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Calculation of Median in Continuous Series | Formula of Median

Last Updated : 12 Apr, 2025
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The median is a measure of central tendency that represents the middle value of a data set when the values are arranged in order, either ascending or descending. In a continuous series (grouped data), where data is presented in class intervals with frequencies, the median divides the distribution such that 50% of the values lie below it and 50% lie above it. It helps identify the central position of the data and is calculated using a formula based on cumulative frequencies.

What is a Continuous Series?

In a continuous series (grouped frequency distribution), the value of a variable is grouped into several class intervals (such as 0-5, 5-10, 10-15) along with the corresponding frequencies. The method used to determine the arithmetic average in a continuous series is the same as that used in a discrete series. The midpoints of several class intervals replace the class interval in a continuous series. When it is done, a continuous series and a discrete series are the same.

Example of Continuous Series

If 10 students of a class score marks between 50-60, 8 students score marks between 60-70, 12 students score marks between 70-80, and 5 students score marks between 80-90, then this information will be shown as:

Marks

No. of Students

50-60

10

60-70

8

70-80

12

80-90

5

Calculation of Median in Continuous Series

The value of the median cannot be easily located for a continuous series. In this situation, the median is located between the lower and upper limits of a class interval. A formula is used to interpolate (guess) the median to obtain the exact value. However, it should be kept in mind that when the median class of a series is first class, then the c.f. in the formula will be taken as zero. The steps required to determine the median of a continuous series are as follows:

Step 1: Arrange the given data in either descending or ascending order.

Step 2: Determine the cumulative frequency, i.e., cf.

Step 3: Calculate the median item using the following formula:

Median(M)=Size~of~[\frac{N}{2}]^{th}~item     

Where N = Total of Frequency

Step 4: Now inspect the cumulative frequencies and find out the cf which is either equal to or just greater than the value determined in the previous step.

Step 5: Now, find the class corresponding to the cumulative frequency equal to or just greater than the value determined in the third step. This class is known as the median class. 

Step 6: Now, apply the following formula for the median:

Median=l_1+\frac{\frac{N}{2}-c.f.}{f}\times{i}     

Where,

l1 = lower limit of the median class

c.f. = cumulative frequency of the class preceding the median class

f = simple frequency of the median class

i = class size of the median group or class

Note: While calculating the median of a given distribution, we have to assume that every class of the distribution is uniformly distributed in the class interval.

Solved Examples: Calculation of Median in Continuous Series

Example 1:

Median in Continuous Series

Solution:

Median in Continuous Series

Median(M)=Size~of~[\frac{N}{2}]^{th}~item

=Size~of~[\frac{30}{2}]^{th}~item=Siz~of~15^{th}~item

Hence, the median lies in the class 15-20.

l1 = 15, f = 8, i = 5, c.f. = 12

Now apply the following formula:
Median=l_{1}+\frac{\frac{N}{2}-c.f.}{f}\times{i}
Median=15+\frac{\frac{30}{2}-12}{8}\times{5}

Median= 16.875

Example 2:

The distribution of income among employees has been observed in a study performed within an organisation. Determine the median wage of the organization's employees.

  • 5 men are paid less than ₹100.
  • 15 men are paid less than ₹200.
  • 23 men are paid less than ₹300.
  • 35 men are paid less than ₹400.
  • 50 men are paid less than ₹500.

Solution:

The above frequencies are the cumulative frequencies (c.f.) of the workers. Thus to calculate the median, first, we have to convert it into simple frequency and presented the data in tabular form.

Median in Continuous Series

Median(M)=Size~of~[\frac{N}{2}]^{th}~item

=Size~of~[\frac{50}{2}]^{th}~item=Size~of~25^{th}~item

Hence, the median lies in the class 300-400.

l1 = 300, f = 12, i = 100, c.f. = 23

Now apply the following formula:
Median=l_{1}+\frac{\frac{N}{2}-c.f.}{f}\times{i}
Median=300+\frac{\frac{50}{2}-23}{12}\times{100}

Median = 316.67

Example 3:

The weekly expenditures of 100 families are listed in the following table. Calculate the weekly expenditure's median.

Median in Continuous Series

Solution:

Median in Continuous Series

Median(M)=Size~of~[\frac{N}{2}]^{th}~item

=Size~of~[\frac{100}{2}]^{th}~item=Size~of~50^{th}~item

Hence, the median lies in the class 1500-3000.

l1 = 1500, f = 25, i = 1500, c.f. = 30

Now apply the following formula:
Median=l_{1}+\frac{\frac{N}{2}-c.f.}{f}\times{i}
Median=1500+\frac{\frac{100}{2}-30}{25}\times{1500}

Median = 2700

Calculation of Median in Continuous Series Practice Problems

Question 1: Given the following continuous series, find the median

Class IntervalFrequency (f)
0 - 105
10 - 208
20 - 3012
30 - 4015
40 - 5010

Question 2: Calculate the median for the following data

Class IntervalFrequency (f)
50 - 606
60 - 709
70 - 8013
80 - 9011
90 - 1007

Question 3: Find the median for this continuous series

Class IntervalFrequency (f)
10 - 204
20 - 306
30 - 4014
40 - 5016
50 - 6010

Question 4: Determine the median for the following data set

Class IntervalFrequency (f)
100 - 1105
110 - 12010
120 - 13015
130 - 14020
140 - 15025

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