JEE Mains · Maths · STD 11 - 13. statistics
Let the mean and variance of \(8\) numbers \(x , y , 10\), \(12,6,12,4,8\), be \(9\) and \(9.25\) respectively. If \(x > y\), then \(3 x-2 y\) is equal to \(...........\).
- A \(24\)
- B \(25\)
- C \(23\)
- D \(22\)
Answer & Solution
Correct Answer
(B) \(25\)
Step-by-step Solution
Detailed explanation
\(\frac{x+y+52}{8}=9 \Rightarrow x+y=20\) For variance \(x-9, y-9,3,3,1,-5,-1,-3\) \(\bar{x}=0\) \(\therefore \frac{(x-9)^2+(y-9)^2+54}{8}-0^2=9.25\) \((x-9)^2+(11-x)^2=20\) \(x=7 \text { or } 13 \therefore y=13,7\) \(3 x-2 y=3 \times 13-2 \times 7=25\)
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