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

$\alpha =\frac{1-0.95}{2}$

$=0.025$ 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, $CI=\stackrel{\u2015}{X}\pm M.E$

$=\stackrel{\u2015}{X}\pm Z\times \frac{\sigma}{\sqrt{n}}$ Substitute values in the above expression. $CI=83.43\pm 1.96\times \frac{14.02}{\sqrt{58}}$

$=83.43\pm 3.60$

$=(83.43-3.60,83.43+3.60)$

$(79.83,87.03)$ 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, $1-2\alpha =0.90$

$\alpha =\frac{1-0.90}{2}$

$=0.050$ 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, $CI=\stackrel{\u2015}{X}\pm M.E$

$=\stackrel{\u2015}{X}\pm Z\times \frac{\sigma}{\sqrt{n}}$ Substitute values in the above expression. $CI=83.43\pm 1.96\times \frac{14.02}{\sqrt{58}}$

$=83.43\pm 3.02$

$=(83.43-3.60,83.43+3.02)$

$=(80.41,86.45)$ 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.