1. **Problem statement:** We have a normally distributed random variable $X$ representing the time for super glue to set, with mean $\mu = 30$ seconds. We want to find the standard deviation $\sigma$ given that $P(X > 39.2) = 0.20$. 2. **Formula and concepts:** For a normal distribution, the standardized variable $Z = \frac{X - \mu}{\sigma}$ follows a standard normal distribution $N(0,1)$. 3. We know $P(X > 39.2) = 0.20$, so equivalently $P\left(Z > \frac{39.2 - 30}{\sigma}\right) = 0.20$. 4. From standard normal tables or using inverse normal function, $P(Z > z) = 0.20$ corresponds to $z \approx 0.8416$. 5. Set up the equation: $$\frac{39.2 - 30}{\sigma} = 0.8416$$ 6. Solve for $\sigma$: $$\sigma = \frac{39.2 - 30}{0.8416} = \frac{9.2}{0.8416} \approx 10.92$$ 7. **Answer:** The standard deviation $\sigma$ is approximately $10.92$ seconds.