Answer:
84.134%
Explanation:
First, determine the value of the z-score.
[tex]\begin{gathered} Z=\frac{X-\mu}{\sigma} \\ =\frac{113-101}{12} \\ =\frac{12}{12} \\ z-score=1 \end{gathered}[/tex]Next, we determine the probability that a value selected at random is at most 113:
[tex]\begin{gathered} P(X\le113)=P(x\le1)_{} \\ =0.84134 \\ =84.134\% \end{gathered}[/tex]Thus, the probability that a value selected at random is in the given interval is 84.134%.
