play?

How many pupils play more than two sports?

A group of pupils were asked how many sports they played.This graph shows the results.

What is the modal number of sports played?

How many pupils took part in the survey?

0

2

4

6

8

10

12

14

0 1 2 3 4 5 6Numbers of sports played

Fre

quen

cy

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This graph represents Year Ten girls’ times for a 100m sprint race.

Grouped data

0

2

4

6

8

10

Fre

que

ncy

Times in seconds12 13 14 15 16 17 18 19 20

What is the modal time interval?

How many girls are in this interval?

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When the mode is not appropriate

94

35

103

6

2

1

0

Numbers of sports played

2

15

17

20

Frequency

Another survey is carried out among university students.The results are represented in this table:

A newspaper reporter writes:

“You may be surprised to learn that the average number of sports played by university students is 0.”

Should the reporter say which average has been used?

Why is the mode a misleading average in this example?

Do you think this is a fair representation of the data?

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Data that is heavily weighted towards one end of the data set is said to be skewed. When data is skewed, the mode is not an appropriate average.

Data that is heavily weighted towards one end of the data set is said to be skewed. When data is skewed, the mode is not an appropriate average.

Skewed data

0

5

10

15

20

25

1 2 3 4 5 6 7Numbers of sports played

Fre

quen

cy

Negatively skewed data

0

2

4

6

8

10

12

14

1 2 3 4 5 6 7

Numbers of sports played

Fre

quen

cy

Positively skewed data

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Contents

A

A

A

A

A

D2.2 The mean

D2.5 Comparing data

D2.1 The mode

D2 Averages and range

D2.3 Calculating the mean from frequency tables

D2.4 The median

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St Clement Danes School holds an inter-form athletics competition for Year 10. Each class must select their five best boys and five best girls for each event.

Comparing data

Here are the times in seconds for 100 metres sprint for the two best classes.

Which class should win and why?

13.116.514.315.4

13.116.413.815.4

12.815.913.415.3

12.215.212.914.7

12.014.911.512.8

10C boys10C girls10B boys10B girls

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The mean

The mean is the most commonly used average.

To calculate the mean of a set of values we add together the values and divide by the total number of values.

Mean =Sum of values

Number of values

For example, the mean time for Class 10B girls is:

12.8 + 14.7 + 15.3 + 15.4 + 15.4

5= 14.72

73.6

5=

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The mean

Calculate the mean times for the other three groups.

mean time

10C boys10C girls10B boys10B girls

14.72 13.18 15.78 12.64

Now calculate means for Class 10B and Class 10C (with girls and boys combined).

mean time

Class 10CClass 10B

13.95 14.21

Based on these results, who should win?

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Calculating the mean

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Calculating a missing data item

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The school athletics team take part in an inter-schools competition. James’s shot results (in metres) are below.

Outliers and their effect on the mean

9.46 9.25 8.77 10.25 10.35 9.59 4.02

A data item that is significantly higher or lower than the other items is called an outlier. Outliers affect the mean, by reducing or increasing it.

A data item that is significantly higher or lower than the other items is called an outlier. Outliers affect the mean, by reducing or increasing it.

Discuss:

What is the mean throw?

Is this a fair representation of James’s ability? Explain.

What would be a fair way for the competition to operate?

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Here are some 1500 metre race results in minutes.

Outliers and their effect on the mean

It may be appropriate in research or experiments to remove an outlier before carrying out analysis of results.

Discuss:

6.26 6.28 6.30 6.39 5.38 4.54 10.59 6.35 7.01

Are there any outliers?

Will the mean be increased or reduced by the outlier?

Calculate the mean with the outlier.

Now calculate the mean without the outlier. How much does it change?

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Contents

A

A

A

A

A

D2.3 Calculating the mean from frequency tables

D2.5 Comparing data

D2.1 The mode

D2 Averages and range

D2.2 The mean

D2.4 The median

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Calculating the mean from a frequency table

Numbers of sports played

Frequency

0 20

1 17

2 15

3 10

4 9

5 3

6 2

Here are the results of a survey carried out among university students.

If you were to write out the whole list of results, what

would it look like?

What do you think the mean will be?

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Calculating the mean from a frequency table

26

3

9

10

15

17

20

Frequency Number of sports

× frequency

4

5

3

2

1

0

Numbers of sports played

TOTAL

0 × 20 = 0

1 × 17 = 17

2 × 15 = 30

3 × 10 = 30

4 × 9 = 36

5 × 3 = 15

6 × 2 = 12

Mean = 140 ÷ 76 =

14076

2 sports (to the nearest whole)

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Because the data is grouped, we do not know individual scores. It is not possible to add up the scores.

