JEE Mains · Maths · STD 11 - 13. statistics
Let the mean and variance of four numbers \(3,7, x\) and \(y(x>y)\) be \(5\) and \(10\) respectively. Then the mean of four numbers \(3+2 \mathrm{x}, 7+2 \mathrm{y}, \mathrm{x}+\mathrm{y}\) and \(x-y\) is ..... .
- A \(10\)
- B \(11\)
- C \(12\)
- D \(48\)
Answer & Solution
Correct Answer
(C) \(12\)
Step-by-step Solution
Detailed explanation
\(5=\frac{3+7+x+y}{4} \Rightarrow x+y=10\) \(\operatorname{Var}(x)=10=\frac{3^{2}+7^{2}+x^{2}+y^{2}}{4}-25\) \(140=49+9+x^{2}+y^{2}\) \(x^{2}+y^{2}=82\) \(x+y=10\) \(\Rightarrow(x, y)=(9,1)\) Four numbers are \(21,9,10,8\) \(\text { Mean }=\frac{48}{4}=12\)
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