1. **Stating the problem:** We have a normal distribution $X \sim N(400, 64)$ with mean $\mu = 400$ and variance $\sigma^2 = 64$, so standard deviation $\sigma = 8$. We want to find values $u$ and $L$ such that: $$P(X < u) = 0.75 \quad \text{and} \quad P(X < L) = 0.25$$ 2. **Formula and explanation:** For a normal distribution, the cumulative distribution function (CDF) is: $$P(X < x) = \Phi\left(\frac{x - \mu}{\sigma}\right)$$ where $\Phi$ is the standard normal CDF. To find $u$ and $L$, we use the inverse CDF (quantile function) of the standard normal distribution: $$z_u = \Phi^{-1}(0.75), \quad z_L = \Phi^{-1}(0.25)$$ Then convert back to $X$ scale: $$u = \mu + z_u \sigma, \quad L = \mu + z_L \sigma$$ 3. **Find $z_u$ and $z_L$:** From standard normal tables or calculator: $$z_u = \Phi^{-1}(0.75) = 0.674$$ $$z_L = \Phi^{-1}(0.25) = -0.674$$ 4. **Calculate $u$ and $L$:** $$u = 400 + 0.674 \times 8 = 400 + 5.392 = 405.392$$ $$L = 400 + (-0.674) \times 8 = 400 - 5.392 = 394.608$$ 5. **Interpretation:** The value $u = 405.392$ is the 75th percentile, and $L = 394.608$ is the 25th percentile of the distribution. **Final answer:** $$u = 405.39, \quad L = 394.61$$