Standard deviation of the first 2n + 1 natural numbers is equal to

Question

Standard deviation of the first 2n + 1 natural numbers is equal to (A) √(n(n+1)/2) (B) √(n(n+1)(2n+1)/3) (C) √(n(n+1)/3) (D) √(n(n-1)/2)

Tardigrade Admin · Accepted Answer

barx = mean =( 1+2+3+...+(2n+1)/(2n+1)) = (∑(2n+1)/(2n+1)) = ((2n+1)(2n+2)/2(2n+1)) = n+1 Varience = σ2 = (1/2n+1) ∑(xi- barx)2 = (2/2n+1) [n2+(n-1)2+.........+(1)2] = (2/2n+1) ∑ n2 = (2/2n+1)⋅(n(n+1)(2n+1)/6) = (n(n+1)/3) S.D. = √Varience = √(n(n+1)/3).

Q. Standard deviation of the first $2 n + 1$ natural numbers is equal to

$\frac{n ( n + 1 )}{2}$

$\frac{n ( n + 1 ) ( 2 n + 1 )}{3}$

$\frac{n ( n + 1 )}{3}$

$\frac{n ( n - 1 )}{2}$

Q. Standard deviation of the first 2n+1 natural numbers is equal to

2n(n+1)​​

3n(n+1)(2n+1)​​

3n(n+1)​​

2n(n−1)​​

xˉ= mean =(2n+1)1+2+3+...+(2n+1)​ =(2n+1)∑(2n+1)​ =2(2n+1)(2n+1)(2n+2)​ =n+1 Varience =σ2=2n+11​∑(xi​−xˉ)2 =2n+12​[n2+(n−1)2+.........+(1)2] =2n+12​∑n2=2n+12​⋅6n(n+1)(2n+1)​ =3n(n+1)​ S.D.=Varience​ =3n(n+1)​​.

Q. Standard deviation of the first $2 n + 1$ natural numbers is equal to

$\frac{n ( n + 1 )}{2}$

$\frac{n ( n + 1 ) ( 2 n + 1 )}{3}$

$\frac{n ( n + 1 )}{3}$

$\frac{n ( n - 1 )}{2}$