1. Given data

• Number of students: $n=7$
• Mean marks: $\mu=62$
• Variance: $\sigma^2=20$

So, $\sum_{i=1}^7 x_i = 7\cdot 62 = 434$ and $\frac{1}{7}\sum_{i=1}^7 (x_i-62)^2 = 20$ which gives $\sum_{i=1}^7 (x_i-62)^2 = 140.$

A student fails if his/her marks are less than $50$.

2. How many students can fail at most?

If a student fails, then $x_i<50$. Hence for every failing student, $x_i-62 < -12.$ So its squared deviation from the mean is $(x_i-62)^2 > 12^2 = 144.$

That means each failing student contributes more than $144$ to the sum $\sum (x_i-62)^2.$

But the total sum of squared deviations is only $140.$

This is impossible even for one failing student, because one failing student alone would already require $\sum (x_i-62)^2 > 144,$ which contradicts the given value $140$.

3. Conclusion

Therefore, in the worst case, the maximum number of students who can fail is $0.$

4. Comparison with stored answer

Stored correct answer: $0$

Our derived answer is also $0$, so they agree.