1. **Problem statement:** We have a normal random variable $X$ with mean $\mu = 30$ and unknown variance $\sigma^2$. Given that $P(X < 35) = 0.69$, we need to find the variance $\sigma^2$. 2. **Formula and concepts:** For a normal distribution, the standardized variable $Z = \frac{X - \mu}{\sigma}$ follows the standard normal distribution $N(0,1)$. 3. Using the given probability, we write: $$P(X < 35) = P\left(Z < \frac{35 - 30}{\sigma}\right) = 0.69$$ 4. Let $z_0 = \frac{35 - 30}{\sigma} = \frac{5}{\sigma}$. We need to find $z_0$ such that $P(Z < z_0) = 0.69$. 5. From standard normal distribution tables or using an inverse normal function, $z_0 \approx 0.496$ because $\Phi(0.496) \approx 0.69$. 6. Equate and solve for $\sigma$: $$0.496 = \frac{5}{\sigma} \implies \sigma = \frac{5}{0.496} \approx 10.08$$ 7. Finally, variance is: $$\sigma^2 = (10.08)^2 \approx 101.6$$ **Answer:** The variance of $X$ is approximately $101.6$.