1. **State the problem:** We need to estimate the number of slow worms whose length is more than one interquartile range (IQR) above the mean length. 2. **Recall the formula:** We calculate $\bar{x} + \text{IQR}$, where $\bar{x}$ is the mean length and IQR is the interquartile range. 3. **Given data:** Length intervals (mm) and frequencies: 125–139: 4 140–154: 4 155–169: 2 170–184: 7 185–199: 20 200–214: 24 215–229: 10 4. **Calculate total number of worms:** $$N = 4 + 4 + 2 + 7 + 20 + 24 + 10 = 71$$ 5. **Estimate the mean $\bar{x}$:** Use midpoints of intervals: Midpoints: 132, 147, 162, 177, 192, 207, 222 Calculate weighted sum: $$\sum f x = 4\times132 + 4\times147 + 2\times162 + 7\times177 + 20\times192 + 24\times207 + 10\times222$$ $$= 528 + 588 + 324 + 1239 + 3840 + 4968 + 2220 = 13707$$ Mean: $$\bar{x} = \frac{13707}{71} \approx 193.0$$ 6. **Estimate the interquartile range (IQR):** - Find cumulative frequencies: 4, 8, 10, 17, 37, 61, 71 - Q1 position: $\frac{71}{4} = 17.75$th value - Q3 position: $\frac{3\times71}{4} = 53.25$th value Q1 lies in 170–184 interval (cumulative 17 to 37) Q3 lies in 200–214 interval (cumulative 37 to 61) Interpolate Q1: $$Q1 = 170 + \frac{17.75 - 17}{37 - 17} \times 15 = 170 + \frac{0.75}{20} \times 15 = 170 + 0.5625 = 170.56$$ Interpolate Q3: $$Q3 = 200 + \frac{53.25 - 37}{61 - 37} \times 15 = 200 + \frac{16.25}{24} \times 15 = 200 + 10.16 = 210.16$$ IQR: $$\text{IQR} = Q3 - Q1 = 210.16 - 170.56 = 39.6$$ 7. **Calculate $\bar{x} + \text{IQR}$:** $$193.0 + 39.6 = 232.6$$ 8. **Determine which class interval 232.6 falls in:** The highest interval is 215–229, so 232.6 is above all intervals. 9. **Estimate number of worms longer than $\bar{x} + \text{IQR}$:** Since 232.6 is above the highest interval, no worms fall in a higher interval. Therefore, estimate is 0 worms. **Final answer:** There are approximately 0 slow worms whose length is more than one interquartile range above the mean.