1. **State the problem:** We want to find the probability that more than 20 homes sold for above market value out of 120 homes, given that 19% of homes sell above market value. 2. **Identify the distribution:** This is a binomial probability problem where: - Number of trials $n = 120$ - Probability of success $p = 0.19$ - We want $P(X > 20)$ where $X$ is the number of homes sold above market value. 3. **Use normal approximation to binomial:** Since $n$ is large, approximate $X$ by a normal distribution with: - Mean $\mu = np = 120 \times 0.19 = 22.8$ - Standard deviation $\sigma = \sqrt{np(1-p)} = \sqrt{120 \times 0.19 \times 0.81} = \sqrt{18.468} \approx 4.296$ 4. **Apply continuity correction:** We want $P(X > 20)$, so approximate $P(X > 20.5)$ for continuity correction. 5. **Calculate the z-score:** $$ z = \frac{20.5 - 22.8}{4.296} = \frac{-2.3}{4.296} \approx -0.535 $$ 6. **Find the probability:** $$ P(X > 20) \approx P(Z > -0.535) = 1 - P(Z \leq -0.535) = 1 - \Phi(-0.535) $$ Using symmetry of normal distribution: $$ 1 - \Phi(-0.535) = \Phi(0.535) $$ From standard normal tables or calculator: $$ \Phi(0.535) \approx 0.704 $$ 7. **Final answer:** The probability that more than 20 homes sold above market value is approximately **0.704**.