1. **State the problem:** We have a normal distribution with mean $\mu = 61.6$ and unknown standard deviation $\sigma$. We know that 75% of the area lies to the right of 60, and we need to find $\sigma$. 2. **Recall the properties of the normal distribution:** The total area under the normal curve is 1. The area to the right of 60 is 0.75, so the area to the left of 60 is 0.25. 3. **Convert the problem to a standard normal distribution problem:** Define the standard normal variable $Z = \frac{X - \mu}{\sigma}$. We want to find $\sigma$ such that $$P(X > 60) = 0.75 \implies P(X \leq 60) = 0.25$$ This means $$P\left(Z \leq \frac{60 - 61.6}{\sigma}\right) = 0.25$$ 4. **Find the z-score corresponding to the cumulative probability 0.25:** From standard normal tables or using inverse normal function, $$z_{0.25} = -0.674$$ 5. **Set up the equation:** $$\frac{60 - 61.6}{\sigma} = -0.674$$ 6. **Solve for $\sigma$:** $$\frac{60 - 61.6}{\sigma} = -0.674 \implies \cancel{\sigma} \times -0.674 = 60 - 61.6 \implies -0.674 \sigma = -1.6$$ $$\implies \sigma = \frac{-1.6}{-0.674} = \frac{1.6}{0.674} \approx 2.374$$ **Final answer:** $$\boxed{\sigma \approx 2.37}$$