First, find the z-score of 86
[tex]\begin{gathered} z=\frac{x-\mu}{\sigma} \\ z=\frac{86-84.28}{6.71} \\ z=\frac{1.72}{6.71} \\ z=0.2563 \end{gathered}[/tex]Rounding to the nearest hundredth, that is 0.26
Next, find z = 0.26 to the left of z in the z-table and we have the following
Multiply by 100%, and we have
[tex]\begin{gathered} P(z<0.26)=0.60257 \\ \\ 0.60257\cdot100\%=60.257\% \end{gathered}[/tex]Therefore, the percentage of students who scored lower than 86 is 60.257%.
