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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.

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Answer:

P(X < 985) = 5.5%

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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:

[tex]\mu = 1476, \sigma = 307[/tex]

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{985 - 1476}{307}[/tex]

[tex]Z = -1.599[/tex]

[tex]Z = -1.599[/tex] has a pvalue of 0.055.

So

P(X < 985) = 5.5%