Answer
$\mu$ is between -22.1 and 308.1.
Work Step by Step
The mean can be counted by summing all the data and dividing it by the number of data: $\frac{-85+325+...+36}{12}=143.$
Standard deviation=$\sqrt{\frac{\sum (x-\mu)^2}{n-1}}=\sqrt{\frac{(-85-143)^2+...+(36-143)^2}{11}}=259.77.$
$\alpha=1-0.955=0.05.$ $\sigma$ is unknown, hence we use the t-distribution with $df=sample \ size-1=12-1=11$ in the table. $t_{\alpha/2}=t_{0.025}=2.201.$ Margin of error:$t_{\alpha/2}\cdot\frac{\sigma}{\sqrt {n}}=2.201\cdot\frac{259.77}{\sqrt{12}}=165.1.$ Hence the confidence interval:$\mu$ is between 143-165.1=-22.1 and 143+165.1=308.1.
