Let mathrm{X} be a random variable with distr | vedclass

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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 $(2)$

$a=\frac{1}{10}, b=\frac{1}{6}$

$\sigma^{2}=\Sigma p_{i} x_{i}^{2}-(\bar{X})^{2}$

$\frac{1}{5}(4)+a(1)+\frac{1}{3}(9)+\frac{1}{5}(16)+b(36)-(2.3)^{2}$

$=\frac{4}{5}+a+3+\frac{16}{5}+36 b-(2.3)^{2}$

$=4+a+3+36 b-(2.3)^{2}$

$=7+a+36 b-(2.3)^{2}$

$=7+\frac{1}{10}+6-(2.3)^{2}$

$=13+\frac{1}{10}-\left(\frac{23}{10}\right)^{2}$

$=\frac{131}{10}-\left(\frac{23}{10}\right)^{2}$

$=\frac{1310-(23)^{2}}{100}$

$=\frac{1310-529}{100}$

$=\frac{781}{100}$

$100 \sigma^{2}=781$

Let $\mathrm{X}$ be a random variable with distribution.

$\mathrm{x}$ $-2$ $-1$ $3$ $4$ $6$

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

If the mean of $X$ is $2.3$ and variance of $X$ is $\sigma^{2}$, then $100 \sigma^{2}$ is equal to :

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