The mean deviation for `n` observations `x_1, x_2, ......... , x_n` from their mean ` X ` is given by (a)`sum_(i=1)^n(x_i- X )` (b) `1/nsum_(i=1)^n(x_i- X )` (c) `sum_(i=1)^n(x_i- X )^2` (d) `1/nsum_(i=1)^n(x_i- X )^2`

The mean deviation for n observations x_1, x_2, ......... , x_n from their mean X is given by (a) sum_(i=1)^n(x_i- X ) (b) 1/nsum_(i=1)^n|x_i- X | (c) sum_(i=1)^n(x_i- X )^2 (d) 1/nsum_(i=1)^n(x_i- X )^2

The mean deviation for n observations x_(1),x_(2),.........,x_(n) from their mean X is given by (a)sum_(i=1)^(n)(x_(i)-X)(b)(1)/(n)sum_(i=1)^(n)(x_(i)-X)(c)sum_(i=1)^(n)(x_(i)-X)^(2)(c)(1)/(n)sum_(i=1)^(n)(x_(i)-X)^(2)

The mean deviation for n observations x_(1),x_(2)…….x_(n) from their median M is given by (i) sum_(i=1)^(n)(x_(i)-M) (ii) (1)/(n)sum_(i=1)^(n)|x_(i)-M| (iii) (1)/(n)sum_(i=1)^(n)(x_(i)-M)^(2) (iv) (1)/(n)sum_(i=1)^(n)(x_(i)-M)

Let x_1, x_2, ... ,x_n be n observations and barX be their arithmetic mean. The standard deviation is given by (a) sum_(i=1)^n(x_i- barX )^2 (b) 1/nsum_(i=1)^n(x_i- barX )^2 (c) sqrt(1/nsum_(i=1)^n(x_i- X )^2) (c) 1/nsum_(i=1)^n x_i^2- barX ^2

Let x_(1),x_(2),x_(n) be n observations and X be their arithmetic mean.The standard mean is given by sum_(i=1)^(n)(x_(i)-X)^(2) (b) (1)/(n)sum_(i=1)^(n)(x_(i)-X)^(2) (c) sqrt((1)/(n)sum_(i=1)^(n)(x_(i)-X)^(2))( c) (1)/(n)sum_(i=1)^(n)xi2-X^(2)

If barx=1/n sum_(i=1)^n x_i then prove that sum_(i=1)^n (x_i-barx)=0

The standard deviation of n observations x_(1),x_(2),......,x_(n) is 2. If sum_(i=1)^(n)x_(i)=20 and sum_(i=1)^(n)x_(i)^(2)=100 then n is

Prove that identity : sum_(i=1)^(n) (x_i-bar x)^2 = sum_(i=1)^(n) x_i^2-n bar x^2= sum_(i=1)^(n) x_i^2 -(sum_(i=1)^(n) x_i)^2/n .

If sum_(i=1)^(n)sin x_(i)=n then sum_(i=1)^(n)cot x_(i)=