Statistics And Probability Question 212
Question: If $ {{\bar{x}}_1} $ and $ {{\bar{x}}_2} $ are the means of two distributions such that $ {{\bar{x}}_1}<{{\bar{x}}_2} $ and $ \bar{x} $ is the mean of the combined distribution, then
Options:
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} $
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Answer:
Correct Answer: D
Solution:
Let $ n_1 $ and $ n_2 $ be the number of observations in two groups having means $ {{\bar{x}}_1} $ and $ {{\bar{x}}_2} $ respectively. Then, $ \bar{x}=\frac{n_1{{{\bar{x}}}_1}+n_2{{{\bar{x}}}_2}}{n_1+n_2} $ Now, $ \bar{x}-{{\bar{x}}_1}=\frac{n_1{{{\bar{x}}}_1}+n_2{{{\bar{x}}}_2}}{n_1+n_2}-{{\bar{x}}_1} $ $ =\frac{n_2({{{\bar{x}}}_2}-{{{\bar{x}}}_1})}{n_1+n_2}>0,,[\because {{\bar{x}}_2}>{{\bar{x}}_1}] $
Þ $ \bar{x},>,{{\bar{x}}_1} $ …..(i) and $ \bar{x}-{{\bar{x}}_2}=\frac{n({{{\bar{x}}}_1}-{{{\bar{x}}}_2})}{n_1+n_2}<0 $ , $ [ \because {{{\bar{x}}}_2}>,{{{\bar{x}}}_1} ] $
Þ $ \bar{x}<,{{\bar{x}}_2} $ ……(ii) From (i) and (ii), $ {{\bar{x}}_1}<\bar{x},<{{\bar{x}}_2} $ .
BETA
