1. **State the problem:** We have 500 children with heights averaging 110 cm and a standard deviation of 6 cm. We want to find how many children belong to the upper 15% of the height distribution. 2. **Formula and concept:** We assume the heights are normally distributed. To find the number of children in the upper 15%, we first find the height cutoff $x$ such that $P(X > x) = 0.15$. 3. **Find the z-score for the upper 15%:** The upper 15% corresponds to the 85th percentile. Using standard normal distribution tables or a calculator, the z-score for 0.85 is approximately $z = 1.04$. 4. **Convert z-score to height cutoff:** Use the formula $$x = \mu + z \sigma$$ where $\mu = 110$, $\sigma = 6$, and $z = 1.04$. Calculate: $$x = 110 + 1.04 \times 6 = 110 + 6.24 = 116.24 \text{ cm}$$ 5. **Find the number of children above this height:** Since 15% of 500 children are above this height, $$\text{Number} = 0.15 \times 500 = 75$$ **Final answer:** 75 children belong to the upper 15% of the group.