1. We are given a normal distribution for IQ scores with mean $\mu = 100$ and standard deviation $\sigma = 15$. 2. The problem states that the area to the right of the score $x$ is 0.0912. This means $P(X > x) = 0.0912$. 3. To find $x$, first find the corresponding z-value $z$ such that $P(Z > z) = 0.0912$ for the standard normal distribution. 4. Using standard normal distribution tables or a calculator, the $z$-value for the upper tail probability 0.0912 is approximately $z = 1.34$ (since $P(Z > 1.34) \approx 0.0918$ which is close, rounding is acceptable). 5. Convert the z-value to the original IQ score $x$ by using the formula: $$x = \mu + z\sigma = 100 + 1.34 \times 15 = 100 + 20.1 = 120.1$$ 6. Rounding to the nearest whole number, $x = 120$. Therefore, the indicated IQ score is **120**.