5 the median of grouped data

8/12/2019 5 the Median of Grouped Data

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The Median of

Grouped Data

Christine Crisp

Teach A Level Maths

Statistics 1


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The Median of Grouped Data

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Statistics 1AQA

EDEXCEL

MEI/OCR

OCR


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Finding the median of raw data is easy.

e.g.1 Find the median of3231172879233413

3432312823171397

The data must be put in numerical order:

The median is the middle value, which we can see is the5thvalue, so,

median = 23

The formula telling us which value we want is ,where nis the number of data items.

2

1n

If nis an even number, we average the 2middle values.


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e.g.2 Find the median of

x 1 2 3 4 5f 4 7 9 6 2

There are 28observations so using we need:2

1n

th5142

128

Accumulating the frequencies:

20114Cu. f54321x

so, the 14thand 15thobservations are both 3.

The median is3.

We need to average the 14thand 15thnumbers.


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With a grouped distribution, we can only estimatethe median.

36543Frequency, f

41 - 5031 - 4021 - 3011 - 201 - 10Length (cm)e.g. 3

There are 21observations so we want to estimate thesize of the 11thone. It lies in the 3rdclass.


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36543Frequency, f

41 - 5031 - 4021 - 3011 - 201 - 10Length (cm)e.g. 3

The first2classes have a cumulative frequency of

7,




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36543Frequency, f

41 - 5031 - 4021 - 3011 - 201 - 10Length (cm)e.g. 3

so to reach 11, we need 4more.The first2classes have a cumulative frequency of

7,




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36543Frequency, f

41 - 5031 - 4021 - 3011 - 201 - 10Length (cm)e.g. 3


7,so to reach 11, we need 4more.


The 3rdclass has afrequency of 5so we need to go part-way along this class.



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36543Frequency, f

41 - 5031 - 4021 - 3011 - 201 - 10Length (cm)e.g. 3







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36543Frequency, f

41 - 5031 - 4021 - 3011 - 201 - 10Length (cm)e.g. 3

Coming up with our own method, needing 4out of thefrequency of 5, we would go 4/5thalong the class.

The class is 10wide







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36543Frequency, f

41 - 5031 - 4021 - 3011 - 201 - 10Length (cm)e.g. 3


The class is 10wide and 4/5thof 10is 8.

The first 2classes have a cumulative frequency of 7,so to reach 11, we need 4more.





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36543Frequency, f

41 - 5031 - 4021 - 3011 - 201 - 10Length (cm)e.g. 3


The class is 10wideWe would go 8along the class, which starts at 205,

and 4/5thof 10is 8.






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36543Frequency, f

41 - 5031 - 4021 - 3011 - 201 - 10Length (cm)e.g. 3


The class is 10wide and 4/5thof 10is 8.We would go 8along the class, which starts at 205,




205



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36543Frequency, f

41 - 5031 - 4021 - 3011 - 201 - 10Length (cm)e.g. 3


The class is 10wide

giving 28 5.

and 4/5thof 10is 8.We would go 8along the class, which starts at 205,

This is a reasonable estimate but not quite the accepted

method which is called Linear Interpolation.






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36543Frequency, f

41 - 5031 - 4021 - 3011 - 201 - 10Length (cm)e.g. 3

Our reasoning was:

The median is the 11thobservation.

There are 7 in class 1 and class 2.We need to add 4 to reach the median.

We need to assume the data are evenly distributed inthe 3rdclass and it can be shown that this means themedian is found at 35along the class not 4.

You dont need to know the reason for this but Ive putan explanation at the end of the presentation.

you will get the correct answer.2

n

2

1n

If you liked our reasoning to get to the estimate, stick to

it, but in locating the median use instead of and



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If you prefer to use a formula to find the estimate ofthe median, the formula is given by

where,wf

F

n

2median l.c.b.

2

n

l.c.b.

Ffw

520 510

7

5

10

105

7510

520

527 median

36543Frequency, f

41 - 5031 - 4021 - 3011 - 201 - 10Length (cm)e.g. 3



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To use linear interpolation to find an estimate of themedian for a grouped frequency distribution, we

locate the class containing the median usingtotal frequency divided by 2,

Fis the cumulative frequencies up to the classcontaining the median,

fis the frequency of the class containing the median,

wis the width of the class containing the median.

( think of n/2 Fas the distance along the class to the

median ),

usew

f

Fn

2median l.c.b. where,

SUMMARY

or, use reasoning to save the need to remember

the formula.


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58107Frequency, f16 - 2011 - 156 - 101 - 5Height (cm)

e.g.1 Estimate the median for the following:.

Solution:

30n

2

n15

The median is in the 2ndclass.Method 1: Without the formula,

The 1stclass has 7. . .


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Solution:

30n

2

n15



There are 10in the 2ndclass . . .

so we need to go 157 = 8along the 2ndclass.