Grouped data

Javelin distances in

metres

Frequency

5 ≤ d < 10 1

10 ≤ d < 15 8

15 ≤ d < 20 12

20 ≤ d < 25 10

25 ≤ d < 30 3

30 ≤ d < 35 1

35 ≤ d < 40 1

36

Here are the Year Ten boys’ javelin scores.

How could you calculate the mean from this data?

How is the data different from the previous examples you

have calculated with?

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It is possible to find an estimate for the mean.

This is done by finding the midpoint of each group.

To find the midpoint of the group 10 ≤ d < 15:

10 + 15 = 25

25 ÷ 2 =

Midpoints

12.5 m

Javelin distances in

metres

Frequency

5 ≤ d < 10 1

10 ≤ d < 15 8

15 ≤ d < 20 12

20 ≤ d < 25 10

25 ≤ d < 30 3

30 ≤ d < 35 1

35 ≤ d < 40 1

Find the midpoints of the other groups.

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135 ≤ d < 40

1

3

10

12

8

1

Frequency Midpoint

30 ≤ d < 35

Frequency × midpoint

25 ≤ d < 30

20 ≤ d < 25

15 ≤ d < 20

10 ≤ d < 15

5 ≤ d < 10

Javelin distances in

metres

Estimating the mean from grouped data

1 × 7.5 = 7.5

8 × 12.5 = 100

12 × 17.5 = 210

10 × 22.5 = 225

3 × 27.5 = 82.5

1 × 32.5

Estimated mean = 695 ÷ 36

1 × 37.5

= 32.5

7.5

12.5

17.5

22.5

27.5

32.5

37.5 = 37.5

36 695TOTAL

= 19.3 m (to 1 d.p.)

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How accurate is the estimated mean?

35.00 31.05 28.89 25.60 25.33 24.11 23.50 21.82 21.78

21.77 21.60 21.00 20.70 20.20 20.00 19.50 19.50 18.82

17.35 17.31 16.64 15.79 15.75 15.69 15.52 15.25 15.00

14.50 12.80 12.50 12.00 12.00 12.00 11.85 10.00 9.50

Here are the javelin distances thrown by Year 10 before the data was grouped.

Work out the mean from the original data above and compare it with the estimated mean found from the grouped data.

How accurate was the estimated mean?

The estimated mean is 19.3 metres (to 1 d.p.).

The actual mean is 18.7 metres (to 1 d.p.).

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A

A

A

A

A

Contents

D2.4 The median

D2.5 Comparing data

D2.1 The mode

D2 Averages and range

D2.2 The mean

D2.3 Calculating the mean from frequency tables

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6.26 6.28 6.30 6.39 5.38 4.54 10.59 6.35 7.01

The median is the middle number when all numbers are in order.The median is the middle number when all numbers are in order.

The median

Calculate the median of the 1500 m results.

Why is this a more appropriate average than the mean for these results?

4.54 5.38 6.26 6.28 6.30 6.35 6.39 7.01 10.59

Write the results in order and find the middle value:

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Choosing the most appropriate average

What are the mean and median for these sets of attendance figures for three lunchtime activities?

Explain your answers.

To decide which of the three activities is the most popular, which average is a better one to use? Why?

23222120191817Choir

20202020202020Drama club

29282520191818Orchestra

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Outliers and the median and mean

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If there are two middle numbers, you need to find what is halfway between them.

2.15 2.21 2.40 2.55 2.80 3.32 3.46 3.63 3.83 4.74

When there are two middle numbers

Here are 10B girls’ long jump results in metres.

How could you work out the median jump?

2.80 m + 3.32 m = 6.12 m

6.12 m ÷ 2 = 3.06 m

If the numbers are far apart, a quick way to find the middle of those two numbers is to add them up and divide by two.

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Finding halfway between two numbers

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One or two middle numbers?

1 2 3 4 6 6 7 8 9 11

If there are 9 numbers in a list, will there be 1 or 2 middle numbers?

2 3 4 6 6 7 8 9 11

If there are 10 numbers in a list, will there be 1 or 2 middle numbers?

If there is an even number of numbers in a list, there will be two middle numbers.

If there is an odd number of numbers in a list, there will be one middle number.

