1. **Problem statement:** Calculate the probability that a sample mean is within $200 of the population mean $\mu=16642$ for sample sizes $n=30, 50, 100, 400$ given population standard deviation $\sigma=2400$. 2. **Formula and explanation:** The sampling distribution of the sample mean $\bar{x}$ is approximately normal with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$. We want $P(|\bar{x} - \mu| < 200)$. This is equivalent to $P(-200 < \bar{x} - \mu < 200)$. Standardize using $Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}$: $$P\left(-\frac{200}{\sigma/\sqrt{n}} < Z < \frac{200}{\sigma/\sqrt{n}}\right) = P(-z < Z < z)$$ where $$z = \frac{200}{\sigma/\sqrt{n}} = \frac{200}{2400/\sqrt{n}} = \frac{200\sqrt{n}}{2400} = \frac{\sqrt{n}}{12}$$ 3. **Calculate $z$ for each $n$:** - For $n=30$: $$z = \frac{\sqrt{30}}{12} = \frac{5.477}{12} \approx 0.456$$ - For $n=50$: $$z = \frac{\sqrt{50}}{12} = \frac{7.071}{12} \approx 0.589$$ - For $n=100$: $$z = \frac{\sqrt{100}}{12} = \frac{10}{12} = 0.833$$ - For $n=400$: $$z = \frac{\sqrt{400}}{12} = \frac{20}{12} = 1.667$$ 4. **Find probabilities using standard normal distribution:** $$P(-z < Z < z) = 2\Phi(z) - 1$$ where $\Phi(z)$ is the cumulative distribution function (CDF) of the standard normal. Using standard normal tables or calculator: - $\Phi(0.456) \approx 0.675$ - $\Phi(0.589) \approx 0.722$ - $\Phi(0.833) \approx 0.797$ - $\Phi(1.667) \approx 0.952$ 5. **Calculate final probabilities:** - For $n=30$: $$P = 2(0.675) - 1 = 0.35$$ - For $n=50$: $$P = 2(0.722) - 1 = 0.444$$ - For $n=100$: $$P = 2(0.797) - 1 = 0.594$$ - For $n=400$: $$P = 2(0.952) - 1 = 0.904$$ 6. **Answer to part (2):** A larger sample size decreases the standard error $\frac{\sigma}{\sqrt{n}}$, which makes the sampling distribution narrower. This increases the probability that the sample mean is close to the population mean, improving the accuracy and reliability of the estimate. **Final answers:** - $n=30$: Probability $\approx 0.35$ - $n=50$: Probability $\approx 0.44$ - $n=100$: Probability $\approx 0.59$ - $n=400$: Probability $\approx 0.90$ - Larger sample size advantage: reduces variability of sample mean, increasing estimation precision.