Statistics
Mean Deviation
Grade 11
Question:
<p>Let \(\bar{X}\) and MD be the mean and the mean deviation about \(\bar{X}\) of \(n\) observations \(x_i,\ i = 1, 2, \ldots, n\). If each of the observations is increased by 5, then the new mean and the mean deviation about the new mean, respectively, are</p>
<p>\(\bar{X},\ \text{MD}\)</p>
<p>\(\bar{X} + 5,\ \text{MD}\)</p>
<p>\(\bar{X},\ \text{MD} + 5\)</p>
<p>\(\bar{X} + 5,\ \text{MD} + 5\)</p>
Step-by-Step Solution
Key Concept: Mean deviation is a measure of absolute deviations from the center; it depends only on the distances from the mean, not on the location of the data. When data shifts by a constant, the mean shifts by that constant, but relative distances remain unchanged.
<p><strong>Step 1:</strong> Original mean is <strong>X̄</strong> and original mean deviation is <strong>MD</strong>.</p><p><strong>Step 2:</strong> When each observation is increased by 5, the new observations become (x₁ + 5), (x₂ + 5), ..., (xₙ + 5).</p><p><strong>Step 3:</strong> New mean = (1/n)Σ(xᵢ + 5) = (1/n)Σxᵢ + 5 = <strong>X̄ + 5</strong></p><p><strong>Step 4:</strong> New mean deviation = (1/n)Σ|xᵢ + 5 - (X̄ + 5)| = (1/n)Σ|xᵢ - X̄| = <strong>MD</strong></p><p><strong>Step 5:</strong> The mean deviation depends only on deviations from the mean. Since (xᵢ + 5) - (X̄ + 5) = xᵢ - X̄, the absolute deviations remain identical.</p><p>∴ <strong>New mean = X̄ + 5, New mean deviation = MD</strong></p>
Correct Answer: B