#### Answer

μ is between 1034.4 and 1226. The population is all daily receipts of the movie. The interval seems to be good, because it is between 950 and 1800.

#### Work Step by Step

The mean can be counted by summing all the data and dividing it by the number of data: $\frac{963+1027+...+1204}{10}=1130.2.$
Standard deviation=$\sqrt{\frac{\sum (x-\mu)^2}{n-1}}=\sqrt{\frac{(963-1130.2)^2+...+(1204-1130.2)^2}{13}}=117.45.$
α=1−0.99=0.01. σ is 16.36, hence we use the z-distribution with df=sample size−1=10−1=9 in the table. $z_{\alpha/2}=z_{0.005}=2.58.$ Margin of error:$z_{\alpha/2}\cdot\frac{\sigma}{\sqrt {n}}=2.58\cdot\frac{117.45}{\sqrt{10}}=95.8.$ Hence the confidence interval:μ is between 1130.2-95.8=1034.4 and 1130.2+95.8=1226. The population is all daily receipts of the movie. The interval seems to be good, because it is between 950 and 1800.