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Answer :
Answer:
69.15% of students from this school earn scores that satisfy the admission requirement.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set 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]
The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Mean of 1527 and a standard deviation of 295.
This means that [tex]\mu = 1527, \sigma = 295[/tex]
The local college includes a minimum score of 1380 in its admission requirements. What percentage of students from this school earn scores that satisfy the admission requirement?
The proportion is 1 subtracted by the pvalue of Z when X = 1380. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1380 - 1527}{295}[/tex]
[tex]Z = -0.5[/tex]
[tex]Z = -0.5[/tex] has a pvalue of 0.3085
1 - 0.3085 = 0.6915
0.6915*100% = 69.15%
69.15% of students from this school earn scores that satisfy the admission requirement.
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