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Sagot :
Using the normal distribution, it is found that approximately 40% of the scores are greater than 413.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
In this problem, we have that the mean and the standard deviation of the scores are given by:
[tex]\mu = 400, \sigma = 50[/tex]
Approximately 40% of the scores are greater than the 60th percentile, which is X when Z = 0.253.
Then:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.253 = \frac{X - 400}{50}[/tex]
X - 400 = 50(0.253)
X = 412.65.
Rounding up, approximately 40% of the scores are greater than 413.
More can be learned about the normal distribution at https://brainly.com/question/24663213
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