Solution
Step 1
Write an expression for the mean


Step 2
Find the standard deviation
![SD=\sqrt[]{\frac{\sum |x-\bar{x}|^2}{n}}](https://img.qammunity.org/2023/formulas/mathematics/college/m3i1hzqri7z9xwc1cf9lnwy4or3ja03bst.png)
![SD=\sqrt[]4-2.6](https://img.qammunity.org/2023/formulas/mathematics/college/qeuvdld7xezo29zdlz597lachqc9kr6pw4.png)
![\begin{gathered} SD=\sqrt[]{(0.36+0.36+2.56+1.96+0.16+0.16+0.16+0.16+1.96+2.56)/(10)} \\ =\sqrt[]{(10.4)/(10)}\text{ =}\sqrt[]{1.04} \\ =1.02 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/34uie5lbrox5pw8a22p0yduj2077lj4a99.png)
Find the Z (confidence interval level) value for 90%

Find the confidence interval
![\begin{gathered} CI=\bar{x}\pm_{}_{}z*\frac{s}{\sqrt[]{n}} \\ \bar{x}\Rightarrow\operatorname{mean} \\ s\Rightarrow\text{standard deviation} \\ n\Rightarrow\text{sample size} \\ z\Rightarrow\text{confidence level} \end{gathered}]()
![\begin{gathered} CI=2.6\pm1.645*\frac{1.02}{\sqrt[]{10}} \\ =2.6\pm1.645*0.32 \\ =2.6\pm0.53 \\ \text{Thus,} \\ CI=2.6+0.53\text{ 0r 2.6-0.53} \\ CI=3.13\text{ or 2.07} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/pr3vtjpbphxud44gfjvlgd44dq8etfcfhw.png)
Hence, a 90% confidence interval estimate of the mean lies between 2 and 3
Therefore, either Michelangelo or Donatello