Show that the two formulae for the standard deviation of ungrouped data: `sigma=sqrt(1/nsum(x_i- barX )^2)` `a n d\ sigma^(prime)=sqrt(1/nsumx_i^2- barX ^2)` are equivalent where `barX =1/nsumx_i`.

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

If standard deviation of 1,2,3,4,….,10 is sigma then standard deviation of 11, 12, …20 is

For a frequency distribution standard deviation is computed by applying the formula (a) sigma=sqrt((sumfd^2)/(sumf)-((sumfd)/(sumf))^2) (b) sigma=sqrt(((sumfd)/(sumf))^2-(sumfd^2)/(sumf)) (c) sigma=sqrt((sumfd^2)/(sumf)-(sumfd)/(sumf)) (d) sigma=sqrt(((sumfd)/(sumf))^2-(sumfd^2)/(sumf))

For a frequency distribution standard deviation is computed by applying the formula (a) sigma=sqrt((sumfd^2)/(sumf)-((sumfd)/(sumf))^2) (b) sigma=sqrt(((sumfd^2)/(sumf))-(sumfd^2)/(sumf)) (c) sigma=sqrt((sumfd^2)/(sumf)-(sumfd)/(sumf)) (d) sqrt(((sumfd)/(sumf))^2-(sumfd^2)/(sumf))

Express the following complex numbers in the standard form a+i b :(5+sqrt(2)i)/(1-sqrt(2)i)

The standard deviation of n observations x_(1), x_(2), x_(3),…x_(n) is 2. If sum_(i=1)^(n) x_(i)^(2) = 100 and sum_(i=1)^(n) x_(i) = 20 show the values of n are 5 or 20

If standard deviation of 1,2,3,4….,10 is sigma then, show that standard deviation of 11, 12,…. , 20 is sigma

The standard deviation (sigma) of variate x is the square root of the A.M. of the squares of all deviations of x from the A.M. observations. If x_i//f_i , i=1,2,… n is a frequency distribution then sigma=sqrt(1/N Sigma_(i=1)^(n) f_i(x_i-barx)^2), N=Sigma_(i=1)^(n) f_i and variance is the square of standard deviation. Coefficient of dispersion is sigma/x and coefficient of variation is sigma/x xx 100 The standard deviation for the set of numbers 1,4,5,7,8 is 2.45 then coefficient of dispersion is

The standard deviation (sigma) of variate x is the square root of the A.M. of the squares of all deviations of x from the A.M. observations. If x_i//f_i , i=1,2,… n is a frequency distribution then sigma=sqrt(1/N Sigma_(i=1)^(n) f_i(x_i-barx)^2), N=Sigma_(i=1)^(n) f_i and variance is the square of standard deviation. Coefficient of dispersion is sigma/x and coefficient of variation is sigma/x xx 100 For a given distribution of marks mean is 35.16 and its standard deviation is 19.76 then coefficient of variation is

Let x_(1),x_(2),x_(3),……..,x_(n) be n observations with mean barx and standard deviation sigma . The mean the standard deviation of kx_(1),kx_(2),……,kx_(n) respectively are (i) barx,ksigma (ii) kbarx,sigma (iii) kbarx,ksigma (iv) barx,sigma