Find the Gini index for this function. Recall that \( \mathrm{L}(\mathrm{x})=\frac{5 \mathrm{x}^{2}}{6}+\frac{\mathrm{x}}{6} \). \( \mathrm{G}=\square \) (Type an integer or a fraction.)
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Bonus Knowledge
To find the Gini index for the given function \( \mathrm{L}(\mathrm{x})=\frac{5 \mathrm{x}^{2}}{6}+\frac{\mathrm{x}}{6} \), we need to calculate the area between the line of equality (45° line) and the Lorenz curve (which is defined by \( L(x) \)). The Gini index \( G \) is given by: \[ G = \frac{A}{A + B} \] where \( A \) is the area between the line of equality and the Lorenz curve, and \( B \) is the area under the Lorenz curve. First, we calculate the area under the Lorenz curve \( L(x) \) from 0 to 1: \[ B = \int_0^1 L(x) \, dx = \int_0^1 \left( \frac{5x^2}{6} + \frac{x}{6} \right) dx = \left[ \frac{5x^3}{18} + \frac{x^2}{12} \right]_0^1 = \frac{5}{18} + \frac{1}{12} \] The LCM of 18 and 12 is 36: \[ B = \frac{10}{36} + \frac{3}{36} = \frac{13}{36} \] Next, the area under the line of equality (the triangle under the line from 0 to 1) is: \[ A + B = \frac{1}{2} = \frac{18}{36} \] Thus, the area \( A \): \[ A = \frac{1}{2} - B = \frac{18}{36} - \frac{13}{36} = \frac{5}{36} \] Finally, we can find the Gini index: \[ G = \frac{A}{A + B} = \frac{\frac{5}{36}}{\frac{18}{36}} = \frac{5}{18} \] Therefore, the Gini index \( G \) is \( \frac{5}{18} \).