1. **Problem statement:** We have battery pack runtimes that are Normally distributed with mean $\mu=2$ hours and standard deviation $\sigma=20$ minutes = $\frac{1}{3}$ hours. 2. **Key formulas and rules:** - To find probabilities and percentiles for a Normal distribution, we use the standard normal variable $Z = \frac{X-\mu}{\sigma}$. - Use standard normal tables or a calculator to find probabilities and quantiles. 3. **Part (a): Percentage lasting longer than 3 hours** - Convert 3 hours to $Z$-score: $$Z = \frac{3 - 2}{\frac{1}{3}} = \frac{1}{\frac{1}{3}} = 3$$ - Find $P(X > 3) = P(Z > 3)$. - From standard normal tables, $P(Z > 3) \approx 0.00135$ or 0.135%. 4. **Part (b): Third quartile (75th percentile)** - The 75th percentile $Q_3$ corresponds to $Z$ such that $P(Z \leq z) = 0.75$. - From standard normal tables, $z_{0.75} \approx 0.674$. - Convert back to $X$: $$Q_3 = \mu + z_{0.75} \sigma = 2 + 0.674 \times \frac{1}{3} = 2 + 0.2247 = 2.2247 \text{ hours}$$ 5. **Part (c): Runtime for top 10% (90th percentile)** - Find $z$ such that $P(Z \leq z) = 0.90$. - From tables, $z_{0.90} \approx 1.282$. - Convert to $X$: $$X = 2 + 1.282 \times \frac{1}{3} = 2 + 0.427 = 2.427 \text{ hours}$$ **Final answers:** - (a) About 0.135% last longer than 3 hours. - (b) Third quartile is approximately 2.22 hours. - (c) Battery packs lasting at least 2.43 hours are in the top 10%.