Step 1 Given Data: The sample size is: \(\displaystyle{n}={58}\) The mean temperature of the city is: \(\displaystyle\mu={83.43}^{{\circ}}{F}\) The standard deviation of the temperature is: \(\sigma=14.02^{\circ}F\) 95% Population: The expression calculate the 95% of the Population is, \(\displaystyle{1}-{2}\alpha={0.95}\)

\(\displaystyle\alpha={\frac{{{1}-{0.95}}}{{{2}}}}\)

\(\displaystyle={0.025}\) Corresponds to probability 0.025, the z value from the right tail end of the z-table 1.96. Step 2 The expression to calculate the interval of the population is, \(\displaystyle{C}{I}=\overline{{{X}}}\pm{M}.{E}\)

\(\displaystyle=\overline{{{X}}}\pm{Z}\times{\frac{{\sigma}}{{\sqrt{{{n}}}}}}\) Substitute values in the above expression. \(\displaystyle{C}{I}={83.43}\pm{1.96}\times{\frac{{{14.02}}}{{\sqrt{{{58}}}}}}\)

\(\displaystyle={83.43}\pm{3.60}\)

\(\displaystyle={\left({83.43}-{3.60},{83.43}+{3.60}\right)}\)

\(\displaystyle{\left({79.83},{87.03}\right)}\) Thus, the interval width of the 95% level is 3.6. and the interval is (79.83, 87.03). Step 3 90% Population: The expression calculate the 95% of the Population is, \(\displaystyle{1}-{2}\alpha={0.90}\)

\(\displaystyle\alpha={\frac{{{1}-{0.90}}}{{{2}}}}\)

\(\displaystyle={0.050}\) Corresponds to probability 0.050, the z value from the right tail end of the z-table is 1.64. The expression to calculate the interval of the Population is, \(\displaystyle{C}{I}=\overline{{{X}}}\pm{M}.{E}\)

\(\displaystyle=\overline{{{X}}}\pm{Z}\times{\frac{{\sigma}}{{\sqrt{{{n}}}}}}\) Substitute values in the above expression. \(\displaystyle{C}{I}={83.43}\pm{1.96}\times{\frac{{{14.02}}}{{\sqrt{{{58}}}}}}\)

\(\displaystyle={83.43}\pm{3.02}\)

\(\displaystyle={\left({83.43}-{3.60},{83.43}+{3.02}\right)}\)

\(\displaystyle={\left({80.41},{86.45}\right)}\) Thus, the interval width of 90% population level is 3.02, and the interval is (80.41, 59.45). Thus, 95% population has higher interval width.