\( X 1\) and \( X..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep.">
\( X... - Solved | MCA Prep" />
\( X 1\) and \( X..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep." />
\( X... - Solved | MCA Prep" />
\( X 1\) and \( X..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep." />
If <span class="math-tex">\(\rm \overline{X}_1\)</span> and <span class="math-tex">\(\rm \overline{X}_2\)</span> are the means of two distributions such that <span class="math-tex">\(\rm \overline{X}_1 < \rm \overline{X}_2\)</span>, and <span class="math-tex">\(\rm \overline{X}\)</span> is the mean of the combined distribution, then: Step-by-step Solution: If \(\overline{X}_1\) and \(\overline{X}_2\) are the means of two distributions with \(\overline{X}_1 < \overline{X}_2\), and \(\overline{X}\) is the mean of the combined distribution, then:
The combined mean \(\overline{X}\) is given by:
\[
\overline{X} = \frac{n_1\overline{X}_1 + n_2\overline{X}_2}{n_1 + n_2}
\]
where \(n_1\) and \(n_2\) are the number of observations in the two distributions.
\[
Key Points:
\]
- Since \(\overline{X}_1 < \overline{X}_2\), the combined mean \(\overline{X}\) will always lie between \(\overline{X}_1\) and \(\overline{X}_2\).
- The exact position of \(\overline{X}\) depends on the sizes \(n_1\) and \(n_2\):
- If \(n_1 = n_2\), then \(\overline{X}\) is exactly the average of \(\overline{X}_1\) and \(\overline{X}_2\).
- If \(n_1 > n_2\), \(\overline{X}\) will be closer to \(\overline{X}_1\).
- If \(n_2 > n_1\), \(\overline{X}\) will be closer to \(\overline{X}_2\).
\[
Final Result:
\]
\[
\overline{X}_1 < \overline{X} < \overline{X}_2
\]Question 16
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