1. **State the problem:** We have a frequency distribution of earthworm lengths with one unknown frequency $K$ for the interval $68 \leq L < 72$. The mean length is given as 71.9 mm. We need to find $K$. 2. **Identify the intervals and frequencies:** - $60 \leq L < 64$: frequency = 3 - $64 \leq L < 68$: frequency = 7 - $68 \leq L < 72$: frequency = $K$ - $72 \leq L < 76$: frequency = 8 - $76 \leq L < 80$: frequency = 6 - $80 \leq L < 84$: frequency = 4 3. **Calculate midpoints of each interval:** - Midpoint for $60 \leq L < 64$ is $\frac{60+64}{2} = 62$ - Midpoint for $64 \leq L < 68$ is $66$ - Midpoint for $68 \leq L < 72$ is $70$ - Midpoint for $72 \leq L < 76$ is $74$ - Midpoint for $76 \leq L < 80$ is $78$ - Midpoint for $80 \leq L < 84$ is $82$ 4. **Use the formula for mean of grouped data:** $$\text{Mean} = \frac{\sum (f \times x)}{\sum f}$$ where $f$ is frequency and $x$ is midpoint. 5. **Set up the equation with known mean 71.9:** $$71.9 = \frac{3 \times 62 + 7 \times 66 + K \times 70 + 8 \times 74 + 6 \times 78 + 4 \times 82}{3 + 7 + K + 8 + 6 + 4}$$ 6. **Calculate the sums without $K$:** - Numerator without $K$: $3 \times 62 = 186$ - $7 \times 66 = 462$ - $8 \times 74 = 592$ - $6 \times 78 = 468$ - $4 \times 82 = 328$ Sum = $186 + 462 + 592 + 468 + 328 = 2036$ 7. **Rewrite the equation:** $$71.9 = \frac{2036 + 70K}{28 + K}$$ 8. **Multiply both sides by denominator:** $$71.9(28 + K) = 2036 + 70K$$ 9. **Expand left side:** $$2013.2 + 71.9K = 2036 + 70K$$ 10. **Bring terms with $K$ to one side and constants to the other:** $$71.9K - 70K = 2036 - 2013.2$$ $$1.9K = 22.8$$ 11. **Solve for $K$:** $$K = \frac{22.8}{1.9} = 12$$ **Final answer:** The value of $K$ is 12.