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Question
Find the standard deviation of the first n natural numbers.
Solution
| `x_i` | 1 | 2 | 3 | 4 | 5 | – | – | n |
| `x_i^2` | 1 | 4 | 9 | 16 | 25 | – | – | n2 |
`sumx_i = 1 + 2 + 3 + 4 + 5 + ... + n = (n(n + 1))/2`
`sumx_i^2 = 1^2 + 2^2 + 3^2 + ... + n^2 = (n(n + 1)(2n + 1))/6`
∴ S.D. `(sigma) = sqrt((sumx_i^2)/n - ((sumx_i)/n)^2`
= `sqrt((n(n + 1)(2n + 1))/(6n) - (n^2(n + 1)^2)/(4n^2))`
= `sqrt(((n + 1)(2n + 1))/6 - (n + 1)^2/4`
= `sqrt((2n^2 + 3n + 1)/6 - (n^2 + 2n + 1)/4)`
= `sqrt((4n^2 + 6n + 2 - 3n^2 - 6n - 3)/12`
= `sqrt((n^2 - 1)/12)`
Hence, the required S.D. = `sqrt((n^2 - 1)/12`
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