1. Problem statement: We have sugar bags filled with an average (mean) of 1004 g and a standard deviation of 2.4 g, assuming a normal distribution. 2. We want to find: a) The percentage of bags containing less than 1000 g. b) The number of bags (out of 890) containing between 1000 g and 1005 g. 3. Formula and rules: For a normal distribution, the z-score is calculated as: $$z = \frac{x - \mu}{\sigma}$$ where $x$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation. 4. Part a) Calculate the percentage of bags with less than 1000 g: $$z = \frac{1000 - 1004}{2.4} = \frac{-4}{2.4} = -1.6667$$ 5. Using standard normal distribution tables or a calculator, find the cumulative probability for $z = -1.6667$: $$P(Z < -1.6667) \approx 0.0478$$ This means approximately 4.78% of bags contain less than 1000 g. 6. Part b) Calculate the number of bags with sugar between 1000 g and 1005 g: Calculate z-scores for both limits: $$z_1 = \frac{1000 - 1004}{2.4} = -1.6667$$ $$z_2 = \frac{1005 - 1004}{2.4} = 0.4167$$ 7. Find cumulative probabilities: $$P(Z < -1.6667) \approx 0.0478$$ $$P(Z < 0.4167) \approx 0.6612$$ 8. Probability that $Z$ is between $z_1$ and $z_2$: $$P(-1.6667 < Z < 0.4167) = 0.6612 - 0.0478 = 0.6134$$ 9. Number of bags in this range out of 890: $$890 \times 0.6134 \approx 546$$ Final answers: - a) 4.78% - b) 546 bags