Question: The diameters of the trees in a particular region are normally distributed with a mean of $27.1$ cm and a standard deviation of $7.2$ cm. a) What is the probability that, if you randomly select one tree, it will have a diameter under $17.7$ cm? b) If you took a sample of $25$ tree diameters, what is the probability that the sample mean diameter is over $31.0$ cm? c) If you took a sample of $100$ tree diameters, what is the probability that the sample mean diameter is over $31.0$ cm?

1. **Problem Statement:** We have tree diameters normally distributed with mean $\mu = 27.1$ cm and standard deviation $\sigma = 7.2$ cm. We want to find probabilities for: a) A single tree diameter under $17.7$ cm. b) Sample mean diameter over $31.0$ cm for sample size $n=25$. c) Sample mean diameter over $31.0$ cm for sample size $n=100$. --- 2. **Formula and Important Rules:** For a normal distribution $X \sim N(\mu, \sigma^2)$, the standardized variable (Z-score) is: $$Z = \frac{X - \mu}{\sigma}$$ For sample means $\bar{X}$ of size $n$, the distribution is: $$\bar{X} \sim N\left(\mu, \frac{\sigma^2}{n}\right)$$ and the standard deviation of the sample mean (standard error) is: $$\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$$ We use the Z-score for sample means as: $$Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}}$$ Probabilities are found using the standard normal distribution tables or a calculator. --- 3. **Part a) Probability single tree diameter under $17.7$ cm:** Calculate Z: $$Z = \frac{17.7 - 27.1}{7.2} = \frac{-9.4}{7.2} = -1.3056$$ Using standard normal tables or calculator, $P(Z < -1.3056) \approx 0.0959$. --- 4. **Part b) Probability sample mean diameter over $31.0$ cm for $n=25$:** Calculate standard error: $$\sigma_{\bar{X}} = \frac{7.2}{\sqrt{25}} = \frac{7.2}{5} = 1.44$$ Calculate Z: $$Z = \frac{31.0 - 27.1}{1.44} = \frac{3.9}{1.44} = 2.7083$$ Find $P(\bar{X} > 31.0) = P(Z > 2.7083)$. From tables or calculator, $P(Z > 2.7083) = 1 - P(Z \leq 2.7083) \approx 1 - 0.9966 = 0.0034$. --- 5. **Part c) Probability sample mean diameter over $31.0$ cm for $n=100$:** Calculate standard error: $$\sigma_{\bar{X}} = \frac{7.2}{\sqrt{100}} = \frac{7.2}{10} = 0.72$$ Calculate Z: $$Z = \frac{31.0 - 27.1}{0.72} = \frac{3.9}{0.72} = 5.4167$$ Find $P(\bar{X} > 31.0) = P(Z > 5.4167)$. This is extremely small, practically zero, but from tables or calculator, approximately $0.0000$ (rounded to 4 decimals). --- **Final answers:** a) $0.0959$ b) $0.0034$ c) Approximately $0.0000$ (very close to zero)