 Hello and welcome to the session. In this session we will learn about general equations of lines of regression. Now we know that the electric form of line of regression of y on x is given as y is equal to n x plus c is given as x is equal to n y plus c where in the first equation the coefficient of x that is m is called the regression coefficient of y on x and in the second equation the coefficient of y that is m is called the regression coefficient of x on y and c is a constant. Now the regression equation is obtained by fitting a straight line equal to n y plus c values y and x by the method least squares. Now in the method of least squares a regression line is so chosen that the sum of the squares of the deviations parallel to the x's of x is minimized and then it is called the line of regression of x and y. Now let n y1 and sum up to x and yn be the n pairs of observations and let the line that is the regression line by the method of least squares bx is equal to n y plus c. Now for the observed value y1 of y we got which are that is the observed value printed value. So for the observed value y1 of y we are getting two values of x that is the observed value x1 and the s2 and y1 plus c. Now the difference between these two values minus x1 is the regression which is made. Now the line AB is of the form that is the equation of the line AB is x is equal to n y plus c and this is one kind of observation which famous on x1 y1. So for the observed value y1 of y we are getting that is the observed value which is x1 and the estimation value which is n y1 plus c minus x1 which is called the deviation d1 that is for the clear record. For the observed value y2 of y we are getting the observed value x2 and the estimate value of the deviation and the differences d4 and so on up to so we can write similarly d2 d3 and so on up to plus c minus x2 3 plus c minus x3 plus c minus x the values d1 square plus d2 square minimum. The values of m of deviations parallel to the axis of x is minimum d1 d2 and so at d1 square plus d2 square plus so on up to dm square is minimum that is 7 plus c minus x1 whole square plus c minus x2 whole square plus c minimum. Now let z is equal to m y1 plus c minus x1 whole square plus m y2 plus c minus x2 whole square that is minimum. Now for the differential values we write the differential coefficient equal to 0 then a minimum value of z. Now this is the v is equal to 2 into m y1 plus c minus x1 whole plus 2 plus c minus x2 whole plus c minus x1 whole is equal to c minus x1 whole plus c minus x2 whole c minus x1 whole plus y2 plus so on up to n the whole plus n c minus the whole is equal to 0. 7 square plus y2 square yn square the whole plus y2 plus so on up to n the whole y2 plus so on up to yn will be equal to summation y we have m into summation y square to summation y minus summation of x into y is equal to 0 into summation y plus summation y square plus c into summation y. Now let us see equation a with b thus c is the equation of line of regression then the constant novel equation b summation x into summation y to summation y square minus summation y which is further equal to x into y minus summation x into summation y minus summation y summation x into summation y square minus summation of x into summation y y square minus summation y y plus c that is the coefficient of y represent the regression coefficient which is the regression coefficient coefficient of x and y is equal to summation y over n summation y square minus summation y for the regression c plus m into summation x is equal to c into summation x plus and solving these two equations x into y minus summation x into summation y over n summation x square minus summation x whole square which is equal to summation of x into y minus to summation y over n rather than into summation x square minus summation into summation x into summation x square minus summation x now we know that in the equation equation of line of regression of y on x that is y is equal to mx plus c which is m represents the regression coefficient that is by x now here we have got this as the value of m therefore by x is equal to summation x y minus summation x into summation y over n x square minus summation x in the normal equations which we have obtained earlier x and y indicate the observed values of the respective variables which we have obtained we have this we have used a simple notation for the sake of simplicity where summation x is equal to summation xi where i varies from 1 to n is equal to summation into yi where i varies from 1 to n etc we have learnt about of regression lines so this completes our session and if you all have enjoyed the session