Respuesta :
Given that the sample has the following information;
[tex]\begin{gathered} \text{size(n)}=541 \\ S.d(\sigma)=10.4 \\ \operatorname{mean}(\mu)=21.9 \\ z=\text{ probability of 80\% confidence interval} \end{gathered}[/tex]We can use the formula below to find the confidence interval;
[tex]\mu\pm z(\frac{\sigma}{\sqrt[]{n}})[/tex]The 80% confidence interval for a sample is usually given to be 1.28
We can then substitute all the values into the formula above.
[tex]\begin{gathered} 21.9\pm1.28(\frac{10.4}{\sqrt[]{541}}) \\ 21.9\pm0.572 \end{gathered}[/tex]We would then have the answer finalized as
Answer:
[tex]21.328<\mu<22.472[/tex]