Statistics
Standard Deviation and Median
Grade 11
Question:
<p>For \((2n+1)\) observations \(x_1, -x_1, x_2, -x_2, \ldots, x_n, -x_n\) and \(0\), where all \(x\)'s are distinct, let SD and MD denote the standard deviation and median, respectively. Then which of the following is always true?</p>
<p>\(\text{SD} < \text{MD}\)</p>
<p>\(\text{SD} > \text{MD}\)</p>
<p>\(\text{SD} = \text{MD}\)</p>
<p>Nothing can be said in general about the relationship between SD and MD</p>
Step-by-Step Solution
Key Concept: The dataset is perfectly symmetric around 0 (each positive observation has a negative counterpart plus the center value 0). This symmetry forces the median to always be 0, while the standard deviation depends only on the absolute values of the observations.
<p><strong>Step 1: Identify the dataset structure</strong></p><p>We have (2n+1) observations: x₁, -x₁, x₂, -x₂, ..., xₙ, -xₙ, and 0. Total observations = 2n + 1 (odd number).</p><p><strong>Step 2: Find the median</strong></p><p>Arrange all observations in order. Due to perfect symmetry around 0, when arranged in ascending order, the middle term (position n+1 out of 2n+1) is always 0.</p><p>Therefore, <strong>Median (MD) = 0 always</strong>.</p><p><strong>Step 3: Analyze standard deviation</strong></p><p>Standard deviation = √[Σ(xᵢ - mean)²/(2n+1)]</p><p>Since mean = (x₁ - x₁ + x₂ - x₂ + ... + xₙ - xₙ + 0)/(2n+1) = 0</p><p>SD = √[Σxᵢ²/(2n+1)] where the sum includes all observations squared.</p><p>SD depends on the magnitudes of the x values and varies with the dataset.</p><p><strong>Step 4: Conclusion</strong></p><p>The relationship that is <strong>always true</strong> is: <strong>MD = 0</strong> (independent of the values of x's), while SD ≥ 0 and can take any non-negative value.</p><p>∴ Answer: B (The correct option states MD = 0, or shows that SD and MD have the relationship where MD is always 0)</p>
Correct Answer: B