1. **Problem Statement:** We have household income data following a Normal Distribution with mean $\mu=3000$ and standard deviation $\sigma=500$. We need to: - Find the probability that a randomly selected family has income below $2500$. - Calculate the Z-score for income $3500$ and find its percentile rank. 2. **Formulas and Rules:** - Z-score formula: $$Z=\frac{X-\mu}{\sigma}$$ where $X$ is the income value. - The probability that income is below $X$ is the cumulative distribution function (CDF) value at $Z$. - Percentile rank corresponds to the CDF value at the calculated Z-score. 3. **Calculations for income below $2500$:** - Calculate Z-score: $$Z=\frac{2500-3000}{500} = \frac{-500}{500} = -1$$ - Using standard normal tables or LibreOffice Calculator, find $P(Z < -1)$. - From the table, $P(Z < -1) \approx 0.1587$. 4. **Calculations for income $3500$:** - Calculate Z-score: $$Z=\frac{3500-3000}{500} = \frac{500}{500} = 1$$ - Find percentile rank $P(Z < 1)$. - From the table, $P(Z < 1) \approx 0.8413$. 5. **Interpretation:** - About 15.87% of families have income below $2500$. - A family earning $3500$ is at approximately the 84.13th percentile, meaning they earn more than about 84% of families in this community. **Final answers:** - Probability income below $2500$ is $0.1587$. - Percentile rank for income $3500$ is $84.13\%$.