1. **Problem Statement:** We are given a normal distribution of weights for 9-ounce bags of potato chips with mean $\mu = 9.12$ ounces and standard deviation $\sigma = 0.15$ ounces. We want to find: A) The probability that a randomly selected bag weighs exactly 8.97 ounces. 2. **Important Note:** For continuous distributions like the normal distribution, the probability of a variable taking any exact value is 0. Instead, we find probabilities over intervals. 3. **Step for A:** Since the probability at exactly 8.97 ounces is 0, we interpret this as finding the probability that the weight is very close to 8.97 ounces or use the probability density function (PDF) value at 8.97. 4. **Calculate the z-score for 8.97:** $$z = \frac{X - \mu}{\sigma} = \frac{8.97 - 9.12}{0.15} = \frac{-0.15}{0.15} = -1$$ 5. **Interpretation:** The z-score of -1 means 8.97 ounces is 1 standard deviation below the mean. 6. **Probability for exact value:** $P(X=8.97) = 0$ for continuous variables. --- 1. **Problem Statement:** B) Find the probability that a bag weighs between 8.97 and 9.27 ounces. 2. **Calculate z-scores:** $$z_1 = \frac{8.97 - 9.12}{0.15} = -1$$ $$z_2 = \frac{9.27 - 9.12}{0.15} = 1$$ 3. **Use empirical rule or standard normal table:** The probability between $z = -1$ and $z = 1$ is approximately 68%. 4. **Answer:** $P(8.97 < X < 9.27) = 0.68$ or 68%. --- 1. **Problem Statement:** C) Find the probability that a bag weighs between 8.82 and 9.57 ounces. 2. **Calculate z-scores:** $$z_1 = \frac{8.82 - 9.12}{0.15} = \frac{-0.30}{0.15} = -2$$ $$z_2 = \frac{9.57 - 9.12}{0.15} = \frac{0.45}{0.15} = 3$$ 3. **Use empirical rule:** Probability between $z = -2$ and $z = 3$ is approximately the area from $-2\sigma$ to $+3\sigma$. From the empirical rule: - Between $-2\sigma$ and $+2\sigma$ is about 95% - Between $+2\sigma$ and $+3\sigma$ adds about 2.35% So total probability: $$0.95 + 0.0235 = 0.9735$$ 4. **Answer:** $P(8.82 < X < 9.57) \approx 0.9735$ or 97.35%. --- **Summary for problem 11:** - A) $P(X=8.97) = 0$ - B) $P(8.97 < X < 9.27) = 0.68$ - C) $P(8.82 < X < 9.57) \approx 0.9735$ --- **Note:** The user asked multiple questions but per instructions, only the first problem (11) is fully solved here.