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Median

Median for a given set of observations may be defined as the middle most value when the observations are arranged either in an ascending order or in a descending order of their magnitudes. Middle-most value means that half of the observations lie above this value and the other half lie below this value. It is also known as the positional average of the given data since it is based on the position of a given observation when the series is arranged in an ascending or descending order.

 

To find the median, first the given observations have to be arranged in ascending or descending order. Let there be n observations, then the median of the observations can be found using the relation:

 

Description: 92933.png 

Description: 92980.png 
 

Example
What is the median of the following observations?
5, 8, 6, 9, 11, 4, 15
Solution
The given observations have to be arranged in ascending or descending order. Arranging the observations in ascending order, we have the observations as 4, 5, 6, 8, 9, 11, 15
Since n = 7 is odd, Description: 92987.png
Description: 92994.png 
The 4th value in the above observations after arranging them is 8.
Hence, the median of the given observations is 8.

 

Example
What is the median of the observations 5, 8, 6, 9, 11, 4?
Solution
The given observations have to be arranged in ascending or descending order.
Arranging the observations in ascending order, we have the observations as 4, 5, 6, 8, 9, 11.
 
Since n = 6 is even, Description: 93004.png
Description: 93010.png Description: 93041.png 
 
In case of a simple frequency distribution, median is calculated using
 
Description: 93050.png 
  • When a continuous or grouped frequency distribution is given, we can find the median by the following relation

Description: 93066.png 
 

Example
Compute the median for the given distribution.
 
Marks 5–14 15–24 25–34 35–44 45–54 55–64
Students 10 18 32 26 14 10
Solution
We have to first find the less than cumulative frequency distribution which is shown in the table below.
 
Class Interval Frequency Less than Cumulative Frequency
4.5 – 14.5 10 10
14.5 – 24.5 18 28
24.5 – 34.5 32 60
34.5 – 44.5 26 86
44.5 – 54.5 14 100
54.5 – 64.5 10 110
 
Description: 93099.png 
55 lies between 28 and 60, i.e., 28 < 55 < 60
We choose the next higher value than 55. Hence, the class interval 24.5 – 34.5 becomes the median class.
Thus, l = 24.5, f = 32, Cf = 28 and C = 34.5 - 24.5 = 10
Description: 93137.png 
Description: 93172.png 

Properties of Median

  1. Median is affected by change in origin as well as change in scale.
     
    Let y = a + bx be the relation between two variables x and y, for any two constants a and b, then the median of y is given by
     
    Me (y) = a + b × Me (x) (where, Me (x) is the median of x).
  2. For a set of observations, the sum of absolute values of the deviations is minimum when the deviations are taken from the median, i.e.Description: 93189.png is minimum, if we choose A as the median.

Merits

  1. It is easily understood.
  2. It is not affected by extreme values.
  3. It can be located graphically.
  4. It is the best measure of qualitative data, such as beauty, intelligence, honesty, etc.
  5. It can be easily located even if the class-intervals in the series are unequal.
  6. It can be determined even by inspection in many cases.

Demerits

  1. It is not subject to algebraic treatments.
  2. It cannot represent the irregular distribution series.
  3. It is a positional average and is based on the middle item.
  4. It does not have sampling stability.
  5. It is an estimate in case of a series containing even number of items.
  6. It does not take in account the values of all items in the series.
  7. It is not suitable in those cases where due importance and weight should be given to extreme n values.

Partition values

Partition values may be defined as the values dividing a given set of observations into equal number of parts.

 

Median divides the given set of observations into two equal parts when the observations are arranged in ascending or descending order.

  1. Quartiles are partition values which divide the given set of observations into four equal parts when the observations are arranged in ascending or descending order.
     
    This implies that there are three quartile values, i.e., first quartile or lower quartile (Q1), second quartile or the median (Q2) and third quartile or upper quartile (Q3).
     
    First quartile is the value for which one-fourth of the observations are less than or equal to Q1 and the remaining three-fourth are more than or equal to Q1. Similarly, the second and third quartiles can be defined.
     
