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devlian [24]
2 years ago
12

The combined SAT scores for the students at a local high school are normally distributed with a mean of 1479 and a standard devi

ation of 302. 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
Mathematics
1 answer:
kotykmax [81]2 years ago
6 0

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 \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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 \mu = 1479, \sigma = 302

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

Z = \frac{X - \mu}{\sigma}

Z = \frac{2294 - 1479}{302}

Z = 2.7

Z = 2.7 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.

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