Respuesta :
Using the normal distribution, it is found that a student has to score 0.6433 standard deviations above the mean to be publicly recognized.
Normal Probability Distribution
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
The top 26% of the scores are given by scores above the 74th percentile, which corresponds to Z = 0.6433, hence, a student has to score 0.6433 standard deviations above the mean to be publicly recognized.
More can be learned about the normal distribution at https://brainly.com/question/24663213
