1. **Problem statement:** A random variable $X$ representing the life in hours of a radio tube is normally distributed with mean $\mu = 200$ and variance $\sigma^2$. We want to find the largest value of $\sigma$ such that at least 90% of the tubes have lives exceeding 150 hours. 2. **Understanding the problem:** We want $P(X > 150) \geq 0.90$. Since $X$ is normal, we use the standard normal variable $Z = \frac{X - \mu}{\sigma}$. 3. **Express the probability in terms of $Z$:** $$P(X > 150) = P\left(Z > \frac{150 - 200}{\sigma}\right) = P\left(Z > \frac{-50}{\sigma}\right)$$ 4. **Use the complement rule:** $$P\left(Z > \frac{-50}{\sigma}\right) = 1 - P\left(Z \leq \frac{-50}{\sigma}\right)$$ We want this to be at least 0.90, so: $$1 - P\left(Z \leq \frac{-50}{\sigma}\right) \geq 0.90$$ 5. **Rearranging:** $$P\left(Z \leq \frac{-50}{\sigma}\right) \leq 0.10$$ 6. **Find the z-value for 0.10 cumulative probability:** From standard normal tables or using inverse CDF, $z_{0.10} \approx -1.28155$. 7. **Set the equation:** $$\frac{-50}{\sigma} = z_{0.10} = -1.28155$$ 8. **Solve for $\sigma$:** $$\sigma = \frac{50}{1.28155} \approx 39.02$$ **Final answer:** The largest value of $\sigma$ is approximately **39.02** hours to satisfy the purchaser's requirement.