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The marks obtained by 50 students of class 10 out of 80 marks are given in the following frequency distribution. Find the median.
Class0-1010-2020-3030-4040-5050-6060-7070-80
Frequency258169532


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Last updated date: 15th Jul 2024
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Answer
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Hint: We will first find the total number of students (N) by summing the frequencies of different classes in the given distribution. Now we will identify the class in which the value $\dfrac{N}{2}$ lies and define that class as Median Class. Now the value of Median is obtained from the below formula
Median $ = l + \left[ {\dfrac{{\dfrac{N}{2} - cf}}{f}} \right]h$

Complete step-by-step answer:
The frequency table with cumulative frequency is,
Class IntervalFrequencyCumulative Frequency
0-1022
10-2057
20-30815
30-401631
40-50940
50-60545
60-70348
70-80250

Here the sum of the frequencies is $N = 50$.
Now the value of $\dfrac{N}{2}$ is,
$ \Rightarrow \dfrac{N}{2} = \dfrac{{50}}{2} = 25$
The value of $\dfrac{N}{2}$ lies in the interval 30-40.
So, the median class is 30-40.
The lower limit of the median class is,
$ \Rightarrow l = 30$
Cumulative frequency of class preceding the median class is,
$ \Rightarrow cf = 15$
The frequency of the median class is,
$ \Rightarrow f = 16$
The height of the class is,
$ \Rightarrow h = 40 - 30 = 10$
Then the value of the median is given by,
Median $ = l + \left[ {\dfrac{{\dfrac{N}{2} - cf}}{f}} \right]h$
Substitute the values,
$ \Rightarrow $ Median $ = 30 + \left[ {\dfrac{{25 - 15}}{{16}}} \right] \times 10$
Subtract the value in the numerator and multiply with 10,
$ \Rightarrow $ Median $ = 30 + \dfrac{{100}}{{16}}$
Divide numerator by the denominator,
$ \Rightarrow $ Median $ = 30 + 6.25$
Add the terms,
$\therefore $ Median $ = 36.25$

Hence the median is 36.25.

Note: Here you need to know what is the median. Median is the Middle Most value of the data and it separates the higher half of the data set from the lower half of the data set. To find the median you need to arrange the data in an ascending order or descending order.