We are given the marks obtained by 8 students in Mathematics and Physics tests. We are asked to calculate the Spearman's rank correlation coefficient.

1. Problem Description

We are given the marks obtained by 8 students in Mathematics and Physics tests. We are asked to calculate the Spearman's rank correlation coefficient.

2. Solution Steps

First, we create a table to rank the marks for Mathematics and Physics separately.

| Mathematics | Physics | Rank (Math) | Rank (Physics) | d = Rank(Math) - Rank(Physics) | d^2 |

|---|---|---|---|---|---|

| 8 | 2 | 1 | 7 | -6 | 36 |

| 2 | 6 | 7 | 3 | 4 | 16 |

| 7 | 4 | 2 | 5 | -3 | 9 |

| 6 | 5 | 3 | 4 | -1 | 1 |

| 4 | 3 | 5 | 6 | -1 | 1 |

| 1 | 8 | 8 | 1 | 7 | 49 |

| 3 | 7 | 6 | 2 | 4 | 16 |

| 5 | 1 | 4 | 8 | -4 | 16 |

Next, we calculate the sum of the squared differences, $\sum d^2 = 36 + 16 + 9 + 1 + 1 + 49 + 16 + 16 = 144$.

The formula for Spearman's rank correlation coefficient is:

$r_s = 1 - \frac{6 \sum d^2}{n(n^2 - 1)}$

where $n$ is the number of pairs of data, which in our case is $n=8$.

Plugging in the values, we get:

$r_s = 1 - \frac{6 \times 144}{8(8^2 - 1)} = 1 - \frac{864}{8(64 - 1)} = 1 - \frac{864}{8(63)} = 1 - \frac{864}{504} = 1 - \frac{12}{7} = \frac{7-12}{7} = -\frac{5}{7}$.

3. Final Answer

The Spearman's rank correlation coefficient is $-\frac{5}{7}$.

$r_s = -5/7 \approx -0.714$

Final Answer: The final answer is $\boxed{-5/7}$