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 $\mu$ and standard deviation $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$. 3. Calculate the standard error: $$\sigma_{\bar{x}} = \frac{2.4}{\sqrt{100}} = \frac{2.4}{10} = 0.24$$ 4. Convert each sample mean to a $z$-score using: $$z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}}$$ 5. (a) For $\bar{x} = 47.6$: $$z = \frac{47.6 - 48.0}{0.24} = \frac{-0.4}{0.24} = -1.67$$ Find $P(z < -1.67)$ from standard normal table or calculator: $$P(z < -1.67) \approx 0.0475$$ 6. (b) For $\bar{x} = 47.8$: $$z = \frac{47.8 - 48.0}{0.24} = \frac{-0.2}{0.24} = -0.83$$ Find $P(z > -0.83)$: $$P(z > -0.83) = 1 - P(z < -0.83)$$ From table, $P(z < -0.83) \approx 0.2033$, so $$P(z > -0.83) = 1 - 0.2033 = 0.7967$$ 7. (c) For $47.5 < \bar{x} < 48.5$: Calculate $z$ for 47.5: $$z_1 = \frac{47.5 - 48.0}{0.24} = \frac{-0.5}{0.24} = -2.08$$ Calculate $z$ for 48.5: $$z_2 = \frac{48.5 - 48.0}{0.24} = \frac{0.5}{0.24} = 2.08$$ Find $P(-2.08 < z < 2.08)$: From table, $P(z < 2.08) \approx 0.9812$ and $P(z < -2.08) \approx 0.0188$ So, $$P(-2.08 < z < 2.08) = 0.9812 - 0.0188 = 0.9624$$ **Final answers:** (a) $0.0475$ (b) $0.7967$ (c) $0.9624$