The mean and variance of seven observations a | vedclass

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(A) Let the unknown observations be $x$ and $y$.Given mean $\bar{x} = 8$ for $n = 7$ observations.$\frac{2+4+10+12+14+x+y}{7} = 8$$42 + x + y = 56 \Ri

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$4$

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(A) Let the unknown observations be $x$ and $y$.Given mean $\bar{x} = 8$ for $n = 7$ observations.$\frac{2+4+10+12+14+x+y}{7} = 8$$42 + x + y = 56 \Rightarrow x + y = 14 \dots (i)$Given variance $\sigma^2 = 16$.$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$$16 = \frac{2^2+4^2+10^2+12^2+14^2+x^2+y^2}{7} - 8^2$$16 + 64 = \frac{4+16+100+144+196+x^2+y^2}{7}$$80 	imes 7 = 460 + x^2 + y^2$$560 = 460 + x^2 + y^2 \Rightarrow x^2 + y^2 = 100 \dots (ii)$From $(i)$, $y = 14 - x$. Substituting in $(ii)$:$x^2 + (14 - x)^2 = 100$$x^2 + 196 - 28x + x^2 = 100$$2x^2 - 28x + 96 = 0 \Rightarrow x^2 - 14x + 48 = 0$$(x - 6)(x - 8) = 0$. Thus, $x = 6$ or $x = 8$.If $x = 6$, $y = 8$. If $x = 8$, $y = 6$.The product $xy = 48$.The square root of the product is $\sqrt{48} = \sqrt{16 	imes 3} = 4 \sqrt{3}$.

The mean and variance of seven observations are $8$ and $16$ respectively. If $5$ of the observations are $2, 4, 10, 12, 14$, then the square root of the product of the remaining two observations is (in $\sqrt{3}$)

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