Answer :
Final answer:
The standardized weight representing the 5th percentile in a normally distributed set of data is a Z-score of -1.64, which corresponds to option A.
Explanation:
The standardized weight that represents the 5th percentile for a normally distributed set of data can be found by looking at a standard normal distribution table or using a Z-score calculator.
The 5th percentile falls in the left tail of the normal distribution. This is equivalent to a Z-score that marks the point below which 5% of the data fall.
The correct Z-score that corresponds to the 5th percentile of a normal distribution is -1.64.
Therefore, the answer to your question is A. -1.64.
This Z-score indicates that the weight of the zucchini squash that corresponds to the 5th percentile is 1.64 standard deviations below the mean weight of the zucchini squash at Fisher Farm.
