A data set is normally distributed with a mean of 18 and astandard deviation of 2.7. Approximately what percent of the data are greaterthan 12.6?

Solution:
Given;
[tex]x=12.8,\mu=18,\sigma=2.7[/tex]The z-score is;
[tex]\begin{gathered} z=\frac{x-\mu}{\sigma} \\ \\ z=\frac{12.8-18}{2.7} \\ \\ z=-1.9259 \end{gathered}[/tex]Thus, the approximate percent of the data greater than 12.6 is;
[tex]P(x>12.8)=97.5\text{ \%}[/tex]ANSWER: 97.5%