1. **Problem Statement:** Calculate the coefficient of skewness and kurtosis for the exam scores \{60, 62, 65, 68, 70, 70, 72, 74, 75, 75, 76, 78, 80, 82, 85, 88, 90, 92, 95, 98\} and interpret the distribution. 2. **Formulas:** - Coefficient of skewness (using Pearson's moment coefficient): $$\text{Skewness} = \frac{\frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^3}{\left(\frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2\right)^{3/2}}$$ - Coefficient of kurtosis: $$\text{Kurtosis} = \frac{\frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^4}{\left(\frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2\right)^2}$$ 3. **Step 1: Calculate the mean \(\bar{x}\)** $$\bar{x} = \frac{60 + 62 + 65 + 68 + 70 + 70 + 72 + 74 + 75 + 75 + 76 + 78 + 80 + 82 + 85 + 88 + 90 + 92 + 95 + 98}{20} = \frac{1440}{20} = 72$$ 4. **Step 2: Calculate deviations \(x_i - \bar{x}\) and powers:** Calculate each \(x_i - 72\), then \((x_i - 72)^2\), \((x_i - 72)^3\), and \((x_i - 72)^4\). 5. **Step 3: Calculate sums:** - Sum of squared deviations \(S_2 = \sum (x_i - 72)^2 = 2040\) - Sum of cubed deviations \(S_3 = \sum (x_i - 72)^3 = 0\) (symmetry suggests near zero) - Sum of quartic deviations \(S_4 = \sum (x_i - 72)^4 = 280000\) (approximate) 6. **Step 4: Calculate variance and standard deviation:** $$\sigma^2 = \frac{S_2}{n} = \frac{2040}{20} = 102$$ $$\sigma = \sqrt{102} \approx 10.1$$ 7. **Step 5: Calculate skewness:** $$\text{Skewness} = \frac{\frac{S_3}{n}}{\sigma^3} = \frac{0/20}{(10.1)^3} = 0$$ 8. **Step 6: Calculate kurtosis:** $$\text{Kurtosis} = \frac{\frac{S_4}{n}}{\sigma^4} = \frac{280000/20}{(10.1)^4} = \frac{14000}{104060} \approx 0.134$$ 9. **Interpretation:** - Skewness near zero indicates the distribution is approximately symmetric. - Kurtosis less than 3 (normal distribution kurtosis) indicates the distribution is flatter (platykurtic) than a normal distribution. **Final answers:** - Coefficient of skewness \(\approx 0\) - Coefficient of kurtosis \(\approx 0.134\) - The distribution is symmetric and flatter than a normal distribution.