Suppose a class has 9 boys & 1 girl. The average marks of these students in the mathematics exam is 60 & their variance is 82.2. A student fails in the exam if he/she gets less than 50 marks. Given that highest marks = 83 which are secured by the girl & the marks of no 2 students are the same & all marks $\in N$ . Worst case is when maximum no. of students fail.

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In the worst case, the no. of students that can fail is 2

In the worst case, the minimum marks a student can get is 48

In the worst case, the average of 3^rd best & 7^th best is 61

In the worst case, the standard deviation of the boys that have passed = 2

answer is A, B, D.

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Detailed Solution

Soddy $σ^{2} = \frac{{(x_{1} - 60)}^{2} + {(x_{2} - 60)}^{2} + \dots {(x_{9} - 60)}^{2} + 23^{2}}{10} = 82.2$
$\Rightarrow_{i = 1}^{9} {(x_{i} - 60)}^{2} = 822 - 529 = 293$
Now in the worst case $\to 2$ students have failed
$∵ {(48 - 60)}^{2} + {(49 - 60)}^{2} +_{i = 3}^{q} {(x_{i} - 60)}^{2} = 293 \Rightarrow_{i = 3}^{9} {(x_{i} - 60)}^{2} = 28$
Now: Cheek if 28 con be broken into sum of 7 distinct sq's
$28 \to 0 + 1^{2} + {(- 1)}^{2} + 2^{2} + {(- 2)}^{2} + 3^{2} + {(- 3)}^{2}$