Let `x_1, x_2, ... ,x_n` be n observations and `barX ` be their arithmetic mean. The standard deviation is given by

Options

A. (a) `sum_(i=1)^n(x_i- barX )^2`
B. (b) `1/nsum_(i=1)^n(x_i- barX )^2`
C. (c) `sqrt(1/nsum_(i=1)^n(x_i- X )^2)`
D. (d) `1/nsum_(i=1)^n x_i^2- barX ^2`

Answer

(c) `sqrt(1/nsum_(i=1)^n(x_i- X )^2)`

Explanation

We know,

Variance is a mean of sum of square of deviation of mean

Variance `= 1/nsum_(i=1)^n(x_i- X )`

Standard deviation is square root of variance
Standard deviation `= sqrt(1/nsum_(i=1)^n(x_i- X )^2)`

Hence, correct option is option 3