Central Tendency & DispersionPYQ July 21Question 3132 of 473

The SD of 1 to 9 natural number is:

\displaystyle 6.65

\displaystyle 2.58

\displaystyle 6.75

\displaystyle 5.62

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Correct Answer

✅ Option b — $\displaystyle 2.58$

**Formula for SD of first

\displaystyle n

natural numbers:**

\sigma = \sqrt{\frac{n^2 - 1}{12}}

**Given:**

\displaystyle n = 9

**Step 1:** Apply the formula.

\sigma = \sqrt{\frac{9^2 - 1}{12}} = \sqrt{\frac{81 - 1}{12}} = \sqrt{\frac{80}{12}}

**Step 2:** Simplify.

\sigma = \sqrt{\frac{80}{12}} = \sqrt{6.\overline{6}} = \sqrt{6.667} \approx 2.582

**Verification by direct computation:** - Data: 1, 2, 3, 4, 5, 6, 7, 8, 9 -

\displaystyle n = 9

\displaystyle \bar{x} = \frac{1+2+...+9}{9} = \frac{45}{9} = 5

\displaystyle \sum x_i^2 = 1+4+9+16+25+36+49+64+81 = 285

\displaystyle \sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2 = \frac{285}{9} - 25 = 31.\overline{6} - 25 = 6.\overline{6}

\displaystyle \sigma = \sqrt{6.\overline{6}} \approx 2.58

Hence, **Option B** is the correct answer.

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