Answer: 0.1039

Using the omitted variable formula, we know that: $$ \hat{\beta}_1 = \beta_1 + \beta_2 \delta $$ Where: $$ \delta = \frac{\text{Cov}(X_1, X_2)}{\text{Var}(X_1)} = \frac{\rho_{X_1X_2} \sigma_{X_1} \sigma_{X_2}}{\text{Var}(X_1)} $$ Given: - $\hat{\beta}_1 = 2.4$ - $\beta_1 = 2.2$ - $\beta_2 = 1.1$ - $\sigma^2_{X_1} = 16 \Rightarrow \sigma_{X_1} = 4$ - $\sigma^2_{X_2} = 49 \Rightarrow \sigma_{X_2} = 7$ Substitute into the formula: $$ 2.4 = 2.2 + 1.1 \cdot \delta \Rightarrow \delta = \frac{2.4 - 2.2}{1.1} = \frac{0.2}{1.1} \approx 0.1818 $$ Now solve for $\rho_{X_1X_2}$: $$ \delta = \frac{\rho_{X_1X_2} \cdot 4 \cdot 7}{16} = \frac{28 \rho_{X_1X_2}}{16} = 1.75 \rho_{X_1X_2} $$ So: $$ 0.1818 = 1.75 \rho_{X_1X_2} \Rightarrow \rho_{X_1X_2} = \frac{0.1818}{1.75} \approx 0.1039 $$ Thus, the correct answer is **D. 0.1039**.

Author: Nikitesh Somanthe