Suppose a class has 7 students. The average marks of these students in the mathematics examination is 62, and their variance is 20. A student fails in the examination if he/she gets less than 50 marks, then in worst case, the number of students can fail is ____

Answer

To solve the problem, we need to determine the worst-case scenario for the number of students who can fail the mathematics examination, given the average marks and variance.

### Step-by-Step Solution:

1. **Understanding the Given Data**:
- Number of students, \( n = 7 \)
- Average marks, \( \bar{x} = 62 \)
- Variance, \( \sigma^2 = 20 \)

2. **Using the Variance Formula**:
The formula for variance is given by:
\[ \sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2 \]
Substituting the known values:
\[ 20 = \frac{1}{7} \sum_{i=1}^{7} (x_i - 62)^2 \]
Multiplying both sides by 7:
\[ 140 = \sum_{i=1}^{7} (x_i - 62)^2 \]

3. **Setting Up for the Worst Case**:
To find the worst-case scenario for the number of students failing (i.e., scoring less than 50), we need to consider how many students can score below 50 while still satisfying the variance condition.

4. **Assuming Students Fail**:
Let's assume \( k \) students score below 50. The highest score for these students can be 49 (the maximum failing score). The remaining \( 7 - k \) students must then score higher to maintain the average of 62.

5. **Calculating Contribution to Variance**:
If \( k \) students score 49, their contribution to the variance will be:
\[ (49 - 62)^2 = 169 \quad \text{(for each failing student)} \]
Therefore, for \( k \) students:
\[ k \cdot 169 \]

6. **Calculating the Remaining Students' Scores**:
Let the scores of the remaining \( 7 - k \) students be \( x_1, x_2, \ldots, x_{7-k} \). Their average must be such that the overall average remains 62:
\[ \frac{k \cdot 49 + (7 - k) \cdot \bar{x}_{remaining}}{7} = 62 \]
Rearranging gives:
\[ k \cdot 49 + (7 - k) \cdot \bar{x}_{remaining} = 434 \]

7. **Finding the Maximum \( k \)**:
To maximize \( k \), we need to ensure that the total variance does not exceed 140:
\[ k \cdot 169 + \sum_{i=1}^{7-k} (x_i - 62)^2 \leq 140 \]
The remaining students must score high enough to balance the variance. The maximum score for the remaining students must be significantly higher than 62 to keep the variance in check.

8. **Testing Values for \( k \)**:
- If \( k = 6 \):
\[ 6 \cdot 169 = 1014 \quad \text{(exceeds 140)} \]
- If \( k = 5 \):
\[ 5 \cdot 169 = 845 \quad \text{(exceeds 140)} \]
- If \( k = 4 \):
\[ 4 \cdot 169 = 676 \quad \text{(exceeds 140)} \]
- If \( k = 3 \):
\[ 3 \cdot 169 = 507 \quad \text{(exceeds 140)} \]
- If \( k = 2 \):
\[ 2 \cdot 169 = 338 \quad \text{(exceeds 140)} \]
- If \( k = 1 \):
\[ 1 \cdot 169 = 169 \quad \text{(exceeds 140)} \]
- If \( k = 0 \):
\[ 0 \cdot 169 = 0 \quad \text{(valid)} \]

9. **Conclusion**:
The maximum number of students that can fail while keeping the average and variance intact is **3**.