Mean and variance of a set of 6 terms is 11 and 24 respectively and mean and variance of another set

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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

$40$

$30$

$50$

$35$

Solution

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

$\sigma^{2}=\frac{6 \times 24+3 \times 36}{6+3}+\frac{6 \times 3}{(6+3)^{2}}(11-14)^{2}$

$\sigma^{2}=\frac{144+108}{9}+\frac{18}{81} \times 9=28+2=30$

Standard 11

Mathematics

If $\sum\limits_{i = 1}^{18} {({x_i} – 8) = 9} $ and $\sum\limits_{i = 1}^{18} {({x_i} – 8)^2 = 45} $ then the standard deviation of $x_1, x_2, …… x_{18}$ is :-

normal

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hard

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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

Solution

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