1. **State the problem:** We want to find the probability that in 200 tosses of a fair coin, the proportion of heads is between 55% and 65%. 2. **Identify the distribution:** The number of heads in 200 tosses follows a binomial distribution with parameters $n=200$ and $p=0.5$ (since the coin is fair). 3. **Sampling distribution of the sample proportion:** The sample proportion $\hat{p}$ of heads is $\hat{p} = \frac{X}{n}$ where $X$ is the number of heads. 4. **Mean and standard deviation of $\hat{p}$:** $$\mu_{\hat{p}} = p = 0.5$$ $$\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.5 \times 0.5}{200}} = \sqrt{\frac{0.25}{200}} = \sqrt{0.00125} \approx 0.03536$$ 5. **Convert the problem to a probability about $\hat{p}$:** We want $P(0.55 \leq \hat{p} \leq 0.65)$. 6. **Standardize using the normal approximation:** $$Z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}}$$ Calculate the Z-scores for 0.55 and 0.65: $$Z_1 = \frac{0.55 - 0.5}{0.03536} = \frac{0.05}{0.03536} \approx 1.414$$ $$Z_2 = \frac{0.65 - 0.5}{0.03536} = \frac{0.15}{0.03536} \approx 4.243$$ 7. **Find the probability using the standard normal distribution:** $$P(0.55 \leq \hat{p} \leq 0.65) = P(1.414 \leq Z \leq 4.243) = \Phi(4.243) - \Phi(1.414)$$ Where $\Phi(z)$ is the cumulative distribution function (CDF) of the standard normal. 8. **Look up or calculate values:** $$\Phi(4.243) \approx 1$$ (very close to 1) $$\Phi(1.414) \approx 0.9213$$ 9. **Calculate final probability:** $$P = 1 - 0.9213 = 0.0787$$ **Answer:** The probability that between 55% and 65% of the outcomes are heads in 200 tosses of a fair coin is approximately **0.0787** or **7.87%**.