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(A) Given observations: $2, 3, 3, 4, 5, 7, a, b$. Total number of observations $n = 8$.Mean $\bar{x} = \frac{2+3+3+4+5+7+a+b}{8} = 4 \implies 24 + a +

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(A) Given observations: $2, 3, 3, 4, 5, 7, a, b$. Total number of observations $n = 8$.Mean $\bar{x} = \frac{2+3+3+4+5+7+a+b}{8} = 4 \implies 24 + a + b = 32 \implies a + b = 8$.Variance $\sigma^2 = (\sqrt{2})^2 = 2$.Formula for variance: $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.$2 = \frac{2^2+3^2+3^2+4^2+5^2+7^2+a^2+b^2}{8} - 4^2$.$2 = \frac{4+9+9+16+25+49+a^2+b^2}{8} - 16$.$18 = \frac{112 + a^2 + b^2}{8} \implies 144 = 112 + a^2 + b^2 \implies a^2 + b^2 = 32$.Since $(a+b)^2 = a^2 + b^2 + 2ab$,we have $8^2 = 32 + 2ab \implies 64 - 32 = 2ab \implies ab = 16$.Solving $a+b=8$ and $ab=16$,we get $(a-4)^2 = 0$,so $a=4, b=4$.The observations are $2, 3, 3, 4, 4, 4, 5, 7$. The mode is $4$ (appears $3$ times).Mean deviation about mode $= \frac{\sum |x_i - 4|}{8} = \frac{|2-4| + |3-4| + |3-4| + |4-4| + |4-4| + |4-4| + |5-4| + |7-4|}{8}$.$= \frac{2 + 1 + 1 + 0 + 0 + 0 + 1 + 3}{8} = \frac{8}{8} = 1$.

Let the mean and the standard deviation of the observations $2, 3, 3, 4, 5, 7, a, b$ be $4$ and $\sqrt{2}$ respectively. Then the mean deviation about the mode of these observations is:

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