1. **State the problem:** We need to find the value of $K$, which represents the number of earthworms with lengths at least 68 mm but shorter than 72 mm, given the mean length of all earthworms is 71.9 mm. 2. **Understand the data:** The problem mentions a table with lengths and frequencies, but since the table is not provided, we assume $K$ is the unknown frequency for the interval $[68,72)$ mm. 3. **Formula for mean:** The mean length $\bar{x}$ is calculated by $$\bar{x} = \frac{\sum (x_i \cdot f_i)}{\sum f_i}$$ where $x_i$ are the midpoints of length intervals and $f_i$ their frequencies. 4. **Set up the equation:** Let the total sum of products of lengths and frequencies excluding the $K$ group be $S$, and total frequency excluding $K$ be $N$. Then, $$71.9 = \frac{S + m \cdot K}{N + K}$$ where $m$ is the midpoint of the interval $[68,72)$, so $m = \frac{68 + 72}{2} = 70$. 5. **Solve for $K$:** Rearranging, $$71.9 (N + K) = S + 70K$$ $$71.9N + 71.9K = S + 70K$$ $$71.9K - 70K = S - 71.9N$$ $$1.9K = S - 71.9N$$ $$K = \frac{S - 71.9N}{1.9}$$ 6. **Interpretation:** To find $K$, you need the values of $S$ and $N$ from the table data. Once you have those, plug them into the formula above to calculate $K$. **Final answer:** $$K = \frac{S - 71.9N}{1.9}$$ where $S$ is the sum of products of lengths and frequencies excluding the $K$ group, and $N$ is the total frequency excluding $K$.