1. **Problem statement:** Given that 7.7% of passengers fly first-class, and a random sample of size 158 is taken, we want to describe the probability distribution of the sample proportion $\hat{p}$ using the Central Limit Theorem (CLT). 2. **Formula and explanation:** The CLT states that for large samples, the sampling distribution of the sample proportion $\hat{p}$ is approximately normal with mean $\mu_{\hat{p}} = p$ and standard deviation $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$, where $p$ is the population proportion and $n$ is the sample size. 3. **Calculate parameters:** - Population proportion $p = 0.077$ - Sample size $n = 158$ Mean: $$\mu_{\hat{p}} = p = 0.0770$$ Standard deviation: $$\sigma_{\hat{p}} = \sqrt{\frac{0.077 \times (1 - 0.077)}{158}} = \sqrt{\frac{0.077 \times 0.923}{158}} = \sqrt{\frac{0.071071}{158}} = \sqrt{0.0004498} = 0.0212$$ 4. **Distribution:** $$\hat{p} \sim N(0.0770, 0.0212)$$ 5. **Find the probability that between 3% and 5% of the sample fly first-class:** We want $P(0.03 \leq \hat{p} \leq 0.05)$. Convert to $z$-scores: $$z_1 = \frac{0.03 - 0.077}{0.0212} = \frac{-0.047}{0.0212} = -2.2170$$ $$z_2 = \frac{0.05 - 0.077}{0.0212} = \frac{-0.027}{0.0212} = -1.2736$$ Using standard normal tables or a calculator: $$P(z_1 \leq Z \leq z_2) = P(-2.2170 \leq Z \leq -1.2736) = \Phi(-1.2736) - \Phi(-2.2170)$$ From standard normal CDF values: $$\Phi(-1.2736) = 0.1017$$ $$\Phi(-2.2170) = 0.0133$$ Therefore: $$P = 0.1017 - 0.0133 = 0.0884$$ **Final answers:** - Distribution: $N(0.0770, 0.0212)$ - Probability between 3% and 5%: $0.0884$