1. **State the problem:** We have a normal distribution $X \sim N(400,64)$ with mean $\mu=400$ and variance $\sigma^2=64$, so standard deviation $\sigma=8$. 2. We want to find values $L$ and $U$ such that: - $P(X < L) = 0.25$ - $P(X < U) = 0.75$ These correspond to the 25th percentile (lower quartile) and 75th percentile (upper quartile) of the distribution. 3. **Formula and rules:** For a normal distribution, the standardized variable is $$Z = \frac{X - \mu}{\sigma}$$ which follows $N(0,1)$. We find $z$-scores $z_L$ and $z_U$ such that: $$P(Z < z_L) = 0.25, \quad P(Z < z_U) = 0.75$$ From standard normal tables or symmetry: $$z_L = -0.6745, \quad z_U = 0.6745$$ 4. **Calculate $L$ and $U$:** $$L = \mu + z_L \sigma = 400 + (-0.6745)(8) = 400 - 5.396 = 394.604$$ $$U = \mu + z_U \sigma = 400 + 0.6745 \times 8 = 400 + 5.396 = 405.396$$ 5. **Interpretation:** The middle 50% of the data lies between approximately 394.6 and 405.4. **Final answer:** $$L \approx 394.6, \quad U \approx 405.4$$