1. **Problem statement:** A vegetable distributor has tomato weights normally distributed with mean $\mu = 0.61$ lb and standard deviation $\sigma = 0.15$ lb. We need to solve: a. What percent of tomatoes weigh less than 0.76 lb? b. In 6000 tomatoes, how many weigh more than 0.31 lb? c. In 4500 tomatoes, how many weigh between 0.31 lb and 0.91 lb? 2. **Step a: Percent weighing less than 0.76 lb** - Calculate number of standard deviations above mean for 0.76 lb: $$z = \frac{0.76 - 0.61}{0.15} = \frac{0.15}{0.15} = 1$$ - Using Empirical Rule, approximately 84% of data lie below $\mu + 1\sigma$ (50% below mean + 34% between mean and $1\sigma$). - Therefore, about 84% of tomatoes weigh less than 0.76 lb. 3. **Step b: Number weighing more than 0.31 lb out of 6000** - Calculate $z$ for 0.31 lb: $$z = \frac{0.31 - 0.61}{0.15} = \frac{-0.30}{0.15} = -2$$ - From the Empirical Rule, 95% of data lie between $\mu \pm 2\sigma$. - Since 2.5% lie below $\mu - 2\sigma$, then the percent above 0.31 lb is: $$100\% - 2.5\% = 97.5\%$$ - Number of tomatoes: $$(0.975)(6000) = 5850$$ 4. **Step c: Number weighing between 0.31 lb and 0.91 lb out of 4500** - Calculate $z$ for 0.91 lb: $$z = \frac{0.91 - 0.61}{0.15} = \frac{0.30}{0.15} = 2$$ - Percentage of tomatoes between $\mu - 2\sigma$ and $\mu + 2\sigma$ is 95%. - Number of tomatoes: $$(0.95)(4500) = 4275$$ **Final answers:** a. 84% of tomatoes weigh less than 0.76 lb. b. 5850 tomatoes weigh more than 0.31 lb. c. 4275 tomatoes weigh between 0.31 lb and 0.91 lb.