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प्रश्न
Find the standard deviation of the first 21 natural numbers.
बेरीज
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उत्तर
Here n = 21
The standard deviation of the first ‘n’ natural numbers,
= `sqrt(("n"^2 - 1)/12)`
1, 2, 3, 4, ..., 21 = `sqrt((21^2 - 1)/12)`
= `sqrt((441 - 1)/12)`
= `sqrt(440/12)`
= `sqrt(36.666)`
= `sqrt(36.67)`
= 6.055
= 6.06
The standard deviation of the first 21 natural numbers = 6.06
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