1. **Problem Statement:** Given tree circumferences data, find statistical measures: mean, median, mode, midrange, range, standard deviation, variance, Q1, Q3, and P10. Data: 1.8, 1.9, 1.8, 2.4, 5.1, 3.1, 5.5, 5.1, 8.3, 13.7, 5.3, 4.9, 3.7, 3.8, 4.0, 3.4, 5.2, 4.1, 3.7, 3.9 2. **Step 1: Organize data in ascending order:** $$1.8, 1.8, 1.9, 2.4, 3.1, 3.4, 3.7, 3.7, 3.8, 3.9, 4.0, 4.1, 4.9, 5.1, 5.1, 5.2, 5.3, 5.5, 8.3, 13.7$$ 3. **Step 2: Calculate mean:** $$\text{mean} = \frac{\sum x_i}{n} = \frac{1.8+1.8+1.9+2.4+3.1+3.4+3.7+3.7+3.8+3.9+4.0+4.1+4.9+5.1+5.1+5.2+5.3+5.5+8.3+13.7}{20} = \frac{90.8}{20} = 4.54$$ 4. **Step 3: Calculate median:** Since $n=20$ (even), median is average of 10th and 11th values: $$\frac{3.9 + 4.0}{2} = 3.95$$ 5. **Step 4: Calculate mode:** Most frequent values are 1.8, 3.7, and 5.1 each appearing twice. So modes are: $$1.8, 3.7, 5.1$$ 6. **Step 5: Calculate midrange:** $$\text{midrange} = \frac{\text{min} + \text{max}}{2} = \frac{1.8 + 13.7}{2} = 7.75$$ 7. **Step 6: Calculate range:** $$\text{range} = 13.7 - 1.8 = 11.9$$ 8. **Step 7: Calculate variance and standard deviation:** Calculate squared deviations: $$\sum (x_i - 4.54)^2 = 94.252$$ Sample variance: $$s^2 = \frac{94.252}{20 - 1} = 4.96$$ Standard deviation: $$s = \sqrt{4.96} = 2.23$$ 9. **Step 8: Calculate Q1 (first quartile):** Position: $0.25 \times (20 + 1) = 5.25$th value Interpolate between 5th (3.1) and 6th (3.4): $$Q1 = 3.1 + 0.25 \times (3.4 - 3.1) = 3.175$$ 10. **Step 9: Calculate Q3 (third quartile):** Position: $0.75 \times (20 + 1) = 15.75$th value Interpolate between 15th (5.1) and 16th (5.2): $$Q3 = 5.1 + 0.75 \times (5.2 - 5.1) = 5.175$$ 11. **Step 10: Calculate P10 (10th percentile):** Position: $0.10 \times (20 + 1) = 2.1$th value Interpolate between 2nd (1.8) and 3rd (1.9): $$P10 = 1.8 + 0.1 \times (1.9 - 1.8) = 1.81$$ --- 12. **Problem 2a: Convert 13.7 ft circumference to z-score:** $$z = \frac{13.7 - 4.54}{2.23} = \frac{9.16}{2.23} = 4.11$$ 13. **Problem 2b: Is 13.7 ft unusual?** A z-score above 3 is generally considered unusual. Since $z=4.11$, 13.7 ft is unusual. 14. **Problem 2c: Using range rule of thumb:** Unusual values are outside $\text{mean} \pm 2 \times \text{std dev}$: $$4.54 \pm 2 \times 2.23 = (0.08, 10.99)$$ Values outside this range: 13.7 only. --- 15. **Problem 3: Frequency distribution with 7 classes, lower limit 1.0, width 2.0:** Classes: 1.0-2.9, 3.0-4.9, 5.0-6.9, 7.0-8.9, 9.0-10.9, 11.0-12.9, 13.0-14.9 Frequencies: 1.0-2.9: 4 3.0-4.9: 10 5.0-6.9: 5 7.0-8.9: 1 9.0-10.9: 0 11.0-12.9: 0 13.0-14.9: 1 --- 16. **Problem 4: Histogram and distribution shape:** Histogram shows most data concentrated in 3.0-4.9 class, with a long right tail due to 8.3 and 13.7. Distribution is right-skewed. --- 17. **Problem 5: Box plot and 5-number summary:** Minimum: 1.8 Q1: 3.175 Median: 3.95 Q3: 5.175 Maximum: 13.7 --- 18. **Problem 6: Empirical rule for men heights (mean=176, std=7):** a. Between 169 and 183 cm is within 1 std dev, approx 68%. b. Between 155 and 197 cm is within 3 std devs, approx 99.7%. --- 19. **Problem 7: Coefficient of variation (CV):** Men heights: mean=69.17, std=2.16 $$CV = \frac{2.16}{69.17} = 0.0312 = 3.12\%$$ Cuckoo eggs: mean=22.14, std=1.34 $$CV = \frac{1.34}{22.14} = 0.0605 = 6.05\%$$ Egg lengths have higher relative variability. --- 20. **Problem 8: Chebyshev rule (mean=176, std=7):** a. Between 162 and 190 cm is $\pm 2$ std devs, at least $1 - \frac{1}{2^2} = 75\%$. b. Between 155 and 197 cm is $\pm 3$ std devs, at least $1 - \frac{1}{3^2} = 88.9\%$. Final answers are summarized above.