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 outside the range 9800 to 10200 inclusive, i.e., the scrap proportion. 2. **Convert the limits to standard normal variable $Z$:** $$Z = \frac{X - \mu}{\sigma}$$ Calculate $Z$ for the lower limit 9800: $$Z_{low} = \frac{9800 - 10000}{100} = \frac{-200}{100} = -2$$ Calculate $Z$ for the upper limit 10200: $$Z_{high} = \frac{10200 - 10000}{100} = \frac{200}{100} = 2$$ 3. **Find the probability that $Z$ lies between $-2$ and $2$:** Using standard normal distribution tables or a calculator, $$P(-2 \leq Z \leq 2) = P(Z \leq 2) - P(Z \leq -2)$$ From tables, $P(Z \leq 2) \approx 0.9772$ and $P(Z \leq -2) \approx 0.0228$ So, $$P(-2 \leq Z \leq 2) = 0.9772 - 0.0228 = 0.9544$$ 4. **Calculate the scrap proportion:** The scrap proportion is the probability that $X$ is outside the range 9800 to 10200, which is $$1 - 0.9544 = 0.0456$$ 5. **Adjust for rounding to nearest 50:** Since measurements are recorded to the nearest 50, the effective limits are 9800 and 10200 inclusive, so the scrap proportion remains approximately $0.0456$. 6. **Compare with given options:** Closest option to $0.0456$ is d. 0.0542. **Final answer:** d. 0.0542