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To find out where a middle number in a very long list, call the number of numbers n. Then the middle number is then

When there are two middle numbers

(n + 1) ÷ 2

101 ÷ 2 = 50.5th number in the list (halfway between the 50th and the 51st).

38 ÷ 2 = 19th number in the list.

For example,

There are 100 numbers in a list. Where is the median?

There are 37 numbers in a list. Where is the median?

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Where is the median?

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Contents

A

A

A

A

A

D2.5 Comparing data

D2.1 The mode

D2 Averages and range

D2.2 The mean

D2.3 Calculating the mean from frequency tables

D2.4 The median

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The range

Here are the high jump scores for two girls in metres.

Joanna 1.62 1.41 1.35 1.20 1.15

Kirsty 1.59 1.45 1.41 1.30 1.30

Find the range for each girl’s results and use this to find out who is consistently better.

Joanna’s range = 1.62 – 1.15 = 0.47

Kirsty’s range = 1.59 – 1.30 = 0.29

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The range

The highest and lowest scores can be useful in deciding who is more consistent.

If the scores are close together then the range will be lower and the scores more consistent.

The lowest score subtracted from the highest score is called the range.

Remember that the range is not an average, but a measure of spread.

If the scores are spread out then the range will be higher and the scores less consistent.

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The range

Joanna Kirsty

Mean

Range

1.35 m

0.47 m

1.41 m

0.29 m

Joanna 1.62 1.41 1.35 1.20 1.15

Kirsty 1.59 1.45 1.41 1.30 1.30

Calculate the mean and the range for each girl.

Use these results to decide which one you would enter into the athletics competition and why.

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Calculating the mean, median and range

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Comparing sets of data

Chris Rob

Mean 24.8 seconds 25.0 seconds

Range 1.4 seconds 0.9 seconds

Here is a summary of Chris and Rob’s performance in the 200 metres over a season. They each ran 10 races.

Which of these conclusions are correct?

Robert is more reliable. Robert is better because his mean is higher. Chris is better because his range is higher. Chris must have run a better time for his quickest race. On average, Chris is faster but he is less consistent.

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Comparing sets of data

Chris Rob

Mean 24.8 seconds 25.0 seconds

Range 1.4 seconds 0.9 seconds

24.424.524.424.524.624.925.025.025.125.8

24.324.925.025.025.025.125.125.125.225.2

Here is the original data for Chris and Rob.

Use the summary table above to decide which data set is Chris’s and which is Rob’s?

Who has the best time?

Who has the worst time?

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Year 912.114.015.315.415.415.615.715.716.116.717.0

Comparing hurdles scores

Year 9 Year 10

Mean

Range

Year 1012.313.715.515.515.615.916.016.116.117.122.9

15.4 16.1

4.9 10.6

Here are the top eleven hurdles scores in seconds for Year 9 and Year 10.

Work out the mean and range.

Which year group do you think is better and why?

Why might Year 10 feel the comparison is unfair?

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Finding the interquartile range

When there are outliers in the data, it is more appropriate to calculate the interquartile range.

The time of 22.9 seconds is an outlier.

The interquartile range is the range of the middle half of the data.

The lower quartile is the data value that is quarter of the way along the list.

The upper quartile is the data value that is three quarters of the way along the list.

interquartile range = upper quartile – lower quartileinterquartile range = upper quartile – lower quartile

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Locating the upper and lower quartiles

There are 11 times in each list.

16.1 – 15.3 = 0.8

16.1 – 15.5 = 0.6

Year 9

12.114.015.315.415.415.615.715.716.116.717.0

Year 10

12.313.715.515.515.615.916.016.116.117.122.9

Interquartile range for Year 9:

Interquartile range for Year 10:

Where is the median in each list?

Where is the lower quartile in each list?

Where is the upper quartile in each list?

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The location of quartiles in an ordered data set

When there are n values in an ordered data set:

The lower quartile =n + 1

4th value

The median =n + 1

2th value

The upper quartile =3(n + 1)

4th value

The interquartile range = the upper quartile – the lower quartile

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Finding the interquartile range

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Review

To review the work you have covered in this topic:

2) Make up challenges involving sets of data for your partner, such as working out the mean.

3) Make a list of possible mistakes to avoid in this topic.

Write out the key words on cards.

Shuffle the cards.

Describe the word on each card to your partner.

Your partner must guess the word.

Do as many as you can in one minute, then swap over.

1) Play “Guess the word”.