Question

The combined SAT scores for the students at a local high school are normally distributed with a mean of 1504 and a standard deviation of 300. The local college includes a minimum score of 1954 in its admission requirements. What percentage of students from this school earn scores that satisfy the admission requirement

Answer 1

Answer:

6.68% of students from this school earn scores that satisfy the admission requirement

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

$\mu = 1504, \sigma = 300$

The local college includes a minimum score of 1954 in its admission requirements. What percentage of students from this school earn scores that satisfy the admission requirement

This is 1 subtracted by the pvalue of Z when X = 1954. So

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

$Z = \frac{1954 - 1504}{300}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

1 - 0.9332 = 0.0668

6.68% of students from this school earn scores that satisfy the admission requirement