The mean and variance of 5 observations are 5 | vedclass

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(A) Let the five observations be $1, 3, 5, a, b$. Given mean $\bar{x} = 5$,so $\frac{1+3+5+a+b}{5} = 5$. $9 + a + b = 25 \implies a + b = 16$. Given v

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$1072$

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(A) Let the five observations be $1, 3, 5, a, b$. Given mean $\bar{x} = 5$,so $\frac{1+3+5+a+b}{5} = 5$. $9 + a + b = 25 \implies a + b = 16$. Given variance $\sigma^2 = 8$,so $\frac{1^2+3^2+5^2+a^2+b^2}{5} - (5)^2 = 8$. $\frac{1+9+25+a^2+b^2}{5} = 8 + 25 = 33$. $35 + a^2 + b^2 = 165 \implies a^2 + b^2 = 130$. We have $(a+b)^2 = a^2 + b^2 + 2ab$,so $16^2 = 130 + 2ab$. $256 = 130 + 2ab \implies 2ab = 126 \implies ab = 63$. The values $a$ and $b$ are roots of $x^2 - 16x + 63 = 0$. $(x-7)(x-9) = 0$,so the remaining observations are $7$ and $9$. The sum of cubes is $7^3 + 9^3 = 343 + 729 = 1072$.

The mean and variance of $5$ observations are $5$ and $8$ respectively. If $3$ observations are $1, 3, 5$,then the sum of cubes of the remaining two observations is

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