1. The problem asks to find the z-score such that the probability that a standard normal variable $Z$ is greater than this z-score is 0.3186. 2. We use the property of the standard normal distribution: $P(Z > z) = 1 - P(Z \leq z)$. 3. Given $P(Z > z) = 0.3186$, then $P(Z \leq z) = 1 - 0.3186 = 0.6814$. 4. We need to find the z-score $z$ such that the cumulative distribution function (CDF) $\Phi(z) = 0.6814$. 5. Using standard normal distribution tables or a calculator, $\Phi(0.47) \approx 0.6814$. 6. Therefore, the z-score corresponding to $P(Z > z) = 0.3186$ is approximately $z = 0.47$. Final answer: $z = 0.47$