The given question asks for the median class with the following information:

• $\Sigma f = 50$ (total frequency)
• The less than cumulative frequency of the class 20–30 is 25.

Step 1: Understanding the problem

The median class is determined by the formula: $L_m = L + \left( \frac{\frac{N}{2} - CF}{f_m} \right) \times h$ Where:

• $L_m$ is the median value (but we will find the median class first).
• $L$ is the lower boundary of the median class.
• $N$ is the total frequency (here $N = 50$).
• $CF$ is the cumulative frequency just before the median class (this will be needed to find the median).
• $f_m$ is the frequency of the median class.
• $h$ is the class width (in this case, the width of the class intervals).

Step 2: Finding the median class

To find the median class, we calculate $\frac{N}{2}$: $\frac{N}{2} = \frac{50}{2} = 25$

The class whose cumulative frequency just crosses 25 will be the median class. It is given that the less than cumulative frequency of the class 20–30 is 25. This means the median class is 20–30.

Thus, the median class is 20–30.

Do you want to calculate the median value using this information? Let me know if you need further details!

Here are some related questions for further exploration:

1. How do you calculate the cumulative frequency for a frequency distribution table?
2. What is the significance of the median class in statistics?
3. How would you calculate the median if the total frequency was odd?
4. What role does the class width play in determining the median value?
5. Can you explain how the median formula adjusts for the position of the data?

Tip:

When solving for the median class, always verify if you are given the cumulative frequency or individual class frequencies, as this changes the approach to the problem.