ધારો કે X={11,12,13, ldots, 40,41} અને Y={61, | vedclass

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Question

$\overline{ x }=\frac{\sum \limits_{ i =11}^{41} i }{31}=\frac{11+41}{2}=26 \quad(31 	ext { elements) }$

$\overline{ y }=\frac{\sum \limits_{ j =6

Vedclass · Accepted Answer

$603$

Vedclass · Answer

$\overline{ x }=\frac{\sum \limits_{ i =11}^{41} i }{31}=\frac{11+41}{2}=26 \quad(31 	ext { elements) }$

$\overline{ y }=\frac{\sum \limits_{ j =61}^{91} j }{31}=\frac{61+91}{2}=76 \quad 	ext { (31 elements) }$

$	ext { Combined mean, }$

$\mu =\frac{31 	imes 26+31 	imes 76}{31+31}$

$=\frac{26+76}{2}=51$

$\sigma^2=\frac{1}{62} 	imes\left(\sum_{i=1}^{31}\left(x_i-\muight)^2+\sum_{i=1}^{31}\left(y_i-\muight)^2ight)=705$

Since, $x _{ i } \in X$ are in $A.P.$ with $31$ elements and common difference $1$,same is $y _{ i } \in y$, when written 

in increasing order.

$	herefore \sum \limits_{i=1}^{31}\left(x_i-\muight)^2=\sum \limits_{i=1}^{31}\left(y_i-\muight)^2$

$=10^2+11^2+\ldots . .+40^2$

$=\frac{40 	imes 41 	imes 81}{6}-\frac{9 	imes 10 	imes 19}{6}=21855$

$	herefore \left|\bar{x}+\bar{y}-\sigma^2ight|=|26+76-705|=603$

$\mathrm{x}$	$-2$	$-1$	$3$	$4$	$6$
$\mathrm{P}(\mathrm{X}=\mathrm{x})$	$\frac{1}{5}$	$\mathrm{a}$	$\frac{1}{3}$	$\frac{1}{5}$	$\mathrm{~b}$

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