First slide

Introduction to statistics

Question

$If x_{1}, x_{2} a r e t h e m e a n s o f t w o d i s t r i b u t i o n s s u c h t h a t x_{1} < x_{2} a n d x i s t h e m e a n o f t h e c o m b i n e d d i s t r i b u t i o n, t h e n$

Moderate

A

$x < x_{1}$
B

$x > x_{2}$
C

$x = \frac{x_{1} + x_{2}}{2}$
D

$x_{1} < x_{} < x_{2}$

Solution

$Let n_{1}, a n d n_{2} b e t h e n u m b e r o f o b s e r v a t i o n s i n t w o g r o u p s h a v i n g m e a n s x_{1} a n d x_{2} r e s p e c t i v e l y . T h e n,$
$\begin{array}{l} \bar{x} = \frac{n_{1} {\bar{x}}_{1} + n_{2} {\bar{x}}_{2}}{n_{1} + n_{2}} \\ \bar{x} - {\bar{x}}_{1} = \frac{n_{1} {\bar{x}}_{1} + n_{2} {\bar{x}}_{2}}{n_{1} + n_{2}} - {\bar{x}}_{1} \\ \bar{x} - {\bar{x}}_{1} = \frac{n_{2} ({\bar{x}}_{2} - {\bar{x}}_{1})}{n_{1} + n_{2}} > 0 [∵ {\bar{x}}_{2} > {\bar{x}}_{1}] \\ \begin{array}{l} \overset{}{\Rightarrow x} > {\bar{x}}_{1}, - - - - - - (1) \\ a n d \bar{x} - {\bar{x}}_{2} = \frac{n_{1} ({\bar{x}}_{1} - {\bar{x}}_{2})}{n_{1} + n_{2}} < 0 [∵ {\bar{x}}_{2} > {\bar{x}}_{1}] \\ \bar{x} < {\bar{x}}_{2} - - - - - - (2) \\ f r o m e q u a t i o n s (1) a n d (2) \\ x_{1} < x_{} < x_{2} \end{array} \end{array}$

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