The combined SAT scores for the students at a local high school are normally distributed with a mean of 1454 and a standard deviation of 310. The local college includes a minimum score of 1795 in its admission requirements.

What percentage of students from this school earn scores that satisfy the admission requirement?

\[ P(X > 1795) = \]

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.

The z-score is calculated as (1795 - 1454) / 310 = 1.103. Looking up the corresponding area in the standard normal distribution table, we find that the percentage is approximately 86.6%.

To find the percentage of students from the local high school who earn scores that satisfy the admission requirement of 1795, we need to calculate the area under the normal distribution curve that corresponds to scores greater than 1795.

First, we need to convert the admission requirement score into a z-score. The formula for calculating a z-score is:
z = (x - mean) / standard deviation

So, for an admission requirement score of 1795, the z-score would be:
z = (1795 - 1454) / 310
z = 1.103

Now, we need to find the area to the right of the z-score of 1.103 on the standard normal distribution table. This area represents the percentage of students who have scores greater than 1795.

Looking up the z-score of 1.103 on the standard normal distribution table, we find that the area to the left is 0.8621. To find the area to the right, we subtract this value from 1:
1 - 0.8621 = 0.1379

So, approximately 13.8% of students from this school earn scores that satisfy the admission requirement of 1795.

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