We need o find the 99% confidence interval for a sample with:
[tex]\begin{gathered} n=52 \\ \\ \overline{x}=85.6\text{ in} \\ \\ \sigma=3.8\text{ in} \end{gathered}[/tex]A 99% confidence interval has a z-value z = 2.576.
And the confidence interval is given by:
[tex]\overline{x}\pm z\times\frac{\sigma}{\sqrt{n}}[/tex]Thus, we obtain:
[tex]\begin{gathered} 85.6\text{ in}\pm2.576\times\frac{3.8\text{ in}}{\sqrt{52}} \\ \\ \cong85.6\text{ }\imaginaryI\text{n}\pm1.4\text{ in} \end{gathered}[/tex]Approximating to the nearest tenth, the 99% confidence interval is:
Answer
[tex]\lbrack84.2\text{ in},87.0\text{ in}\rbrack[/tex]