The combined SAT scores for the students at a local high school are normally distributed with a mean of 1497 and a standard deviation of 302. The local college requires a minimum score of 2463 for admission.

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

[tex]P(X\ \textgreater \ 2463)=\square[/tex]

Enter your answer as a percent accurate to 1 decimal place (do not enter the "%" sign).

To determine what percentage of students from a local high school earn scores that satisfy the minimum admission requirement of 2463, we need to use the properties of the normal distribution.

1. Understand the Problem:
- The SAT scores are normally distributed with a mean of 1497 and a standard deviation of 302.
- We need to find the percentage of students with scores higher than 2463.

2. Calculate the Z-score:
The Z-score tells us how many standard deviations an element is from the mean. It is calculated using the formula:

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

Where [tex]\(X\)[/tex] is the score of interest (2463), [tex]\(\mu\)[/tex] is the mean (1497), and [tex]\(\sigma\)[/tex] is the standard deviation (302).

[tex]\[
Z = \frac{2463 - 1497}{302} \approx 3.20
\][/tex]

3. Find the Probability Below the Score:
Using the Z-score, we look up the cumulative probability from the standard normal distribution, which gives the probability that a score is less than 2463.

- [tex]\[ P(X < 2463) \approx 0.9993 \][/tex]

This means approximately 99.93% of students score below 2463.

4. Find the Probability Above the Score:
To find the probability that a student's score is above 2463, subtract the cumulative probability from 1:

[tex]\[
P(X > 2463) = 1 - 0.9993 \approx 0.0007
\][/tex]

5. Convert to a Percentage:
To express this probability as a percentage, multiply by 100:

[tex]\[
P(X > 2463) \approx 0.07\%
\][/tex]

Approximately 0.07% of the students meet or exceed the score requirement of 2463 for admission.