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

0.35% of students from this school earn scores that satisfy the admission requirement.

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 = (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 = (X - \mu)/(\sigma)`

`Z = (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.