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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. \(\rm \overline{X}_1 < \overline{X} < \rm \overline{X}_2\).

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