અહી mathrm{X} એ વિતરણનું યાર્દચ્છિક ચલ છે.
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Question

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

Vedclass · Accepted Answer

$781$

Vedclass · Answer

$\mathrm{x}$
			$-2$
			$-1$
			$3$
			$4$
			$6$
		
		
			$\mathrm{P}(\mathrm{X}=\mathrm{x})$
			$\frac{1}{5}$
			$\mathrm{a}$
			$\frac{1}{3}$
			$\frac{1}{5}$
			$\mathrm{~b}$
			 
		
	


$\bar{X}=2.3$

$-a+6 b=\frac{9}{10} \ldots (1)$

$\sum P_{i}=\frac{1}{5}+a+\frac{1}{3}+\frac{1}{5}+b=1$

$a+b=\frac{4}{15} \ldots (2)$

From equation $(1)$ and $(2)$

$a=\frac{1}{10}, b=\frac{1}{6}$

$\sigma^{2}=\Sigma p_{i} x_{i}^{2}-(\bar{X})^{2}$

$\frac{1}{5}(4)+a(1)+\frac{1}{3}(9)+\frac{1}{5}(16)+b(36)-(2.3)^{2}$

$=\frac{4}{5}+a+3+\frac{16}{5}+36 b-(2.3)^{2}$

$=4+a+3+36 b-(2.3)^{2}$

$=7+a+36 b-(2.3)^{2}$

$=7+\frac{1}{10}+6-(2.3)^{2}$

$=13+\frac{1}{10}-\left(\frac{23}{10}\right)^{2}$

$=\frac{131}{10}-\left(\frac{23}{10}\right)^{2}$

$=\frac{1310-(23)^{2}}{100}$

$=\frac{1310-529}{100}$

$=\frac{781}{100}$

$100 \sigma^{2}=781$

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

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