The mean and variance of 7 observations are 8 | vedclass

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

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$2$

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(A) Let the two remaining observations be $x$ and $y$.Given the mean $\bar{x} = 8$ for $n = 7$ observations:$\frac{2 + 4 + 10 + 12 + 14 + x + y}{7} = 8$$42 + x + y = 56 \implies x + y = 14$  $(i)$Given the variance $\sigma^2 = 16$:$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$$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$$560 = 460 + x^2 + y^2 \implies x^2 + y^2 = 100$  $(ii)$We know that $(x - y)^2 = 2(x^2 + y^2) - (x + y)^2$$(x - y)^2 = 2(100) - (14)^2 = 200 - 196 = 4$$|x - y| = \sqrt{4} = 2$

The mean and variance of $7$ observations are $8$ and $16,$ respectively. If five observations are $2, 4, 10, 12, 14,$ then the absolute difference of the remaining two observations is

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