how to calculate median. merits and demerits of Median

1. MEDIAN
Median is a value which divides the series into two equal parts.
It is position which is exactly in the center, equal number of terms lie on
either side of it, when terms are arranged in ascending or descending
order.
Definition:
“The median is that value of the variable which divides the group into
two equal parts, one part comprising all values greater and the other all
values less than median.”
OR
“Median of a series is the value of the item actual or estimated when a
series is arranged in order of magnitude which divides the distribution
into two parts.”

2. Calculation of Median:
A. Individual Series:
To find the value of Median, in this case, the terms are arranged in
ascending or descending order first; and then the middle term taken is
called Median.
Two cases arise in individual type of series:
(a) When number of terms is odd:
The terms are arranged in ascending or descending order and then are taken
as Median.
Example 1. Find Median from following data:
17 19 21 13 16 18 24 22 20
13 16 17 18 19 20 21 22 24
Solution: Arranging the terms in ascending order

3. n = Total number of terms 9
Now = (n+1)/2 = (9+1)/2 = 5
Median = 5th term = 19.
(b) When number of terms is even:
In this case also, the terms are arranged in ascending or descending
order and then mean of two middle terms is taken as Median.
Example 2. From the following figures of ages of some students,
calculate the median age:
Age in
years:
18 16 14 11 13 10 9 20
Solution: Arranging the terms in ascending order
Age in
years:
9 10 11 13 14 16 18 20

4. Now, median age =[(n/2)th term + ((n/2)+1)th term]/2
=[(8/2th term + ((8/2)+1)th term]/2
=[4th term + (4+1)th term ]/2
=[4th term +5th term ]/2
=[13+14]/2
=27/2
Md =13.5 years
(B) Discrete Series:
Here also the data is arranged in ascending or descending order and the (n+1)/2
term is taken after finding cumulative frequencies. Value of variable
corresponding to that term is the value of Median.
Calculation of Median in Discrete Series:
After arranging the terms, taking cumulative frequencies, we take (N/2) and then
calculate.

5. Steps to Calculate:
(1) Arrange the data in ascending or descending order.
(2) Find cumulative frequencies.
(3) Find the value of the middle item by using the formula
Median = value of (N/2)th term
(4) Find that total in the cumulative frequency column which is equal to (N/2)th or nearer
to that value.
(5) Locate the value of the variable corresponding to that cumulative frequency This is the
value of Median.

6. Example: Compute median for following data.
X: 10 20 30 40 50 60 70
f: 4 7 21 34 25 12 3
Solution:
X f Cf
10 4 =4
20 7 (4+7)=11
30 21 (11+21)=32
40 34 (32+34)=66
50 25 (66+25)=91
60 12 (91+12)=103
70 3 (3+103)=106
N=106

7. Now, median = (N/2)th term
= (106/2)th term
= 53th term
= 40
(C) Continuous Series:
In this case cumulative frequencies is taken and then the value from
the class-interval in which (N/2)th term lies is taken. Using the
formula.
Md= L+ [{(N/2)-Cf}/f]× i
Where N is the sum of given frequency,
L is lower limit of class interval in which frequency lies.
Cf is the cumulative frequency,
f the frequency of that interval and
i is the length of class interval.

8. Median can also be calculated from formula given below:
Md= L–[(Cf-N1)/f ]× I
Where L is upper limit of median class.
Important Working Hints:
1. Take N1=N/2
2. Looking from the lowest to upper side in the Cf column we get the value which
is firstly numerically less than N1. It is value of Cf.
3. We come one step down, make the terms in column f and also the class
interval to get f and L. L is taken from lower limit of class interval.
Note: We often use first formula; although second formula has also
been applied in the example given below:

9. Example 1. Calculate median for the following data.
Class
interval:
0-10 10-20 20-30 30-40 40-50 50-60 60-70
Frequency: 7 18 34 50 35 20 6
Solution:
Class interval f Cf
0-10 7 7
10-20 18 25
20-30 34 59
30-40 50 109
40-50 35 144
50-60 20 164
60-70 6 170
N=170

