JEE Mains · Maths · STD 11 - 13. statistics
The mean and variance of the data \(4, 5,6,6,7,8, x\), \(y\) where \(x < y\) are \(6\) , and \(\frac{9}{4}\) respectively. Then \(x^{4}+y^{2}\) is equal to
- A \(162\)
- B \(320\)
- C \(674\)
- D \(420\)
Answer & Solution
Correct Answer
(B) \(320\)
Step-by-step Solution
Detailed explanation
mean \(\bar{x}=\frac{4+5+6+6+7+8+x+y}{8}=6\) \(\Rightarrow x+y=48-36=12\) Variance \(=\frac{1}{8}\left(16+25+36+36+49+64+x^{2}+y^{2}\right)-36=\frac{9}{4}\) \(\Rightarrow x^{2}+y^{2}=80\) \(\therefore x=4 ; y=8\) \(x^{4}+y^{2}=256+64=320\)
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