we can use the following formula for the margin of error:
[tex]ME=1.96\sqrt[]{\frac{p(1-p)}{n}}[/tex]where p is the proportion and n the sample size
p = 58% = 0.58
ME = 1.5% = 0.015
therefore:
[tex]0.015=1.96\sqrt[]{\frac{0.58(1-0.58)}{n}}[/tex]now, we need to solve for n
[tex]\begin{gathered} \frac{0.015}{1.96}=\sqrt[]{\frac{0.58(1-0.58)}{n}} \\ \frac{0.58(1-0.58)}{n}=(\frac{0.015}{1.96})^2 \\ 0.58(1-0.58)=n\cdot(\frac{0.015}{1.96})^2 \\ 0.58\cdot\: 0.42=n\cdot0.00765^2 \\ n=\frac{0.58\cdot0.42}{0.00765^2}=4159.17226 \end{gathered}[/tex]