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 for the sample mean $\bar{X}$: (a) $P(\bar{X} < 47.6)$ (b) $P(\bar{X} > 47.8)$ (c) $P(47.5 < \bar{X} < 48.5)$ 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) $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{2.4}{\sqrt{100}}=0.24$. 3. **Convert to standard normal variable $Z$:** $$Z=\frac{\bar{X}-\mu}{\sigma_{\bar{X}}}$$ 4. **Calculate probabilities:** (a) For $\bar{X} < 47.6$: $$Z=\frac{47.6-48.0}{0.24} = \frac{-0.4}{0.24} = -1.67$$ Using standard normal tables or calculator: $$P(Z < -1.67) \approx 0.0475$$ (b) For $\bar{X} > 47.8$: $$Z=\frac{47.8-48.0}{0.24} = \frac{-0.2}{0.24} = -0.83$$ $$P(Z > -0.83) = 1 - P(Z \leq -0.83) = 1 - 0.2033 = 0.7967$$ (c) For $47.5 < \bar{X} < 48.5$: Calculate $Z$ for both bounds: $$Z_1=\frac{47.5-48.0}{0.24} = -2.08$$ $$Z_2=\frac{48.5-48.0}{0.24} = 2.08$$ Probability: $$P(-2.08 < Z < 2.08) = P(Z < 2.08) - P(Z < -2.08) = 0.9812 - 0.0188 = 0.9624$$ **Final answers:** (a) $0.0475$ (b) $0.7967$ (c) $0.9624$