If X1 ¯ and X2¯ the means of two distributions such that X1 <X2¯ and X¯ is the mean of the combined distribution then

If X1 ¯ and X2¯ the means of two distributions such that X1

A
$\bar{X} < {\bar{X}}_{1}$
B
$\bar{X} > {\bar{X}}_{2}$
C
$\bar{X} = \frac{{\bar{X}}_{1} + {\bar{X}}_{2}}{2}$
D
${\bar{X}}_{1} < \bar{X} < {\bar{X}}_{2}$

Solution:

Let $n_{1}$ and $n_{2}$ be the number of observations in two groups having means $X_{1}$ and $X_{2}$ respectively. Then

$\bar{X} = \frac{n_{1} {\bar{X}}_{1} + n_{2} {\bar{X}}_{2}}{n_{1} + n_{2}}$

Now,

$\begin{array}{l} \bar{X} - {\bar{X}}_{1} = \frac{n_{1} {\bar{X}}_{1} + n_{2} {\bar{X}}_{2}}{n_{1} + n_{2}} - {\bar{X}}_{1} \\ \Rightarrow \bar{X} - {\bar{X}}_{1} = \frac{n_{2} ({\bar{X}}_{2} - {\bar{X}}_{1})}{n_{1} + n_{2}} > 0 \\ \Rightarrow \bar{X} > {\bar{X}}_{1} \\ And, \bar{X} - {\bar{X}}_{2} = \frac{n ({\bar{X}}_{1} - {\bar{X}}_{2})}{n_{1} + n_{2}} < 0 \\ \Rightarrow \bar{X} < {\bar{X}}_{2} \end{array}$

If $\bar{X_{1}} a n d \bar{X_{2}}$ the means of two distributions such that $X_{1} < \bar{X_{2}} a n d \bar{X}$ is the mean of the combined distribution then

Solution:

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