The mean and variance of a set of 15 numbers are 12 and 14 respectively.

Question

The mean and variance of a set of 15 numbers are 12 and 14 respectively. The mean and variance of another set of 15 numbers are 14 and $$\sigma^{2}$$

Accepted Answer

We know that if ${n}_{1},{n}_{2}$ are the sizes, ${\vec{X}}_{1},{\vec{X}}_{2}$ are the means and ${\sigma }_{1},{\sigma }_{2}$ are the standard deviation of the series, then the combine variance of the series.

$\begin{gathered} {\sigma }^{2}=\frac{{n}_{1}{\sigma }_{2}^{1}+{n}_{2}{\sigma }_{2}^{2}}{{n}_{1}+{n}_{2}}+\frac{{n}_{1}\cdot {n}_{2}}{{({n}_{1}+{n}_{2})}^{2}}{({\vec{X}}_{1}-{\vec{X}}_{2})}^{2} \ \Rightarrow 13=\frac{15	imes 14+15	imes {\sigma }^{2}}{15+15}+\frac{15	imes 15}{(15+15{)}^{2}}(12-14{)}^{2} \ \Rightarrow 13=\frac{14+{\sigma }^{2}}{2}+\frac{1}{4}	imes 4 \ \Rightarrow 14+{\sigma }^{2}=2	imes 12 \ \Rightarrow {\sigma }^{2}=10 \end{gathered}$

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