Final answer:
The z-score associated with the 63rd percentile in a standard normal distribution is approximately 0.31, rounded to the nearest hundredth.
Step-by-step explanation:
If x is in the 63rd percentile of a standard normal distribution, we want to find the associated z-score. The z-score tells us how many standard deviations a value is from the mean. By definition, a z-score is the number representing the position of a raw score in terms of how many standard deviations it is above or below the mean. To find the z-score corresponding to the 63rd percentile, we can use a standard normal distribution table or a calculator with statistical functions.
In the standard normal distribution, the mean is 0 and the standard deviation is 1. Checking a standard normal distribution table or using a z-score calculator, we find that the z-score that roughly corresponds to the 63rd percentile is approximately 0.31 (this value may slightly differ depending on the table or calculator precision). Rounding this to the nearest hundredth, we obtain a z-score of 0.31.