Central Tendency & DispersionPYQ Sep 24Question 3148 of 473
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The Standard Deviation of the series is 3,6,9,12,15\displaystyle 3, 6, 9, 12, 15 is:

Options

A6.36\displaystyle 6.36
B4.24\displaystyle 4.24
C4.12\displaystyle 4.12
D3.28\displaystyle 3.28
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Correct Answer

✅ Option b — 4.24\displaystyle 4.24

All Options:

  • A6.36\displaystyle 6.36
  • B4.24\displaystyle 4.24
  • C4.12\displaystyle 4.12
  • D3.28\displaystyle 3.28

Detailed Solution & Explanation

**Given data:** 3, 6, 9, 12, 15; n=5\displaystyle n = 5 **Step 1: Calculate Mean.** xˉ=3+6+9+12+155=455=9\bar{x} = \frac{3+6+9+12+15}{5} = \frac{45}{5} = 9 **Step 2: Calculate squared deviations.** xi(xi−xˉ)(xi−xˉ)23−6366−39900123915636∑=90\begin{array}{|c|c|c|} \hline x_i & (x_i - \bar{x}) & (x_i - \bar{x})^2 \\ \hline 3 & -6 & 36 \\ 6 & -3 & 9 \\ 9 & 0 & 0 \\ 12 & 3 & 9 \\ 15 & 6 & 36 \\ \hline & & \sum = 90 \\ \hline \end{array} **Step 3: Calculate Variance.** σ2=905=18\sigma^2 = \frac{90}{5} = 18 **Step 4: Calculate SD.** σ=18=32≈3×1.4142=4.243≈4.24\sigma = \sqrt{18} = 3\sqrt{2} \approx 3 \times 1.4142 = 4.243 \approx 4.24 Alternatively: the data 3,6,9,12,15 is 3 × (1,2,3,4,5). SD of (1,2,3,4,5) = 2\displaystyle \sqrt{2} (formula: (n2−1)/12\displaystyle \sqrt{(n^2-1)/12} = 24/12\displaystyle \sqrt{24/12} = 2\displaystyle \sqrt{2}). So SD = 32≈4.24\displaystyle 3\sqrt{2} \approx 4.24. But option C is 4.12, not 4.24. Let me verify: 18=4.2426...\displaystyle \sqrt{18} = 4.2426..., so Option B (4.24) is the accurate answer. Hence, **Option B** is the correct answer.

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