Question

find standard deviation of first 21 nnatural number?

Answer 1

Answer:

The standard deviation of first 21 natural number is √36.66

We know that б =  √ [Σxi²/n - (Σxi/n)²]

Σxi = [n * (n + 1)] / 2 .........(i)

Σxi² =  [n * (n + 1) * (2n + 1)] / 6   .........(ii)

here, n = 21

now we have to put the value of  n in equation (i) and (ii)

Σxi = ( 21 * 22 ) / 2 = 231

Σxi² = ( 21 * 22 * 43 ) / 6 = 3311

now ,

б =  √ [ (3311 / 21) - (231/21)² ]

б =  √ [ 157.66 - 11² ]

б =  √  [ 157.66 - 121 ] = √36.66

So the standard deviation of first 21 natural number is √36.66

Answer 2

Answer:

√n^2/12

√21^2-1/12

√441-1/12

√440/12

√36.6

6.05