1. **Problem statement:** A researcher samples 200 MU students to find the proportion who prefer chocolate ice cream. The population proportion is $p=0.26$. We want to find probabilities about the sample proportion $\hat{p}$. 2. **Formula and rules:** The sample proportion $\hat{p}$ is approximately normal for large $n$ with mean $p$ and standard deviation $\sigma_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}$. 3. **Calculate standard deviation:** $$\sigma_{\hat{p}}=\sqrt{\frac{0.26(1-0.26)}{200}}=\sqrt{\frac{0.26\times0.74}{200}}=\sqrt{0.000962}=0.0310$$ 4. **(A) Probability $\hat{p}>0.30$:** Calculate the z-score: $$z=\frac{0.30-0.26}{0.0310}=\frac{0.04}{0.0310}=1.29$$ Find $P(Z>1.29)$ from standard normal table: $$P(Z>1.29)=1-P(Z\leq1.29)=1-0.9015=0.0985$$ 5. **(B) Probability $0.25<\hat{p}<0.30$:** Calculate z-scores: $$z_1=\frac{0.25-0.26}{0.0310}=-0.32$$ $$z_2=\frac{0.30-0.26}{0.0310}=1.29$$ Find probabilities: $$P(0.25<\hat{p}<0.30)=P(-0.32