 Hello and welcome to the session. In this session, we will discuss the question which says that the lines of regression of a set of data are x is equal to 0.51y minus 1.12, y is equal to 0.97x plus 7.15. The values of x is equal to 6.78. Find first part new value of x and y. Second part, coefficient B yx and B xy. Third part, standard deviation of y. Fourth part, the value of y for x is equal to 2. And fifth part, y is equal to 3. Now, before starting, we should know some results of regression y bar that is, secondly, the regression coefficient of y on x that is given by the formula r into, that is, is given by the formula over sigma y. Where r is the correlation coefficient, sigma y is the standard deviation of y. And next, for the regression equation on x, that is, y is equal to mx plus c, the regression coefficient of y on x, that is, B yx is equal to, let it be equation number 1. So, here the coefficient of x in this equation is m. So, B yx will be equal to m, that is, the coefficient of x in this equation. So, regression equation plus c, let it be equation number 2, y, that is, B xy of y in equation number 2, coefficient of y in, so the regression, this will work out as a key idea. We will start with the solution. Our equations of the lines of regression are given to us, but we have to find the mean value of x and y. Now, given the equations of the lines of regression, phi 1y minus 1.12 and y is equal to 7.15. Now, let us name it as equals, which is given in the key idea that I have given the equation number 1 and equation number 2, x bar, that is, the mean value of x and y bar, that is, the mean value of y. Now, also I am giving the given equations, let it be equation c, y is equal to, and to simultaneously equal to 12, this is a point of intersection of lines 1 and 2 of y bar, that is, the equal to 5 of y is equal to 5. Now, this is the equation number 1, equation number 2, phi 1y minus 1.12. Now, this is the line. Now, using this result, which is given in the key idea, the regression coefficient of x and y, that is, which is equal to 0.51. Now, from 2 equal to 0.97 x, which is given in the the regression coefficient of y on x, that is, so this is the coefficient of x, so this will be equal to 0.97. Now, using these results, which I have given in the key idea, now, v y x is equal to 0.95, which will be equal to, by the formula, sigma y over sigma x is equal to 0.51. So, this will be equal to r into sigma y. Therefore, into b x y is equal to 0.97 into 0.51 is equal to r into sigma y over sigma x into r into sigma x over sigma y, which implies into b x y is equal to 0.4947 plus minus 0.7 approximately. Now, as the value of b y x and b x y y both are positive, then r is also positive, y both are negative, then r is also. Using the positive value here, this will be 0.7 approximately. Sending deviation of y also given to us is equal to 2.6. Now, v y x is equal to 0.97, which is equal to r into sigma y by sigma x, sigma y over sigma x is equal to 0.97. Now, r is 0.7, sigma x is equal to 0.97, 0.97 into 2.6 over 0.7 approximately. Now, the line of deviation of y nominated value of y for the given value of x equation number 2, 0.97 into 9 is equal to 8.735, which is equal to, we are getting y is equal to the value of x for y is equal to 3. Now, the line of deviation of y gives the best estimated value of h for a given value of y. y is equal to 12 in equation number 1 and we will get the value of h equal to 12 in equation number 1, v equal to 0.51 into 12 minus 1.12, which is equal to 6.12 minus 1.12, which is equal to i is equal to 12, x is equal to, that is x bar is equal to 5 and mean value of r is equal to 12, v x y is equal to 0.51 and by regression coefficient v y x is equal to 0.97, then in the third part of y, that is sigma y is equal to 3.6 approximately, x is equal to 12, x is equal to, that is the given question, we have enjoyed the session.