Question

The combined SAT scores for the students at a local high school are normally distributed with a mean of 1476 and a standard deviation of 307. The local college includes a minimum score of 985 in its admission requirements. What percentage of students from this school earn scores that fail to satisfy the admission requirement? P(x < 985) = % Enter your answer as a percent accurate to 1 decimal place (do not enter the "%" sign). Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

P(X < 985) = 5.5%

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 = (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 = 1476, \sigma = 307`

What percentage of students from this school earn scores that fail to satisfy the admission requirement? P(x < 985) = %

This is the pvalue of Z when X = 985. So

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

`Z = (985 - 1476)/(307)`

`Z = -1.599`

`Z = -1.599` has a pvalue of 0.055.

So

P(X < 985) = 5.5%