Given:
a.) Approximately normal with a mean of 460 and a standard deviation of 31.
b.) The probability that the score of a randomly selected examinee is more than 530.
Step 1: Let's determine the z-score.
[tex]\text{ z-score = }\frac{x-\mu}{\sigma}[/tex][tex]\text{ = }\frac{530\text{ - 460}}{31}[/tex][tex]\text{ = 2.258064516129032}[/tex][tex]\text{ z-score }\approx\text{ 2.26}[/tex]Step 2: Let's use the z-score table to determine its equivalent probability.
From the given chart, it appears that at z-score of 2.26, the probability is 0.9881
Therefore, the answer is 0.9881 or 98.81%
