# If $\rm \overline{X}_1$ and $\rm \overline{X}_2$ are the means of two distributions such that $\rm \overline{X}_1 < \rm \overline{X}_2$, and $\ 0 votes 16 views closed If \(\rm \overline{X}_1$ and $\rm \overline{X}_2$ are the means of two distributions such that $\rm \overline{X}_1 < \rm \overline{X}_2$, and $\rm \overline{X}$ is the mean of the combined distribution, then:
1. $\rm \overline{X} < \rm \overline{X}_1$
2. $\rm \overline{X} > \rm \overline{X}_2$
3. $\rm \overline{X}= \dfrac{\overline{X}_1+\overline{X}_2}{2}$
4. $\rm \overline{X}_1 < \overline{X} < \rm \overline{X}_2$

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Correct Answer - Option 4 : $\rm \overline{X}_1 < \overline{X} < \rm \overline{X}_2$

Concept:

• Average/Mean: Mean of 'n' observations = $\rm \overline x=\dfrac{Sum\ of\ Observations}{n}$.
• Combined\Weighted Average/Mean: If p sets with number of elements n1, n2, n3, ..., np have averages $\rm \overline x_1,\overline x_2,\overline x_3,\ ...,\overline x_p$ respectively, then the combined average of all the elements of all the p sets is: $\rm \overline x=\dfrac{n_1\overline x_1+n_2\overline x_2+n_3\overline x_3+\ ...\ +n_p\overline x_p}{n_1+n_2+n_3+\ ...\ +n_p}$.
• The combined average of two sets always lies between the individual averages.

Calculation:

Let's say that there are two sets with number of elements n1 and n2 and averages $\rm \overline{X}_1$ and $\rm \overline{X}_2$.

The sum of the values of both the sets will be $\rm n_1\overline{X}_1+n_2\overline{X}_2$ and the number of elements combined will be n1 + n2.

Since $\rm \overline{X}_1 < \rm \overline{X}_2$, we can say that:

$\rm (n_1+n_2)\overline{X}_1 <n_1\overline{X}_1+n_2\overline{X}_2< (n_1+n_2)\overline{X}_2$

⇒ $\rm \dfrac{(n_1+n_2)\overline{X}_1}{n_1+n_2} <\dfrac{n_1\overline{X}_1+n_2\overline{X}_2}{n_1+n_2}<\dfrac{(n_1+n_2)\overline{X}_2}{n_1+n_2}$

⇒ $\rm \overline{X}_1 < \overline{X} < \rm \overline{X}_2$.

∴ The combined average of two sets always lies between the individual averages, i.e. $\rm \overline{X}_1 < \overline{X} < \rm \overline{X}_2$.