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Answer:
0.35% 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 and standard deviation
, the zscore of a measure X is given by:
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 pvalue, we get the probability that the value of the measure is greater than X.
The combined SAT scores for the students at a local high school are normally distributed with a mean of 1479 and a standard deviation of 302.
This means that
The local college includes a minimum score of 2294 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 = 2294. So
has a pvalue of 0.9965
1 - 0.9965 = 0.0035
0.0035*100% = 0.35%
0.35% of students from this school earn scores that satisfy the admission requirement.