    When the given observations contain ungrouped/discrete data, the quartile values can be found using the relation Description: 93204.png
     
    where, i = 1, 2, 3

Example
What is the value of the first quartile for the observations 15, 18, 10, 20, 23, 28, 12, 16?
Solution
First, let us arrange the observations in ascending order. The observations after arranging will be 10, 12, 15, 16, 18, 20, 23, 28
Description: 93222.png 
Description: 93303.png 
If the given observations show a continuous frequency distribution, then
Description: 93309.png 

 

Example
Find the value of third quartile for the given distribution.
 
Class Interval Less than 10 10–19 20–29 30–39 40–49 50–59
Frequency 5 18 38 20 9
2
Solution
This is an example of an open-ended distribution. We find that the lower limit of the first class is not given. It can assume any value between 0 and 10. Let us denote the first Lower Class Boundary as L and construct the cumulative frequency distribution converting the inclusive table into exclusive.
 
Class Interval Frequency Less than Cumulative Frequency
L–9.5 5 5
9.5–19.5 18 23
19.5–29.5 38 61
29.5–39.5 20 81
39.5–49.5 9 90
49.5–59.5 2 92
 
 
From the above table, N = 92.
 
For calculating the 3rd quartile, we start with finding the value of 3 N/4.
 
Description: 93321.png 
 
Description: 93351.png 
Description: 93357.png
  1. Deciles are the partition values that divide a given set of observations into 10 equal parts. This implies that there must be 9 deciles for a set of observations.
     
    When the given observations contain ungrouped/discrete data, the decile values can be found using the relationDescription: 93372.png, where, i = 1, 2, 3... 9.

Example
What is the value of the sixth decile for the observations 15, 18, 10, 20, 23, 28, 12, 16?
Solution
First, let us arrange the observations in ascending order. The observations after arranging will be 10, 12, 15, 16, 18, 20, 23, 28
 
Description: 93380.png 
 
If the given observations show a continuous frequency distribution, then
 
Description: 93401.png 

 

Example
Find the value of fifth decile for the given distribution.
 
Class Interval Less than 10 10–19 20–29 30–39 40–49 50–59
Frequency 5 18 38 20 9
2
Solution
This is an example of an open-ended distribution. We will proceed in the same way as done in Example.
 
Class Boundary Frequency Less than Cumulative Frequency
L–9.5 5 5
9.5–19.5 18 23
19.5–29.5 38 61
29.5–39.5 20 81
39.5–49.5 9 90
49.5–59.5 2 92
 
 
From the above table, N = 92.
 
For calculating the 5th decile, we start with finding the value of 5 N/10.
 
Description: 93427.png 
 
Description: 93439.png 
Description: 93533.png
  1. Percentiles are partition values that divide the given set of observations into hundred equal parts. Hence, there are 99 percentiles for a set of observations.
     
    When the given observations contain ungrouped/discrete data, the percentile values can be found using the relation
     
    Description: 93545.png where, i = 1, 2, 3 ... 99.

Example
What is the value of the f ifty-ninth percentile for the observations 15, 18, 10, 20, 23, 28, 2, 16?
Solution
First, let us arrange the observations in ascending order. The observations after arranging will be 2, 10, 15, 16, 18, 20, 23, 28.
 
Description: 93567.png 
If the given observations show a continuous frequency distribution, then
 
Description: 93578.png 
 
Description: 93584.png 

 

Example
Find the value of seventieth percentile for the given distribution.
 
Class Interval Less than10 10–19 20–29 30–39 40–49 50–59
Frequency 5 18 38 20 9
2
Solution
This is an example of an open-ended distribution. We will proceed in the same way as done in Example.
 
Class Boundary Frequency Less than Cumulative Frequency
L–9.5 5 5
9.5–19.5 18 23
19.5–29.5 38 61
29.5–39.5 20 81
39.5–49.5 9 90
49.5–59.5 2 92
 
From the above table, N = 92.
 
For calculating the 70th percentile, we start with finding the value of 70 N/100.
 
Description: 93637.png 
 
Description: 93647.png 





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