1. **State the problem:** Calculate the standard deviation and coefficient of skewness from the given frequency distribution table. 2. **Recall formulas:** - Mean $\bar{x} = \frac{\sum f x}{\sum f}$ - Variance $\sigma^2 = \frac{\sum f x^2}{\sum f} - \bar{x}^2$ - Standard deviation $\sigma = \sqrt{\sigma^2}$ - Coefficient of skewness (Pearson's first coefficient) $\text{Skewness} = \frac{\bar{x} - \text{Mode}}{\sigma}$ 3. **Calculate total frequency $n$:** $$n = 7 + 9 + 13 + 12 + 5 + 5 + 2 + 2 + 2 + 3 = 60$$ 4. **Calculate mean $\bar{x}$:** $$\bar{x} = \frac{\sum f x}{n} = \frac{17.5 + 67.5 + 162.5 + 210 + 112.5 + 137.5 + 65 + 75 + 85 + 142.5}{60} = \frac{1072.5}{60} = 17.875$$ 5. **Calculate variance $\sigma^2$:** $$\frac{\sum f x^2}{n} = \frac{43.75 + 506.25 + 2031.25 + 3675 + 2531.25 + 3781.25 + 2112.5 + 2812.5 + 3612.5 + 6768.75}{60} = \frac{27075}{60} = 451.25$$ $$\sigma^2 = 451.25 - (17.875)^2 = 451.25 - 319.14 = 132.11$$ 6. **Calculate standard deviation $\sigma$:** $$\sigma = \sqrt{132.11} \approx 11.49$$ 7. **Find mode:** Mode class is the class with highest frequency, which is $[10, 15)$ with frequency 13. 8. **Calculate mode using formula:** $$\text{Mode} = L + \frac{(f_1 - f_0)}{(2f_1 - f_0 - f_2)} \times h$$ Where: - $L = 10$ (lower class boundary of modal class) - $f_1 = 13$ (frequency of modal class) - $f_0 = 9$ (frequency of class before modal) - $f_2 = 12$ (frequency of class after modal) - $h = 5$ (class width) $$\text{Mode} = 10 + \frac{(13 - 9)}{(2 \times 13 - 9 - 12)} \times 5 = 10 + \frac{4}{(26 - 21)} \times 5 = 10 + \frac{4}{5} \times 5 = 10 + 4 = 14$$ 9. **Calculate coefficient of skewness:** $$\text{Skewness} = \frac{17.875 - 14}{11.49} = \frac{3.875}{11.49} \approx 0.337$$ **Final answers:** - Standard deviation $\approx 11.49$ - Coefficient of skewness $\approx 0.337$