The variance sigma^2 of the data is . . . . | vedclass

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Question

$x_i$
			$f_i$
			$f_ix_i$
			$f_ix_i^2$
		
		
			$0$
			$3$
			$0$
			$0$
		
		
			$1$
			$2$
			$2$
			$2$
		
		
			$5

Vedclass · Accepted Answer

$29$

Vedclass · Answer

$x_i$
			$f_i$
			$f_ix_i$
			$f_ix_i^2$
		
		
			$0$
			$3$
			$0$
			$0$
		
		
			$1$
			$2$
			$2$
			$2$
		
		
			$5$
			$3$
			$15$
			$75$
		
		
			$6$
			$2$
			$12$
			$72$
		
		
			$10$
			$6$
			$60$
			$600$
		
		
			$12$
			$3$
			$36$
			$432$
		
		
			$17$
			$3$
			$51$
			$867$
		
		
			 
			
			$\sum f_i = 22$
			
			 
			$\sum f_ix_i^2 = 2048$
		
	


$ 	herefore \quad \sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}=176$

$ 	ext { So } \overline{\mathrm{x}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}=\frac{176}{22}=8 $

$ 	ext { for } \sigma^2=\frac{1}{\mathrm{~N}} \sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2-(\overline{\mathrm{x}})^2 $

$ \quad=\frac{1}{22} 	imes 2048-(8)^2$

$ \quad=93.090964 $

$\quad=29.0909$

Diameters	$33-36$	$37-40$	$41-44$	$45-48$	$49-52$
No. of circles	$15$	$17$	$21$	$22$	$25$

The variance $\sigma^2$ of the data is $ . . . . . .$

$x_i$ $0$ $1$ $5$ $6$ $10$ $12$ $17$

$f_i$ $3$ $2$ $3$ $2$ $6$ $3$ $3$

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