1. **State the problem:** We want to find the percentage of light bulbs that last less than 1160 hours given a normal distribution with mean $\mu = 1200$ hours and standard deviation $\sigma = 60$ hours. 2. **Formula and rules:** For a normal distribution, we convert the value to a standard normal variable $Z$ using the formula: $$Z = \frac{X - \mu}{\sigma}$$ where $X$ is the value of interest. 3. **Calculate the Z-score:** $$Z = \frac{1160 - 1200}{60} = \frac{-40}{60} = -\frac{2}{3} \approx -0.6667$$ 4. **Find the cumulative probability:** We look up the cumulative probability for $Z = -0.6667$ in the standard normal distribution table or use a calculator. 5. **Using symmetry and tables:** The cumulative probability for $Z = -0.6667$ is approximately 0.2525. 6. **Interpretation:** This means about 25.25% of light bulbs last less than 1160 hours. 7. **Final answer:** To the nearest tenth, the percentage is **25.3%**.