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
Class      Frequency
010        10
1040       15
4050       5
5070       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 ${\sum}_{}$Fx = 222
SO mean of that data is $\frac{222}{20}$ = 11.1 , So our median class in 11  15 ,
And median = L + $\frac{\left({\displaystyle \frac{n}{2}}\right)c{f}_{b}}{{f}_{m}}\times w$
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 ) cf_{b} is the cumulative frequency of the groups before the median group.( Here cf_{b} = 9 )
4 ) f_{m} is the frequency of the median group. ( here f_{m} = 8 )
5 ) w is the group width. ( Here w = 5 )
So,
Median = 11 + $\frac{\left({\displaystyle \frac{20}{2}}\right)9}{8}\times 5=11+\frac{109}{8}\times 5=11+\frac{5}{8}=11+0.625$ = 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 ${\sum}_{}$Fx = 320
SO mean of that data is $\frac{320}{20}$ = 16 , So our median class in 16  20 ,
And median = L + $\frac{\left({\displaystyle \frac{n}{2}}\right)c{f}_{b}}{{f}_{m}}\times w$
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 ) cf_{b} is the cumulative frequency of the groups before the median group.( Here cf_{b} = 9 )
4 ) f_{m} is the frequency of the median group. ( here f_{m} = 8 )
5 ) w is the group width. ( Here w = 5 )
So,
Median = 16 + $\frac{\left({\displaystyle \frac{20}{2}}\right)9}{8}\times 5=16+\frac{109}{8}\times 5=16+\frac{5}{8}=16+0.625$ = 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 
1620 
5 
2125 
6 
2630 
12 
3135 
14 
3640 
26 
4145 
12 
4650 
16 
5155 
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 f_{i}
Cumulative frequency (c.f.)
Midpoint x_{i}
x_{i} – Med.
f_{i} x_{i} – Med.
15.520.5
5
5
18
20
100
20.525.5
6
11
23
15
90
25.530.5
12
23
28
10
120
30.535.5
14
37
33
5
70
35.540.5
26
63
38
0
0
40.545.5
12
75
43
5
60
45.550.5
16
91
48
10
160
50.555.5
9
100
53
15
135
100
735
The class interval containing theor 50^{th} 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 
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  MidPoint  D From median 
fD 
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 
13^{th} item whose class interval is 10 – 20
Thus, the value of Mean Deviation calculated from median is 7.16