The mean is given by:
![\begin{gathered} \mu=(23.8+17+29.1+28+15.3+6.8+29.3+16.8+1.3+21.3)/(10) \\ so: \\ \mu=(188.7)/(10) \\ \mu=18.87 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ziaa32n64ecqmi6a7azp4dluewd6979hwv.png)
The median is:
![{}{}\lbrace1.3,6.8,15.3,16.8,17,21.3,23.8,28,29.1,29.3\rbrace](https://img.qammunity.org/2023/formulas/mathematics/college/ep6ykkxdm23kx3xvw5amjkzjo7dmisvqpj.png)
The median is the middle number in a list of values, first we had to sort the list in increasing order. Since we have an even number of values we have to average the middle two numbers, so:
![Me=(17+21.3)/(2)=(38.3)/(2)=19.15](https://img.qammunity.org/2023/formulas/mathematics/college/2fu9mwb9kzi9yoefpgpbmiur8lzv6ahdzg.png)
Finally, the standard deviation is:
![\begin{gathered} \sigma=\sqrt{(1)/(N)\sum_{n\mathop{=}1}^N(x_i-\mu)^2} \\ so: \\ \sigma=\sqrt{(1)/(10)\sum_{n\mathop{=}1}^(10)(x_i-18.87)^2} \\ \\ \sigma=√(80.1921) \\ \sigma\approx8.9550 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/o4dlqoqd3b34n05f1kuas7ky0htiagpj7l.png)