Answer
$\mu$ is between 5.21 and 7.79. The interval doesn't contain 4 years, but there is an outlier so the result might not be correct.
Work Step by Step
The mean can be counted by summing all the data and dividing it by the number of data: $\frac{4+4+...+15}{20}=6.5.$ Standard deviation=$\sqrt{\frac{\sum (x-\mu)^2}{n-1}}=\sqrt{\frac{(4-6.5)^2+...+(15-6.5)^2}{19}}=3.51.$ $\alpha=1-0.9=0.1.$ $\sigma$ is 3.51 hence we use the z-distribution with $df=sample \ size-1=20-1=19$ in the table. $z_{\alpha/2}=z_{0.05}=1.645.$ Margin of error:$z_{\alpha/2}\cdot\frac{\sigma}{\sqrt {n}}=1.645\cdot\frac{3.51}{\sqrt{20}}=1.29.$ Hence the confidence interval:$\mu$ is between 6.5-1.29=5.21 and 6.5+1.29=7.79. The interval doesn't contain 4 years, but there is an outlier so the result might not be correct.
