1. **State the problem:** We are given class intervals and their corresponding frequencies, and we need to find the median of the data. 2. **List the data:** Class intervals: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60 Frequencies: 8, 10, 12, 22, 30, 18 3. **Calculate the cumulative frequency (CF):** - CF for 0-10 = 8 - CF for 10-20 = 8 + 10 = 18 - CF for 20-30 = 18 + 12 = 30 - CF for 30-40 = 30 + 22 = 52 - CF for 40-50 = 52 + 30 = 82 - CF for 50-60 = 82 + 18 = 100 4. **Find the total number of observations, $n$:** $$n = 8 + 10 + 12 + 22 + 30 + 18 = 100$$ 5. **Find the median class:** Median position = $\frac{n}{2} = \frac{100}{2} = 50$ The median class is the class whose cumulative frequency is just greater than or equal to 50. From the CF, 52 is the first CF greater than 50, so the median class is 30-40. 6. **Identify median class parameters:** - Lower boundary, $l = 30$ - Frequency of median class, $f = 22$ - Cumulative frequency before median class, $CF = 30$ - Class width, $h = 10$ 7. **Apply the median formula:** $$\text{Median} = l + \left(\frac{\frac{n}{2} - CF}{f}\right) \times h$$ Substitute values: $$= 30 + \left(\frac{50 - 30}{22}\right) \times 10 = 30 + \left(\frac{20}{22}\right) \times 10$$ $$= 30 + 0.9091 \times 10 = 30 + 9.091 = 39.091$$ 8. **Final answer:** The median of the data is approximately **39.09**.