1. **Problem statement:** We have exam scores normally distributed with mean $\mu=70$ and standard deviation $\sigma=5$. We want to find the proportion of scores between 68 and 73. 2. **Formula and rules:** For a normal distribution, we use the standard normal variable $Z=\frac{X-\mu}{\sigma}$ to find probabilities. The proportion between two values $a$ and $b$ is $P(a < X < b) = P\left(\frac{a-\mu}{\sigma} < Z < \frac{b-\mu}{\sigma}\right)$. 3. **Calculate Z-scores:** $$Z_1 = \frac{68-70}{5} = \frac{\cancel{68-70}}{5} = \frac{-2}{5} = -0.4$$ $$Z_2 = \frac{73-70}{5} = \frac{\cancel{73-70}}{5} = \frac{3}{5} = 0.6$$ 4. **Find probabilities from Z-table or calculator:** $$P(Z < -0.4) = 0.3446$$ $$P(Z < 0.6) = 0.7257$$ 5. **Calculate proportion between 68 and 73:** $$P(-0.4 < Z < 0.6) = P(Z < 0.6) - P(Z < -0.4) = 0.7257 - 0.3446 = 0.3811$$ 6. **Final answer:** Approximately **0.3811** or 38.11% of exam scores lie between 68 and 73 points.