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Q.
If the combined mean of two groups is $\frac{40}{3}$ and if the mean of one group with $10$ observations is $15$, then the mean of the other group with $8$ observations is equal to
Let $\bar{x}_{1}$ and $\bar{x}_{2}$ be the mean of the two observations with $n_{1}=10$ and $n_{2}=8$. Then, their combined mean $\bar{X}$ can be given as
$\bar{X}=\frac{n_{1} \bar{X}_{1}+n_{2} \bar{X}_{2}}{n_{1}+n_{2}}$
$\Rightarrow \frac{40}{3}=\frac{10 \times 15+8 \times \bar{x}_{2}}{10+8}$
$\Rightarrow \frac{40}{3}=\frac{150+8 \bar{x}_{2}}{18}$
$\Rightarrow \bar{x}_{2}=\frac{45}{4}$