The coefficients of rank correlation between marks in Mathematics and marks in Physics obtained by a certain group of students is 0.8. If the sum of the squars of the difference in the rank is given to be 33, then find the number of students in the group.

Answer

To solve the problem, we will use the formula for the coefficient of rank correlation (denoted as \( r_s \)):

\[ r_s = 1 - \frac{6 \sum d_i^2}{n(n^2 - 1)} \]

Where:
- \( d_i \) is the difference in ranks,
- \( n \) is the number of students,
- \( \sum d_i^2 \) is the sum of the squares of the differences in ranks.

Given:
- The coefficient of rank correlation \( r_s = 0.8 \)
- The sum of the squares of the differences in ranks \( \sum d_i^2 = 33 \)

We need to find \( n \).

### Step 1: Substitute the known values into the formula

Substituting the known values into the formula:

\[ 0.8 = 1 - \frac{6 \times 33}{n(n^2 - 1)} \]

### Step 2: Simplify the equation

Rearranging the equation gives:

\[ 0.8 = 1 - \frac{198}{n(n^2 - 1)} \]

Subtracting 1 from both sides:

\[ 0.8 - 1 = -\frac{198}{n(n^2 - 1)} \]

This simplifies to:

\[ -0.2 = -\frac{198}{n(n^2 - 1)} \]

Multiplying both sides by -1:

\[ 0.2 = \frac{198}{n(n^2 - 1)} \]

### Step 3: Cross-multiply to eliminate the fraction

Cross-multiplying gives:

\[ 0.2 n(n^2 - 1) = 198 \]

### Step 4: Simplify the equation

Expanding the left side:

\[ 0.2n^3 - 0.2n = 198 \]

### Step 5: Rearranging the equation

Rearranging gives:

\[ 0.2n^3 - 0.2n - 198 = 0 \]

### Step 6: Multiply through by 5 to eliminate the decimal

Multiplying the entire equation by 5 to simplify:

\[ n^3 - n - 990 = 0 \]

### Step 7: Solve the cubic equation

Now we can solve the cubic equation \( n^3 - n - 990 = 0 \). We can use trial and error or synthetic division to find integer solutions.

Testing \( n = 10 \):

\[ 10^3 - 10 - 990 = 1000 - 10 - 990 = 0 \]

Thus, \( n = 10 \) is a solution.

### Conclusion

The number of students in the group is:

\[ \boxed{10} \]