548. The knowledge of 10 students was tested in two tests: $A$ and $B$. The scores on a 100-point scale were as follows (the first row shows the scores for test $A$, and the second row shows the scores for test $B$):

| 95 | 90 | 86 | 84 | 75 | 70 | 62 | 60 | 57 | 50 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 92 | 93 | 83 | 80 | 55 | 60 | 45 | 72 | 62 | 70 |

Find the sample coefficient of rank correlation Kendall between the scores of the two tests.
Solution. When solving problem 540, the condition of which coincides with the condition of the current problem, two rank sequences were obtained (in the first row, the ranks by test $A$ are given, in the second row - by test $B$):

$$
\begin{array}{rrrrrrrrrrr}
x_{i} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
y_{i} & 2 & 1 & 3 & 4 & 9 & 8 & 10 & 5 & 7 & 6
\end{array}
$$

To the right of $y_{1}=2$, there are $R_{1}=8$ ranks $(3,4,9,8,10,5,7,6)$ greater than $y_{1}=2$; to the right of $y_{2}=1$, there are $R_{2}=8$ ranks greater than $y_{2}=1$. Similarly, we obtain: $R_{3}=7, R_{4}=6, R_{5}=1, R_{6}=1, R_{7}=0$, $R_{8}=2, R_{9}=0$. Therefore, the sum of the ranks

$$
R=R_{1}+R_{2}+\ldots+R_{9}=8+8+7+6+1+1+2=33
$$

We will find the Kendall rank correlation coefficient, considering that $R=33, n=10$:

$$
\tau_{\mathrm{B}}=\frac{4 R}{n(n-1)}-1=\frac{4 \cdot 33}{10 \cdot 9}-1=0.47
$$

Thus, $\tau_{\mathrm{R}}=0.47$