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Sagot :
Using the normal distribution, it is found that a student has to score 0.675 standard deviations above the mean to be publicly recognized.
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 25% is at least the 100 - 25 = 75th percentile, which is X when z has a p-value of 0.75.
- Looking at the z-table, z = 0.675 has a p-value of 0.75.
Hence, a student has to score 0.675 standard deviations above the mean to be publicly recognized.
A similar problem is given at https://brainly.com/question/25784380
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