First, calculate the mean of the scores. Since there are 10 scores, then the mean is given by:
![\begin{gathered} \bar{s}=(96+91+85+89+83+85+94+95+90+92)/(10) \\ =(900)/(10) \\ =90 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/eeyofp1g5litchtfsf0nh6qwwetfiwnal1.png)
Calculate the absolute value of the difference between the mean and each of the scores:
![\begin{gathered} |s_1-\bar{s}|=|96-90|=6 \\ |s_2-\bar{s}|=|91-90|=1 \\ |s_3-\bar{s}|=|85-90|=5 \\ |s_4-\bar{s}|=|89-90|=1 \\ |s_5-\bar{s}|=|83-90|=7 \\ |s_6-\bar{s}|=|85-90|=5 \\ |s_7-\bar{s}|=|94-90|=4 \\ |s_8-\bar{s}|=|95-90|=5 \\ |s_9-\bar{s}|=|90-90|=0 \\ |s_(10)-\bar{s}|=|92-90|=2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/2hbkueo3rtnr3mnbhe6fyy2adeaqgxkoa8.png)
The mean absolute deviation is the average of the absolute deviations from each score to the mean. Calculate the mean from the deviations:
![\begin{gathered} (6+1+5+1+7+5+4+5+0+2)/(10) \\ =(36)/(10) \\ =3.6 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/c5b7eql9mtmxp9gjflecvoqafx9k7oi2w9.png)
Therefore, the mean absolute deviation of Taylor's scores, is 3.6