1. **Problem statement:** We have a population with variance $\sigma^2 = 1$ (so standard deviation $\sigma = 1$). A sample of size 10 is taken: $-0.7518, 1.4977, 1.7274, 1.8371, -0.3193, 0.7773, 1.0900, 0.7659, 0.3623, 1.7205$. We want to find a 95% confidence interval for the population mean $\mu$. 2. **Formula for confidence interval when population variance is known:** $$\bar{x} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$$ where $\bar{x}$ is the sample mean, $z_{\alpha/2}$ is the critical z-value for confidence level $1-\alpha$, $\sigma$ is population standard deviation, and $n$ is sample size. 3. **Calculate sample mean $\bar{x}$:** $$\bar{x} = \frac{\sum x_i}{n} = \frac{-0.7518 + 1.4977 + 1.7274 + 1.8371 - 0.3193 + 0.7773 + 1.0900 + 0.7659 + 0.3623 + 1.7205}{10}$$ Calculate sum: $$-0.7518 + 1.4977 = 0.7459$$ $$0.7459 + 1.7274 = 2.4733$$ $$2.4733 + 1.8371 = 4.3104$$ $$4.3104 - 0.3193 = 3.9911$$ $$3.9911 + 0.7773 = 4.7684$$ $$4.7684 + 1.0900 = 5.8584$$ $$5.8584 + 0.7659 = 6.6243$$ $$6.6243 + 0.3623 = 6.9866$$ $$6.9866 + 1.7205 = 8.7071$$ So, $$\bar{x} = \frac{8.7071}{10} = 0.87071$$ 4. **Find critical z-value for 95% confidence:** For 95% confidence, $\alpha = 0.05$, so $z_{\alpha/2} = z_{0.025} \approx 1.96$. 5. **Calculate margin of error:** $$E = z_{\alpha/2} \frac{\sigma}{\sqrt{n}} = 1.96 \times \frac{1}{\sqrt{10}} = 1.96 \times 0.3162 = 0.6195$$ 6. **Construct confidence interval:** $$\left(\bar{x} - E, \bar{x} + E\right) = (0.87071 - 0.6195, 0.87071 + 0.6195) = (0.2512, 1.4902)$$ --- 7. **Part b: Find sample size for 95% confidence interval with total length 0.5** The length $L$ of the confidence interval is: $$L = 2E = 2 z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$$ We want $L = 0.5$, so: $$0.5 = 2 \times 1.96 \times \frac{1}{\sqrt{n}}$$ Solve for $n$: $$\frac{0.5}{2 \times 1.96} = \frac{1}{\sqrt{n}}$$ $$\frac{0.5}{3.92} = \frac{1}{\sqrt{n}}$$ $$0.12755 = \frac{1}{\sqrt{n}}$$ Invert both sides: $$\sqrt{n} = \frac{1}{0.12755} = 7.843$$ Square both sides: $$n = 7.843^2 = 61.5$$ Since sample size must be an integer, round up: $$n = 62$$ **Final answers:** - 95% confidence interval for $\mu$ is approximately $$(0.251, 1.490)$$ - Required sample size for interval length 0.5 is $$62$$