sum(xi-bar(x))=

Find the mean bar(x) of first 8 natural numbers. Also verify sum(x-bar(x))=0

variance of 6 observation is 250 then sum(x_i-barx)^2=1200.

If barx is the mean of x_1, x_2, x_3, ... .. , x_n , then sum_(i=1)^(n)(x_i-barx)=

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

Fill in the blanks If (sumx_i)/n= barx then (sum(x_i-a))/n -_______.

If bar(X) is the mean of a distribution of X, then under usual notation sum sum_(i=1)^(n)f_(i)(x_(2)-bar(x)) 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 for distribution of 18 observations sum(x_i-5)=3a n dsum(x_i-5)^2=43 , find the mean and standard deviation.

If for distribution of 18 observations sum(x_i-5)=3a n dsum(x_i-5)^2=43 , find the mean and standard deviation.