Statistics
Combined Mean
Grade 11
Question:
<p>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</p>
<p>\(\bar{x} < \bar{x}_1\)</p>
<p>\(\bar{x} > \bar{x}_2\)</p>
<p>\(\bar{x} = \dfrac{\bar{x}_1 + \bar{x}_2}{2}\)</p>
<p>\(\bar{x}_1 < \bar{x} < \bar{x}_2\)</p>
Step-by-Step Solution
Key Concept: When combining two distributions, the overall mean is the weighted average of individual means, where weights are the respective frequencies. Use this relationship to find the combined mean or establish constraints on individual means.
<p><strong>Step 1:</strong> Write the formula for combined mean of two distributions.</p><p>The combined mean is: <strong>x̄ = (n₁x̄₁ + n₂x̄₂)/(n₁ + n₂)</strong></p><p><strong>Step 2:</strong> Use the given constraint x̄₁ < x̄.</p><p>Since x̄₁ < x̄, we have: x̄₁ < (n₁x̄₁ + n₂x̄₂)/(n₁ + n₂)</p><p>Multiply both sides by (n₁ + n₂):</p><p>x̄₁(n₁ + n₂) < n₁x̄₁ + n₂x̄₂</p><p>n₁x̄₁ + n₂x̄₁ < n₁x̄₁ + n₂x̄₂</p><p>n₂x̄₁ < n₂x̄₂</p><p><strong>Step 3:</strong> Conclude the relationship.</p><p>Since n₂ > 0 (positive frequency), we can divide by n₂:</p><p><strong>x̄₁ < x̄₂</strong></p><p>Similarly, x̄ < x̄₂ implies x̄₁ < x̄₂.</p><p>∴ <strong>Answer: D</strong> (x̄₁ < x̄₂)</p>
Correct Answer: D