<p>Consider any set of observations \(x_1, x_2, x_3, \ldots, x_{101}\). It is given that \(x_1 < x_2 < x_3 < \ldots < x_{100} < x_{101}\); then the mean deviation of this set of observations about a point \(k\) is minimum when \(k\) equals</p>
Step-by-Step Solution
Key Concept: When a constant is added to all observations, the mean increases by that constant while variance remains unchanged. Use the relationship between variance, mean, and sum of squares.
<p><strong>Step 1:</strong> Let the original set be {x₁, x₂, ..., x₁₀₁} with mean μ and variance σ².</p><p><strong>Step 2:</strong> Given: x₁ < x₂ < ... < x₁₀₁ with mean = 20, variance = 5.</p><p><strong>Step 3:</strong> The new set after adding 50 to each observation is {x₁+50, x₂+50, ..., x₁₀₁+50}.</p><p><strong>Step 4:</strong> New mean = μ + 50 = 20 + 50 = 70</p><p><strong>Step 5:</strong> Variance property: Var(X + c) = Var(X) for any constant c. This is because variance measures dispersion/spread around the mean, which doesn't change when all points shift by the same amount.</p><p><strong>Step 6:</strong> New variance = σ² = 5</p><p><strong>Step 7:</strong> Therefore, mean = 70 and variance = 5.</p><p>∴ Answer: B</p>
Correct Answer: B