a. A random sample of 43 cars in the drive-thru of a popular fast food restaurant revealed an average bill of $18.58 per car. The population standard deviation is $6.22. Estimate the mean bill for all cars from the drive-thru with 97% confidence. Round intermediate and final answers to two decimal places.

Given[tex]\begin{gathered} n=43 \\ Mean\text{ = \$18.58} \\ \sigma=\text{ \$6.22} \end{gathered}[/tex]SolutionFormula[tex]\text{Confident interval =M }\pm\frac{Z\sigma}{\sqrt[]{n}}[/tex]where[tex]\begin{gathered} M=\text{ mean or Average} \\ Z-score=Z_{97}=2.17 \\ n=43 \end{gathered}[/tex]Substitute the parameters into the Confident Interval formula[tex]\text{Confident interval =18.58}\pm\frac{2.17\times6.22}{\sqrt[]{43}}[/tex]Then we calculate the Addition and subtractionFirst the Addition[tex]\begin{gathered} \text{Confident interval =18.58+}\frac{2.17\times6.22}{\sqrt[]{43}} \\ \\ \text{Confident interval =18.58+}\frac{13.4974}{\sqrt[]{43}} \\ \\ \text{Confident interval =18.58+}\frac{13.4974}{6.5574} \\ \text{Confident interval =18.58+}2.05833 \\ \text{Confident interval =}20.63833342 \\ \\ \text{Confident interval =}20.64\text{ two decimal places} \end{gathered}[/tex]Then now for subtraction[tex]\begin{gathered} \text{Confident interval =18.58-}\frac{2.17\times6.22}{\sqrt[]{43}} \\ \\ \text{Confident interval =18.58-}\frac{13.4974}{\sqrt[]{43}} \\ \\ \text{Confident interval =18.58-}\frac{13.4974}{6.5574} \\ \text{Confident interval =18.58-}2.05833 \\ \text{Confident interval =}16.5216658 \\ \\ \text{Confident interval =16.52 two decimal places} \end{gathered}[/tex]The final answer[tex](16.52,\text{ 20.64)}[/tex]