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Answer :
The percentage of students from this school who earn scores that satisfy the admission requirement is given as follows:
53.98%.
How to obtain probabilities using the normal distribution?
The z-score of a measure X of a variable that has mean symbolized by [tex]\mu[/tex] and standard deviation symbolized by [tex]\sigma[/tex] is obtained by the rule presented as follows:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The z-score represents how many standard deviations the measure X is above or below the mean of the distribution, depending if the obtained z-score is positive or negative.
- Using the z-score table, the p-value associated with the calculated z-score is found, and it represents the percentile of the measure X in the distribution.
The mean and the standard deviation for this problem are given as follows:
[tex]\mu = 1488, \sigma = 309[/tex]
The proportion of scores above 1457 is one subtracted by the p-value of Z when X = 1457, hence:
Z = (1457 - 1488)/309
Z = -0.1
Z = -0.1 has a p-value of 0.4602.
Hence:
1 - 0.4602 = 0.5398 = 53.98%.
More can be learned about the normal distribution at https://brainly.com/question/25800303
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