Percentage of Students Scoring Above a Given Score

In a recent analysis of a new educational program, researchers recorded the test scores of 100 students who participated. The scores are subject to further statistical examination. The mean score was found to be 75, with a standard deviation of 12. The researchers decided to identify what percentage of students scored above 87.

To solve this, use the properties of the normal distribution, assuming the scores are normally distributed. You will need to calculate the z-score for the score of 87 using the formula:

$$ z = \f\frac{X - \mu}{\sigma} $$

where $X$ is the score being analyzed, $\mu$ is the mean, and $\sigma$ is the standard deviation. After calculating the z-score, refer to the standard normal distribution table (z-table) to determine the corresponding percentile. What percentage of students scored above 87?