1. Problem 18: Given import data from various Asian countries for years 1984 and 1985 (in million US dollars), compute the requested values a_3 and a_4. The data points (import values for 1984) are: \[ a = [297.2, 266.1, 820.1, 101.8, 431.9, 19.5, 8.4, 113.2, 3.7] \] a_3 is the third element in the array (since indexing starts at 1): $$a_3 = 820.1$$ Similarly, a_4 is the fourth element: $$a_4 = 101.8$$ 2. Problem 20: From data in Problem 18 (using 1984 values), calculate: a. Variance: - Find the mean \(\bar{a}\): $$\bar{a} = \frac{1}{9}\sum_{i=1}^9 a_i = \frac{297.2 + 266.1 + 820.1 + 101.8 + 431.9 + 19.5 + 8.4 + 113.2 + 3.7}{9} = \frac{2061.9}{9} = 229.1$$ - Calculate variance: $$\sigma^2 = \frac{1}{9-1} \sum_{i=1}^9 (a_i - \bar{a})^2$$ Calculate each squared deviation: $$(297.2 - 229.1)^2 = 4624.81$$ $$(266.1 - 229.1)^2 = 1369$$ $$(820.1 - 229.1)^2 = 348102.4$$ $$(101.8 - 229.1)^2 = 16214.1$$ $$(431.9 - 229.1)^2 = 41498.2$$ $$(19.5 - 229.1)^2 = 43715.2$$ $$(8.4 - 229.1)^2 = 48700.1$$ $$(113.2 - 229.1)^2 = 13466.4$$ $$(3.7 - 229.1)^2 = 50616.2$$ Sum of squared deviations: $$4624.81 + 1369 + 348102.4 + 16214.1 + 41498.2 + 43715.2 + 48700.1 + 13466.4 + 50616.2 = 566206.4$$ Thus, $$\sigma^2 = \frac{566206.4}{8} = 70775.8$$ b. Standard deviation: $$\sigma = \sqrt{70775.8} \approx 266.0$$ c. Coefficient of skewness (sample skewness): $$g_1 = \frac{n}{(n-1)(n-2)} \sum \left( \frac{a_i - \bar{a}}{\sigma} \right)^3$$ Calculate each \( (a_i - \bar{a})/\sigma \) and cube: $$(\frac{297.2 - 229.1}{266.0})^3 = (0.254)^3 = 0.0164$$ $$(\frac{266.1 - 229.1}{266.0})^3 = (0.137)^3 = 0.0026$$ $$(\frac{820.1 - 229.1}{266.0})^3 = (2.21)^3 = 10.75$$ $$(\frac{101.8 - 229.1}{266.0})^3 = (-0.479)^3 = -0.110$$ $$(\frac{431.9 - 229.1}{266.0})^3 = (0.765)^3 = 0.447$$ $$(\frac{19.5 - 229.1}{266.0})^3 = (-0.791)^3 = -0.495$$ $$(\frac{8.4 - 229.1}{266.0})^3 = (-0.826)^3 = -0.564$$ $$(\frac{113.2 - 229.1}{266.0})^3 = (-0.443)^3 = -0.087$$ $$(\frac{3.7 - 229.1}{266.0})^3 = (-0.848)^3 = -0.610$$ Sum: $$0.0164 + 0.0026 + 10.75 - 0.11 + 0.447 - 0.495 - 0.564 - 0.087 - 0.610 = 9.349$$ Then, $$g_1 = \frac{9}{8 \times 7} \times 9.349 = \frac{9}{56} \times 9.349 \approx 1.503$$ d. Coefficient of kurtosis (excess kurtosis): $$g_2 = \frac{n(n+1)}{(n-1)(n-2)(n-3)} \sum \left( \frac{a_i - \bar{a}}{\sigma} \right)^4 - \frac{3(n-1)^2}{(n-2)(n-3)}$$ Calculate each \( (a_i - \bar{a})/\sigma \) to the fourth power: $$(0.254)^4=0.0042$$ $$(0.137)^4=0.0004$$ $$(2.21)^4=23.76$$ $$(-0.479)^4=0.053$$ $$(0.765)^4=0.342$$ $$(-0.791)^4=0.391$$ $$(-0.826)^4=0.465$$ $$(-0.443)^4=0.039$$ $$(-0.848)^4=0.517$$ Sum: $$0.0042 + 0.0004 + 23.76 + 0.053 + 0.342 + 0.391 + 0.465 + 0.039 + 0.517 = 25.57$$ Plug into formula: Numerator coefficient: $$\frac{9 \times 10}{8 \times 7 \times 6} = \frac{90}{336} = 0.2679$$ First term: $$0.2679 \times 25.57 = 6.85$$ Second term: $$\frac{3 \times (8)^2}{7 \times 6} = \frac{3 \times 64}{42} = 4.57$$ Thus, $$g_2 = 6.85 - 4.57 = 2.28$$ Final answers: - $$a_3 = 820.1$$ - $$a_4 = 101.8$$ - Variance: $$70775.8$$ - Standard deviation: $$266.0$$ - Coefficient of skewness: $$1.503$$ - Coefficient of kurtosis: $$2.28$$