
The median class of the frequency distribution given below is
Class $0 - 10$ $10 - 20$ $20 - 30$ $30 - 40$ $40 - 50$ Frequency $7$ $15$ $13$ $17$ $10$
A. $40 - 50$
B. $30 - 40$
C. $20 - 30$
D. $10 - 20$
Class | $0 - 10$ | $10 - 20$ | $20 - 30$ | $30 - 40$ | $40 - 50$ |
Frequency | $7$ | $15$ | $13$ | $17$ | $10$ |
Answer
414.6k+ views
Hint : First, we shall analyze the given information so that we can able to solve the problem.
The given frequency distribution is grouped data. Here we are asked to calculate the median class for the given grouped data. The median class is the class interval whose cumulative frequency is just greater than or equal to $\dfrac{N}{2}$ where $N$ is the total frequency of the given grouped data. First, we need to prepare a frequency distribution table containing the cumulative frequency.
Complete step-by-step solution:
Here we are asked to calculate the median class for the given grouped data. First, we need to prepare a frequency distribution table containing the cumulative frequency.
We shall add the frequencies of all the classes preceding the given class to obtain the cumulative frequency of a class.
Here, the total frequency is $N = 7 + 15 + 13 + 17 + 10$
$N = 62$
Now, we shall calculate the value of $\dfrac{N}{2}$
Thus, we get $\dfrac{N}{2} = \dfrac{{62}}{2}$
$\dfrac{N}{2} = 31$
The median class is the class interval whose cumulative frequency is just greater than or equal to $\dfrac{N}{2}$ where $N$ is the total frequency of the given grouped data.
Thus, we need to obtain the class interval whose cumulative frequency is just greater than $\dfrac{N}{2}$
Hence, the cumulative frequency just greater than $\dfrac{N}{2}$ is $35$ .
Now, we shall check the class interval whose cumulative frequency is $35$
Therefore, the required median class is $20 - 30$
Hence, option C) is correct.
Note:We can understand that the median class is the class interval whose cumulative frequency is just greater than or equal to half the total frequency of the given grouped data. We usually denote the total frequency by the letter $N$. Hence, the cumulative frequency just greater than $\dfrac{N}{2}$ is $35$ . Therefore, we got the required median class $20 - 30$.
The given frequency distribution is grouped data. Here we are asked to calculate the median class for the given grouped data. The median class is the class interval whose cumulative frequency is just greater than or equal to $\dfrac{N}{2}$ where $N$ is the total frequency of the given grouped data. First, we need to prepare a frequency distribution table containing the cumulative frequency.
Complete step-by-step solution:
Here we are asked to calculate the median class for the given grouped data. First, we need to prepare a frequency distribution table containing the cumulative frequency.
We shall add the frequencies of all the classes preceding the given class to obtain the cumulative frequency of a class.
Class interval | Frequency | Cumulative frequency |
$0 - 10$ | $7$ | $7$ |
$10 - 20$ | $15$ | $22$ |
$20 - 30$ | $13$ | $35$ |
$30 - 40$ | $17$ | $52$ |
$40 - 50$ | $10$ | $62$ |
Here, the total frequency is $N = 7 + 15 + 13 + 17 + 10$
$N = 62$
Now, we shall calculate the value of $\dfrac{N}{2}$
Thus, we get $\dfrac{N}{2} = \dfrac{{62}}{2}$
$\dfrac{N}{2} = 31$
The median class is the class interval whose cumulative frequency is just greater than or equal to $\dfrac{N}{2}$ where $N$ is the total frequency of the given grouped data.
Thus, we need to obtain the class interval whose cumulative frequency is just greater than $\dfrac{N}{2}$
Hence, the cumulative frequency just greater than $\dfrac{N}{2}$ is $35$ .
Now, we shall check the class interval whose cumulative frequency is $35$
Therefore, the required median class is $20 - 30$
Hence, option C) is correct.
Note:We can understand that the median class is the class interval whose cumulative frequency is just greater than or equal to half the total frequency of the given grouped data. We usually denote the total frequency by the letter $N$. Hence, the cumulative frequency just greater than $\dfrac{N}{2}$ is $35$ . Therefore, we got the required median class $20 - 30$.
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