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(D) Given,number of students $n = 7$,mean $\bar{x} = 62$,and variance $\sigma^2 = 20$.The formula for variance is $\sigma^2 = \frac{1}{n} \sum_{i=1}^{

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(D) Given,number of students $n = 7$,mean $\bar{x} = 62$,and variance $\sigma^2 = 20$.The formula for variance is $\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2$.Substituting the values: $20 = \frac{1}{7} \sum_{i=1}^{7} (x_i - 62)^2$.$\sum_{i=1}^{7} (x_i - 62)^2 = 20 	imes 7 = 140$.$A$ student fails if $x_i If a student fails,the minimum contribution to the sum of squares is $(49 - 62)^2 = (-13)^2 = 169$.Since the total sum of squares is only $140$,and $169 > 140$,it is impossible for even one student to have marks less than $50$ while maintaining a variance of $20$ with a mean of $62$.Therefore,the number of students who can fail is $0$.

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 they get less than $50$ marks. In the worst-case scenario,what is the maximum number of students who can fail?

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