1. **Problem statement:** We want to find the probability that a randomly chosen battery lasts at least 135 hours, given that the battery life is normally distributed with mean $\mu = 160$ hours and standard deviation $\sigma = 30$ hours. 2. **Formula and rules:** For a normal distribution, the probability that a value $X$ is greater than or equal to some value $x$ is given by: $$P(X \geq x) = 1 - P(X < x) = 1 - \Phi\left(\frac{x - \mu}{\sigma}\right)$$ where $\Phi$ is the cumulative distribution function (CDF) of the standard normal distribution. 3. **Calculate the z-score:** $$z = \frac{135 - 160}{30} = \frac{-25}{30} = -\frac{5}{6} \approx -0.8333$$ 4. **Find the cumulative probability for $z$:** Using standard normal tables or a calculator, find $\Phi(-0.8333)$. Since $\Phi(-z) = 1 - \Phi(z)$, we first find $\Phi(0.8333) \approx 0.7977$. Therefore, $$\Phi(-0.8333) = 1 - 0.7977 = 0.2023$$ 5. **Calculate the desired probability:** $$P(X \geq 135) = 1 - \Phi\left(\frac{135 - 160}{30}\right) = 1 - 0.2023 = 0.7977$$ **Final answer:** The probability that a randomly chosen battery lasts at least 135 hours is approximately **0.798** or **79.8%**.