The weights of bags of baby carrots are normally distributed, with a mean of 33 ounces and a standard deviation of 0.32 ounce. Bags in the upper $4.5 \%$ are too heavy and must be repackaged. What is the most a bag of baby carrots can weigh and not need to be repackaged? Click here to view Page 1 of the Standard Normal Table Click here to view Page 2 of the Standard Normal Table. A bag of baby carrots can weigh at most ounces without needing to be repackaged. (Round to two decimal places as needed.)

Step 1 ：The weights of bags of baby carrots are normally distributed, with a mean of 33 ounces and a standard deviation of 0.32 ounce. Bags in the upper 4.5% are too heavy and must be repackaged. We need to find the weight that corresponds to the 95.5th percentile of the distribution (since the upper 4.5% are too heavy and need to be repackaged).

Step 2 ：This is a z-score problem, where we need to find the z-score that corresponds to the 95.5th percentile and then convert that z-score back into a weight in ounces.

Step 3 ：Given that the mean weight is 33 ounces and the standard deviation is 0.32 ounce, we can calculate the z-score for the 95.5th percentile, which is approximately 1.695.

Step 4 ：Using the formula for converting a z-score to a raw score (X = μ + Zσ), where X is the raw score, μ is the mean, Z is the z-score, and σ is the standard deviation, we can find the maximum weight a bag of baby carrots can be without needing to be repackaged.

Step 5 ：Substituting the given values into the formula, we get X = 33 + 1.695 * 0.32, which simplifies to approximately 33.54 ounces.

Step 6 ：Final Answer: A bag of baby carrots can weigh at most \(\boxed{33.54}\) ounces without needing to be repackaged.