Answer
$\mu$ is between 134.252 and 210.698.
Work Step by Step
The mean can be counted by summing all the data and dividing it by the number of data: $\frac{1+0+...+173}{40}=172.475.$
Standard deviation=$\sqrt{\frac{\sum (x-\mu)^2}{n-1}}=\sqrt{\frac{(1-172.475)^2+...+(173-172.475)^2}{39}}=119.4983.$
$\alpha=1-0.95=0.05.$ $\sigma$ is unknown, hence we use the t-distribution with $df=sample \ size-1=40-1=39$ in the table. $t_{\alpha/2}=t_{0.025}=2.023.$ Margin of error:$z_{\alpha/2}\cdot\frac{s}{\sqrt {n}}=2.023\cdot\frac{119.4983}{\sqrt{40}}=38.223.$ Hence the confidence interval:$\mu$ is between 172.475-38.223=134.252 and 172.475+38.223=210.698.
