Let in a series of 2 n observations, half of | vedclass

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Question

Let observations are denoted by $x _{i}$ for $1 \leq i< 2 n$

$\bar{x}=\frac{\sum x_{i}}{2 n}=\frac{(a+a+\ldots+a)-(a+a+\ldots+a)}{2 n}$

$\Rig

Vedclass · Accepted Answer

$425$

Vedclass · Answer

Let observations are denoted by $x _{i}$ for $1 \leq i< 2 n$

$\bar{x}=\frac{\sum x_{i}}{2 n}=\frac{(a+a+\ldots+a)-(a+a+\ldots+a)}{2 n}$

$\Rightarrow \overline{ x }=0$

and $\sigma_{ x }^{2}=\frac{\sum x _{i}^{2}}{2 n }-(\overline{ x })^{2}=\frac{ a ^{2}+ a ^{2}+\ldots+ a ^{2}}{2 n }-0= a ^{2}$

$\Rightarrow \sigma_{x}=a$

Now, adding a constant $b$ then $\overline{ y }=\overline{ x }+ b =5$

$\Rightarrow b =5$

and $\sigma_{y}=\sigma_{x}$ (No change in S.D.) $\Rightarrow a=20$

$\Rightarrow a^{2}+b^{2}=425$

$X$	$c$	$2c$	$3c$	$4c$	$5c$	$6c$
$f$	$2$	$1$	$1$	$1$	$1$	$1$

$X$	$1$	$3$	$5$	$7$	$9$
$(f)$	$4$	$24$	$28$	$\alpha$	$8$

Let in a series of $2 n$ observations, half of them are equal to $a$ and remaining half are equal to $-a.$ Also by adding a constant $b$ in each of these observations, the mean and standard deviation of new set become $5$ and $20 ,$ respectively. Then the value of $a^{2}+b^{2}$ is equal to ....... .

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