Central Tendency & DispersionMTP Nov 18Question 2887 of 473
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If the mean of the set of observations x1,x2,x3,…,xn\displaystyle x_1, x_2, x_3, \dots, x_n is xˉ\displaystyle \bar{x}, then the mean of the observation xi+ki\displaystyle x_i + ki, where i=1,2,3,…,n\displaystyle i = 1, 2, 3, \dots, n

Options

Axˉ+k(n+1)\displaystyle \bar{x} + k(n+1)
Bxˉ+kn\displaystyle \bar{x} + kn
Cxˉ+kn\displaystyle \bar{x} + \frac{k}{n}
Dxˉ+k2(n+1)\displaystyle \bar{x} + \frac{k}{2}(n+1)
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Correct Answer

✅ Option d — xˉ+k2(n+1)\displaystyle \bar{x} + \frac{k}{2}(n+1)

All Options:

  • Axˉ+k(n+1)\displaystyle \bar{x} + k(n+1)
  • Bxˉ+kn\displaystyle \bar{x} + kn
  • Cxˉ+kn\displaystyle \bar{x} + \frac{k}{n}
  • Dxˉ+k2(n+1)\displaystyle \bar{x} + \frac{k}{2}(n+1)

Detailed Solution & Explanation

**Step 1: Compute the new mean.** New observations are yi=xi+ki\displaystyle y_i = x_i + ki for i=1,2,…,n\displaystyle i = 1, 2, \ldots, n. yˉ=∑i=1n(xi+ki)n=∑xi+k∑i=1nin\bar{y} = \frac{\sum_{i=1}^{n}(x_i + ki)}{n} = \frac{\sum x_i + k\sum_{i=1}^{n} i}{n} **Step 2: Substitute known values.** ∑xin=xˉand∑i=1ni=n(n+1)2\frac{\sum x_i}{n} = \bar{x} \quad \text{and} \quad \sum_{i=1}^{n} i = \frac{n(n+1)}{2} yˉ=xˉ+k⋅n(n+1)2n=xˉ+k(n+1)2\bar{y} = \bar{x} + \frac{k \cdot \frac{n(n+1)}{2}}{n} = \bar{x} + \frac{k(n+1)}{2} **Note:** The derived answer is xˉ+k(n+1)2\displaystyle \bar{x} + \frac{k(n+1)}{2} (Option D), not Option A. Option A says xˉ+k(n+1)\displaystyle \bar{x} + k(n+1), which is twice the correct value. The correct derivation gives **Option D**. Hence, **Option D** is the correct answer.

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