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Solution:

30n

2

n15



There are 10in the 2ndclass . . .45

10

8


so we want 8/10thof the class width:


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Solution:

30n

2

n15



There are 10in the 2ndclass . . .45

10

8

The l.c.b. is 55, so the estimate of the median is 95.


so we want 8/10thof the class width:


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587Frequency, f16 - 2011 - 151 - 5Height (cm)


Solution:

30n

2

n15

where,wf

Fn

2median l.c.b.

distance along class: Fn2

106 - 10

Method 2: Using the formula,


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106 - 10

587Frequency, f16 - 2011 - 151 - 5Height (cm)


Solution:

30n

2

n15

distance along class:

wf

Fn


Fn2 15



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Solution:

30n

2

n15


wf

Fn


Fn2 715



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Solution:

30n

2

n15


wf

Fn


Fn2 7 8frequency of class,

15



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Solution:

30n

2

n15


wf

Fn


Fn2 7 810ffrequency of class,

15



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Solution:



30n

2

n15


wf

Fn


Fn2 7 810f

555510 wfrequency of class,

width of class,

15



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Solution: 2

n15


wf

Fn


Fn2 7 8

510

8

55 median 59

10ffrequency of class,

15

555510 wwidth of class,



30n



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In the next example and the solutions to the exercise,

Ive used the formula.

However, if you choose to use the formula, you will needto memorize it.

If you find it easy to work each problem out usingreasoning, just stick to that. Its all the formula isdoing anyway.

h d f G d


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11n

Solution:

The median is in the 3rdclass.

2

n55

wf

Fn

2median l.c.b.


50555

5

4

50510 median 111

class width= 5510515

Th M d f G d D


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Exercise

61075Frequency, f

41 - 5036 - 4031 - 3521 - 30Length(cm)

Use linear interpolation to estimate the median of thefollowing:

1.

2.

162420Frequency, f

16 - 1813 - 1510 - 12Age (yrs)

Th M di f G d D


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wf

Fn

2median l.c.b.

28n 142n

536510

2535 median

distance along class:Fn

2

21214


Solutions:

61075Frequency, f

41 - 5036 - 4031 - 3521 - 30Length(cm)1.

Th M di f G d D t


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Solution:

The median is in the 2ndclass.

314324

1013 median

As the data giveages, the boundariesare13 and16, not 125and155.

2.

162420Frequency, f

16 - 1813 - 1510 - 12Age (yrs)

31316 class width=

60n 302n

wf

Fn

2median l.c.b.


102030

Th M di f G d D t


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The next 4slides show you how the linear interpolationformula is derived.

You are not expected to know the derivation so youcan skip over them unless you are interested.

SKIP

Th M di f G d D t


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Well start with the example we used before.

36543Frequency, f41 - 5031 - 4021 - 3011 - 201 - 10Length (cm)e.g. 3


Since we have 7 observations in the first 2 classes, themedian is the 4thvalue in the 3rdclass.The observations could be anywhere in the class but onaverage we expect them to be evenly spaced so we assume

that the 3rd

class looks like this:

The estimate of the median is 275.

x x x xx

215 235 255 275 295

205 305

Th M di f G d D t


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205 305x x x xx

215 235 255 275 295

36543Frequency, f

41 - 5031 - 4021 - 3011 - 201 - 10Length (cm)

3rdclass:

The diagram shows we want to add 7to the l.c.b.

05 1 1 1

This is 7tenths of the class width or 1010

7

The 7 tenths comes from35parts of the 5parts given bythe class frequency.

We need to express this as a formula.

Th M di f G d D t


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205 305x x x xx

215 235 255 275 295

36543Frequency, f

41 - 5031 - 4021 - 3011 - 201 - 10Length (cm)

3rdclass:

So, the estimate is: 527105

53520

This is 7tenths of the class width or 1010

7

The 7 tenths comes from35parts of the 5parts given bythe class frequency.

The diagram shows we want to add 7to the l.c.b.

We need to express this as a formula.

Th M di f G d D t


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205 305x x x xx

215 235 255 275 295

36543Frequency, f

41 - 5031 - 4021 - 3011 - 201 - 10Length (cm)

3rdclass:

The estimate of the median is: 527105

53

520

In general, we have

wf

Fn

2median l.c.b.

We can think of n/2 Fas the distance along the classto the median.

fis the frequency and wthe width, both for the classcontaining the median.


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The following slides contain repeats ofinformation on earlier slides, shown withoutcolour, so that they can be printed and

photocopied.

For most purposes the slides can be printedas Handouts with up to 6slides per sheet.


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To use linear interpolation to find an estimate of themedian for a grouped frequency distribution, we

locate the class containing the median usingtotal frequency divided by 2,

Fis the cumulative frequencies up to the classcontaining the median,

fis the frequency of the class containing the median,wis the width of the class containing the median.

( think of n/2 Fas the distance along the class to the

median ),

usew

f

Fn


SUMMARY

or, use reasoning to save the need to remember theformula.


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Solution:

The median is in the 2ndclass.

2

n15


where,

Fn2 7 8

510

855 median 59

10ffrequency of class,

15

555510 wwidth of class,

58107Frequency, f

16 - 2011 - 156 - 101 - 5Height (cm)


30n

wf

Fn

2median l.c.b.


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11n

Solution:


2

n55


50555

550

510 median 111


wf

Fn

2median l.c.b.

5 the median of grouped data

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