1. **State the problem:** We have a random variable (RV) with a normal distribution, mean $\mu = 62.4$, and we want to find the standard deviation $\sigma$ given that $P(X > 79.2) = 0.20$. 2. **Recall the normal distribution properties:** For a normal distribution $X \sim N(\mu, \sigma^2)$, the standardized variable $Z = \frac{X - \mu}{\sigma}$ follows the standard normal distribution $N(0,1)$. 3. **Translate the probability:** We want $P(X > 79.2) = 0.20$, which is equivalent to $P\left(Z > \frac{79.2 - 62.4}{\sigma}\right) = 0.20$. 4. **Find the z-score corresponding to the upper tail probability 0.20:** From standard normal tables or using inverse normal functions, $P(Z > z) = 0.20$ implies $z \approx 0.8416$. 5. **Set up the equation:** $$ \frac{79.2 - 62.4}{\sigma} = 0.8416 $$ 6. **Solve for $\sigma$:** $$ \sigma = \frac{79.2 - 62.4}{0.8416} = \frac{16.8}{0.8416} \approx 19.95 $$ 7. **Interpretation:** The standard deviation $\sigma$ is approximately 19.95. **Final answer:** $\boxed{19.95}$