Let a, b in R. Let the mean and the variance | vedclass

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(A) Given the observations are $-3, 4, 7, -6, a, b$. The number of observations $N = 6$. Mean $\overline{x} = \frac{-3 + 4 + 7 - 6 + a + b}{6} = 2 \im

Vedclass · Accepted Answer

$\frac{13}{3}$

Vedclass · Answer

(A) Given the observations are $-3, 4, 7, -6, a, b$. The number of observations $N = 6$. Mean $\overline{x} = \frac{-3 + 4 + 7 - 6 + a + b}{6} = 2 \implies 2 + a + b = 12 \implies a + b = 10$. Variance $\sigma^2 = \frac{\sum x_i^2}{N} - (\overline{x})^2 = 23$. $\frac{(-3)^2 + 4^2 + 7^2 + (-6)^2 + a^2 + b^2}{6} - 2^2 = 23$. $\frac{9 + 16 + 49 + 36 + a^2 + b^2}{6} = 23 + 4 = 27$. $110 + a^2 + b^2 = 162 \implies a^2 + b^2 = 52$. Since $(a + b)^2 = a^2 + b^2 + 2ab$,we have $100 = 52 + 2ab \implies 2ab = 48 \implies ab = 24$. The values $a$ and $b$ are roots of $t^2 - 10t + 24 = 0$,which are $t = 4, 6$. So,the observations are $-3, 4, 7, -6, 4, 6$. Mean deviation about mean = $\frac{\sum |x_i - \overline{x}|}{6} = \frac{|-3-2| + |4-2| + |7-2| + |-6-2| + |4-2| + |6-2|}{6}$. $= \frac{|-5| + |2| + |5| + |-8| + |2| + |4|}{6} = \frac{5 + 2 + 5 + 8 + 2 + 4}{6} = \frac{26}{6} = \frac{13}{3}$.

Let $a, b \in R$. Let the mean and the variance of $6$ observations $-3, 4, 7, -6, a, b$ be $2$ and $23$,respectively. The mean deviation about the mean of these $6$ observations is:

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