Central Tendency & DispersionMTP June 24 Series IIQuestion 3210 of 473
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For a set of 100\displaystyle 100 observations, taking assumed mean as 4\displaystyle 4, the sum of the deviations is −11\displaystyle -11 cm, and the sum of the squares of these deviations is 257\displaystyle 257 cm2\displaystyle ^2. The coefficient of variation is:

Options

A41.13%\displaystyle 41.13\%
BNone of these
C40.13%\displaystyle 40.13\%
D40.13%\displaystyle 40.13\%
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Correct Answer

✅ Option a — 41.13%\displaystyle 41.13\%

All Options:

  • A41.13%\displaystyle 41.13\%
  • BNone of these
  • C40.13%\displaystyle 40.13\%
  • D40.13%\displaystyle 40.13\%

Detailed Solution & Explanation

**Given:** n=100\displaystyle n = 100, Assumed Mean A=4\displaystyle A = 4, ∑di=−11\displaystyle \sum d_i = -11, ∑di2=257\displaystyle \sum d_i^2 = 257 (where di=xi−A\displaystyle d_i = x_i - A) **Step 1: Calculate actual mean.** xˉ=A+∑din=4+−11100=4−0.11=3.89\bar{x} = A + \frac{\sum d_i}{n} = 4 + \frac{-11}{100} = 4 - 0.11 = 3.89 **Step 2: Calculate Variance.** σ2=∑di2n−(∑din)2=257100−(−11100)2\sigma^2 = \frac{\sum d_i^2}{n} - \left(\frac{\sum d_i}{n}\right)^2 = \frac{257}{100} - \left(\frac{-11}{100}\right)^2 =2.57−0.0121=2.5579= 2.57 - 0.0121 = 2.5579 **Step 3: Calculate SD.** σ=2.5579≈1.5993\sigma = \sqrt{2.5579} \approx 1.5993 **Step 4: Calculate CV.** CV=σxˉ×100=1.59933.89×100≈41.11%CV = \frac{\sigma}{\bar{x}} \times 100 = \frac{1.5993}{3.89} \times 100 \approx 41.11\% This is approximately 41.13%\displaystyle 41.13\% (Option A). Hence, **Option A** is the correct answer.

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