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The mean and variance of the data $\mathrm{4,5,6,6,7,8, x, y}$ , where $\mathrm{x< y}$ , are $\mathrm{6\: and\: \frac{9}{4}}$ respectively. Then $\mathrm{x^{4}+y^{2}}$ is equal to
Option: 1
$162$

Option: 2
$320$

Option: 3
$674$

Option: 4
$420$

Answers (1)

best_answer

$\mathrm{Mean =\frac{4+5+6+6+7+8+x+y}{8}=6 \Rightarrow x+y=12}\\$ ......(1)

$\mathrm{Varciance =\frac{\sum x_{i}^{2}}{x}-(\bar{x})^{2} \Rightarrow \frac{4^{2}+5^{2}+6^{2}+6^{2}+7^{2}+8^{2}+x^{2}+y^{2}-(6)^{2}}{8}=\frac{9}{4}}$

$\mathrm{\Rightarrow 16+25+36+36+4 y+64+x^{2}+y^{2} =18+288} \\$

$\mathrm{\Rightarrow x^{2}+y^{2} =306-226=80}$ ........(2)

$(1)^{2}$ $\mathrm{\Rightarrow x^{2}+y^{2}+2 x y=144 \Rightarrow 2 xy=144-80=64\Rightarrow x y=32}\\$ ......(3)

From eqn (1) and (2) $\mathrm{ x=4, \quad y=8}\\$

Hence the correct answer is option 2.

$\mathrm{x^{4}+y^{2}=256+64=320}$

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