Question:
<p>Let observations be \(a_1, a_2, \ldots, a_n, a_{n+1}, a_{n+2}, \ldots, a_{2n}\). If \((a_1+5), (a_2+5), \ldots, (a_n+5)\) and \((a_{n+1}-3), (a_{n+2}-3), \ldots, (a_{2n}-3)\) are the new observations, then the new mean \(\bar{x}'\) equals:</p>
<p>\(\bar{x}\)</p>
<p>\(\bar{x} + 1\)</p>
<p>\(\bar{x} + 2\)</p>
<p>\(\bar{x} - 1\)</p>
Step-by-Step Solution
Key Concept: When each observation in a dataset is shifted by a constant, the mean shifts by that same constant. Split the 2n observations into two groups with different shifts, then use the weighted average principle for combined means.
<p><strong>Step 1:</strong> Let the original mean of all 2n observations be x̄. Then: (a₁ + a₂ + ... + a₂ₙ) = 2nx̄</p><p><strong>Step 2:</strong> Split into two groups. Let x̄₁ = mean of (a₁, a₂, ..., aₙ) and x̄₂ = mean of (aₙ₊₁, ..., a₂ₙ). Then: x̄ = (nx̄₁ + nx̄₂)/(2n) = (x̄₁ + x̄₂)/2</p><p><strong>Step 3:</strong> New observations: (a₁+5), ..., (aₙ+5) have mean (x̄₁ + 5), and (aₙ₊₁-3), ..., (a₂ₙ-3) have mean (x̄₂ - 3)</p><p><strong>Step 4:</strong> New combined mean: x̄' = [n(x̄₁+5) + n(x̄₂-3)]/(2n) = (x̄₁ + x̄₂ + 5 - 3)/2 = [(x̄₁ + x̄₂)/2] + 1 = x̄ + 1</p><p>∴ Answer: B</p>
Correct Answer: B