Properties of Median in Statistics

The value which occupies the centre position among the observations when arranged in ascending or descending order is the median. Fifty per cent of scores are above or below the median. Hence, it is named as 50^th 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 an even and an odd number of observations.

Median Properties

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

• The 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.
• The median cannot be manipulated algebraically. It cannot be weighed and combined.
• In a grouping procedure, the median is stable.
• The median is not applicable to qualitative data.
• The values must be grouped and ordered for computation.
• The 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 the 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 to find the median for a continuous distribution is:

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

Video Lesson: Mean, Median and Mode

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 4^th item + value of 5^th 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 observations 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 = 5^th 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

= 4^th 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,….. x_n, − x_n 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 the 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 the mean is 39.6. Then the median of the distribution is given by _____________.

Solution:

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

i.e., 0.32 = [39.6 − M_o] / 6.5

M_o = 37.52, and also we know that, M_o = 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.

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

Here, the total number of items is 41, i.e., an odd number. Hence, the median is [41 + 1] / 2, i.e., the 21st item. From the cumulative frequency table, we find that the median, i.e., the 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

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

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, on arranging in ascending order, becomes as follows: 11, 15, 19, 20, 24, 27, 31

Median = value of the 4th observation

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