The variance of first $20$ natural number is

\frac{133}{4}

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\frac{279}{12}

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\frac{133}{2}

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\frac{399}{4}

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Solution

The correct option is D $\frac{133}{4}$
Variance of $n$ natural numbers
$\frac{n^{2} - 1}{12} = \frac{(20)^{2} - 1}{12}$ $[∵ n = 20]$

$\frac{400 - 1}{12} = \frac{399}{12} = \frac{133}{4}$

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Similar questions

Q. The variance of first

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natural numbers is

Q. Variance of first

20

natural number is

Q. The variance of first n natural numbers is

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

Q. Explain the method of mathematical induction and use it to show that

11^{n + 2} + 12^{2 n + 1}

where n is natural number is divisible by 133

Verify each of the following :

(i) $(\frac{5}{7} \times \frac{12}{13}) \times \frac{7}{18} = \frac{5}{7} \times (\frac{12}{13} \times \frac{7}{18})$
(ii) $\frac{- 13}{24} \times (\frac{- 12}{5} \times \frac{35}{36}) = (\frac{- 13}{24} \times \frac{- 12}{5}) \times \frac{35}{36}$
(iii) $(\frac{- 9}{5} \times \frac{- 10}{3}) \times \frac{21}{- 4} = \frac{- 9}{5} \times (\frac{- 10}{3} \times \frac{21}{- 4})$

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The variance of first 20 natural number is

The correct option is D 1334Variance of n natural numbersn2−112=(20)2−112 [∵n=20]400−112=39912=1334

The variance of first $20$ natural number is

The correct option is D $\frac{133}{4}$
Variance of $n$ natural numbers
$\frac{n^{2} - 1}{12} = \frac{(20)^{2} - 1}{12}$ $[∵ n = 20]$

$\frac{400 - 1}{12} = \frac{399}{12} = \frac{133}{4}$