Step 1

From the provided information,

Margin of error \(\displaystyle{\left({E}\right)}=\${130}\)

Confidence level \(\displaystyle={95}\%\)

Population standard deviation \(\displaystyle{\left(\sigma\right)}=\${591}\)

Step 2

The z value at \(\displaystyle{95}\%\) confidence level from the standard normal table is 1.96.

The required sample size can be obtained as:

\(\displaystyle{n}={\left({\frac{{{z}_{{\frac{\alpha}{{2}}}}\sigma}}{{{E}}}}\right)}^{{{2}}}\)

\(\displaystyle={\left({\frac{{{1.96}\times{591}}}{{{130}}}}\right)}^{{{2}}}\)

\(\displaystyle={79.3963}\)

\(\displaystyle\approx{80}\)

Thus, the required sample size is 80.

From the provided information,

Margin of error \(\displaystyle{\left({E}\right)}=\${130}\)

Confidence level \(\displaystyle={95}\%\)

Population standard deviation \(\displaystyle{\left(\sigma\right)}=\${591}\)

Step 2

The z value at \(\displaystyle{95}\%\) confidence level from the standard normal table is 1.96.

The required sample size can be obtained as:

\(\displaystyle{n}={\left({\frac{{{z}_{{\frac{\alpha}{{2}}}}\sigma}}{{{E}}}}\right)}^{{{2}}}\)

\(\displaystyle={\left({\frac{{{1.96}\times{591}}}{{{130}}}}\right)}^{{{2}}}\)

\(\displaystyle={79.3963}\)

\(\displaystyle\approx{80}\)

Thus, the required sample size is 80.