Mean and variance of a set of 6 terms is 11 a | vedclass

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Question

$\sigma^{2}=\frac{n_{1} \sigma_{1}^{2}+n_{2} \sigma_{2}^{2}}{n_{1}+n_{2}}+\frac{n_{1} n_{2}}{\left(n_{1}+n_{2}\right)^{2}}\left(\bar{x}_{1}-\bar{x}_{2

Vedclass · Accepted Answer

$30$

Vedclass · Answer

$\sigma^{2}=\frac{n_{1} \sigma_{1}^{2}+n_{2} \sigma_{2}^{2}}{n_{1}+n_{2}}+\frac{n_{1} n_{2}}{\left(n_{1}+n_{2}ight)^{2}}\left(\bar{x}_{1}-\bar{x}_{2}ight)^{2}$

$\sigma^{2}=\frac{6 	imes 24+3 	imes 36}{6+3}+\frac{6 	imes 3}{(6+3)^{2}}(11-14)^{2}$

$\sigma^{2}=\frac{144+108}{9}+\frac{18}{81} 	imes 9=28+2=30$

Mean and variance of a set of $6$ terms is $11$ and $24$ respectively and the mean and variance of another set of $3$ terms is $14$ and $36$ respectively. Then variance of all $9$ terms is equal to

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