To calculate the median for the given grouped data, we can use the following steps:

1. Identify the cumulative frequencies to locate the median class.
2. Apply the median formula for grouped data: $\text{Median} = L + \left( \frac{\frac{N}{2} - F}{f} \right) \times h$ where:
   • $L$ = Lower boundary of the median class.
   • $N$ = Total frequency.
   • $F$ = Cumulative frequency before the median class.
   • $f$ = Frequency of the median class.
   • $h$ = Class width.

Step 1: Data Overview

Step 2: Find Cumulative Frequencies

Total frequency $N = 40$.

Step 3: Locate the Median Class

We calculate $\frac{N}{2} = \frac{40}{2} = 20$.

The median class is the class where the cumulative frequency first exceeds 20, which is the class interval 20 - 30.

Step 4: Apply the Median Formula

• $L = 20$ (lower boundary of the median class)
• $N = 40$
• $F = 12$ (cumulative frequency before the median class)
• $f = 12$ (frequency of the median class)
• $h = 10$ (class width)

Substitute these into the formula: $\text{Median} = 20 + \left( \frac{20 - 12}{12} \right) \times 10$ $\text{Median} = 20 + \left( \frac{8}{12} \right) \times 10$ $\text{Median} = 20 + \left( 0.6667 \right) \times 10$ $\text{Median} = 20 + 6.67 = 26.67$

Conclusion:

The median of the given data distribution is approximately 26.67.

Would you like further details on this process or have any questions?

Here are 5 related questions:

1. How do we calculate the mode for this grouped data?
2. What is the difference between mean and median?
3. Can you explain how to calculate the quartiles for grouped data?
4. How do outliers affect the median compared to the mean?
5. How would you calculate the mean for this frequency distribution?

Tip: The median is less sensitive to outliers than the mean, making it a better measure for skewed distributions.