The yield strength of an aerospace aluminum alloy is normally distributed with a mean $\mu = 450\text{ MPa}$ and standard deviation $\sigma = 20\text{ MPa}$. Using the standard normal distribution table where the cumulative probability $\Phi(2.0) \approx 0.9772$, the probability that a randomly tested specimen has a yield strength less than $410\text{ MPa}$ is closest to:

(A) 0.0228

(B) 0.0456

(C) 0.9772

(D) 0.0013

Answer Key:

Verified in official key

### 1. Strategy & Theoretical Principles Transform the normal variable $X$ to the standard normal score $Z$: $$Z = \frac{X - \mu}{\sigma}$$ By symmetry of the standard normal distribution: $$P(Z < -z) = P(Z > z) = 1 - \Phi(z)$$ ### 2. Analytical Formulation & Calculation Given: - $\mu = 450\text{ MPa}$ - $\sigma = 20\text{ MPa}$ - $X = 410\text{ MPa}$ $$Z = \frac{410 - 450}{20} = \frac{-40}{20} = -2.0$$ Evaluating the cumulative tail probability: $$P(X < 410) = P(Z < -2.0) = 1 - \Phi(2.0) = 1 - 0.9772 = 0.0228$$ ### 3. Distractor Forensics - **Option A (0.0228):** Correct tail probability ($2.28\%$). - **Option B (0.0456):** Two-tailed probability ($2 \times 0.0228$). - **Option C (0.9772):** $P(X < 490)$. ### 4. Exam Trap & Standard Reference - **Trap Alert:** Distinguish one-tailed lower probability from two-tailed rejection regions. - **Standard Reference:** Kreyszig, E., *Advanced Engineering Mathematics*, Chapter 24.

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