Here’s the English translation of your solution: --- | Class Interval | Frequency (\(f_i\)) | Midpoint (\(x_i\)) | \(f_i x_i\) | |----------------|---------------------|---------------------|-------------| | 0–20 | 7 | 10 | 70 | | 20–40 | 11 | 30 | 330 | | 40–60 | \(k\) | 50 | \(50k\) | | 60–80 | 9 | 70 | 630 | | 80–100 | 13 | 90 | 1170 | | **Total** | \(\sum f_i = 40 + k\) | | \(\sum f_i x_i = 2200 + 50k\) | Given: The combined (weighted) mean is 54 \[ \frac{\sum f_i x_i}{\sum f_i} = 54 \Rightarrow \frac{2200 + 50k}{40 + k} = 54 \] Cross-multiplying: \[ 2160 + 54k = 2200 + 50k \Rightarrow 54k - 50k = 2200 - 2160 \Rightarrow 4k = 40 \Rightarrow k = 10 \] Therefore, the required value is \(k = 10\).