Median Middle value in a distribution is known as Median. Calculation of median. 1. Calculation of median in a series of individual observations or Calculation of median for ungrouped data 2. Calculation of median for grouped data a) Calculation of median in a discrete series. b) Calculation of median in a continuous series. c) Calculation of median in unequal class intervals. d) Calculation of median in open-end classes. Merits and Demerits of Median.

1. BIOSTATISTICS AND RESEARCH METHODOLOGY
Unit-1: median
PRESENTED BY
Gokara Madhuri
B. Pharmacy IV Year
UNDER THE GUIDANCE OF
Gangu Sreelatha M.Pharm., (Ph.D)
Assistant Professor
CMR College of Pharmacy, Hyderabad.
email: sreelatha1801@gmail.com

2. MEDIAN
• The median by definition refers to the middle value in a distribution. As the
name itself suggests, median is the value of the middle item of a series,
arranged in either ascending or descending order of magnitude.
CALCULATION OF MEDIAN
• The data are arranged in ascending order of magnitude to find out the value
of the median. If the data set contain an odd number of values , the middle
of the array is the median. If there is an even number of items, the median
is the average of the middle items.
Types:
1) Calculation of median in a series of individual observations or
Calculation of median for ungrouped data
2) Calculation of median for grouped data
a) Calculation of median in a discrete series.
b) Calculation of median in a continuous series.
c) Calculation of median in unequal class intervals.
d) Calculation of median in open-end classes.

3.  For calculating median in a series of individual observations, the following
steps should be considered.
i. Arrange the data in ascending or descending order. The answer will be the
same both ways.
ii. Median is located by finding the size of n+1th item.
Symbolically; M=Size of the n+1th item
where M=median
n= number of observations
Example: Systolic blood pressure(in mm of Hg) of eight
male patients are given below. Find out the value of median.
(100,110,120,90,130,140,160,150)
Solution: Arrange the values in ascending order of
magnitude.
S.N
O
Data arranged in
ascending order
1 90
2 100
3 110
4 120
5 130
6 140
7
8
150
160
There are 8 values i.e., even number of values, therefore,
there are 2 middle values at 4th and 5th positions whose values
are 120 and 130 respectively. Hence the value of median is
calculated as Median =120+130/2 = 250/2=125
i.e., the median B.P is 125mm of Hg

4. For the computation of the median in discrete series, one should take into
consideration the following steps:
1. Data should be arranged in ascending or descending order of magnitude.
2. Find out the cumulative frequencies.
3. Median=Size of (n+1)/2th item
4. Find out the value of (n+1)/2th item. It can be found by first locating the
cumulative frequency which is equal to (n+1)/2 or the next higher to this and the
determine the value corresponding to it. This will be the value of the median.
Example: Calculate median for the blood samples off or HDL of 43 patients.
No. of
patient
s
6 4 16 7 8 2
HDL(in
mg/dl)
20 9 25 50 40 80
Solution: Arranging the marks in ascending order and preparing the following table.

5. Here n=43 Median=M= (n+1/2)th value= (43+1/2)th value=22nd value
The above table shows that all items from 11 to 26 have their values 25.
Since 22nd item lies. This interval, therefore, it value is 25
Hence Median =25 mg/dl
HDL(in mg/dl) No of patients (f) Cumulative
frequency(C)
9 4 4
20 6 10
25 16 26
40 8 34
50 7 41
80 2 43
n=sum of f=43

6.  The value of the median in a continuous series,first determine the particular
class in which the value of the median lies.
 Use n/2 which will divide the area of the curve into two parts.
 The following formula is used for determining the exact value of the median.
Median = 𝑀 = 𝐿 +
𝑛
2
−𝑐𝑓
𝑓
× 𝑖
where L = Lower limit of the median class
cf = cumulative frequency of the class preceding the median class
f = frequency of the median class
i = class-interval
However, it should be remembered that instead of the lower limit of the median
class, one can take the upper limit of the median class also. In this case class-
intervals should be in descending order and frequencies are to be cumulative from
top to bottom

7.  Example: Find the median of blood samples for triglyceride test of the
following data is
 Solution: let us prepare the following computation table. Let x = the range of
triglycerides and f = number of patients
 Computation of median
Median = size of (n/2)th item =
Size of 43/2th item
= 21.5th item, which lies in the class 40 -50
Therefore Median class = 40 – 50, L = 40,
n/2 = 21.5, c.f = 8, f = 20, i = 10
𝑀 = 𝐿 +
𝑛
2
−𝑐𝑓
𝑓
× 𝑖
= 40 +
21.5−8
20
× 10 = 40 +
135
20
= 40 + 6.75 = 46.75 𝑚𝑔/𝑑𝑙
Glyceride
(in mg/dl)
20 - 30 30 -40 40 - 50 50 - 60 60 - 70
Patients 5 10 20 5 3
x f C . F.
20 - 30 3 3
30 - 40 5 8
40 - 50 20 28
50 - 60 10 38
60 - 70 5 43
n=43

8.  In unequal class-intervals, frequencies need not be adjusted to make the class
intervals equal .
 The formula given in the continuous series can be used here.
 Median = 𝐿 +
𝑛
2
−𝑐𝑓
𝑓
× 𝑖
 Example: Data recorded on the production of tablets during 6 consecutive days is
presented below. Calculate the median.
Classes 0 -
10
10 -
30
30 –
60
60 -
80
80 -
90
90 -
100
Freque
ncy
5 16 30 12 6 1
Classes Frequenc
y
Cumulative
Frequency(cf)
0 - 10 5 5
10 - 30 16 21
30 - 60 30 51
60 - 80 12 63
80 - 90 6 69
90 - 100 1 70
n=70
Median = 𝐿 +
𝑛
2
−𝑐𝑓
𝑓
× 𝑖
Median = size of n/2 = 70/2 = 35th item
and median lies in the class = 30 – 60
L = 30, n/2 = 35, cf = 21, f = 30, i = 30
Median = 30 +
14
30
× 30 = 30 + 14 = 44
Median = 44

9.  Since the median is not affected by the values of extreme ends, we are not
concerned with extreme values for calculation of the median in open-end
classes.
 Example: Systolic blood pressure in (mm of hg) of 5 female patients are given
below. Find out value of median.
Value Less than 10 10 - 20 20 - 30 30 - 40 40 and above
Frequency 4 8 14 6 4
Size of item
classes
Frequ
ency
Cumulative
frequency(cf)
Less than 10 4 4
10 - 20 8 12
20 - 30 14 26
30 - 40 6 32
40 and above 4 36
Median = Size of n/2 = 36/2 = 18th item
Median lies in the class = 20 – 30
Median = 𝐿 +
𝑛
2
−𝑐𝑓
𝑓
× 𝑖
L = 20 , n/2 = 18, cf = 12, f = 14, i = 10
Median = 20 +
18−12
14
× 10 = 20 +
6
14
×
10 = 20 + 4.29
Median = 24.29 mm of hg.

10.  It is easily understood.
 It is not affected by extreme values.
 It can be located graphically.
 It is the best measure for qualitative data such as beauty,
intelligence, honesty etc.
 It can be easily located even if the class-intervals in the
series are unequal.
DEMERITS OF MEDIAN
 It is not subject to algebraic treatments.
 It cannot represent the irregular distribution series.
 It is a positional average and is based on the middle item.
 It is an estimate in case of a series containing even
number of items.
 It does not have sampling stability.