1. The problem asks for the 95% confidence interval for the true population proportion $p$. 2. The general formula for a confidence interval for a population proportion is: $$p \pm z \times \sigma(p)$$ where $z$ is the z-score corresponding to the confidence level, and $\sigma(p)$ is the standard error of the proportion. 3. For a 95% confidence level, the z-score is approximately 1.96. 4. Therefore, the 95% confidence interval is: $$p \pm 1.96 \times \sigma(p)$$ 5. The options involving $\mu$ are incorrect because $\mu$ represents a population mean, not a proportion. 6. The option with $\pm 1.0 \sigma(p)$ is incorrect because the z-score for 95% confidence is 1.96, not 1.0. Final answer: $p \pm 1.96 \sigma(p)$