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Question: If \({\bar{x}}_{1}\) and \({\bar{x}}_{2}\) are the means of two distributions such that \({\bar{x}}_...

If xˉ1{\bar{x}}_{1} and xˉ2{\bar{x}}_{2} are the means of two distributions such that xˉ1<xˉ2{\bar{x}}_{1} < {\bar{x}}_{2} and xˉ\bar{x} is the mean of the combined distribution, then

A

xˉ<xˉ1\bar{x} < {\bar{x}}_{1}

B

xˉ>xˉ2\bar{x} > {\bar{x}}_{2}

C

Xˉ=Xˉ1+Xˉ22\bar{X} = \frac{{\bar{X}}_{1} + {\bar{X}}_{2}}{2}

D

xˉ1<xˉ<xˉ2{\bar{x}}_{1} < \bar{x} < {\bar{x}}_{2}

Answer

xˉ1<xˉ<xˉ2{\bar{x}}_{1} < \bar{x} < {\bar{x}}_{2}

Explanation

Solution

Let n1n_{1}and n2n_{2} be the number of observations in two groups having means xˉ1{\bar{x}}_{1}and xˉ2{\bar{x}}_{2} respectively. Then,

xˉ=n1xˉ1+n2xˉ2n1+n2\bar{x} = \frac{n_{1}{\bar{x}}_{1} + n_{2}{\bar{x}}_{2}}{n_{1} + n_{2}}

Now, xˉxˉ1=n1xˉ1+n2xˉ2n1+n2xˉ1\bar{x} - {\bar{x}}_{1} = \frac{n_{1}{\bar{x}}_{1} + n_{2}{\bar{x}}_{2}}{n_{1} + n_{2}} - {\bar{x}}_{1}

=n2(xˉ2xˉ1)n1+n2>0,[xˉ2>xˉ1]= \frac{n_{2}({\bar{x}}_{2} - {\bar{x}}_{1})}{n_{1} + n_{2}} > 0,\lbrack\because{\bar{x}}_{2} > {\bar{x}}_{1}\rbrack

xˉ>xˉ1\bar{x} > {\bar{x}}_{1} .....(i)

and xˉxˉ2=n(xˉ1xˉ2)n1+n2<0\bar{x} - {\bar{x}}_{2} = \frac{n({\bar{x}}_{1} - {\bar{x}}_{2})}{n_{1} + n_{2}} < 0, [xˉ2>xˉ1]\left\lbrack \because{\bar{x}}_{2} > {\bar{x}}_{1} \right\rbrack

xˉ<xˉ2\bar{x} < {\bar{x}}_{2} ......(ii)

From (i) and (ii), xˉ1<xˉ<xˉ2{\bar{x}}_{1} ⥂ < \bar{x} < {\bar{x}}_{2}.