Let a, b, c in N and a

Question

(D) Given observations are $9, 25, a, b, c$ with $a < b < c$. Mean $\bar{x} = \frac{9 + 25 + a + b + c}{5} = 18 \implies a + b + c = 56$. Mean deviati

Vedclass · Accepted Answer

$33$

Vedclass · Answer

(D) Given observations are $9, 25, a, b, c$ with $a Mean $\bar{x} = \frac{9 + 25 + a + b + c}{5} = 18 \implies a + b + c = 56$. Mean deviation about mean $= \frac{\sum |x_i - \bar{x}|}{5} = 4 \implies |9-18| + |25-18| + |a-18| + |b-18| + |c-18| = 20$. $9 + 7 + |a-18| + |b-18| + |c-18| = 20 \implies |a-18| + |b-18| + |c-18| = 4$. Variance $= \frac{\sum (x_i - \bar{x})^2}{5} = \frac{136}{5} \implies (9-18)^2 + (25-18)^2 + (a-18)^2 + (b-18)^2 + (c-18)^2 = 136$. $81 + 49 + (a-18)^2 + (b-18)^2 + (c-18)^2 = 136 \implies (a-18)^2 + (b-18)^2 + (c-18)^2 = 6$. Let $x = a-18, y = b-18, z = c-18$. We have $|x| + |y| + |z| = 4$ and $x^2 + y^2 + z^2 = 6$. Since $a The integer solutions for $x^2 + y^2 + z^2 = 6$ are $\{-1, 1, 2\}$ or $\{-2, -1, 1\}$ etc. Testing $x = -1, y = 1, z = 2$: $|-1| + |1| + |2| = 4$ (Matches). $a-18 = -1 \implies a = 17$. $b-18 = 1 \implies b = 19$. $c-18 = 2 \implies c = 20$. Check sum: $17 + 19 + 20 = 56$ (Correct). $2a + b - c = 2(17) + 19 - 20 = 34 + 19 - 20 = 33$.

Let $a, b, c \in N$ and $a < b < c$. Let the mean,the mean deviation about the mean,and the variance of the $5$ observations $9, 25, a, b, c$ be $18, 4$,and $\frac{136}{5}$,respectively. Then $2a + b - c$ is equal to:

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