If the S.D. of the 1^{st} n natural numbers is sqrt(30) then the value of n is

Option a: 19. Solution: The standard deviation (S.D.) of the first n natural numbers is given by the formula: text{S.D.} = sqrt(n^2 - 1/12) We are given the standard deviation is sqrt(30). Set up the equation: sqrt(30) = sqrt(n^2 - 1/12) Square both sides: 30 = n^2 - 1/12 360 = n^2 - 1 n^2 = 361 implies n = 19 Hence, Option A is the correct answer.

If the S.D. of the $1^{st}$ $n$ natural numbers is $\sqrt{30}$ then the value of $n$ is

The correct answer is Option a: $19$. The standard deviation (S.D.) of the first $n$ natural numbers is given by the formula: $$\text{S.D.} = \sqrt{\frac{n^2 - 1}{12}}$$ We are given the standard deviation is $\sqrt{30}$. Set up the equation: $$\sqrt{30} = \sqrt{\frac{n^2 - 1}{12}}$$ Square both sides: $$30 = \frac{n^2 - 1}{12}$$ $$360 = n^2 - 1$$ $$n^2 = 361 \implies n = 19$$ Hence, **Option A** is the correct answer.

Central Tendency & DispersionPYQ May 18Question 3115 of 473

All Questions

If the S.D. of the $\displaystyle 1^{st}$ $\displaystyle n$ natural numbers is $\displaystyle \sqrt{30}$ then the value of $\displaystyle n$ is

Options

\displaystyle 19

\displaystyle 20

\displaystyle 21

DNone of these

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Correct Answer

✅ Option a — $\displaystyle 19$

All Options:

A $\displaystyle 19$
B $\displaystyle 20$
C $\displaystyle 21$
DNone of these

Detailed Solution & Explanation

The standard deviation (S.D.) of the first

\displaystyle n

natural numbers is given by the formula:

\text{S.D.} = \sqrt{\frac{n^2 - 1}{12}}

We are given the standard deviation is

\displaystyle \sqrt{30}

. Set up the equation:

\sqrt{30} = \sqrt{\frac{n^2 - 1}{12}}

Square both sides:

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

360 = n^2 - 1

n^2 = 361 \implies n = 19

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

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If the S.D. of the 1st\displaystyle 1^{st}1st n\displaystyle nn natural numbers is 30\displaystyle \sqrt{30}30​ then the value of n\displaystyle nn is