Question

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. 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

Answer 1

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.