1. **State the problem:** We have a metal component's tensile strength $X$ that is normally distributed with mean $\mu = 10000$ and standard deviation $\sigma = 100$. We want to find the proportion of components with tensile strength exceeding 10150. 2. **Convert the problem to a standard normal distribution:** We calculate the $z$-score for 10150 using the formula: $$ z = \frac{X - \mu}{\sigma} = \frac{10150 - 10000}{100} = \frac{150}{100} = 1.5 $$ 3. **Find the probability:** The proportion exceeding 10150 is $P(X > 10150) = P(Z > 1.5)$ where $Z$ is standard normal. 4. **Use standard normal tables or a calculator:** $$ P(Z > 1.5) = 1 - P(Z \leq 1.5) $$ From standard normal tables, $P(Z \leq 1.5) \approx 0.9332$. 5. **Calculate the final answer:** $$ P(Z > 1.5) = 1 - 0.9332 = 0.0668 $$ 6. **Interpretation:** Approximately 6.68% of the components have tensile strength exceeding 10150 kilograms per square centimeter. Note: Since measurements are recorded to the nearest 50, the rounding does not affect the calculation of the proportion exceeding 10150 significantly.