[tex]\begin{gathered} \mu=13 \\ \sigma=1.1 \\ \end{gathered}[/tex]
So:
[tex]\begin{gathered} P(\frac{X1-\mu}{\sigma}\leq Z\leq\frac{X2-\mu}{\sigma}) \\ P(\frac{14.1-13}{1.1}\leq Z\leq\frac{15.2-13}{1.1}) \\ P(1\leq Z\leq2)=0.1359 \end{gathered}[/tex]
Therefore:
[tex]0.1352\times3000\approx408[/tex]
Answer:
408 phones
