1. **Problem statement:** Find the point estimate and 90% confidence interval for the mean price of all such college textbooks given a sample mean of 145, population standard deviation 35, and sample size 25. 2. **Point estimate:** The point estimate of the population mean $\mu$ is the sample mean $\bar{x}$. $$\text{Point estimate} = \bar{x} = 145$$ 3. **Confidence interval formula:** Since the population standard deviation $\sigma$ is known and the population is normal, use the z-distribution: $$\bar{x} \pm z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}$$ where $z_{\alpha/2}$ is the critical z-value for the confidence level, $\sigma=35$, and $n=25$. 4. **Find critical z-value for 90% confidence:** $$\alpha = 1 - 0.90 = 0.10$$ $$\alpha/2 = 0.05$$ From z-tables, $z_{0.05} = 1.645$. 5. **Calculate margin of error:** $$ME = 1.645 \times \frac{35}{\sqrt{25}} = 1.645 \times \frac{35}{5} = 1.645 \times 7 = 11.515$$ 6. **Construct confidence interval:** $$145 \pm 11.515 = (145 - 11.515, 145 + 11.515) = (133.485, 156.515)$$ 7. **Interpretation:** We are 90% confident that the true mean price of all such college textbooks lies between 133.485 and 156.515. **Final answers:** - Point estimate: $145$ - 90% confidence interval: $\boxed{(133.485, 156.515)}$