1. **State the problem:** We are given a frequency distribution of marks obtained by 100 candidates in a Biology exam. We need to calculate the mean mark, given the frequencies and the class intervals, with one unknown frequency $x$. 2. **Given data:** | Marks Interval | Frequency | | 15-24 | 6 | | 25-34 | 14 | | 35-44 | 24 | | 45-54 | 14 | | 55-64 | $x$ | | 65-74 | 10 | | 75-84 | 6 | | 85-94 | 4 | Total candidates = 100 3. **Find $x$:** Sum of frequencies must be 100. $$6 + 14 + 24 + 14 + x + 10 + 6 + 4 = 100$$ $$78 + x = 100$$ $$x = 100 - 78 = 22$$ 4. **Calculate midpoints ($m$) of each class interval:** $$15-24: \frac{15+24}{2} = 19.5$$ $$25-34: \frac{25+34}{2} = 29.5$$ $$35-44: \frac{35+44}{2} = 39.5$$ $$45-54: \frac{45+54}{2} = 49.5$$ $$55-64: \frac{55+64}{2} = 59.5$$ $$65-74: \frac{65+74}{2} = 69.5$$ $$75-84: \frac{75+84}{2} = 79.5$$ $$85-94: \frac{85+94}{2} = 89.5$$ 5. **Calculate the product of midpoints and frequencies ($m \times f$):** $$19.5 \times 6 = 117$$ $$29.5 \times 14 = 413$$ $$39.5 \times 24 = 948$$ $$49.5 \times 14 = 693$$ $$59.5 \times 22 = 1309$$ $$69.5 \times 10 = 695$$ $$79.5 \times 6 = 477$$ $$89.5 \times 4 = 358$$ 6. **Sum of $m \times f$:** $$117 + 413 + 948 + 693 + 1309 + 695 + 477 + 358 = 5010$$ 7. **Calculate the mean:** $$\text{Mean} = \frac{\sum m f}{\sum f} = \frac{5010}{100} = 50.1$$ **Final answer:** The mean mark is **50.1** (correct to 1 decimal place).