1. **State the problem:** We have a normal distribution for replacement times with mean $\mu = 12.6$ years and standard deviation $\sigma = 1.5$ years. We want to find: - The probability $P(X < 8.1)$. - The warranty time $w$ such that only 4.4% of pieces are replaced before $w$ years. 2. **Formula and rules:** For a normal variable $X$, the standardized z-score is: $$z = \frac{X - \mu}{\sigma}$$ The probability $P(X < x)$ corresponds to $P(Z < z)$ where $Z$ is standard normal. 3. **Calculate $P(X < 8.1)$:** $$z = \frac{8.1 - 12.6}{1.5} = \frac{-4.5}{1.5} = -3.0$$ Using standard normal tables or a calculator: $$P(Z < -3.0) \approx 0.0013$$ So, $$P(X < 8.1) = 0.0013$$ 4. **Find warranty time $w$ for 4.4% replacement:** We want $P(X < w) = 0.044$. Find $z$ such that $P(Z < z) = 0.044$. From z-tables or inverse normal function: $$z \approx -1.70$$ Convert back to $w$: $$w = \mu + z \sigma = 12.6 + (-1.70)(1.5) = 12.6 - 2.55 = 10.1$$ 5. **Final answers:** - $P(X < 8.1) = 0.0013$ - Warranty time $w = 10.1$ years These answers are rounded as requested.