Statistics
Mean, Mode and Variance
Grade 11
Question:
<p>Let \(\bar{x}\), \(M\) and \(\sigma^2\) be respectively, the mean, mode and variance of \(n\) observations \(x_1, x_2, ..., x_n\) and \(d_i = -x_i - a\), \(i = 1, 2, ..., n\), where \(a\) is any number.</p><p><strong>Statement-1:</strong> Variance of \(d_1, d_2, ..., d_n\) is \(\sigma^2\).</p><p><strong>Statement-2:</strong> Mean and mode of \(d_1, d_2, ..., d_n\) are \(-\bar{x} - a\) and \(-M - a\), respectively.</p>
<p>Statement-1 and Statement-2 are both true.</p>
<p>Statement-1 and Statement-2 are both false.</p>
<p>Statement-1 is true and Statement-2 is false.</p>
<p>Statement-1 is false and Statement-2 is true.</p>
Step-by-Step Solution
Key Concept: Variance is invariant under linear transformations of the form y = -x + c (only affected by the coefficient of x, not the constant term), while mean and mode transform directly according to the linear transformation applied to each observation.
<p><strong>Step 1:</strong> Analyze how d<sub>i</sub> = -x<sub>i</sub> - a transforms the data.</p><p><strong>Step 2 (Mean):</strong> Mean of d<sub>i</sub> = E[d<sub>i</sub>] = E[-x<sub>i</sub> - a] = -E[x<sub>i</sub>] - a = -x̄ - a ✓</p><p><strong>Step 3 (Mode):</strong> If x<sub>i</sub> = M is the mode (most frequent value), then d<sub>i</sub> = -M - a is the corresponding mode. Mode transforms directly: -M - a ✓</p><p><strong>Step 4 (Variance):</strong> Var(d<sub>i</sub>) = Var(-x<sub>i</sub> - a) = Var(-x<sub>i</sub>) = (-1)² · Var(x<sub>i</sub>) = σ². The constant term -a does not affect variance; only the coefficient (-1) matters. ✓</p><p><strong>Step 5:</strong> Both Statement-1 and Statement-2 are correct. Statement-2 correctly explains why Statement-1 is true (the transformation properties ensure variance preservation).</p><p>∴ Answer: A (Both statements are true, and Statement-2 is the correct explanation of Statement-1)</p>
Correct Answer: A