1. **Problem a:** Find $k$ such that $P(\mu - k\sigma \leq X \leq \mu + k\sigma) = 0.8502$ for a normal distribution. 2. The probability that $X$ lies within $k$ standard deviations of the mean is given by: $$P(\mu - k\sigma \leq X \leq \mu + k\sigma) = P(-k \leq Z \leq k) = 2\Phi(k) - 1$$ where $Z = \frac{X - \mu}{\sigma}$ is the standard normal variable and $\Phi$ is the CDF of $Z$. 3. Set up the equation: $$2\Phi(k) - 1 = 0.8502$$ 4. Solve for $\Phi(k)$: $$\Phi(k) = \frac{0.8502 + 1}{2} = 0.9251$$ 5. Find $k$ such that $\Phi(k) = 0.9251$. Using standard normal tables or inverse CDF, $$k \approx 1.44$$ --- **Final answer:** $$k = 1.44$$