$$57\,\,\,\,\,\,59\,\,\,\,\,\,66\,\,\,\,\,\,66\,\,\,\,\,\,66\,\,\,\,\,\,66\,\,\,\,\,\,69\,\,\,\,\,\,70\,\,\,\,\,\,73\,\,\,\,\,\,73\,\,\,\,\,\,76\,\,\,\,\,\,80\,\,\,\,\,\,89\,\,\,\,\,\,$$

Next, calculate the first quartile. To find the index $i_1$ of the first quartile ($Q=1$), use the following formula:
$$\begin{array}{rcl} i_1 &=& \cfrac{Q}{4}(n-1)+1\\ &=& \cfrac{1}{4}(13 - 1) + 1=4 \end{array}$$
Since $i_1=4$ is an integer, the first quartile is the score located at the $4^{th}$ position of the ordered data:
$$X_{4} = 66$$
Next, calculate the third quartile. To find the index $i_3$ of the third quartile ($Q=3$), use the following formula:
$$\begin{array}{rcl} i_3 &=& \cfrac{Q}{4}(n-1)+1\\ &=& \cfrac{3}{4}(13 - 1) + 1=10 \end{array}$$
Since $i_3=10$ is an integer, the third quartile is the score located at the $10^{th}$ position of the ordered data:
$$X_{10} = 73$$
Calculate the interquartile range:
$$\text{IQR}=Q_3-Q_1=73-66=7$$
According to the Interquartile Range Rule, a score $X$ is considered an outlier if:

• The score lies more than $1.5\cdot IQR\,$ below the first quartile: $X < (Q_1 - 1.5\cdot IQR)$
  $$Q_1 - 1.5\cdot IQR = 66 - 1.5 \cdot 7 = 55.5$$
• The score lies more than $1.5\cdot IQR\,$ above the third quartile: $X > (Q_3 + 1.5\cdot IQR)$
  $$Q_3 + 1.5\cdot IQR = 73 + 1.5 \cdot 7 = 83.5$$

This means that any score $X<55.5$ or $X>83.5$ should be considered an outlier, of which there is $1$ in the sample, namely: $89$.