ISI2015-DCG-2

Question

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

• A. $a <b$
• B. $a>b$
• C. $a=b$
• D. None of these

Comments

Option A

Answer 1

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.

Answer 2

$\underline{\textbf{Answer:}\Rightarrow}\;\;\mathbf{A.}$

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

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.

Answer 3

a=60
b=69
a<b
A)

Answer 4

$a = \sum_{x \ \epsilon \ S } {(x - 9)}^2$

 

$b = \sum_{x \ \epsilon \ S } {(x - 10)}^2$

 

$b = \sum_{x \ \epsilon \ S } {((x - 9) - 1 )}^2$

 

$b = \sum_{x \ \epsilon \ S } {((x - 9)}^2 - 2(x - 9) + 1)$

$\sum_{x \ \epsilon \ S } \ (x - 9)^2 = a$

$\sum_{x \ \epsilon \ S } \ 2(x - 9) = 0$

$\sum_{x \ \epsilon \ S } \ 1 = 9$

 

$b = a + 9$

 

$b > a$

Comments

Can you please explain this step ?

∑x ϵ S 2(x−9)=0

∑x ϵ S 1=9

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$ MiNiPanda

2*[ (6-9) + (10 - 9 ) + (7-9) + (13-9) + (5-9) + (12-9) + (8 - 9) + ( 11 - 9) + (9 - 9 ) ]

= 2*[ -3 +1 -2 + 4 - 4 + 3 -1 + 2 + 0]

= 2*0 = 0

1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 9

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Thank you pankaj_vir