1. **Problem statement:** A company completes projects with mean duration $\mu=120$ days and standard deviation $\sigma=25$ days. A sample of $n=30$ projects is taken. We want to describe the sampling distribution of the sample mean and find the probability that the sample mean is between 115 and 125 days. 2. **Sampling distribution of the sample mean:** According to the Central Limit Theorem, the sampling distribution of the sample mean $\overline{X}$ for sample size $n$ is approximately normal with mean $\mu_{\overline{X}}=\mu$ and standard deviation $\sigma_{\overline{X}}=\frac{\sigma}{\sqrt{n}}$. 3. **Calculate the standard deviation of the sample mean:** $$\sigma_{\overline{X}}=\frac{25}{\sqrt{30}}=\frac{25}{5.477}=4.564$$ 4. **Describe the sampling distribution:** $$\overline{X} \sim N(120, 4.564)$$ 5. **Find the probability $P(115 < \overline{X} < 125)$:** Convert to standard normal variable $Z$: $$Z=\frac{\overline{X}-\mu}{\sigma_{\overline{X}}}$$ Calculate $Z$ values: $$Z_1=\frac{115-120}{4.564} = \frac{-5}{4.564} = -1.096$$ $$Z_2=\frac{125-120}{4.564} = \frac{5}{4.564} = 1.096$$ 6. **Use standard normal distribution table or calculator:** $$P(-1.096 < Z < 1.096) = P(Z < 1.096) - P(Z < -1.096)$$ From standard normal tables: $$P(Z < 1.096) \approx 0.8637$$ $$P(Z < -1.096) = 1 - P(Z < 1.096) = 1 - 0.8637 = 0.1363$$ So, $$P(-1.096 < Z < 1.096) = 0.8637 - 0.1363 = 0.7274$$ **Final answer:** The probability that the sample mean is between 115 and 125 days is approximately $0.727$ or 72.7%.