In this case we want to estimate the mean of the population; since the sample size is larger than 60 and we know the standard deviation of the population the confidence interval is given by:
[tex]\begin{gathered} \bar{x}\pm z\sigma_{\bar{x}} \\ \text{ where} \\ \sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}} \end{gathered}[/tex]Now, from a normal distribution table we know that the z-value for a 95% confidence interval is 1.96; plugging these and the values of the mean and standard deviation given we have that:
[tex]81\pm(1.96)(\frac{8}{\sqrt{60}})=81\pm2.024[/tex]Choosing the negative sign we can calculate the lower limit:
[tex]81-2.024=79[/tex]Choosing the positive sign we can calculate the upper limit:
[tex]81+2.024=83[/tex]Threfore, the lower and upper limits of the confidence interval are 79 and 83, respectively.