Calculate the median for the given frequency distribution.

The total frequency is the sum of all frequencies:

$$N = 6 + 15 + 20 + 7 + 2 = 50$$

The median position is given by $\frac{N}{2}$:

$$\text{Median position} = \frac{50}{2} = 25$$

This means the median is the 25th value in the ordered data.

We need to find the class interval where the 25th value falls. We do this by calculating the cumulative frequencies:

• For class 1-5: Cumulative frequency = 6
• For class 6-10: Cumulative frequency = $6 + 15 = 21$
• For class 11-15: Cumulative frequency = $21 + 20 = 41$ Since the 25th value is greater than 21 and less than or equal to 41, the median class is 11-15.

The formula for the median of grouped data is:

$$\text{Median} = L + \left( \frac{\frac{N}{2} - cf_b}{f_m} \right) \times c$$

Where:

• $L$ = Lower class boundary of the median class. For the class 11-15, the lower boundary is 10.5 (midpoint between 10 and 11).
• $\frac{N}{2}$ = Median position = 25.
• $cf_b$ = Cumulative frequency of the class before the median class. This is the cumulative frequency for the 6-10 class, which is 21.
• $f_m$ = Frequency of the median class. For the 11-15 class, $f_m = 20$.
• $c$ = Class width of the median class. For the class 11-15, the width is $15 - 11 + 1 = 5$. (Alternatively, $15.5 - 10.5 = 5$).

$$\text{Median} = 10.5 + \left( \frac{25 - 21}{20} \right) \times 5$$ $$\text{Median} = 10.5 + \left( \frac{4}{20} \right) \times 5$$ $$\text{Median} = 10.5 + \left( \frac{1}{5} \right) \times 5$$ $$\text{Median} = 10.5 + 1$$ $$\text{Median} = 11.5$$

Comparing this result with the given options:
A. $10 \frac{1}{2} = 10.5$
B. 12
C. 13
D. $11 \frac{1}{2} = 11.5$

The calculated median is 11.5, which corresponds to option D.

The final answer is $\boxed{\text{D. } 11 \frac{1}{2}}$.