1. **Problem statement:** Find the $x$-value in a normal distribution such that 12% of the cases lie above it. 2. **Understanding the problem:** In a normal distribution, the total area under the curve is 1 (or 100%). The problem asks for the $x$-value where the area to the right (above) is 0.12. 3. **Formula and approach:** We use the cumulative distribution function (CDF) of the standard normal distribution, denoted as $\Phi(x)$, which gives the area to the left of $x$. Since 12% is above $x$, the area to the left is $1 - 0.12 = 0.88$. 4. **Find the z-score:** We need to find $z$ such that $$\Phi(z) = 0.88$$ Using standard normal distribution tables or a calculator, the z-score for 0.88 is approximately $$z \approx 1.175$$ 5. **Interpretation:** This means that the $x$-value corresponding to 12% of the cases above it is about 1.175 standard deviations above the mean. 6. **If the distribution has mean $\mu$ and standard deviation $\sigma$, then:** $$x = \mu + z \sigma = \mu + 1.175 \sigma$$ This is the general formula to find the $x$-value for the given percentile in any normal distribution. **Final answer:** The $x$-value with 12% of the cases above it corresponds to a z-score of approximately $1.175$ in the standard normal distribution.