10. N1 = N/2
= 170/2
=85
Now, Cf= 59;
f=50;
L=30
Now, Md = L+[(N1-Cf.)/f]*i
=30+[(85-59)/50]*10
=30+(26/5)
=30+5.2
Md =35.2
2nd Formula (L as upper limit) can also be used. But we often use first formula.
Md = L-[(Cf.-N1)/f]*i
Here Cf =109 will be taken and L is upper limit.
Md = 40-[(109-85)/50]*10
= 40-(24/5)
= 40-4.8
Md = 35.2

11. MEDIAN BY GRAPHICAL METHOD
The median value of a series can be calculated through the
graphic presentation of data in the form of Ogives.
This can be done in 2 ways.
1. Presenting the data graphically in the form of 'less than'
ogive or 'more than' ogive .
2. Presenting the data graphically and simultaneously in
the form of 'less than' and more than' ogives. The two
ogives are drawn together.

12. For example:
Marks inclusive
series
Conversion into
exclusive series
No. of students Commulative
frequency
(X) (f) (Cf)
410-419 409.5-419.5 14 14
420-429 419.5-429.5 20 34
430-439 429.5-439.5 42 76
440-449 439.5-449.5 54 130
450-459 449.5-459.5 45 175
460-469 459.5-469.5 18 193
470-479 469.5-479.5 7 200

13. 1. Less than Ogive approach
Marks Commulative frequency (Cf)
Less than 419.5 14
Less than 429.5 34
Less than 439.5 76
Less than 449.5 130
Less than 459.5 175
Less than 469.5 193
Less than 479.5 200

14. Steps involved in calculating median using less than ogive approach-
1. Convert the series into a ‘less than’ commulative frequency distribution as
shown above.
2. let N be the total number of students who’s data is given. N will also be the
commulative frequency of the last interval. Find the (N/2)th item(student)
and mark it on the y- axis. In this case the (N/2)th item(student) is 200/2=
100th student.
3. Draw a perpendicular from 100 to the right to cut the ogive curve at a point
A.
4. From point A where the ogive curve is cut, draw a perpendicular on the x-
axis. The point at which it touches x-axis will be the median value of the
series as shown in the graph.

15. .
The median turns out to be 443.94..

16. 2. More than Ogive approach
More than marks Cumulative Frequency (C.M)
More than 409.5 200
More than 419.5 186
More than 429.5 166
More than 439.5 124
More than 449.5 70
More than 459.5 25
More than 469.5 7
More than 479.5 0

17. Steps involved in calculating median using more than Ogive approach –
1. Convert the series into a 'more than ' cumulative frequency distribution as
shown above.
2. Let N be the total number of students who's data is given.N will also be the
cumulative frequency of the last interval.Find the (N/2)th item(student) and
mark it on the y-axis.In this case the (N/2)th item (student) is 200/2 = 100th
student.
3. Draw a perpendicular from 100 to the right to cut the Ogive curve at point A.
4. From point A where the Ogive curve is cut, draw a perpendicular on the x-axis.
The point at which it touches the x-axis will be the median value of the series
as shown in the graph.

18. The median turns out to be 443.94..

19. 3.Less than and more than Ogive approach -
Another way of graphical determination of median is through simultaneous graphic
presentation of both the less than and more than Ogives.
1.Mark the point A where the Ogive curves cut each other.
2.Draw a perpendicular from A on the x-axis. The corresponding value on the x-axis
would be the median value.

20. The median turns out to be 443.94.

21. Merits of Median :
1) It is easy to compute and understand.
2) It is well defined an ideal average should be.
3) It can also be computed in case of frequency distribution with
open ended classes.
4) It is not affected by extreme values and also interdependent of
range or dispersion of the data.
5) It can be determined graphically.
6) It is proper average for qualitative data where items are not
measured but are scored.
7) It is only suitable average when the data are qualitative & it is
possible to rank various items according to qualitative characteristics.
8) It can be calculated easily by watching the data.
9) In some cases median gives better result than mean.

22. Demerits :
1) For computing median data needs to be arranged in ascending
or descending order.
2) It is not based on all the observations of the data.
3) It can not be given further algebraic treatment.
4) It is affected by fluctuation of sampling.
5) It is not accurate when the data is not large.
6) In some cases median is determined approximately as the mid
-point of two observations whereas for mean this does not happen.