For a bivariate date, you are given following information:
` sum x = 125 , sum x^(2) = 1650 , sum y = 100 , sum y^(2) = 1500 , sum xy= 50 , n= 25`
Find line of regression.

If sum x = 50, sum y = - 30, sum xy = 115 and n = 10 , the covariance is

If sum x_(1) = 16, sum y_(1) = 48 sum (x_(1) - 3) (y_(1) - 4) = 22 and n = 2 , find the cov (x,y).

In a blvariate data sum x=30, sumy=400, sumx^(2)=196, sumxy=850 and n=10 . Find the regression coefficient of y on x.

Calculate Kari Pearson's coefficient of correlation between the values of x and y for the following data : n = 10, sum x = 55, sum y = 40, sum x^(2) = 385, sum y^(2) = 192 and sum (x + y)^(2) = 947

For a series the information available is n = 10 sum x = 60, sum x^(2) = 1000 . The standard deviation is

x and y are a pair of correlated variables . Ten observations of their values (x_(p)y_(t)) have the following result : sum x = 55, sum y = 55, sum x^(2) = 385, sum xy = 350 . Predict the value of y when the value of x is 6

If xy + yz + zx = 1 then sum (x + y)/(1 - xy)=

If n= 10, sum_(i=1)^(10) x_(i) = 60 and sum_(i=1)^(10) x_(i)^(2) = 1000 then find s.d.

Given r = 0.8, sum xy = 60 , sigma_(y) = 2.5 and sum x^(2) = 90 , find the number of items, where x and y are deviation from their respective means.

For the given data the calculation corresponding to all values of series (x,y) is as following sum(x-barx)^(2)=35, sum(y-bary)^(2)=25, sum(x-barx)(y-bary)=20 Find Karl Pearson's correlation coefficient.