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Let $S=\{6, 10, 7, 13, 5, 12, 8, 11, 9\}$ and $a=\underset{x \in S}{\Sigma} (x-9)^2$ & $b = \underset{x \in S}{\Sigma} (x-10)^2$. Then

1. $a <b$
2. $a>b$
3. $a=b$
4. None of these

recategorized | 113 views

S = {5,6,7,8,9,10,11,12,13}. a = $4^2+3^2+2^2+1^2+0^2+1^2+2^2+3^2+4^2$ and b = $5^2+4^2+3^2+2^2+1^2+0^2+1^2+2^2+3^2$ and it is obvious that b > a (OPTION A). In general the sum of square of differences from the median is minimum when all the numbers are positive.
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+1 vote
$\underline{\textbf{Answer:}\Rightarrow}\;\;\mathbf{A.}$

\begin {align}\textbf{Given:}\;\;\;\;\;\;\;\mathrm a &= \sum_{x=5}^{13}(x-9)^2\\ \text b &= \sum_{x=5}^{13}(x-10)^2 \\&=\sum_{x = 5}^{13}((x-9)-1)^2\\&=\sum_{x = 5}^{13}[(x-9)^2 + 1 -2(x-9)] \\&=\sum_{x = 5}^{13}(x-9)^2 + \sum_{x=5}^{13}1-2\sum _{x = 5}^{13}(x-9)\\&=a + \sum_{x = 5}^{13}1-2 \underbrace {\sum_{x=5}^{x = 13}(x-9)}_\text {= 0 }\\&=a + 9 -2(-4-3-2-1-0+1+2+3+4)\\&=a + 9 \end {align}

Now, we get:

\begin{align} &b = a + 9\\ &\Rightarrow b-a > 0\\ &\Rightarrow b\gt a \\ &\text{or,} ~a \lt b \end {align}

$\therefore \textbf A$ is the correct answer.
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