Question

Assertion :If ${\overline{x}}_1\&{\overline{x}}_2$ are the means of two distributions such that ${\overline{x}}_1<{\overline{x}}_2\&\overline{x}$ is the combined mean of the distribution, then ${\overline{x}}_1<\overline{x}<{\overline{x}}_2$. Reason: If $n_1\&n_2$ be the number of observation of two groups whose means are ${\overline{x}}_1\&{\overline{x}}_2$ respectively, then $\overline{x}=\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2}{n_1+n_2}$

• A. Both (A) &. (R) are individually true & (R) is correct explanation of (A).
• B. Both Assertion & Reason are individually true but Reason is not the correct (proper) explanation of Assertion.
• C. Assertion is true Reason is false.
• D. Assertion is false Reason is true.

Answer 1

The correct option is A Both (A) &. (R) are individually true & (R) is correct explanation of (A).

$\because \overline{x}=\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2}{n_1+n_2}$
$\Rightarrow \overline{x}-{\overline{x}}_1=\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2}{n_1+n_2}-{\overline{x}}_1$
$\Rightarrow \overline{x}-{\overline{x}}_1=\frac{n_2({\overline{x}}_2-{\overline{x}}_1)}{n_1+n_2}$
$\Rightarrow \overline{x}-{\overline{x}}_1>0(\because {\overline{x}}_2>{\overline{x}}_1)$
$\Rightarrow \overline{x}>{\overline{x}}_1$ ......( A )
and $\overline{x}-{\overline{x}}_2=\frac{n_2({\overline{x}}_2-{\overline{x}}_1)}{n_1+n_2}$
$\Rightarrow \overline{x}-{\overline{x}}_2<0(\because {\overline{x}}_2>{\overline{x}}_1)$
$\Rightarrow \overline{x}<{\overline{x}}_2$ ...( B)
From (A) & (B) $\overline{x_1}<\overline{x}<{\overline{x}}_2$