1. **Problem statement:** We have fried chicken dinners with mean weight $\mu=48.0$ hectograms and standard deviation $\sigma=2.4$ hectograms. We take a random sample of size $n=100$. We want to find probabilities about the sample mean $\bar{x}$. 2. **Formula and rules:** The sampling distribution of the sample mean $\bar{x}$ is approximately normal with mean $\mu_{\bar{x}}=\mu=48.0$ and standard deviation (standard error) $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{2.4}{\sqrt{100}}=\frac{2.4}{10}=0.24$. 3. We convert sample mean values to $z$-scores using $$z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$$ 4. (a) Probability $\bar{x}<47.6$: $$z=\frac{47.6-48.0}{0.24}=\frac{-0.4}{0.24}=-1.67$$ Using standard normal tables or calculator, $P(z<-1.67)\approx 0.0475$. 5. (b) Probability $\bar{x}>47.8$: $$z=\frac{47.8-48.0}{0.24}=\frac{-0.2}{0.24}=-0.83$$ So $P(\bar{x}>47.8)=P(z>-0.83)=1-P(z\leq -0.83)=1-0.2033=0.7967$. 6. (c) Probability $47.5 < \bar{x} < 48.5$: Calculate $z$ for 47.5: $$z_1=\frac{47.5-48.0}{0.24}=-2.08$$ Calculate $z$ for 48.5: $$z_2=\frac{48.5-48.0}{0.24}=2.08$$ Probability: $$P(47.5<\bar{x}<48.5)=P(-2.08