1. **Problem Statement:** Find the probability $P(34.7 < X < 63.5)$ for a normally distributed random variable $X$ with mean $\mu = 56.3$ and standard deviation $\sigma = 7.2$. 2. **Formula and Important Rules:** The probability for a range in a normal distribution is found using the standard normal variable $Z = \frac{X - \mu}{\sigma}$. The probability $P(a < X < b)$ is equivalent to $P\left(\frac{a - \mu}{\sigma} < Z < \frac{b - \mu}{\sigma}\right)$. 3. **Calculate the Z-scores:** $$Z_1 = \frac{34.7 - 56.3}{7.2} = \frac{-21.6}{7.2} = -3$$ $$Z_2 = \frac{63.5 - 56.3}{7.2} = \frac{7.2}{7.2} = 1$$ 4. **Find the probabilities from the standard normal table:** - $P(Z < -3) \approx 0.0013$ - $P(Z < 1) \approx 0.8413$ 5. **Calculate the probability for the range:** $$P(34.7 < X < 63.5) = P(-3 < Z < 1) = P(Z < 1) - P(Z < -3) = 0.8413 - 0.0013 = 0.84$$ 6. **Interpretation:** There is approximately an 84% chance that the value of $X$ lies between 34.7 and 63.5. **Final answer:** $$P(34.7 < X < 63.5) \approx 0.84$$