Properties Of Median In Statistics

The value which occupies the centre position amongst the observations when arranged in ascending or descending order is the median. Fifty per cent scores are above or below the median. Hence, it is named as 50th percentile or positional average. The location of the median is dependent on whether the data set consists of an even or odd number of values. The method of finding the median is different for even and an odd number of observations.

Median Properties

In statistics, the properties of the median are explained in the following points.

  • Median is not dependent on all the data values in a dataset.
  • The median value is fixed by its position and is not reflected by the individual value.
  • The distance between the median and the rest of the values is less than the distance from any other point.
  • Every array has a single median.
  • Median cannot be manipulated algebraically. It cannot be weighed and combined.
  • In a grouping procedure, the median is stable.
  • Median is not applicable to qualitative data.
  • The values must be grouped and ordered for computation.
  • Median can be determined for ratio, interval and ordinal scale.
  • Outliers and skewed data have less impact on the median.
  • If the distribution is skewed, the median is a better measure when compared to mean.

Formula to Find Median for Discrete Series

Calculating the median for individual series is as follows:

  • The data is arranged in ascending or descending order.
  • If it is an odd-sized sample, median = value of ([n + 1] / 2)th item.
  • If it is an even-sized sample, median = ½ [ value of (n / 2)th item + value of ([n / 2] + 1)th item]

Calculating the median for discrete series is as follows:

  • Arrange the data in ascending or descending order.
  • The cumulative frequencies need to be computed.
  • Median = (n / 2)th term, n refers to cumulative frequency.

The formula for finding the median for a continuous distribution is:

Median for Discrete Series Formula

Where l = lower limit of the median class

f = frequency of the median class

N = the sum of all frequencies

i = the width of the median class

C = the cumulative frequency of the class preceding the median class

Solved Examples on Median

Example 1: If a variable takes the discrete values α − 4, α − 7 / 2, α − 5 / 2, α − 3, α − 2, α + 1 / 2, α − 1 / 2, α + 5 ( α > 0), then the median is

Solution:

Arrange the data as α − 7 / 2, α − 3, α − 5 / 2, α − 2 , α − 1 / 2, α + 1 / 2, α − 4, α + 5

Median = 1 / 2 [value of 4th item + value of 5th item ]

Median = (α − 2 + [α − 1 / 2]) / [2]

= [2α − 5 / 2] / 2

= α − 5 / 4

Example 2: The median of a set of 9 distinct observations is 20.5. If each of the largest fourth observation of the set is increased by 2, then the median of the new set is ________.

Solution:

Since n = 9, then median term = ([9 + 1] / [2])th = 5th term.

Now, the last four observations are increased by 2.

∵ The median is the 5th observation, which remains unchanged.

There will be no change in the median.

Example 3: The median of 10, 14, 11, 9, 8, 12, 6 is _________.

Solution:

Arrange the items in ascending order, i.e. , 6, 8, 9, 10, 11, 12, 14.

If n is odd then, Median = value of ([n + 1] / 2)th term

Median = ([7 + 1] / 2)th term

= 4th term

= 10

Example 4: For a symmetrical distribution Q1 = 25 and Q3 = 45, the median is _______.

Solution:

As the distribution is symmetrical, therefore,

Q2(Median) = [Q1 + Q3] / 2

= [25 + 45] / 2

= 35

Example 5: For (2n + 1) observations x1, − x1, x2,− x2,….. xn, − xn and 0 where x’s are all distinct. Let S.D. and M.D. denote the standard deviation and median, respectively. Then, what is the relation between standard and median?

Solution:

On arranging the given observations in ascending order,

The median of the given observations = (n + 1)th term = 0

S. D. > M .D.

Example 6: Karl-Pearsons coefficient of skewness of a distribution is 0.32. Its S.D. is 6.5 and mean is 39.6. Then the median of the distribution is given by _____________.

Solution:

We know that Sk = [M − Mo] / σ, where M = Mean, Mo = Mode, s = S.D.

i.e., 0.32 = [39.6 − Mo] / 6.5

Mo = 37.52 and also know that, Mo = 3 median – 2 mean

37.52 = 3 (Median) – 2 (39.6)

Median = 38.81

Example 7: The following data gives the distribution of the height of students

Height (in cm) 160 150 152 161 156 154 155
Number of students 12 8 4 4 3 3 7

The median of the distribution is

A) 154

B) 155

C) 160

D) 161

Solution:

Arranging the data in ascending order of magnitude, we obtain

Height (in cm) 150 152 154 155 156 160 161
Number of students 8 4 3 7 3 12 4
Cumulative frequency 8 12 15 22 25 37 41

Here, the total number of items is 41 i.e., an odd number. Hence, the median is [41 + 1] / 2 th i.e., 21st item. From the cumulative frequency table, we find that median i.e., 21st item is 155, (All items from 16 to 22nd are equal, each 155).

Example 8: In a moderately asymmetrical distribution the mode and mean are 7 and 4 respectively. The median is

A) 4     

B) 5

C) 6     

D) 7

Solution:

For a moderately Skewed distribution,                    

Mode = 3 median –  2 mean            

7  = 3 median – [2 × 4] 

15 = 3 median                    

Median = 5. 

Example 9: If in a frequency distribution, the mean and median are 21 and 22 respectively, then its mode is approximately

A) 25.5

B) 24.0

C) 22.0

D) 20.5

Solution:

We know that,                    

Mode = 3 Median – 2 Mean 

= 3(22) – 2(21)                      

= 66 – 42 

= 24. 

Example 10: Following are the marks obtained by 9 students in a mathematics test: 50, 69, 20, 33, 39, 40, 65, 59 the mean deviation from the median is

A) 9

B) 10.5

C) 12.67

D) 14.76

Solution: 

Marks obtained by 9 students in Mathematics are: 50, 69, 20, 33, 53, 39, 40, 65, 59 Let us rewrite the given data in ascending order. 20, 33, 39, 40, 50, 53, 59, 65, 69.

Here, n = 9

Median = 5th term = 50

xi

di=|xi−Me|
20 30
33 17
39 11
40 10
50 0
53 3
59 9
65 15
69 19
N = 2

Σd1 = 114

M.D.= 114 / 9 = 12.67

Example 11: If the mean deviation about the median of the numbers a, 2a, …….. 50a is 50, then find the value of |a|.

Solution:

Median = 25.5a

xiM=[24.5a+23.5a+..+0.5a+0.5a+..+24.5a]=2a[0.5+1.5+.+24.5]xiM=25×25a50=25×25a50a=4\sum |x_{i}-M|=[24.5a+23.5a+ …….. +0.5a+0.5a+ …….. +24.5a]\\ =2a[0.5+1.5+ ……. + 24.5]\\ \sum |x_{i}-M|=25 \times 25a\\ 50=\frac{25 \times 25a}{50}\\ \Rightarrow a=4

Example 12: Find the median of 11, 15, 13, 27, 19, 24 and 20. If 13 is replaced by 31 then find the new median.

A) 10

B) 20

C) 30

D) 40

Solution:

Arranging the given data in ascending order, 11, 15, 13, 27, 19, 24, 20,

The number of observations = 7 (odd)

Median = 19,

When 13 is replaced by 31, the data becomes as follows on arranging in ascending order. 11, 15, 19, 20, 24, 27, 31

Median = value of 4th observation

Since 4th observation in the new series is 20, the new median is 20.