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. Middlemost 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:
Example
What is the median of the following observations?
5, 8, 6, 9, 11, 4, 15
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,
The 4th value in the above observations after arranging them is 8.
Hence, the median of the given observations is 8.
Since n = 7 is odd,
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,
In case of a simple frequency distribution, median is calculated using
Arranging the observations in ascending order, we have the observations as 4, 5, 6, 8, 9, 11.
 When a continuous or grouped frequency distribution is given, we can find the median by the following relation
Example
Compute the median for the given distribution.
Marks  5â€“14  15â€“24  25â€“34  35â€“44  45â€“54  55â€“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.
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 â€“ 34.5 becomes the median class.
Thus, l = 24.5, f = 32, C_{f} = 28 and C = 34.5  24.5 = 10
Class Interval  Frequency  Less than Cumulative Frequency 
4.5 â€“ 14.5  10  10 
14.5 â€“ 24.5  18  28 
24.5 â€“ 34.5  32  60 
34.5 â€“ 44.5  26  86 
44.5 â€“ 54.5  14  100 
54.5 â€“ 64.5  10  110 
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 â€“ 34.5 becomes the median class.
Thus, l = 24.5, f = 32, C_{f} = 28 and C = 34.5  24.5 = 10
Properties of Median
 Median is affected by change in origin as well as change in scale.
Me (y) = a + b Ã— Me (x) (where, Me (x) is the median of x).
 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., is minimum, if we choose A as the median.
Merits
 It is easily understood.
 It is not affected by extreme values.
 It can be located graphically.
 It is the best measure of qualitative data, such as beauty, intelligence, honesty, etc.
 It can be easily located even if the classintervals in the series are unequal.
 It can be determined even by inspection in many cases.
Demerits
 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 does not have sampling stability.
 It is an estimate in case of a series containing even number of items.
 It does not take in account the values of all items in the series.
 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.
 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.
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
If the given observations show a continuous frequency distribution, then
If the given observations show a continuous frequency distribution, then
Example
Find the value of third quartile for the given distribution.
Class Interval  Less than 10  10â€“19  20â€“29  30â€“39  40â€“49  50â€“59 
Frequency  5  18  38  20  9 
2

Solution
This is an example of an openended 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.
From the above table, N = 92.
For calculating the 3rd quartile, we start with finding the value of 3 N/4.
Class Interval  Frequency  Less than Cumulative Frequency 
Lâ€“9.5  5  5 
9.5â€“19.5  18  23 
19.5â€“29.5  38  61 
29.5â€“39.5  20  81 
39.5â€“49.5  9  90 
49.5â€“59.5  2  92 
 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.
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
If the given observations show a continuous frequency distribution, then
Example
Find the value of fifth decile for the given distribution.
Class Interval  Less than 10  10â€“19  20â€“29  30â€“39  40â€“49  50â€“59 
Frequency  5  18  38  20  9 
2

Solution
This is an example of an openended distribution. We will proceed in the same way as done in Example.
From the above table, N = 92.
For calculating the 5th decile, we start with finding the value of 5 N/10.
Class Boundary  Frequency  Less than Cumulative Frequency 
Lâ€“9.5  5  5 
9.5â€“19.5  18  23 
19.5â€“29.5  38  61 
29.5â€“39.5  20  81 
39.5â€“49.5  9  90 
49.5â€“59.5  2  92 
 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.
Example
What is the value of the f iftyninth 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.
If the given observations show a continuous frequency distribution, then
If the given observations show a continuous frequency distribution, then
Example
Find the value of seventieth percentile for the given distribution.
Class Interval  Less than10  10â€“19  20â€“29  30â€“39  40â€“49  50â€“59 
Frequency  5  18  38  20  9 
2

Solution
This is an example of an openended distribution. We will proceed in the same way as done in Example.
From the above table, N = 92.
For calculating the 70th percentile, we start with finding the value of 70 N/100.
Class Boundary  Frequency  Less than Cumulative Frequency 
Lâ€“9.5  5  5 
9.5â€“19.5  18  23 
19.5â€“29.5  38  61 
29.5â€“39.5  20  81 
39.5â€“49.5  9  90 
49.5â€“59.5  2  92 