1. **Problem statement:** We have a sample of size $n=64$ from a normal distribution with variance $\sigma^2=4$. The $100(1-\alpha)\%$ confidence interval for the mean $\mu$ is given as $[10.4175, 11.5825]$. We need to find the $100(1-\alpha)\%$ lower confidence bound for $\mu$. 2. **Recall the confidence interval formula for the mean when 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 value from the standard normal distribution, $\sigma$ is the population standard deviation, and $n$ is the sample size. 3. **Find the sample mean $\bar{x}$:** The confidence interval is symmetric around $\bar{x}$, so $$\bar{x} = \frac{10.4175 + 11.5825}{2} = \frac{22}{2} = 11$$ 4. **Find the margin of error (ME):** $$ME = 11.5825 - 11 = 0.5825$$ 5. **Calculate the critical value $z_{\alpha/2}$:** Given $ME = z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$, we solve for $z_{\alpha/2}$: $$z_{\alpha/2} = \frac{ME \times \sqrt{n}}{\sigma} = \frac{0.5825 \times 8}{2} = \frac{4.66}{2} = 2.33$$ 6. **Lower confidence bound formula:** The $100(1-\alpha)\%$ lower confidence bound is $$\bar{x} - z_{\alpha} \frac{\sigma}{\sqrt{n}}$$ Note that for a one-sided bound, the critical value is $z_{\alpha}$ (not $z_{\alpha/2}$). 7. **Find $z_{\alpha}$:** Since $z_{\alpha/2} = 2.33$, this corresponds approximately to $\alpha/2 = 0.01$, so $\alpha = 0.02$. For the lower bound, $z_{\alpha}$ corresponds to $\alpha = 0.02$, so $$z_{\alpha} = z_{0.02} = 2.05$$ (approximate value from standard normal tables). 8. **Calculate the lower confidence bound:** $$11 - 2.05 \times \frac{2}{8} = 11 - 2.05 \times 0.25 = 11 - 0.5125 = 10.4875$$ **Final answer:** The $100(1-\alpha)\%$ lower confidence bound for $\mu$ is **10.4875**.