Median

Median is the middle value of a distribution. It divides the data in two equal parts.
Middle number of an un-grouped data : The ungrouped data are x1, x2,x3,…,xn then the middle value after arranging the data either ascending or descending order is the middle number of the data.
Steps to find the mid point of the ungrouped data :
1) Arrange the data either ascending or descending order of their values.
2) Determine the total number of observations, say n.
3) If n is odd then the middle number will be the median. And if n is even then mean of middle two numbers will be the me-dian.

Merits and Demerits of Me-dian
Example 1: Find the median of 12,15,10,18 ,8.
Solution : Data in ascending order : 8,10,12,15,18
From the above middle number 12 is the me-dian.
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Example 2 : Find the median of 23,46,18,32,65,20.
Solution : Data in ascending order : 18,20,23,32,46,65.
From the above we can say that there are two middle numbers 23 and 32.
So, the me-dian = (23 + 32 ) / 2 = 55 / 2 = 27.5
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Example 3 : Numbers 50,42,2x + 10, 2x - 8 ,12,11, 8, 6 are written in descending order and their me-dian is 25, find x.
Solution : 50,42,2x + 10, 2x, 8 ,12,11, 8, 6 from this the median is mean of 2x + 10 and 2x - 8.
Median = ( 2x + 10 + 2x -8) / 2 = 25
⇒ ( 4x + 2 ) / 2 = 25
⇒ 4x + 2 = 50
⇒ 4x = 48
∴ x = 12
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Cumulative Frequency Distribution :
The total of a frequency and all frequencies below it in a frequency distribution.
It is the 'running total' of frequencies.
Example : Scores : 4,4, 5,5,1,1,2,3,3,3,2
Scores Frequency Cumulative frequency (cf)
1
2
2
2
2
2 + 2 = 4
3
3
2 + 2 + 3 = 7
4
2
2 + 2 + 3 + 2 = 9
5
2
2 + 2 + 3 + 2 + 2 = 11
11
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There are two types of cumulative frequency distribution : 1) Less than
2) Greater than.
For Less than cumulative frequency distribution, we add up the frequencies from the above and in Greater than cumulative frequency distribution we add up the frequencies from below.
Example :
Class intervals Frequency Less than type Cumulative frequency Greater than type Cumulative frequency
0 - 10
9
9 50
10 - 20
14
9 + 14 = 23 50 - 9 = 41
20 - 30
8
9 + 14 + 8 = 31 50 - 9 -14 = 27
30 - 40
10
9 + 14 + 8 + 10 = 41 50 - 9 -14 - 8 = 19
40 - 50
9
9 + 14 + 8 + 10 + 9 = 50 50 - 9 -14 - 8 - 10 = 9
50
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Me-dian of a Grouped or Continuous Frequency Distribution :
To find the me-dian of a grouped data, use the following steps :
1) Obtain the frequency distribution and obtain N = Σ fi.
2) Prepare cumulative frequency distribution.
3) Find N / 2.
4) Find the cumulative frequency just greater than N / 2 and determine the corresponding class. This class is known as the Me-dian Class.
5) Use the following formula :

Where, l = Lower limit me-dian class
f = frequency of the me-dian class
h = width of the me-dian class
F = Cumulative frequency of the class preceding the me-dian class.
N = Σ fi.
Some solved examples :
1) Find the me-dian for the following data:
Classes 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80
Frequency 4 8 10 12 10 4 2
Solution :
Classes Frequency (fi) Cumulative frequency
10 - 20
4
4
20 - 30
8
12
30 - 40
10
22
40 - 50
12
34
50 - 60
10
44
60 - 70
4
48
70 - 80
2
50
Total N = Σ fi = 50
N / 2 = 50 / 2 = 25,
So, 40 – 50 is the me-dian class and its lower limit = l = 40
Cumulative frequency preceding the me-dian class = f = 22
Frequency of the me-dian class = F = 12
Class width h = 50 -40 = 10

⇒ Median = 40 +[ ( 25 – 22) / 12 ] x 10
⇒ = 40 + ( 3 / 12 ) x 10
⇒ = 40 + 0.25 x 10
⇒ = 40 + 2.5
∴ Me-dian = 42.5
Statistics

• Statistics
• Pictograph
• Pie chart
• Bar Graph
• Double Bar Graph
• Histogram
• Frequency polygon
• Frequency distribution (Discrete )
• Frequency distribution continuous (or grouped)
• Measures of central tendency (Mean-median and Mode
• Ogive or Frequency curve.

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