Let X be a random variable, and let P(X=x) de | vedclass

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Question

Let equation of line is $y=m x+c$


	
		
			$x$
			$0$
			$1$
			$2$
			$3$
			$4$
			$R -\{0,1,2,3,4\}$
		
		
			$P ( x )$
			$C$

Vedclass · Accepted Answer

$42$

Vedclass · Answer

Let equation of line is $y=m x+c$


	
		
			$x$
			$0$
			$1$
			$2$
			$3$
			$4$
			$R -\{0,1,2,3,4\}$
		
		
			$P ( x )$
			$C$
			$m + c$
			$2 m + c$
			$3 m + c$
			$4 m+c$
			$0$
		
	


$\sum_{ x =0}^4( mx + c )=1 \Rightarrow 10 m +5 c =1 \Rightarrow 2 m + c =\frac{1}{5}$   $. . . (1)$

$	ext { mean }=\sum x _{ i } P _{ i }=\sum_{ i =0}^4\left( mx _{ i }+ c ight) x _{ i }=30 m +10 c =\frac{5}{2}$

$	herefore 3 m + c =\frac{1}{4} \ldots(2)$

$	ext { from (1) and (2) m= } \frac{1}{20}, c =\frac{1}{10}$

$\sum P _{ i } x _{ i }^2=\sum_{ i =0}^4\left( mx _{ i }+ c ight) x _1^2$

$=\sum_{ i =0}^4\left( mx _{ i }^3+ cx _{ i }^2ight) \Rightarrow 100 m +30 c (	ext { Now putting } m 	ext { and } c )$

$\Rightarrow \Sigma P _{ i }^2=5+3=8$

$	ext { Variance }=\Sigma P _{ i } x _{ i }^2-\left(\Sigma P _{ i } x _{ i }ight)^2=8-\left(\frac{5}{2}ight)^2=\frac{7}{4}$

$	herefore 24 \alpha=42$

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