How to find median of discrete frequency distriburion and continuous frequency distribution andFurther how to find m.d.(m) in discreate frquency and in continuous frequency distributiotion

 Example-

Class - - - - - Frequency

0-10 - - - - - - - -10

10-40 - - - - - - -15

40-50 - - - - - - -5

50-70 - - - - - - -5

Solution-

Solution: 




Median of continuous frequecy :

Answer :
Given  :
In a continuous frequency distribution, the median of the data is 21.
And
If each observation is increased by 5 .

Then the new median is also increased by 5 , so new median  =  21  +  5  =  26  ( Ans )

We can understand it As :

We have observations of number of toy for every children  and total number of children   = 20

3, 5, 6, 6, 8, 8, 9, 9, 10, 11, 11, 12, 12, 12, 14, 15, 15, 18, 18, 20

First we form a continues frequency table , As  :

 

Number of toys 1 - 5 6 - 10 11 - 15 16 - 20
Frequency 2 7 8 3

Here frequency is number of children that have toys in between ( 1 -5  ,  6 - 10 , 11- 15  and 16 - 20 )

Here  Fx  =  222

SO mean of that data is  22220  =  11.1  , So our median class in 11 - 15  ,

And median  =  L + n2 - cfbfm× w

Here

1 ) L is the lower class boundary of the group containing the median . ( here L  = 11 )
2 ) n is the total number of data . ( here n  =  20 )
3 ) cfb is the cumulative frequency of the groups before the median group.( Here cfb = 9 )
4 ) fm is the frequency of the median group. ( here fm  = 8 )
5 ) w is the group width. ( Here w  = 5 )

So,

Median  = 11 + 202 -98× 5  = 11 + 10 - 98× 5  = 11 + 58 =  11 + 0.625   =  11.625

Now we observe the change in median after increase its observation by 5 , now our new observations is

8, 10, 11, 11, 13, 13, 14, 14, 15, 16, 16, 17, 17, 17, 19, 20, 20, 23, 23, 25

First we form a continues frequency table , As  :
 
Number of toys 1 - 5 6 - 10 11 - 15 16 - 20  | 20 - 25
Frequency 0 2 7 8  | 3


Here  Fx  =  320

SO mean of that data is  32020  =  16  , So our median class in 16 - 20  ,

And median  =  L + n2 - cfbfm× w

Here

1 ) L is the lower class boundary of the group containing the median . ( here L  = 16 )
2 ) n is the total number of data . ( here n  =  20 )
3 ) cfb is the cumulative frequency of the groups before the median group.( Here cfb = 9 )
4 ) fm is the frequency of the median group. ( here fm  = 8 )
5 ) w is the group width. ( Here w  = 5 )

So,

Median  = 16 + 202 -98× 5  = 16 + 10 - 98× 5  = 16 + 58 =  16 + 0.625   =  16.625

So , we can see that after increment of 5 in observations ,we get our new median increased by 5  .


Example of mean deviation:-

 

Calculate the mean deviation about median age for the age distribution of 100 persons given below:

Age

Number

16-20

5

21-25

6

26-30

12

31-35

14

36-40

26

41-45

12

46-50

16

51-55

9
 



Solution:-

The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.

The table is formed as follows.

Age

Number fi

Cumulative frequency (c.f.)

Mid-point xi

|xiMed.|

fi |xiMed.|

15.5-20.5

5

5

18

20

100

20.5-25.5

6

11

23

15

90

25.5-30.5

12

23

28

10

120

30.5-35.5

14

37

33

5

70

35.5-40.5

26

63

38

0

0

40.5-45.5

12

75

43

5

60

45.5-50.5

16

91

48

10

160

50.5-55.5

9

100

53

15

135

 

100

     

735

The class interval containing theor 50th item is 35.5 – 40.5.

Therefore, 35.5 – 40.5 is the median class.

It is known that,

Here, l = 35.5, C = 37, f = 26, h = 5, and N = 100

Thus, mean deviation about the median is given by,



To calculate mean deviation from median, the following formula is used

 


 
When Mean deviation are taken from Median (M)
For Individual Series
For Discrete and Continuous Series
 
The calculation process of Mean Deviation from median can be better understood with the help of the following example.
Example: For the following data calculate Mean Deviation from median.
X f
0 – 10 5
10 – 20 9
20 – 30 6
30 – 40 3
40 – 50 2











 
X f C.f Mid-Point |D|
From median
f|D|
0 – 10 5 5 5 13.88 69.4
10 – 20 9 14 15 3.88 34.92
20 – 30 6 20 25 6.12 36.72
30 – 40 3 23 35 16.12 48.36
40 – 50 2 25 45 26.12 52.24
  N = 25     66.12 179.18




13th item whose class interval is 10 – 20


Thus, the value of Mean Deviation calculated from median is 7.16 

 

 

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