A simple random sample of size n is drawn from a normally distributed population, and the mean of the sample is (x with a line over it), while the standard deviation is s. What is the 99% confidence interval for the population mean? Use the table below to help you answer the question.Confidence Level. 90%. 95%. 99%z*-score. 1.645. 1.96. 2.58

Answer:Last option [tex]{\displaystyle {\overline {x}}} \± \frac{2.58*s}{\sqrt{n}}[/tex]Step-by-step explanation:[tex]{\displaystyle {\overline {x}}}[/tex]If we extract a sample of size n from a normal population and we want to estimate the population mean then the confidence interval for the mean will be:[tex]{\displaystyle {\overline {x}}} \± Z_{\alpha/ 2}*\frac{s}{\sqrt{n}}[/tex]Where Where [tex]{\displaystyle {\overline {x}}}[/tex]  is an estimator for the mean μ.n is the sample size and s is its standard deviation.[tex]\alpha[/tex] is the level of significance[tex]\alpha[/tex]= 1-Confidence Level[tex]\alpha= 1-0.99[/tex] [tex]\alpha = 0.01[/tex]Then [tex]Z_{\frac{\alpha}{2}} = Z_{\frac{0.01}{2}} = Z_{0.005}=2.58[/tex]Then the 99% confidence interval for the population mean is[tex]{\displaystyle {\overline {x}}} \± \frac{2.58*s}{\sqrt{n}}[/tex]