A) \[\overline{x}=\frac{{{\overline{x}}_{1}}+{{\overline{x}}_{2}}}{2}\]
B) \[\overline{x}>{{\overline{x}}_{2}}\]
C) \[\overline{x}<{{\overline{x}}_{1}}\]
D) \[{{\overline{x}}_{1}}<\overline{x}<{{\overline{x}}_{2}}\]
Correct Answer: D
Solution :
[d]: Let \[{{\overline{x}}_{1}}\]and \[{{\overline{x}}_{2}}\]are the arithmetic means of two distributions with \[{{n}_{1}}\]and \[{{n}_{2}}\] observations respectively. Clearly, the combined mean \[\overline{x}=\frac{{{n}_{1}}{{\overline{x}}_{1}}+{{n}_{2}}{{\overline{x}}_{2}}}{{{n}_{1}}+{{n}_{2}}}\] Now, \[\overline{x}-{{\overline{x}}_{1}}=\frac{{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}({{\overline{x}}_{2}}-{{\overline{x}}_{1}})\] \[\Rightarrow \]\[\overline{x}-{{\overline{x}}_{1}}>0\]as\[{{\overline{x}}_{2}}>{{\overline{x}}_{1}}\] ...(i) Similarly, \[\overline{x}-{{\overline{x}}_{2}}=\frac{{{n}_{1}}}{{{n}_{2}}+{{n}_{1}}}({{\overline{x}}_{1}}-{{\overline{x}}_{2}})\] \[\Rightarrow \]\[\overline{x}-{{\overline{x}}_{2}}<0\]as\[{{\overline{x}}_{2}}>{{\overline{x}}_{1}}\] ?(ii) and (ii)\[\Rightarrow \]\[{{\overline{x}}_{1}}<\overline{x}<{{\overline{x}}_{2}}\]
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