[tex]\begin{gathered} \text{Given} \\ \mu=470 \\ \sigma=120 \\ x=610 \end{gathered}[/tex]
Find the z-score of Sheila's score.
[tex]\begin{gathered} z=\frac{x-\mu}{\sigma} \\ z=\frac{610-470}{120} \\ z=\frac{140}{120} \\ z=1.17 \end{gathered}[/tex]
Since we are looking for the percent of takers that earned a higher score then Sheila, find P( z > 1.17)
Looking at the value to the left of the z-score we have
[tex]\begin{gathered} P(z>1.17)=1-0.87900 \\ P(z>1.17)=0.121 \end{gathered}[/tex]
Multiply by 100% and we get
[tex]0.121\cdot100\%=12.1\%[/tex]
Rounding to the nearest whole number, the percent of takers who earned a higher score than Sheila is 12%
