1. **Problem statement:** Given detection times: 12, 14, 13, 15, 13, 16, 14, 13, 17, 12, we need to compute various statistics. 2. **Sum of all detection times:** $$\text{Sum} = 12 + 14 + 13 + 15 + 13 + 16 + 14 + 13 + 17 + 12 = 139$$ 3. **Mean (exact):** $$\text{Mean} = \frac{\text{Sum}}{n} = \frac{139}{10} = 13.9$$ 4. **Mean rounded to two decimals:** $$13.90$$ 5. **Ordered data (ascending):** $$12, 12, 13, 13, 13, 14, 14, 15, 16, 17$$ 6. **5th value in ordered data:** $$13$$ 7. **6th value in ordered data:** $$14$$ 8. **Median detection time:** Since $n=10$ (even), median is average of 5th and 6th values: $$\text{Median} = \frac{13 + 14}{2} = 13.5$$ 9. **Mode of detection times:** The value appearing most frequently is $13$ (appears 3 times). 10. **Sum of absolute deviations about the mean:** Calculate each absolute deviation: $|12-13.9|=1.9$, $|14-13.9|=0.1$, $|13-13.9|=0.9$, $|15-13.9|=1.1$, $|13-13.9|=0.9$, $|16-13.9|=2.1$, $|14-13.9|=0.1$, $|13-13.9|=0.9$, $|17-13.9|=3.1$, $|12-13.9|=1.9$ Sum: $$1.9 + 0.1 + 0.9 + 1.1 + 0.9 + 2.1 + 0.1 + 0.9 + 3.1 + 1.9 = 13.0$$ 11. **Mean absolute deviation (MAD):** $$\text{MAD} = \frac{13.0}{10} = 1.30$$ 12. **Median absolute deviation (MeD):** Order absolute deviations: $$0.1, 0.1, 0.9, 0.9, 0.9, 1.1, 1.9, 1.9, 2.1, 3.1$$ Median is average of 5th and 6th: $$\frac{0.9 + 1.1}{2} = 1.00$$ 13. **Skewness:** Since mean (13.9) > median (13.5) > mode (13), the distribution is positively skewed (right-skewed). **Final answers:** (a) 139 (b) 13.9 (c) 13.90 (d) 13 (e) 14 (f) 13.5 (g) 13 (i) 13.00 (j) 1.30 (MAD), 1.00 (MeD) (k) Positively skewed because mean > median > mode.