Standard deviation of first n natural number is 2. What is the value of n?

Option a: 7. Solution: Formula for SD of first n natural numbers: sigma = sqrt(n^2 - 1/12) Given: sigma = 2 Step 1: Set up equation. 2 = sqrt(n^2 - 1/12) Step 2: Square both sides. 4 = n^2 - 1/12 Step 3: Solve. n^2 - 1 = 48 n^2 = 49 n = 7 Verification: sigma = sqrt(49-1/12) = sqrt(48/12) = sqrt(4) = 2 ✓ Hence, Option A is the correct answer.

Central Tendency & DispersionMTP Nov 21Question 3179 of 473

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Standard deviation of first $\displaystyle n$ natural number is 2. What is the value of $\displaystyle n$ ?

Options

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Correct Answer

✅ Option a — 7

All Options:

Detailed Solution & Explanation

**Formula for SD of first

\displaystyle n

natural numbers:**

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

**Given:**

\displaystyle \sigma = 2

**Step 1:** Set up equation.

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

**Step 2:** Square both sides.

4 = \frac{n^2 - 1}{12}

**Step 3:** Solve.

n^2 - 1 = 48

n^2 = 49

n = 7

**Verification:**

\displaystyle \sigma = \sqrt{\frac{49-1}{12}} = \sqrt{\frac{48}{12}} = \sqrt{4} = 2

✓ Hence, **Option A** is the correct answer.

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Standard deviation of first n\displaystyle nn natural number is 2. What is the value of n\displaystyle nn?