Let the observations mathrm{x}_{mathrm{i}}(1 | vedclass

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Question

$\sum_{i=1}^{10}\left(x_{i}-5\right)=10$

$\Rightarrow$ Mean of observation $\mathrm{x}_{\mathrm{i}}-5=\frac{1}{10} \sum_{\mathrm{i}=1}^{3}\left(\ma

Vedclass · Accepted Answer

$(3, 3)$

Vedclass · Answer

$\sum_{i=1}^{10}\left(x_{i}-5ight)=10$

$\Rightarrow$ Mean of observation $\mathrm{x}_{\mathrm{i}}-5=\frac{1}{10} \sum_{\mathrm{i}=1}^{3}\left(\mathrm{x}_{\mathrm{i}}-5ight)=1$

$\Rightarrow \mu=$ mean of observation $\left(\mathrm{x}_{\mathrm{i}}-3ight)$

$=\left(	ext { mean of observation }\left(\mathrm{x}_{\mathrm{i}}-5ight)ight)+2$

$=1+2=3$

Variance of observation

$\mathrm{x}_{\mathrm{i}}-5=\frac{1}{10} \sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}-5ight)^{2}-\left(\mathrm{Mean} 	ext { of }\left(\mathrm{x}_{\mathrm{i}}-5ight)ight)^{2}=3$

$\Rightarrow \quad \lambda=$ variance of observation $\left(\mathrm{x}_{\mathrm{i}}-3ight)$

$=$ variance of observation $\left(\mathrm{x}_{\mathrm{i}}-5ight)=3$ $	herefore \quad(\mu, \lambda)=(3,3)$

${x_i}$	$92$	$93$	$97$	$98$	$102$	$104$	$109$
${f_i}$	$3$	$2$	$3$	$2$	$6$	$3$	$3$

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