Answer
$\mu$ is between 6132 and 19664.
Work Step by Step
Given mean:12898.
Given $\sigma$:7718.8.
$\alpha=1-0.95=0.05.$ $\sigma$ is known, hence we use the z-distribution using the table. $z_{\alpha/2}=z_{0.025}=1.96.$ Margin of error:$z_{\alpha/2}\cdot\frac{\sigma}{\sqrt {n}}=1.96\cdot\frac{7718.8}{\sqrt{5}}\approx6766.$ Hence the confidence interval:$\mu$ is between 12898-6766=6132 and 12898+6766=19664.
