1. **State the problem:** We want to find the probability that a randomly chosen battery lasts at least 135 hours, given 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 a certain value $x$ is found using the standard normal variable $Z$: $$Z = \frac{X - \mu}{\sigma}$$ We then use standard normal distribution tables or a calculator to find $P(Z \geq z)$. 3. **Calculate the Z-score:** $$Z = \frac{135 - 160}{30} = \frac{-25}{30} = -0.8333$$ 4. **Find the probability:** We want $P(X \geq 135) = P(Z \geq -0.8333)$. Since the normal distribution is symmetric, $$P(Z \geq -0.8333) = 1 - P(Z < -0.8333) = P(Z > -0.8333) = 1 - \Phi(-0.8333)$$ But $P(Z \geq -0.8333)$ is the same as $P(Z \leq 0.8333)$ because of symmetry. Using standard normal tables or a calculator, $$\Phi(0.8333) \approx 0.7967$$ 5. **Final answer:** The probability that a battery lasts at least 135 hours is approximately **0.7967**.