1. **Problem statement:** We have rods with lengths normally distributed with standard deviation $\sigma = 0.005$ cm. The specification requires lengths between 42.49 cm and 42.51 cm. **Part (a):** Mean $\mu = 42.501$ cm. Find the percentage of rods rejected (lengths outside 42.49 to 42.51). 2. **Formula and rules:** The length $X$ follows $N(\mu, \sigma^2)$. We find rejection probability as $P(X < 42.49) + P(X > 42.51)$. Use the standard normal variable $Z = \frac{X - \mu}{\sigma}$. 3. **Calculations for (a):** Calculate $Z$-scores: $$Z_1 = \frac{42.49 - 42.501}{0.005} = \frac{-0.011}{0.005} = -2.2$$ $$Z_2 = \frac{42.51 - 42.501}{0.005} = \frac{0.009}{0.005} = 1.8$$ 4. **Find probabilities:** Using standard normal tables or calculator: $$P(Z < -2.2) = 0.0139$$ $$P(Z > 1.8) = 1 - P(Z < 1.8) = 1 - 0.9641 = 0.0359$$ 5. **Total rejection percentage:** $$P_{reject} = 0.0139 + 0.0359 = 0.0498 = 4.98\%$$ --- 6. **Part (b):** Find mean $\mu$ so that only 3 rods in 1000 have length less than 42.494 cm. 7. **Set up:** $$P(X < 42.494) = 0.003$$ Using $Z = \frac{X - \mu}{\sigma}$, $$P\left(Z < \frac{42.494 - \mu}{0.005}\right) = 0.003$$ 8. **Find $Z$-value for 0.003:** From standard normal tables, $P(Z < -2.75) \approx 0.003$. 9. **Solve for $\mu$:** $$\frac{42.494 - \mu}{0.005} = -2.75$$ Multiply both sides by 0.005: $$42.494 - \mu = -2.75 \times 0.005 = -0.01375$$ Rewrite: $$\mu = 42.494 + 0.01375 = 42.50775$$ --- 10. **Part (c):** To reduce total rejects (both undersize and oversize), set mean $\mu$ at the midpoint of the specification limits: $$\mu = \frac{42.49 + 42.51}{2} = 42.5$$ This centers the distribution, minimizing rejects on both sides. --- **Final answers:** - (a) Rejection percentage $\approx 4.98\%$ - (b) Mean setting $\mu \approx 42.50775$ cm - (c) Mean setting to reduce rejects $\mu = 42.5$ cm