1. **State the problem:** We are given a frequency distribution table with class intervals and frequencies, and we need to determine the median class. 2. **Recall the formula and rules:** The median class is the class interval where the cumulative frequency reaches or exceeds half of the total frequency. 3. **Given data:** Class intervals and frequencies: - [0,25[: 4 - [25,50[: 5 - [50,75[: 3 - [75,100[: 4 - [100,125[: 6 - [125,150[: 1 Total frequency $N = 23$ 4. **Calculate half of total frequency:** $$\frac{N}{2} = \frac{23}{2} = 11.5$$ 5. **Calculate cumulative frequencies:** - Up to [0,25[: 4 - Up to [25,50[: $4 + 5 = 9$ - Up to [50,75[: $9 + 3 = 12$ - Up to [75,100[: $12 + 4 = 16$ - Up to [100,125[: $16 + 6 = 22$ - Up to [125,150[: $22 + 1 = 23$ 6. **Determine the median class:** The cumulative frequency just before exceeding 11.5 is 9 (up to [25,50[), and the next cumulative frequency is 12 (up to [50,75[), which exceeds 11.5. Therefore, the median class is **[50,75[**. **Final answer:** The median class is **[50,75[**.