Question

Find the variance of first n natural numbers.

Answer 1

Answer:

( n² - 1 ) / 12

Step-by-step explanation:

variance = E(X²) + [E(X)]²

             = (mean of their squares) - (square of their mean)

mean of first n natural numbers

= ( 1 + 2 + ... + n ) / n

= [ n(n+1) / 2 ] / n

= ( n + 1 ) / 2

mean of squares of first n natural numbers

= ( 1² + 2² + ... + n² ) / n

= [ n(n+1)(2n+1) / 6 ] / n

= ( n + 1 ) ( 2n + 1 ) / 6

So the variance is

( n + 1 ) ( 2n + 1 ) / 6   -   [ ( n + 1 ) / 2 ] ²

= [ ( n + 1 ) / 12 ] × [ 2(2n + 1) - 3(n + 1) ]

= [ ( n + 1 ) / 12 ] × [ 4n + 2 - 3n - 3 ]

= [ ( n + 1 ) / 12 ] × [ n - 1 ]

= ( n² - 1 ) / 12

Answer 2

Varience = {n∑(X²)−[ ∑(X)]²} / n²

We know, sum of n natural numbers = n(n+1)/2

Sum of squares of first n natural numbers = n(n+1)(2n+1)/6

Applying this to above equation of varience,

Varience = [n•n(n+1)(2n+1)/6 − n(n+1)/2²] /n²

= n²•(n+1)(2n+1)/6n² − n²(n+1)²/4n²

= [(n+1)(2n+1)/6 − (n+1)²] /4

= (2n²+3n+1)/6 − (n²+2n+1)/4

= (4n²+6n+2 − 3n²−6n−3) /12

= (n²−1)/12

Hence varience of n natural numbers = (n²−1)/12